uni-oldenburg.deoops.uni-oldenburg.de/91/2/predom05.pdf · 2013-01-17 · Abstract For certain...

Domains of convergence ininfinite dimensional

holomorphy

Von der Fakultat fur Mathematik und Naturwissenschaften derCarl von Ossietzky Universitat Oldenburg zur Erlangung desGrades und Titels eines Doktors der Naturwissenschaften

(Dr. rer. nat.) angenommene Dissertation von

Christopher Prengel

geboren am 28.10.1975

in Hannover

http://www.bis.uni-oldenburg.de/publikationen/dissertation/2006/predom05/predom05.html

http://www.bis.uni-oldenburg.de/publikationen/dissertation/2006/predom05/predom05.html

Gutachter: apl. Prof. Dr. A. Defant

Zweitgutachter: Prof. Dr. P. Pflug

Tag der Disputation: 26.10.2005

Abstract

For certain subsets F(R) of holomorphic functions on a Reinhardt domainR in a Banach sequence space X, mainly for H∞(R), Pm(X) and P(X), wegive precise descriptions of the domain of convergence domF(R), i.e. the setof all points of R where the monomial expansion∑

α∈N(N)0

∂αf(0)

α!zα

of every function f ∈ F(R) is (unconditionally) convergent.

The results are obtained by an interplay of complex analysis and local Banachspace theory, improving work of Ryan and Lempert about monomial expan-sions of holomorphic functions on `1 and using methods from the theory ofmultidimensional Bohr radii.

The problem of conditional convergence is solved for polynomials. It is closelyconnected with bases for full and symmetric tensor products and convergencepreserving permutations.

Zusammenfassung

Fur gewisse Teilmengen F(R) holomorpher Funktionen auf einem Reinhardt-gebiet R in einem Banachfolgenraum X, vor allem fur H∞(R), Pm(X) undP(X), geben wir eine genaue Beschreibung des KonvergenzbereichsdomF(R), d.h. der Menge aller Punkte aus R, in denen die Monomial-entwicklung ∑

α∈N(N)0

∂αf(0)

α!zα

einer jeden Funktion f ∈ F(R) (unbedingt) konvergiert.

Die Resultate werden durch ein Zusammenspiel von komplexer Analysis undlokaler Banachraumtheorie erzielt, setzen Arbeiten von Ryan und Lempertuber Monomialentwicklungen holomorpher Funktionen auf `1 fort und bedie-nen sich der Theorie mehrdimensionaler Bohrradien.

Das Problem bedingter Konvergenz wird fur Polynome gelost. Es ist engverknupft mit Basen fur volle und symmetrische Tensorprodukte und kon-vergenzerhaltenden Permutationen.

I thank my supervisor Andreas Defant for his guidance during the preparationof this thesis.

Gracias a Manuel Maestre por las discusiones utiles y por su invitacion a Va-lencia. Tambien a todas las personas que hacıan que mi estancia in Valenciafuera muy agradable, especialmente Pilar y su familia, Pablo y Marta, y porsupuesto Manolo.

Thank you Melissa for correcting my English.

Contents

Introduction 9

Preliminaries 15

1. Norm estimates of identities between symmetric tensor products of Banach se-quence spaces 25

Identities between symmetric and full tensor products . . . . . . . . . . . . . 26

Volume estimates for unit balls of spaces of m-homogeneous polynomials . . . 31

`-norm estimates for the injective and projective tensor product . . . . . . . . 33

Volume ratio of m-fold tensor products . . . . . . . . . . . . . . . . . . . . 44

2. Domains of convergence 51

Domains of convergence of holomorphic functions . . . . . . . . . . . . . . . 51

Lower inclusions for domH∞(R) . . . . . . . . . . . . . . . . . . . . . . . 65

The arithmetic Bohr radius . . . . . . . . . . . . . . . . . . . . . . . . . . 75

Upper inclusions for domH∞(R) . . . . . . . . . . . . . . . . . . . . . . 84

A description of domPm(X) . . . . . . . . . . . . . . . . . . . . . . . . 89

A conjecture . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

The domain of convergence of the polynomials . . . . . . . . . . . . . . . . . 105

The exceptional role of `1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

Domains of convergence, Bohr radii and unconditionality . . . . . . . . . . . 113

3. Tensor and s-tensor orders 119

Orders on Nm . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

Orders for tensor products . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

Bases for symmetric tensor products and spaces of polynomials . . . . . . . . 167

Pointwise convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170

Bibliography 174

Introduction

A holomorphic function f on a Reinhardt domain R in Cn has a power seriesexpansion

f(z) =∑α∈Nn

0

cα(f)zα

with coefficients cα(f) = ∂αf(0)α!

that converges to f in every point of R.

In infinite dimensions, if f is holomorphic on a Reinhardt domain R in aBanach sequence space X and we define cα(f) := ∂αf(0)

α!then

f(z) =∑

α∈N(N)0

cα(f)zα

holds for finite sequences z = (z1, . . . , zn, 0, . . .) ∈ R and we ask: Where onR does this monomial expansion series converge? We call the set dom(f) ofall those points the domain of convergence of f .

In this thesis we give precise descriptions of the domain of convergence

domF(R) =⋂

f∈F(R)

dom(f)

of certain subsets F(R) ⊂ H(R) of holomorphic functions.

Historically the problem that we consider here is connected with the defini-tion of holomorphy in infinite dimensions.

Holomorphy in finite dimensions can be characterized by convergence ofmonomial expansion series in the neighbourhood of every point and it wasthis approach that Hilbert tried at the beginning of the 20th century. In [46]he proposed a theory of analytic functions in infinitely many variables basedon representations by power series∑

α∈N(N)0

cαzα (0.1)

(absolutely) convergent on infinite dimensional polydiscs

|z1| ≤ ε1, |z2| ≤ ε2, |z3| ≤ ε3, . . . (0.2)

10 Introduction

where (εn) is a sequence of positive numbers.

In the following years it became clear by the work of Frechet and Gateauxthat expansions in power series of homogeneous polynomials are more suit-able, which formally can be obtained from (0.1) by building blocks∑

α∈N(N)0

cαzα =

∑m

∑|α|=m

cαzα

This theory was continued by students of Frechet. The current form forfunctions between normed spaces can be found in papers of Graves [43] andTaylor [75, 76] from the mid 1930’s, where the equivalence of Frechet differ-entiability, Gateaux differentiability plus continuity, and representation bya power series of homogeneous polynomials in the neighbourhood of everypoint is proved.

A natural environment where it is possible to work with monomial expansionsare fully nuclear locally convex spaces with basis, as was shown in 1978 byBoland and Dineen [15], see [35], [37] and [36, sec. 3.3, 4.5] for more resultsin this direction.

Matos [60] characterized those holomorphic functions on an open subset ofa Banach space with unconditional basis having monomial series representa-tions in the neighbourhood of every point to be functions of a certain holo-morphy type (see also [61]). He proved that in `1 these are all holomorphicfunctions. As we will see, the space `1 plays a special role in our theory.

In some sense the results we give in this thesis justify the historical devel-opment. One problem with the Hilbert approach is that his definition ismore restrictive than he actually thought. He gives us a criterion to de-cide if a power series (0.1) defines an analytic function in the domain (0.2).He claims that it is sufficient for the series to be bounded in the follow-ing sense: Every finite dimensional section

∑α∈Nn

0cαz

α of (0.1) (putting

the variables zn+1, zn+2, zn+3, . . . equal to zero and identifying (α1, . . . , αn) =

(α1, . . . , αn, 0, 0, . . .) ∈ N(N)0 ) is convergent for |z1| ≤ ε1, . . . , |zn| ≤ εn and

supn∈N

sup|zi|≤εi

∣∣ ∑α∈Nn

0

cαzα∣∣ <∞ (0.3)

It can be seen from an example of Bohnenblust and Hille [12] that this isnot true (see example 2.4 and also p. 12 below). But our results even show,extending Hilbert’s definition of boundedness in an appropriate way (seeremark 2.9), that `1 is almost the only Banach sequence space where thecriterion is true. Nevertheless it is remarkable here that a power series (0.1)satisfying (0.3) with all εn equal, say, to r > 0 defines a bounded holomorphicfunction on rBc0 in the current sense (remark 2.9).

Introduction 11

The roots of our results can be found in the work of Harald Bohr at thebeginning of the 20th century. He studied Dirichlet series∑

n

an

ns(0.4)

where an are complex coefficients and s is a complex variable. The domainswhere such a series converges absolutely, uniformly or conditionally are halfplanes [Re s > σ0] where σ0 = a, u or c is called the abscissa of absolute,uniform or conditional convergence respectively. More precisely σ0 = inf σtaken over all σ ∈ R such that on [Re s > σ] we have convergence of therequested type. Bohr wanted to know the greatest possible width T of thestrip of uniform but non absolute convergence over all Dirichlet series (0.4):

-

6

T = supP anns

a− u

u a

uniform convergence -

absolute convergence -

He attacked this problem using the nowadays so called Bohr method (see[13]). Using the fundamental theorem of arithmetic he established a one toone correspondence between Dirichlet series and power series in infinitelymany variables, given by∑

n

an

ns!

∑α∈N(N)

0

cαzα , cα = apα (0.5)

where p = (pn) is the sequence of prime numbers p1 < p2 < p3 < . . .

Then he defined S to be the least upper bound of all q > 0 such that∑|cα| εα < ∞ for every power series

∑cαz

α bounded in Hilbert sense on|zn| ≤ 1, i.e. with

supn

sup|zi|≤1

∣∣ ∑α∈Nn

0

cα zα∣∣ <∞ (0.6)

and every ε ∈ `q ∩ ] 0 , 1[ N and used the prime number theorem to prove

T =1

S(0.7)

Thus Bohr shifted the problem to the setting of bounded power series ininfinitely many variables and he was able to prove S ≥ 2 ([13, Satz III]):

12 Introduction

Theorem 0.1. Let∑cαz

α be a power series bounded in the domain [|zi| ≤ 1]and ε ∈ `2 ∩ ] 0 , 1[ N. Then ∑

α∈N(N)0

|cα| εα <∞

Bohr did not know if 2 was the exact value of S, not even if S < ∞ (i.e. ifT > 0, in other words: if there is a Dirichlet series (0.4) whose absissa a ofabsolute convergence and u of uniform convergence do not coinside).

Um dies Problem zu erledigen, ist ein tieferes Eindringen in die Theorie derPotenzreihen unendlich vieler Variabeln notig, als es mir in §3 gelungen ist.1

Bohr hoped to solve this problem by studying the relation between the ab-solute value of a power series and the sum of the absolute values of theindividual terms. He went down to dimension one and asked (see [14]):

Given 0 < r < 1, does there always exist a power series∑cnz

n convergentand bounded by one on the unit disc such that

∞∑n=1

|cn| rn > 1

is true ? This question found a satisfying answer.

Theorem 0.2 (Bohr’s power series theorem).Let

∑cnz

n be a power series in C with

∣∣ ∞∑n=1

cnzn∣∣ ≤ 1 (0.8)

for all |z| < 1. Then∞∑

n=1

|cnzn| ≤ 1

for all |z| < 1/3 and the value 1/3 is optimal, i.e. for every r > 1/3 there isa power series

∑cnz

n satisfying (0.8) and∑∞

n=1|cn|rn > 1.

Bohr did not manage to extend this result to power series in several variables.The finiteness of S was then obtained by Toeplitz [77]. He proved S ≤ 4 byconstructing for every δ > 0 an ε ∈ `4+δ ∩ ] 0 , 1[ N and a 2-homogeneouspolynomial

∑|α|=2 cαz

α bounded on [|zi| ≤ 1] with∑

|α|=2|cα|εα =∞. Then

after some time Bohnenblust and Hille [12] proved in 1931 that 2 is in factthe exact value of S. They showed that the upper bound 4 of Toeplitz is bestpossible if one just considers 2-homogeneous polynomials instead of arbitrary

1Bohr [13], p. 446

Introduction 13

power series, and they constructed examples of m-homogeneous polynomialsfor arbitrary m to get the result (see example 2.4).

In 1989 Dineen and Timoney [37] studied the existence of absolute monomialbases for spaces of holomorphic functions on locally convex spaces and saw apossibility to solve their basis problem with the help of a polydisc version ofBohr’s theorem. In [38] they used this polydisc result to reprove the estimateS ≤ 2 in the way that Bohr originally might have thought of, showing that ifε ∈] 0 , 1[ N is such that

∑|cα|εα <∞ for all power series

∑cαz

α bounded on[|zi| ≤ 1], then ε ∈ `2+δ for all δ > 0 (a sort of converse of theorem 0.1). Inorder to prove their multidimensional extension of theorem 0.2 they used aprobabilistic method from [59] which goes back to the mid 70’s and of coursewas not accessible to Bohr.

The initial step for our theory is to see that the considerations above aboutbounded power series in infinitely many variables also work for boundedholomorphic functions on B`∞ . The power series in the definition of S canbe replaced by these functions:

S = supr > 0 | `r ∩B`∞ ⊂ domH∞(B`∞)

and the results above can be summarized in a very simple way as

`2 ∩B`∞ ⊂ domH∞(B`∞) ⊂ `2+ε ∩B`∞ (0.9)

for every ε > 0. Here the lower inclusion is a restatement of Bohr’s theo-rem 0.2 and the upper one reflects the result of Dineen and Timoney. Theexamples of Bohnenblust and Hille give

`2+ε ∩B`∞ 6⊂ domH∞(B`∞)

for every ε > 0.

Another influence for our work are papers of Ryan [70] and Lempert [53]where they studied monomial expansions of holomorphic functions on `1.There it is proved that dom(f) = rB`1 holds for every holomorphic functionf on some multiple rB`1 of the unit ball in `1, again underlining the factthat this Banach space plays a special role in infinite dimensional complexanalysis.

In 1997 a systematic study of multidimensional extensions of Bohr’s powerseries theorem started with the work [11]. Several so called Bohr radii wheredefined and estimated, for example in [3, 2, 4, 5, 10, 11, 24, 25, 62]. Asidefrom techniques of complex analysis a keypoint in these estimates is oftenthe use of probabilistic methods. On the other hand questions related to theone dimensional Bohr phenomenon were investigated, as in [7, 16, 17].

The tools developed in these investigations provide important ideas for ourtheory and made it possible to describe the domain of convergence domH∞(R)

14 Introduction

for bounded Reinhardt domains R in a wide class of Banach sequence spaces,extending (0.9) in a far reaching way. Further we give a description for thedomain of convergence of the spaces Pm(X) of m-homogeneous polynomialsand P(X) of all polynomials. This is done in chapter 2.

Our results arise from an interplay of complex analysis and local Banachspace theory. There is a close connection with unconditionality in spaces ofm-homogeneous polynomials, and during the investigation of the domainsof convergence in chapter 2 the problem occurs to give estimates for theunconditional constant of identity mappings between those spaces which issolved going through full tensor products. That this can be done is justified inchapter 1 with a general theorem, providing a helpful tool to give estimatesfor many important invariants of local Banach space theory, such as typeor cotype constants as well as the `-norm, of identities between symmtrictensor products working in the full tensor product, which in general is easierto handle. As an application, we give in chapter 1 asymptotically correctvolume estimates for unit balls of m-homogeneous polynomials Pm(Xn) andof the volume ratio for the m-fold injective and projective tensor product⊗m

αXn, α = ε or π.

The convergence of the monomial expansion series that we consider in chapter2 is the convergence of the net of finite sums or unconditional convergence.In chapter 3 we investigate the problem of conditional convergence. Thisis solved for polynomials. Here the task is to order the monomials in anappropriate way, and we overcome this problem with a systematic study oforders that produce bases for tensor and s-tensor products of Banach spaceswith basis.

Preliminaries

Tensor products and m-homogeneous polynomials

Let E1, . . . , Em and F be vector spaces over K. We write Lm(E1, . . . , Em;F )for the vector space of all m-linear mappings T :

∏mi=1Ei −→ F and then

Lm(E,F ) if E1 = . . . = Em = E and Lms (E,F ) for the subspace of symmetric

mappings. If F = K we simple write Lm(E1, . . . , Em), Lm(E) and Lms (E).

For normed spaces E1, . . . , Em, F the corresponding spaces of continuousoperators are denoted with L instead of L, provided with the norm

‖T‖ = sup‖xi‖≤1

‖T (x1, . . . , xm)‖

Linearization of m-linear mappings produces the (algebraic) tensor product⊗m

i=1Ei, i.e. we have

Lm(E1, . . . , Em;F ) = L(⊗mi=1Ei, F )

for all vector spaces F via T LT , LT (x1 ⊗ · · · ⊗ xm) = T (x1, . . . , xm).If E = E1 = . . . = Em we write ⊗mE = ⊗m

i=1Ei for the m-fold tensorproduct of E and ⊗m,sE for the symmetric m-fold tensor product of E whichcan be obtained by linearization of symmetric m-linear maps. ⊗m,sE is acomplemented subspace of ⊗mE and we denote ιm : ⊗m,sE −→ ⊗mE thenatural injection and

σm :

⊗mE −→ ⊗mE

x1 ⊗ · · · ⊗ xm 1

m!

∑σ∈Sm

xσ(1) ⊗ · · · ⊗ xσ(m)

the natural projection which may also be considered as a mapping ⊗mE −→⊗m,sE.

Let m and n ∈ N. We will often need the set M(m,n) = 1, . . . , nm andits subset J (m,n) = j ∈M(m,n) | j1 ≤ . . . ≤ jm as well as

J (m) = j ∈ Nm | j1 ≤ . . . ≤ jm

16 Preliminaries

Further the notation xi := xi1 ⊗ · · · ⊗ xim for elements x1, . . . , xn of a vectorspace and i ∈M(m,n) will be very useful.

Now suppose that E is n-dimensional with basis x1, . . . , xn. Then xi | i ∈M(m,n) and σm(xj) | j ∈ J (m,n) are bases of ⊗mE and ⊗m,sE, respec-tively, in particular dim⊗mE = nm and dim⊗m,sE =

(m+n−1

n−1

).

The projective tensor norm π on the tensor product ⊗mi=1Ei of normed spaces

Ei can be defined as the norm generated by the absolute convex hull of

BE1 ⊗ · · · ⊗BEm := x1 ⊗ · · · ⊗ xm | xi ∈ BEi

or by the property that

L(⊗mπ,i=1Ei, F ) = Lm(E1, . . . , Em;F )

isometrically for all normed spaces F . The injective tensor norm ε is thenorm induced by the embedding ⊗m

i=1Ei → (⊗mπ,i=1E

′i)′. In general by a

tensor norm α of order m we mean a method of assigning to each m-tuple(E1, . . . , Em) of normed spaces a norm α = α( · ;E1, . . . , Em) satisfying ε ≤α ≤ π (α is reasonable) and the metric mapping property: For Ti ∈ L(Ei, Fi)

‖⊗mi=1Ti : ⊗m

α,i=1Ei −→ ⊗mα,i=1Fi‖ = ‖T1‖ · · · ‖Tm‖

If α is reasonable but does not necessarily satisfy the metric mapping prop-erty, we speak of a tensor product norm (of order m). We write ⊗m

α,i=1Ei for

the space ⊗mi=1Ei equipped with the norm α and ⊗m

α,i=1Ei for its completion.

An m-homogeneous polynomial P : E −→ F between vector spaces E andF is the restriction of an m-linear mapping T ∈ Lm(E,F ) to the diagonal,i.e. P (x) = T (x, . . . , x) for all x ∈ E. The mapping T can be chosen tobe symmetric, and in this way the space Pm(E,F ) of all m-homogeneouspolynomials P : E −→ F can be identified with Lm

s (E,F ). If E and F arenormed spaces the subspace of continuous m-homogeneous polynomials isdenoted by Pm(E,F ), equipped with norm

‖P‖ = sup‖x‖≤1

‖P (x)‖

In the same way as above the projective s-tensor norm πs is characterized bythe fact that L(⊗m

πsE,F ) = Pm(E,F ) isometrically for all normed spaces E

and F and the injective s-tensor norm εs on ⊗m,sE is the norm induced by⊗m,sE → (⊗m,s

πsE ′)′. Then an s-tensor norm αs of order m assigns to each

normed space E a norm αs = αs( · ;⊗m,sE) on ⊗m,sE such that εs ≤ αs ≤ πs

(αs is reasonable) and the metric mapping property is satisfied:

‖⊗m,sT : ⊗m,sαsE −→ ⊗m,s

αsF‖ = ‖T‖m

for all normed spaces E and F and all operators T : E −→ F , where ⊗m,sT =σm ⊗mT ιm. Again if we only require that αs is reasonable, we call it an

Preliminaries 17

s-tensor product norm. Note that the restriction of every tensor norm tosymmetric tensor products gives an s-tensor norm.

Now let α be a tensor norm and αs an s-tensor norm, both of order m. Wesay that α and αs are equivalent – notation α ∼ αs – if there is a constantcm > 0 such that the natural injection ιm : ⊗m,s

αsE −→ ⊗m

αE and projectionσm : ⊗m

αE −→ ⊗m,sαsE have norm smaller than cm for all normed spaces E.

By the norm extension theorem of Floret [40], given an s-tensor norm αs oforder m, there is always an equivalent tensor norm α, and the constant cmcan be chosen to be smaller than mm/m!.

Now let 1 ≤ p ≤ ∞. The natural norm ∆p of order m on the p-integrablefunctions assigns to each m-tuple (Lp(µ1), . . . , Lp(µm)) of Lp-spaces (overmeasure spaces (Ωi, µi)) the norm ∆p = ∆p( · ;⊗m

i=1Lp(µi)) on ⊗mi=1Lp(µi)

coming from the natural inclusion

⊗mi=1Lp(µi) → Lp(µ1 ⊗ · · · ⊗ µm)

where the element f1 ⊗ · · · ⊗ fm ∈ ⊗mi=1Lp(µi) is assigned to the equivalence

class of the function f defined by f(x1, . . . , xm) = f1(x1) · · · fm(xm).

We denote ∆p,s the symmetric counterpart of ∆p on them-fold tensor productof discreet Lp-spaces. We write ⊗m

p `np := ⊗m

∆p`np and ⊗m,s

p `np := ⊗m,s∆p,s

`np .

So ∥∥ ∑i∈M(m,n)

λiei

∥∥⊗m

p `np

=( ∑

i∈M(m,n)

|λi|p)1/p

∥∥ ∑j∈J (m,n)

λjej

∥∥⊗m,s

p `np

=( ∑

j∈J (m,n)

|λj|p)1/p

i.e. ⊗mp `

np = `n

m

p and ⊗m,sp `np = `dn

p isometrically, where dn =(

m+n−1n−1

)is the

dimension of ⊗m,sp `np .

Note that ∆p and ∆p,s are reasonable on the class of Banach spaces wherethey are defined, and we also treat them as tensor product respectively s-tensor product norms.

For more information on tensor and s-tensor norms as well asm-homogeneouspolynomials see [22], [39], [40] and [36].

Banach sequence spaces

By a Banach sequence space we mean a Banach space X ⊂ CN of sequencesin C such that `1 ⊂ X ⊂ `∞, and satisfying that |x| ≤ |y| implies x ∈ Xand ‖x‖ ≤ ‖y‖ whenever x ∈ CN, y ∈ X. In particular, the canonical unitvectors en = (δnk)k belong to X.

18 Preliminaries

Defined in this way a Banach sequence space X in particular is a complexKothe function space in the sense of [57] (the complexification of the realBanach sequence space XR = x ∈ X | x ∈ RN ) or (the complexification of)a Banach function space as introduced in [8], but lacking the Fatou property.

Note that the unit vectors en in a Banach sequence space build a 1-uncon-ditional basic sequence and in particular every infinite dimensional complexBanach space with 1-unconditional Schauder basis satisfying the maximalitycondition can be considered as a Banach sequence space.

Every Banach sequence space X satisfies

`1 ⊂ X ⊂ `∞ (0.10)

so `1 and `∞ are the extreme cases. By the closed graph theorem, the setinclusion X ⊂ Y between Banach sequence spaces X and Y already impliesthat the inclusion mapping X → Y is continuous.

A Banach sequence space X is symmetric if an element x ∈ CN belongs toX if and only if its decreasing rearrangement x∗ does, and in this case theyhave the same norm. The decreasing rearrangement x∗ of a sequence x ∈ Xcan be defined by

x∗n := inf supi∈N\I|xi| | I ⊂ N , |I| < n

If x is a zero sequence this is nothing else than the sequence |x| rearrangedin some non-increasing order. For a Banach sequence space X with basisbeing symmetric it is equivalent to say that any permutation of an elementof X remains in X with equal norm or equivalently the basis (en) of X issymmetric and ∥∥ ∞∑

n=1

anen

∥∥ =∥∥ ∞∑

n=1

aneπ(n)

∥∥for every permutation π and all an ∈ C such that the series converge.

Concerning the notations that we use for elements x of a Banach sequencespace X always think of functions N −→ C. For example |x| = (|xn|)n, x > 0means xn > 0 for all n, xp = (xp

n)n for x ≥ 0 and p > 0, xy is the sequencewith n-th coordinate xnyn for x, y ∈ X and so on.

Let 0 < p ≤ ∞. A Banach sequence space X is called p-convex if thep-convexity constant of X

M(p)(X) := supn‖id : `np (X) −→Mn

p (X)‖

is finite and p-concave if its p-concavity constant

M(p)(X) := supn‖id : Mn

p (X) −→ `np (X)‖

Preliminaries 19

is finite, where Mnp (X) and `np (X) denote Xn with the quasi norm (norm, if

p ≥ 1)

‖x‖ =

∥∥( n∑

i=1

|xi|p)1/p∥∥

Xp <∞

‖max1≤i≤n

|xi|‖X p =∞

and ‖x‖ = ‖(‖x1‖X , . . . , ‖xn‖X)‖`np, respectively. For 1 ≤ p ≤ ∞, p-convexity

is decreasing and p-concavity increasing in p ([57, 1.d.5]). For p-convexitythis is also true if p < 1 (see [20, lemma 4]). Every Banach sequence space is1-convex and ∞-concave, we say trivially convex and trivially concave, with

M(p)(X) = M(∞)(X) = 1

for p ≤ 1 (see [57, p. 46] and [20, lemma 4]).

Now let X and Y be a Banach sequence space. The Kothe dual of X isdefined to be the space

X× = y ∈ CN | xy ∈ `1

It is again a Banach sequence space with norm ‖y‖X× = sup‖x‖X≤1‖xy‖`1(this number is finite for every y ∈ X× by the closed graph theorem). TheKothe dual X× can be regarded as a subspace of X ′ via

λ [x ∞∑

n=1

λnxn]

and we have X× = X ′ if and only if X = spanen. Further X ⊂ Y impliesY × ⊂ X×.

Symmetry carries over from X to X×.

We are going to use the following sets: For 1 ≤ r <∞ the r-th power of X,defined by

Xr := x ∈ CN | |x|1/r ∈ Xand the product

X · Y := xy | x ∈ X, y ∈ Y of X and Y .

The fundamental function of X is defined by λX(n) = ‖∑n

i=1 ei‖, and wedenote n-th section Xn = spane1, . . . , en of X to be the space generatedby the first n standard unit vectors.

We mention two types of examples for Banach sequence spaces from theliterature. Let 1 = w1 ≥ w2 ≥ . . . ≥ wn ≥ . . . ≥ 0 be a sequence of weightswith

∑∞n=1wn =∞ and 1 ≤ p <∞. Then

d(w, p) := x ∈ CN | supπ

∞∑n=1

wn|xπ(n)|p <∞

20 Preliminaries

together with the norm

‖x‖ = supπ

( ∞∑n=1

wn|xπ(n)|p)1/p

is a Banach space which is called Lorentz sequence space. Here the supis taken over all permutations π. The unit vectors en form a normalized(w1 = 1) basis and by the definition of the norm it is easy to check that thisbasis is 1-unconditional. The norm can be expressed as

‖x‖ =

( ∞∑n=1

wn|x∗n|p)1/p

taking one special permutation instead of all (use e.g. Hardy’s lemma [8,prop. 3.6]) and from this we get that d(w, p) is a symmetric Banach sequencespace in our sense. For more general information on Lorentz sequence spacessee [56, 4.e].

Taking all weights equal to one we get the usual `p-spaces. More general if

we define wn := nqp−1 for 1 ≤ q ≤ p <∞ then d(w, q) = `p,q are the discrete

variants of the Lp,q-spaces which play an important role in interpolationtheory (see e.g. [8, chapter IV] or [57, 2.b.8]). The space `p,q is r-convex if0 < r < p and 0 < r ≤ q (see e.g. [20, p. 159]).

As a second example we name Orlicz sequence spaces. We take the defi-nition from [56, chapter 4]. Let ϕ : R≥0 −→ R≥0 be an Orlicz function,i.e. a continuous, non decreasing and convex function with ϕ(0) = 0 andlimt→∞ ϕ(t) = ∞. ϕ is called a degenerate Orlicz function if ϕ(t) = 0 forsome t > 0. The Orlicz function generates a sequence space

`ϕ = x ∈ CN | ∃ρ>0 :∞∑

n=1

ϕ( |xn|ρ

)<∞

which is a symmetric Banach sequence space with norm

‖x‖ = infρ > 0 |∞∑

n=1

ϕ( |xn|ρ

)≤ 1

(see e.g. [68, theorem 10]).

The unit vectors form a normalized 1-unconditional basis of the subspace

hϕ := x ∈ `ϕ | ∀ ρ > 0 :∞∑

n=1

ϕ( |xn|ρ

)<∞

(see [56, prop. 4.a.2]) which is equal to `ϕ if and only if the Orlicz functionϕ satisfies the ∆2-property

lim supt→∞

ϕ(2t)

ϕ(t)<∞

Preliminaries 21

In this case and under the additional assumption that ϕ is non-degeneratethe following conditions are equivalent (see [49], proposition 14 and lemma5):

(1) `ϕ is p-convex.

(2) ϕ is equivalent to an Orlicz function ϕ which satisfies

ϕ(λx) ≥ λpϕ(x)

for all x ≥ 0 and λ ≥ 1.

(3) ϕ is equivalent to an Orlicz function ϕ such that ϕ( p√· ) is convex.

(4) `ϕ satisfies an upper p-estimate, i.e. there is a constant c > 0 such thatfor all n and all x1, . . . , xn ∈ `ϕ pairwise disjoint

∥∥ n∑i=1

xi

∥∥ ≤ c( n∑

i=1

‖xi‖p)1/p

(Two elements u, v ∈ X are called disjoint if min|u|, |v| = 0, i.e. forall n: un = 0 or vn = 0.)

Here equivalence of ϕ and ϕ means that there exists t0 > 0 and a constantc > 0 such that

c−1 ϕ(t) ≤ ϕ(t) ≤ c ϕ(t)

for all 0 < t ≤ t0.

Reinhardt domains and holomorphic mappings

Let E and F be complex Banach spaces and U be an open subset of E. Afunction f : U −→ F is called Frechet differentiable if for all ξ ∈ U we canfind a continuous linear mapping A : E −→ F such that

limx→ξ

‖f(x)− f(ξ)− A(x− ξ)‖‖x− ξ‖

= 0

where f ′(ξ) := Df(ξ) := A is called the Frechet derivative of f in ξ. Form ≥ 2 the function f is m times Frechet differentiable if f (m−1) : U −→L(E,Lm−2(E,F )) = Lm−1(E,F ) is Frechet differentiable. By Schwarz’ the-orem we have f (m) ∈ Lm

s (E,F ). We denote by Cm(U, F ) the vector spaceof all m-times Frechet differentiable mappings f : U −→ F and Cm(U) :=Cm(U,C).

The function f : U −→ F is Gateaux holomorphic or G-holomorphic if forall ξ ∈ U and all x ∈ E the function λ f(ξ + λx) is holomorphic onU(ξ, x) = λ ∈ C | ξ+λx ∈ U or equivalently if f is weakly holomorphic on

22 Preliminaries

every finite dimensional component of U , i.e. for every functional ϕ ∈ F ′ andevery finite dimensional subspace E0 ⊂ E the function ϕ f is holomorphicon U ∩ E0. Let HG(U, F ) denote the space of all G-holomorphic mappingsf : U −→ F and HG(U) := HG(U,C).

The function f : U −→ F is holomorphic iff it has a Taylor series expansion ateach point ξ of U : For all ξ ∈ U we can find a unique sequence of continuousm-homogeneous polynomials Pmf(ξ) ∈ Pm(E,F ) and a radius r > 0 suchthat

f(ξ + x) =∞∑

m=0

Pmf(ξ)(x) (0.11)

or

f(ξ + x) =∞∑

m=1

am(f)xm

uniformly on rBE, where am(f) ∈ Ls(E,F ) are the symmetric m-linear map-pings corresponding to the m-homogeneous polynomials such thatPmf(ξ)(x) = am(f)xm := am(f)(x, . . . , x). In this case f ∈ C∞(ξ + rBE, F )and am(f) = f (m)(ξ)/m!. We denote H(U, F ) the vector space of all holo-morphic mappings f : U −→ F and H(U) := H(U,C).

A function f : U −→ F is holomorphic if and only if it is Frechet differentiableif and only if it is G-holomorphic and continuous.

If f ∈ H(U, F ), ξ ∈ U and B ⊂ E balanced such that ξ + B ⊂ U then wehave for all x ∈ B

Pmf(ξ)(x) =1

2πi

∫|λ|=1

f(ξ + λx)

λm+1dλ , m = 0, 1, 2, . . . (0.12)

(Cauchy integral formulas), in particular

‖Pmf(ξ)‖B ≤ ‖f‖ξ+B , m = 0, 1, 2, . . . (0.13)

(Cauchy inequalities), see for example [19], p. 175 or [36], p. 148.

We will only consider scalar valued holomorphic mappings. Note that wehave f ∈ H(U) if and only if f is continuous and f is holomorphic on everyfinite dimensional component of U .

We denote H∞(U) the Banach space of all bounded holomorphic functionson U with norm ‖f‖U = supz∈U |f(z)|.

By a Reinhardt domain R ⊂ Cn we mean an open set which is n-circularand balanced, i.e. it satisfies λR ⊂ R for all λ ∈ B`n

∞ .

In the literature one usually only assumes R to be open and n-circular (i.e.z ∈ R and u ∈ Cn with |u| = |z| implies u ∈ R) and the additional assump-tion that λR ⊂ R for every λ ∈ C with |λ| ≤ 1 is called completeness of R.Thus our Reinhardt domains are always complete.

Preliminaries 23

In the same way a Reinhardt domain R in a Banach sequence space X is anopen, circular and balanced subset of X, in other words satisfying λR ⊂ Rfor all λ ∈ B`∞ . Note that if R ⊂ X is a Reinhardt domain then the n-thprojection Rn := R ∩Xn of R is a Reinhardt domain in Cn.

An important example for a Reinhardt domain in a Banach sequence space Xis the open unit ball BX , whose n-th projection BXn is even a logarithmicallyconvex Reinhardt domain in Cn.

1. Norm estimates of identitiesbetween symmetric tensor products

of Banach sequence spaces

An active research area of Banach space theory in recent years is the localtheory which investigates the structure of infinite dimensional Banach spacesby relating it to the structure of their finite dimensional subspaces. It hasits roots in the work of Grothendieck in the 1950s. An important method isto derive properties of a Banach space from the asymptotical behaviour ofisometric invariants of its finite dimensional subspaces when the dimensiongrows to infinity. For example in [28] an asymptotically correct estimate inn of the unconditional basis constant of the space Pm(`np ) of m-homogeneouspolynomials on `np is used to prove that Pm(E) cannot have an unconditionalbasis for every infinite dimensional Banach space E with unconditional basis.Other important invariants of the local theory are for example type andcotype constants or the Gordon-Lewis constant. These constants are definedfor general operators between Banach spaces and we will see that it can beuseful to know their asymptotic behaviour (in the dimension n) for identitymappings

id : ⊗m,sαsXn −→ ⊗m,s

βsYn

between spaces of symmetric tensor products, where X and Y are Banach se-quence spaces. For example in chapter 2 we are confronted with the problemof giving the asymptotic for χM(id : Pm(Xn) −→ Pm(Yn)), a generalizationof the unconditional basis constant of the monomials (see definition 2.59),which by the identification Pm(Xn) = ⊗m,s

εsX ′

n for Banach sequence spacesX is a problem of the above type.

In general it is easier to calculate those estimates for identities between fulltensor products, since there is a well developed theory for the α-tensor prod-uct E⊗αF of two Banach spaces E and F which in many cases can be easilyextended to m-fold tensor products ⊗m

α,i=1Ei (see e.g. Floret [39]). For ex-

ample in the case of α = ε or π the α-tensor product ⊗mα,i=1Ei is associative,

which makes it possible to proceed by induction and to deduce an estimatefor id : ⊗m

αXn −→ ⊗mβ Yn from that of id : Xn −→ Yn.

26 1. Norm estimates of identities between symmetric tensor products

For sequences (an) and (bn) in R≥0 the notation an ≺ bn means an ≤ c bn forsome constant c > 0 independent of n and we write an bn if an ≺ bn andbn ≺ an.

In this chapter we are going to give general estimates which permit in manycases to work in the full tensor product. In particular we prove that

A(id : ⊗m,sαsXn −→ ⊗m,s

βsYn) A(id : ⊗m

αXn −→ ⊗mβ Yn)

if α ∼ αs, β ∼ βs and A is an ideal norm. Here the asymptotic is with respectto n (m is always fixed in this chapter). This generalizes results from [21]and [29]. There it is proved that gl(⊗m,s

εs`np ) gl(⊗m

ε `np ) and C2(⊗m,s

εsXn)

C2(⊗mε Xn) for a large class of symmetric Banach sequence spaces X.

For the notion of a normed operator ideal see e.g. [22, I.9] or [63].

As an application we use this reduction to full tensor products to give asymp-totic descriptions for the volume of the unit ball BPm(Xn) and the volume ratiovr(⊗m

αXn), α = ε or π for a wide class of symmetric Banach sequence spacesX.

Identities between symmetric and full ten-

sor products

As in [21] and [29] for our estimates we use the fact that for α = ε, and infact for an arbitrary tensor norm α with equivalent s-tensor norm αs, thesymmtric tensor product ⊗m,s

αsXn is a complemented subspace of ⊗m

αXn andwith a certain construction the full α-tensor product is also complementedin the symmetric one (more precisely in ⊗m,s

αsXnm).

First we define the algebraic mappings which are involved in the complementconstruction (see also [29, section 3] or [39, 1.10]).

For N, n ∈ N, k = 0, . . . , N and i = 1, . . . , N we define the mappings

Ii : Kn −→ KNn+k ,n∑

j=1

λj ej n∑

j=1

λj en(i−1)+j

Pi : KNn+k −→ Kn ,

Nn+k∑j=1

λj ej n∑

j=1

λn(i−1)+j ej

(1.1)

and Ii := Ii1 ⊗ · · · ⊗ Iim , Pi := Pi1 ⊗ · · · ⊗ Pim for i ∈ 1, . . . , Nm. Then

Ii( ∑

j∈1,...,nm

λjej

)=

∑n(i−1)+1≤j≤ni

λj−n(i−1)ej

Pi

( ∑j∈1,...,Nn+km

λjej

)=

∑j∈1,...,nm

λn(i−1)+jej

(1.2)

Identities between symmetric and full tensor products 27

(where 1 = (1, . . . , 1) ∈ Nm if we write i+ 1 with i ∈ Nm).

Further let idn, ⊗m idn and ⊗m,s idn denote the identity map on Kn, ⊗mKn

and ⊗m,sKn, respectively. Note that Ii is an injection, Pi a projection and

idNn =N∑

i=1

Ii idn Pi

⊗m idNn =∑

i∈1,...,Nm

Ii ⊗m idn Pi

(1.3)

Recall that (ej)j∈1,...,nm is a basis of ⊗mKn and (σm(ej))1≤j1≤...≤jm≤n a basisof ⊗m,sKn. From (1.2) we get

σm Ii( ∑

j∈1,...,nm

λjej

)=

∑n(i−1)+1≤j≤ni

λj−n(i−1)σm(ej) (1.4)

andm!Pi ιm

( ∑j∈1,...Nn+km

λjσm(ej))

=∑

j∈1,...,nm

λn(i−1)+jej (1.5)

since

Pi(σm(ej)) =

1

m!ej−n(i−1) n(i− 1) + 1 ≤ j ≤ ni

0 otherwise

In particular

I := σm I1 ⊗ · · · ⊗ Im : ⊗mKn −→ ⊗m,sKmn+k (1.6)

is an injection,

P := m!P1 ⊗ · · · ⊗ Pm ιm : ⊗m,sKmn+k −→ ⊗mKn (1.7)

a projection and PI = ⊗m idn is the identity on ⊗mKn, i.e. ⊗mKn is acomplemented subspace of ⊗m,sKmn+k and the diagram

⊗mKn ⊗mKn

⊗mKmn+k ⊗mKmn+k

⊗m,sKmn+k ⊗m,sKmn+k-

-

?

6

?

6

⊗m idn

I1⊗···⊗Im m! P1⊗···⊗Pm

σm ιm

⊗m,s idmn+k

(1.8)

commutes.

Note that the norms α, β and αs, βs in the following can be ∆p respectively∆p,s. In this case for example ⊗m

αXn or ⊗m,sαsXn requires X = `p.


Lemma 1.1. Let X and Y be Banach sequence spaces, α, β be tensor productnorms and αs, βs s-tensor product norms. Let

f :⋃n

L(⊗mαXn,⊗m

β Yn) −→ R≥0

g :⋃n

L(⊗m,sαsXn,⊗m,s

βsYn) −→ R≥0

be functions and ci > 0, i = 1, 2, 3 be constants such that for all n

(i) f|L(⊗mα Xn,⊗m

β Yn) and g|L(⊗m,sαs Xn,⊗m,s

βsYn) satisfy the triangle inequality.

(ii) For all N ∈ N and all i ∈ 1, . . . , Nm

f(Ii ⊗m idn Pi : ⊗mαXNn −→ ⊗m

β YNn)

≤ c1 f(⊗m idn : ⊗mαXn −→ ⊗m

β Yn)

(iii) For all M ≤ N

f(⊗m idM : ⊗mαXM −→ ⊗m

β YM) ≤ c2 f(⊗m idN : ⊗mαXN −→ ⊗m

β YN)

(iv) For all k = 0, . . . ,m

f(⊗m idn : ⊗mαXn −→ ⊗m

β Yn)

≤ c3 g(⊗m,s idmn+k : ⊗m,sαsXmn+k −→ ⊗m,s

βsYmn+k)

Then

f(⊗m idn : ⊗mαXn −→ ⊗m

β Yn) ≤ c g(⊗m,s idn : ⊗m,sαsXn −→ ⊗m,s

βsYn)

for all n ≥ m with c ≤ c1 c2 c3 (m+ 1)m.

Proof. Let n ∈ N. Using (1.3), (i) and (ii) we get

f(⊗m idNn) ≤ c1Nm f(⊗m idn)

for all N . Together with (iii) and (iv) this implies

f(⊗m idmn+k) ≤ c2 f(⊗m id(m+1)n) ≤ c1 c2 (m+ 1)m f(⊗m idn)

≤ c1 c2 c3 (m+ 1)m g(⊗m,s idmn+k)

for k = 0, . . . ,m.

Proposition 1.2. Let (A,A) be a normed operator ideal, α, β tensor productnorms and αs, βs s-tensor product norms of one of the following type:

(a) α and β are tensor norms with equivalent s-tensor norms αs and βs ,respectively.

Identities between symmetric and full tensor products 29

(b) α = ∆p, αs = ∆p,s and β is a tensor norm with equivalent s-tensornorm βs.

(c) α is a tensor norm with equivalent s-tensor norm αs and β = ∆q,βs = ∆q,s.

(d) α = ∆p, αs = ∆p,s and β = ∆q, βs = ∆q,s.

Further let X and Y be symmetric Banach sequence spaces. Then

A(id : ⊗mαXn −→ ⊗m

β Yn) ≤ cA(id : ⊗m,sαsXn −→ ⊗m,s

βsYn)

for all n ≥ m where (a) c ≤ (mm)2

m!(m + 1)m, (b) c ≤ (m(m + 1))m, (c)

c ≤ (m(m+1))m

m!and (d) c ≤ (m+ 1)m.

Proof. We are going to check (i) – (iv) of lemma 1.1 for A = f = g. A isan ideal norm, hence satisfies (i) and for i ∈ 1, . . . , Nm and Ii : ⊗m

β Yn −→⊗m

β YNn and Pi : ⊗mαXNn −→ ⊗m

αXn (as defined right before (1.2)) we have

A(Ii ⊗m idn Pi) ≤ ‖Ii‖‖Pi‖A(⊗m idn)

We claim ‖Ii‖ = 1 = ‖Pi‖. Since X and Y are symmetric, the mappingIl : Yn −→ YNn is an isometry and Pl : XNn −→ Xn a norm-1-projection forl = 1, . . . , N (see (1.1)). Thus if α and β are tensor norms then the claimfollows by the metric mapping property. On the other hand if α = ∆p andβ = ∆q then a glance at (1.2) assures that Ii : ⊗m

q `nq −→ ⊗m

q `Nnq is even an

isometry and Pi : ⊗mp `

Nnp −→ ⊗m

p `np a norm-1-projection. Hence (ii) follows

with c1 = 1.

To see (iii) take M ≤ N and let P : XN −→ XM and I : YM −→ YN thenatural projection and injection, respectively. Then

⊗mαXM ⊗m

β YM

⊗mαXN ⊗m

β YN

-

-?

6

⊗m idM

⊗mI ⊗mP

⊗m idN

commutes, and again if α and β are tensor norms then the vertical mappingshave norm 1 by the metric mapping property. For the other cases just notethat (⊗mI)(ej) = ej and (⊗mP )(ej) = ej if j1, . . . , jm ≤ M and zero other-wise which implies that ⊗mI : ⊗m

q `Mq −→ ⊗m

q `Nq is the natural injection and

⊗mP : ⊗mp `

Np −→ ⊗m

p `Mp the natural projection. Hence

A(⊗m idM) ≤ ‖⊗mI‖‖⊗mP‖A(⊗m idN) = A(⊗m idN)

and we can take c2 = 1.


To prove (iv) define I and P as in (1.6) and (1.7). Then

⊗mαXn ⊗m

β Yn

⊗m,sαsXmn+k ⊗m,s

βsYmn+k

-

-?

6

⊗m idn

I P

⊗m,s idmn+k

commutes by (1.8) and we calculate the norms of I and P to get the constantc3. In (1.4) and (1.5) we see that I is an isometry if α = ∆p and αs = ∆p,s

and that P is a norm-1-projection if β = ∆q and βs = ∆q,s. If α and β aretensor norms with equivalent s-tensor norms αs and βs, respectively, thenwe have ‖I‖ and 1

m!‖P‖ bounded by mm/m! since this is a bound for ‖σm‖

and ‖ιm‖. Thus we can take c3 = (mm)2

m!in case (a), c3 = 1 in case (d) and

c3 = mm in the cases (b) and (c).

An application of lemma 1.1 now finishes the proof.

Proposition 1.3. Let (A,A) be a normed operator ideal, α, αs, β, βs as inproposition 1.2 satisfying one of the four conditions (a)–(d) and X and Ysymmetric Banach sequence spaces. Then


βsYn) ≤ cA(id : ⊗m


for all n ≥ m where (a) c ≤(

mm

m!

)2, (b) c ≤ mm

m!, (c) c ≤ mm

m!, (d) c ≤ 1.

Proof. Note that the mapping

ιm,p :⊗m,s

p `np −→ ⊗mp `

np∑

j∈J (m,n) λjσm(ej) ∑

j∈J (m,n) λjej

is an isometry and σm,q : ⊗mq `

nq −→ ⊗m,s

q `q defined by

σm,q(ej) :=

σm(ej) j ∈ J (m,n)

0 otherwise

is a norm-1-projection. All the diagrams

(a)

⊗m,sαsXn ⊗m,s

βsYn

⊗mαXn ⊗m

β Yn

-

-?

6

⊗m idn

ιm σm

⊗m idn

(b)

⊗m,sp `np ⊗m,s

βsYn

⊗mp `

np ⊗m

β Yn

-

-?

6

⊗m idn

ιm,p σm

⊗m idn

(c)

⊗m,sαsXn ⊗m,s

q `nq

⊗mαXn ⊗m

q `nq

-

-?

6

⊗m idn

ιm σm,q

⊗m idn

(d)

⊗m,sp `np ⊗m,s

q `nq

⊗mp `

np ⊗m

q `nq

-

-?

6

⊗m idn

ιm,p σm,q

⊗m idn

Volume estimates for unit balls of spaces of m-homogeneous polynomials 31

are commutative which gives the claim since ‖ιm‖ and ‖σm‖ are bounded bymm/m!.

We have proved the following theorem.

Theorem 1.4. Let (A,A) be a normed operator ideal and α, αs, β, βs betensor product norms satisfying one of the conditions (a)–(d) of proposition1.2. Let X and Y be symmetric Banach sequence spaces. Then


βsYn) A(id : ⊗m


Volume estimates for unit balls of spaces

of m-homogeneous polynomials

For the rest of this chapter all Banach spaces are real.

Several ideas of what follows now are taken from [18], where the volumeratio of the tensor product of two `np -spaces is estimated, a result which wasoriginally obtained by Schutt [73].

By the volume volB of a Borel subsets B ⊂ RN we mean the Lebesguemeasure of B and we identify ⊗mRn = Rnm

and ⊗m,sRn = Rdn , dn =dim⊗m,sRn =

(m+n−1

n−1

)by

⊗mRn −→ Rnm∑i∈M(m,n) λiei (λi)i∈M(m,n)

,⊗m,sRn −→ Rdn∑

j∈J (m,n) λjσm(ej) (λj)j∈J (m,n)

when we write volB for a subset B ⊂ ⊗mRn or B ⊂ ⊗m,sRn.

It is well known that

(volB`np)1/n n−1/p (1.9)

for 1 ≤ p ≤ ∞ (see e.g. [64, p. 11]).

We are going to prove the following theorem.

Theorem 1.5. Let X be a symmetric Banach sequence space. Then

(volBPm(Xn)

)1/dn

(λX(n))m

n(m+1)/2X 2-convex, non trivially concave

λX(n)

nX 2-concave, non trivially convex

where dn =(

m+n−1n−1

)stands for the dimension of Pm(Xn).


In the case of `p-spaces this is for example

(volBPm(`n

p )

)1/dn

n−(m( 1

2− 1

p)+ 1

2) 2 ≤ p <∞

n−(1− 1p) 1 < p ≤ 2

Before we start working on the proof of the theorem let us collect some knownfacts about Banach sequence spaces that we will use in the following.

Let X be a Banach sequence space. Then also the Kothe dual X× is a Banachsequence space and (X×)n = (Xn)′ isometrically. For 1 ≤ p ≤ ∞ we havethat X is p-convex if and only if X× is p′-concave and in this case

M(p)(X) = M(p′)(X×) (1.10)

where p′ is the conjugate exponent satisfying 1/p + 1/p′ = 1 ([57, 1.d.4]).Our interest will be the case p = 2, so we mention that in particular

M(2)(Xn) = M(2)((Xn)′) (1.11)

Note that M(2)(Xn) ≤M(2)(X) and M(2)(Xn) ≤M(2)(X) for all n.

Now let X be symmetric. Then the fundamental functions satisfy

λX(n)λX×(n) = n (1.12)

(see e.g. [56, p. 118]) and we have for 1 < p <∞

‖id : `np −→ Xn‖

1 X p-convex

n−1/p λX(n) X p-concave(1.13)

and dually

‖id : Xn −→ `np‖

n1/p (λX(n))−1 X p-convex

1 X p-concave(1.14)

(see [31, 3.5]). In particular from [29, lemma 6.2] we get

M(2)(Xn) n1/2

λX(n) ‖id : Xn −→ `n2‖ (1.15)

whenever X is 2-convex with finite concavity and

M(2)(Xn) λX(n)

n1/2 ‖id : `n2 −→ Xn‖ (1.16)

if X is 2-concave and non trivially convex.

`-norm estimates for the injective and projective tensor product 33

`-norm estimates for the injective and pro-

jective tensor product

Let γn be the standard Gaussian probability measure on Rn with densitye−1/2

Pni=1 x2

i /(2π)n/2 and let E be a Banach space. Then L(`n2 , E) is a closedsubspace of L2(Rn, γn, E) and the induced norm is called the `-norm onL(`n2 , E), denoted by `. In other words, if g1, . . . , gn are standard Gaussianrandom variables on a probability space (Ω,Σ, µ), i.e. they are independentand all have the standard normal distribution γ1 = e−x2/2/

√2π dx, then

`(u) =(∫

Ω

∥∥ n∑i=1

gi(ω)uei

∥∥2

Eµ(dω)

)1/2

for u ∈ L(`n2 , E), and by the rotational invariance of γn the standard unitvectors e1, . . . , en can be replaced by any orthonormal basis of `n2 (see [33, p.239] for a proof of this fact).

The `-norm can be extended to operators u : E −→ F between arbitraryBanach spaces by

`(u) = sup`(uv) | n ∈ N , v ∈ L(`n2 , E) , ‖v‖ ≤ 1

and all operators u with `(u) <∞ together with ` form the normed operatorideal of Gaussian operators (see [78, p. 81]).

Remark 1.6. There is a constant c > 0 such that for all finite dimensionalBanach spaces E = (RN , ‖ · ‖)

c−1 `(id : `N2 −→ E)−1 ≤ (volBE)1/N ≤ cN−1 `(id : `N2 −→ E ′)

Proof. Let E = (RN , ‖ · ‖) be a Banach space. Then we have(volT ′BE′

volB`N2

)1/N

≤ N−1/2 `(T ) (1.17)

for all linear operators T : `N2 −→ E (see [64, remark 3.14]). Since

c−1 n−1/2 ≤ (volB`n2)1/n ≤ c n−1/2 (1.18)

for some constant c independent of n this implies the upper estimate:

(volBE)1/N ≤ vol(B`N2

)1/N N−1/2 `(id : `N2 −→ E ′) ≤ cN−1 `(id : `N2 −→ E ′)

On the other hand the converse of Santalo’s inequality by Bourgain andMilman gives a constant c′ > 0 such that for all n and all convex symmetricand bounded subsets B ⊂ Rn with non-empty interior(

volB volB

(volB`n2)2

)1/n

≥ c′


(see [64, corollary 7.2]). Thus using again (1.17) with T = id : `N2 −→ E and(1.18) we obtain

c′ ≤(

volBE

volB`N2

)1/N(volBE′

volB`N2

)1/N

≤ (volB`N2

)−1/N N−1/2 `(id : `N2 −→ E) (volBE)1/N

≤ c `(id : `N2 −→ E) (volBE)1/N

Corollary 1.7. Let (En)n be a sequence of Banach spaces (Rn, ‖ · ‖n) suchthat

`(id : `n2 −→ En) `(id : `n2 −→ E ′n) ≺ n

Then

(volBEn)1/n `(id : `n2 −→ En)−1 n−1 `(id : `n2 −→ E ′n)

Now we have Pm(Xn) = ⊗m,sεs

(Xn)′ and (Pm(Xn))′ = ⊗m,sπsXn and an esti-

mate of the form

`(id : ⊗m,s2 `n2 −→ ⊗m,s

πsXn) `(id : ⊗m,s

2 `n2 −→ ⊗m,sεs

(Xn)′) ≺ dn (1.19)

would provide us with an asymptotical description of the desired volume interms of the `-norm. Since dn nm and

`(id : ⊗m,s2 `n2 −→ ⊗m,s

αsYn) `(id : ⊗m

2 `n2 −→ ⊗m

α Yn)

for α ∈ ε, π and any Banach sequence space Y by theorem 1.4 it will beenough to work in the full tensor product. So what we are going to do is togive asymptotically correct estimates for `(id : ⊗m

2 `n2 −→ ⊗m

αXn), α = ε orπ in terms of the fundamental function λX(n). This will imply (1.19) andhence also the estimates for (volBPm(Xn))

1/dn .

We first treat the case α = ε.

Lemma 1.8. There is a constant cm > 0 such that for all n ∈ N, all Banachspaces E and all linear operators u : `n2 −→ E

‖u‖m−1 `(u) ≤ `(⊗mu : ⊗m2 `

n2 −→ ⊗m

ε E) ≤ cm ‖u‖m−1 `(u)

Proof. The result follows by induction on m if we can find a constant c > 0with

max‖u‖ `(v) , ‖v‖ `(u) ≤ `(u⊗ v : `n2 ⊗2 `n2 −→ E ⊗ε F )

≤ c (‖u‖ `(v) + ‖v‖ `(u))


for all n ∈ N, Banach spaces E, F and operators u : `n2 −→ E andv : `n2 −→ F (see also [18, p. 243]).

For the lower bound let x = x1 ∈ `n2 with norm 1 and complete to anorthonormal basis x1, . . . , xn of `n2 . Let gij, i = 1, . . . ,m, j = 1, . . . , n bea family of standard Gaussian random variables. Then by the contractionprinciple (see [33, p. 231])

`(u⊗ v) =

(∫Ω

∥∥∥ m∑i=1

n∑j=1

gij(ω)uxi ⊗ vxj

∥∥∥2

E⊗εFµ(dω)

)1/2

≥ ‖vx1‖(∫

Ω

∥∥∥ m∑i=1

gi1(ω)uxi

∥∥∥2

Eµ(dω)

)1/2

= ‖vx‖ `(u)

implying ‖v‖ `(u) ≤ `(u⊗ v) and with the same argument we get ‖u‖ `(v) ≤`(u⊗ v).

The upper bound is a consequence of Chevet’s inequality (see [78, p. 318])and Kahane’s inequality for Gaussian random variables (see e.g. [33, p.228]).

If X is a Banach sequence space with finite concavity then

`(id : `n2 −→ Xn) λX(n) (1.20)

(see [29, p. 294]). This is due to Kahane’s inequality and the fact that inthis case Gaussian and Rademacher averages can be estimated by each otherwith universal constants (see [78, p. 15 and 16]).


`(id : ⊗m2 `

n2 −→ ⊗m

ε Xn)

λX(n)

X 2-convex,non trivially concave

(λX(n))m

n(m−1)/2X 2-concave

Proof. By lemma 1.8 we have

`(id : ⊗m2 `

n2 −→ ⊗m

ε Xn) ‖id : `n2 −→ Xn‖m−1 `(id : `n2 −→ Xn)

which by (1.13) and (1.20) implies the result.

Now we look at α = π.

We recall the definition of three important constants which we use in thefollowing.


Denote rn = sgn(sin 2nπ( · )) the n-th Rademacher function and for a Banachspace E let Rn

2 (E) be En with the semi-norm ‖x‖ = ‖∑n

i=1 rixi‖L2([0,1],E) andW n

p = En with norm ‖x‖ = supϕ∈BE′‖(ϕ(xi))

ni=1‖`n

p. Now let E and F be

Banach spaces and u : E −→ F be a linear operator. For n ∈ N denoteun : En −→ F n the mapping un(x) = (u(xi))

ni=1. Recall that u has type

p/cotype p/is p-summing if the type p constant

Tp(u) = supn‖un : `np (E) −→ Rn

2 (F )‖

the cotype p constant

Cp(u) = supn‖un : Rn

2 (E) −→ `np (F )‖

and the p-summing constant

πp(u) = supn‖un : W n

p (E) −→ `np (F )‖

of u, respectively, is finite. These definitions give ideal norms (see e.g. [22] or[63]), in particular we can apply theorem 1.4 for A = Tp,Cp, πp. A Banachspace is said to have type p/cotype p/to be p-summing if the identity opertoridE has this property, and one defines A(E) := A(idE) for A = Tp,Cp, πp.In the thesis of Sevilla-Peris [74] the general concept of the A-property isdefined: Given a Banach operator ideal (A,A), a Banach space E is saidto have the A-property, if idE ∈ A, and A(E) := A(idE) is called the A-constant of E.

We will use the following general inequality for the `-norm. There is a con-stant c > 0 such that for all Banach spaces E and every operator u : `2 −→ E

`(u) ≤ cT2(E)π2(u′) (1.21)

holds (see [78, p. 83 and theorem 25.1]).

First we give an asymptotic description of the type 2 constant of ⊗mπ Xn which

is dual to that in [29]. There it is shown that

C2(⊗mε Xn) n(m−1)/2 C2(Xn) (1.22)

holds for every symmetric Banach sequence space which is 2-concave or 2-convex and non trivially concave.

Recall that for Banach sequence spaces (more generally for Banach lattices)there is a close relation between the notions of type 2/cotype 2 and 2-convexity/2-concavity. A Banach sequence space X has type 2 if and onlyif it is 2-convex and non trivially concave (see [33, 16.20]) and it has cotype2 if and only if it is 2-concave ([57, 1.d.6]). From [57, 1.d.6] and Kahane’sinequality it follows that

T2(Xn) M(2)(Xn) (1.23)


andC2(Xn) M(2)(Xn) (1.24)

if X is non trivially concave.

Theorem 1.10. Let X be a symmetric Banach sequence space, 2-convex andnon trivially concave or 2-concave and non trivially convex. Then

T2(⊗mπ Xn) n(m−1)/2 T2(Xn)

Proof. The proof is based on duality. For the lower bound we use (1.22) andthe fact that

C2(E) ≤ T2(E′) (1.25)

for every Banach space E (see e.g. [22, p. 106]). For X we even have(asymptotically) the full duality

T2(Xn) M(2)(Xn) M(2)((Xn)′) C2((Xn)′)

and together with (1.22) we obtain

n(m−1)/2 T2(Xn) C2(⊗mε (Xn)′) = C2((⊗m

π Xn)′) ≤ T2((⊗mπ Xn)′)

by the duality of ε and π.

Now we prove the upper estimate. An inequality reverse to (1.25) is ingeneral not true (see for example [22, p. 105]), and we are going to factorizethrough suitable spaces [Xn]m and [Xn]m which permit an appropriate reverseestimate.

First assume that X is 2-convex and non trivially concave. As in [29] wedefine inductively the space [X]m of all functions f : Nm −→ R satisfyingf(i, ·) ∈ X for all i ∈ Nm−1 and (‖f(i, ·)‖X)i∈Nm−1 ∈ [X]m−1 with norm

‖f‖[X]m = ‖(‖f(i, ·)‖X)i∈Nm−1‖[X]m−1

which satisfies all conditions of a Banach sequence space (choose an order ofNm) except perhaps the maximality condition. But (1.11), (1.23) and (1.24)still hold for this space (since it is a Banach lattice with finite concavity: Byinduction it is easy to check that

M(p)([X]m) ≤M(p)(X)m

is true).

For the subspace [Xn]m of all functions f ∈ [X]m which are zero outside1, . . . , nm we have ([Xn]m)′ = [(X×)n]m isometrically and from [29, p.304] we get

C2([(X×)n]m) ≺ M(2)((X

×)n)m


hence together with (1.23), (1.11) and (1.24)

T2([Xn]m) M(2)([Xn]m) M(2)(([Xn]m)′)

C2([(X×)n]m) ≺M(2)((X

×)n)m M(2)(Xn)m(1.26)

This implies T2([Xn]m) 1 since X is 2-convex. Note that we have ⊗mXn =[Xn]m naturally by x1⊗· · ·⊗xm f with f(i) = x1(i1) · · ·xm(im) and [Xn]m

induces a norm α on ⊗mXn satisfying

α(x1 ⊗ · · · ⊗ xm) = ‖x1‖ · · · ‖xm‖

and hence α ≤ π. Thus factorizing the identity id : ⊗mπ Xn −→ ⊗m

π Xn

through [Xn]m we obtain

T2(⊗mπ Xn) ≤ ‖id : ⊗m

π Xn −→ [Xn]m‖T2([Xn]m) ‖id : [Xn]m −→ ⊗mπ Xn‖

≺ ‖id : [Xn]m −→ ⊗mπ Xn‖

Now in [29, lemma 5.5] it is proved that

‖id : ⊗mε Yn −→ [Yn]m‖ ≤ Km−1

G M(2)(Yn)m−1 n(m−1)/2

for every Banach sequence space Y , where KG denotes the Grothendieckconstant. Hence by duality

‖id : [Xn]m −→ ⊗mπ Xn‖ = ‖id : ⊗m

ε (X×)n −→ [(X×)n]m‖ ≺ n(m−1)/2

and thus T2(⊗mπ Xn) ≺ n(m−1)/2 n(m−1)/2 T2(Xn) since X has type 2.

Now letX be 2-concave and non trivially convex. Then we use the space [X]mof functions f : Nm −→ R satisfying f(·, i) ∈ X and (‖f(·, i)‖X)i∈Nm−1 ∈[X]m−1 with norm

‖f‖[X]m = ‖(‖f(·, i)‖X)i∈Nm−1‖[X]m−1

By induction we see that [X]m is 2-concave with

M(2)([X]m) ≤M(2)(X)m

As above we have ([Xn]m)′ = [(X×)n]m isometrically for the subspace [Xn]mof all functions f ∈ [X]m which are zero outside 1, . . . , nm and from [29,p. 307] we get

C2([(X×)n]m) ≺ M(2)((X

×)n)m

Again using (1.23), (1.11) and (1.24) we obtain

T2([Xn]m) M(2)([Xn]m) M(2)(([Xn]m)′)

C2([(X×)n]m) ≺M(2)((X

×)n)m


and with [29, lemma 6.2] it follows

T2([Xn]m) ≺M(2)((X×)n)

(n1/2

λX×(n)

)m−1

(1.27)

Here we use that X is 2-concave with non trivial convexity. Again we have⊗mXn = [Xn]m naturally by x1⊗ · · ·⊗xm f with f(i) = x1(i1) · · ·xm(im)and the norm β induced on ⊗mXn satisfies

β(x1 ⊗ · · · ⊗ xm) = ‖x1‖ · · · ‖xm‖

implying β ≤ π. Hence factorizing through [Xn]m gives

T2(⊗mπ Xn) ≤ T2([Xn]m) ‖id : [Xn]m −→ ⊗m

π Xn‖

and since‖id : ⊗m

ε Yn −→ [Yn]m‖ ≤ λY (n)m−1

for any Banach sequence space Y by [29, lemma 6.4], using (1.27), duality,(1.11) and (1.23) we conclude

T2(⊗mπ Xn) ≺M(2)((X

×)n)

(n1/2

λX×(n)

)m−1

λX×(n)m−1 n(m−1)/2 T2(Xn)

In view of (1.21) our next aim is to calculate the 2-summing normπ2(id : ⊗m

ε Xn −→ `nm

2 ).

The following definition we have taken from [30, p. 10]. By O(n) we denotethe orthogonal group.

Definition 1.11. An n-dimensional Banach space E = (Rn, ‖ · ‖) is said tohave enough symmetries in O(n) if there is a compact subgroup G of O(n)such that

(1) Every g ∈ G is an isometry on E.

(2) Every linear operator u : Rn −→ Rn which commutes with G is amultiple of the identity.

If E = (Rn, ‖ · ‖) has enough symmetries in O(n) then

EE = ‖id : `n2 −→ E‖−1B`n2

(1.28)

for the ellipsoid EE of maximal volume in BE (see e.g. [30, p. 11]) and inparticular

d(`n2 , E) = ‖id : `n2 −→ E‖ ‖id : E −→ `n2‖ (1.29)

since in this case the Banach Mazur distance to the Hilbert space equals‖id : E −→ Emax‖ where Emax = Rn with the norm generated by EE (see [9,lemma 4.6]). Another consequence of (1.28) is the following lemma whichcan be found in [30, lemma 3].


Lemma 1.12. Let E = (Rn, ‖ · ‖1) and F = (Rn, ‖ · ‖2) be n-dimensionalBanach spaces with enough symmetries in O(n). Then

π2(id : E −→ F ) = n1/2‖id : `n2 −→ F‖‖id : `n2 −→ E‖

We will prove an analogue for tensor products (proposition 1.14) which isalready included in [74]. This is done for the convenience of the reader andalso with a different proof, which is shorter and for us seems to be moretransparent.

Let S(Rn) be the group of bijections generated by

Mε : Rn −→ Rn , x n∑

k=1

εkxkek , ε ∈ −1, 1n

Tσ : Rn −→ Rn , x n∑

k=1

xσ(k)ek , σ ∈ Πn

where Πn are the permutations on 1, . . . , n. Then

S(⊗mRn) = ⊗mS(Rn) = S1 ⊗ · · · ⊗ Sm | Si ∈ S(Rn)

is a group of bijections on ⊗mRn. If X is a symmetric Banach sequencespace then all S ∈ S(Rn) are isometries on Xn. A norm α on ⊗mXn is calledsymmetrically invariant if all symmetries S ∈ S(⊗mRn) are isometries on⊗m

αXn. Every m-fold tensor norm α is symmetrically invariant on ⊗mXn:For all S = S1⊗ · · · ⊗ Sm ∈ S(⊗mRn) we have, writing E = ⊗m

αXn for short

‖S ′ : E ′ −→ E ′‖ = ‖S : E −→ E‖ =m∏

i=1

‖Si : Xn −→ Xn‖ = 1

by the metric mapping property of α and hence also

‖(S ′)−1‖L(E′) = ‖(S−1)′‖L(E′) = 1

This implies S ′BE′ = BE′ , thus

‖Sz‖E = supz′∈BE′

|〈z′, Sz〉| = supz′∈BE′

|〈S ′z′, z〉| = ‖z‖E

A proof of the following result can be found in [29]. We use a differentargument from [42, p. 35] which was given there for m = 2 and can bedirectly applied to the general case.

Lemma 1.13. Let X be a symmetric Banach sequence space and α be asymmetrically invariant norm on ⊗mXn. Then ⊗m

αXn = (Rnm, ‖ · ‖α) has

enough symmetries in O(nm), the identification being given by

I :⊗mXn −→ Rnm∑λi ei (λi)

and ‖ · ‖α = α (I−1( · )).


Proof. Every S ∈ S(Rn) can be written as S = MεTσ with some ε ∈ −1, 1nand σ ∈ Πn, hence in the same way S = MεTσ for S ∈ S(⊗mRn) with ε =(ε1, . . . , εm) ∈ (−1, 1n)m and σ = σ1 × · · · × σm a bijection on 1, . . . , nmwith σi ∈ Πn, abbreviating Mε = Mε1 ⊗ · · · ⊗Mεm and Tσ = Tσ1 ⊗ · · · ⊗Tσm .Note that Mεei = ε1(i1) · · · εm(im) ei and

Tσ(ei) = eσ−1(i1) ⊗ · · · ⊗ eσ−1(im) = eσ−1(i)

for i ∈ 1, . . . , nm. Thus we have

S( ∑

i∈M(m,n)

λiei

)=

∑i∈M(m,n)

ε1(i1) · · · εm(im)λσ(i) ei

which implies that S is an isometry on ⊗m2 `

n2 and by the assumption on α,

S is also an isometry on ⊗mαXn.

Now assume that the linear operator u : ⊗mRn −→ ⊗mRn commutes witheach element of S(⊗mRn). Then in particular

〈e′i, uej〉 = 〈e′i,MεuMεej〉 = 〈e′i, uej〉 ε1(i1) · · · εm(im) ε1(j1) · · · εm(jm)

and〈e′i, uej〉 = 〈e′i, TσuTσ−1ej〉 = 〈T ′σe′i, uTσ−1ej〉 = 〈e′σ(i), ueσ(j)〉

for all i, j ∈ M(m,n), ε = (ε1, . . . , εm) ∈ (−1, 1n)m and products σ =σ1 × · · · × σm of permutations σi ∈ Πn, where e′i = e′i1 ⊗ · · · ⊗ e′im , i ∈M(m,n) are the coefficient functionals of the basis ei, i ∈M(m,n) of ⊗mRn,generated from the coefficient functionals e′1, . . . , e

′n of e1, . . . , en. This implies

〈e′i, uej〉 = 0 for i 6= j and c := 〈e′1, ue1〉 = 〈e′i, uei〉 for all i ∈ M(m,n) sincewe can choose σ1, . . . , σm ∈ Πn independently. Thus u = c id⊗mRn and

G := ISI−1 | S ∈ S(⊗mRn)

is the requested subgroup of O(nm).

Lemma 1.13 in particular proves that (1.28), (1.29) and lemma 1.12 hold forcertain tensor products. The first two statements of the next proposition willbe used in section 4.

Proposition 1.14. Let X and Y be symmetric Banach sequence spaces andα, β be symmetrically invarivant norms on ⊗mXn and ⊗mYn, respectively.Then

(1) E⊗mα Xn = ‖id : `n

m

2 −→ ⊗mαXn‖−1B`nm

2

(2) d(`nm

2 ,⊗mαXn) = ‖id : `n

m

2 −→ ⊗mαXn‖ ‖id : ⊗m

αXn −→ `nm

2 ‖

(3) π2(id : ⊗mαXn −→ ⊗m

β Yn) = nm/2‖id : `n

m

2 −→ ⊗mβ Yn‖

‖id : `nm

2 −→ ⊗mαXn‖


It is not hard to estimate the norm of the identity id : `nm

2 −→ ⊗mε Xn.

Remark 1.15. Let X be a symmetric Banach sequence space. Then

‖id : `nm

2 −→ ⊗mε Xn‖

1 X 2-convex

(λX(n)/n1/2)m X 2-concave

Proof. By factorization and the metric mapping property of ε we have

‖id : `nm

2 −→ ⊗mε Xn‖ ≤ ‖id : ⊗m

ε `n2 −→ ⊗m

ε Xn‖ = ‖id : `n2 −→ Xn‖m

= sup‖x‖`n

2≤1

‖x‖mXn= sup

‖x‖`n2≤1

‖⊗mx‖⊗mε Xn

≤ ‖id : ⊗m2 `

n2 −→ ⊗m

ε Xn‖ sup‖x‖`n

2≤1

‖⊗mx‖⊗m2 `n

2

= ‖id : `nm

2 −→ ⊗mε Xn‖

hence ‖id : `nm

2 −→ ⊗mε Xn‖ = ‖id : `n2 −→ Xn‖m and a glance at (1.16)

finishes the proof.

Now we are prepared to give our `-norm estimate for the projective tensorproduct.


`(id : ⊗m2 `

n2 −→ ⊗m

π Xn)

n(m−1)/2 (λX(n))m X 2-convex,

non trivially concave

nm−1 λX(n)X 2-concave,non trivially convex

Proof. Using (1.21), theorem 1.10 and proposition 1.14 we obtain

`(id : `nm

2 −→ ⊗mπ Xn) ≤ T2(⊗m

π Xn)π2(id : ⊗mε (X×)n −→ `n

m

2 )

n(m−1)/2 T2(Xn)nm/2 ‖id : `nm

2 −→ ⊗mε (X×)n‖−1

which implies the upper estimate, since

T2(Xn)

1 X 2-convex, non trivially concave

λX(n)/n1/2 X 2-concave, non trivially convex

by (1.23) and (1.16) and

‖id : `nm

2 −→ ⊗mε (X×)n‖−1

(λX(n)/n1/2)m X 2-convex

1 X 2-concave


by remark 1.15, (1.12) and the remark following it.

Then with the help of theorem 1.9 we get

`(id : `nm

2 −→ ⊗mπ Xn) `(id : `n

m

2 −→ ⊗mε (Xn)′) ≺ nm

and this by corollary 1.7 and theorem 1.9 finishes the proof.

Now by theorems 1.9 and 1.16 and the duality of ε and π it follows

`(id : `nm

2 −→ ⊗mαXn) `(id : `n

m

2 −→ (⊗mαXn)′) nm (1.30)

for α = ε or π if X is 2-convex with non trivial concavity or 2-concave withnon trivial convexity, hence(

volB⊗mα Xn

)1/nm

`(id : `nm

2 −→ ⊗mαXn)−1

by corollary 1.7 and together with the remarks that follow it we have provedthe following result.

Theorem 1.17. Let X be a symmetric Banach sequence space and α = ε orπ. Then(

volB⊗m,sαs Xn

)1/dn (volB⊗m

α Xn

)1/nm

λX(n)−1 X 2-convex,


n(m−1)/2/(λX(n))m X 2-concave,non trivially convex

α = ε

(n(m−1)/2(λX(n))m)−1 X 2-convex,


(nm−1 λX(n))−1 X 2-concave,non trivially convex

α = π

This in particular implies theorem 1.5.

Example 1.18. Let 1 < p <∞.

(a)(volB⊗m,s

εs `np

)1/dn (volB⊗m

ε `np

)1/nm

nm( 1

2− 1

p)− 12 p ≤ 2

n−1p p ≥ 2

(b)(volB⊗m,s

πs `np

)1/dn (volB⊗m

π `np

)1/nm

n1− 1

p−m p ≤ 2

n12−( 1

2+ 1

p)m p ≥ 2


(c) `(id : ⊗m,s2 `n2 −→ ⊗m,s

εs`np ) `(id : ⊗m

2 `n2 −→ ⊗m

ε `np )

nm( 1

p− 1

2)+ 12 p ≤ 2

n1p p ≥ 2

(d) `(id : ⊗m,s2 `n2 −→ ⊗m,s

πs`np ) `(id : ⊗m

2 `n2 −→ ⊗m

π `np )

nm−1+ 1

p p ≤ 2

n( 12+ 1

p)m− 12 p ≥ 2

(e) T2(⊗mπ `

np ) n(m−1)/2 T2(`

np )

n

m2

+ 1p−1 p ≤ 2

nm−1

2 p ≥ 2

Volume ratio of m-fold tensor products

The calculations of the last section permit to give asymptotically correctestimates for the volume ratio of the m-fold injective and projective tensorproduct.

The volume ratio of a Banach space E = (Rn, ‖ · ‖) can be defined by

vr(E) =

(volBE

vol EE

)1/n

where EE is the ellipsoid of maximal volume in BE. See [64], [78] or [33] formore information.

Remark 1.19. Let (En)n be a sequence of Banach spaces (Rn, ‖ · ‖n) withenough symmetries in O(n) such that

`(id : `n2 −→ En) `(id : `n2 −→ E ′n) ≺ n

Then

vr(En) n1/2 ‖id : `n2 −→ En‖`(id : `n2 −→ En)

Proof. We have

vr(En) = n1/2 ‖id : `n2 −→ En‖ (volBEn)1/n n1/2 ‖id : `n2 −→ En‖`(id : `n2 −→ En)

by (1.28), (1.9) and corollary 1.7.

Lemma 1.13 and (1.30) then imply

Volume ratio of m-fold tensor products 45

Corollary 1.20. Let X be a symmetric Banach sequence space which is 2-convex with non trivial concavity or 2-concave with non trivial convexity.Then

vr(⊗mαXn) nm/2 ‖id : `n

m

2 −→ ⊗mαXn‖

`(id : `nm

2 −→ ⊗mαXn)

for α ∈ ε, π.

In particular, using (1.15), (1.16) and (1.20) we get

vr(Xn)

n1/2 λX(n) X 2-convex, non trivially concave

1 X 2-concave, non trivially convex

Thus to express vr(⊗mαXn) asymptotically in terms of the fundamental func-

tion λX(n) it remains to prove an appropriate estimate for‖id : `n

m

2 −→ ⊗mπ Xn‖, since we already have theorems 1.9 and 1.16 and

remark 1.15. The desired estimate follows from the next result by duality.

Proposition 1.21. Let X be a symmetric Banach sequence space. Then

‖id : ⊗mε Xn −→ `n

m

2 ‖

nm/2

λX(n)X 2-convex, non trivially concave

nm−1/2

(λX(n))mX 2-concave, 2m

2m−1-convex

Proof. By (1.22) and factorization we have

n(m−1)/2 C2(Xn) = C2(⊗mε Xn)

≤ ‖id : ⊗mε Xn −→ `n

m

2 ‖C2(`nm

2 ) ‖id : `nm

2 −→ ⊗mε Xn‖

= ‖id : ⊗mε Xn −→ `n

m

2 ‖ ‖id : `nm

2 −→ ⊗mε Xn‖

This by remark 1.15, (1.15) and (1.24) gives the lower estimates.

If X is 2-convex then for all λ ∈ `nm

2

‖λ‖`nm2

=(∑

i1

∑(i2,...,im)

|λi|2)1/2

≤ n1/2 maxi1

( ∑(i2,...,im)

|λi|2)1/2

≤ n(m−1)/2 ‖id : Xn −→ `n2‖ maxi1,...,im−1

‖(λ(i1,...,im))nim=1‖Xn

≤ n(m−1)/2 ‖id : Xn −→ `n2‖∥∥∥∑λi ei

∥∥∥⊗m

ε Xn

which implies the upper bound in case of finite concavity (for m = 2 andX = `p this argument was used in [72, p. 126]).


Now let pm := 2m2m−1

and suppose X to be 2-concave and pm-convex. Thenby the multilinear version of a Hardy-Littlewood theorem [45] by Praciano-Pereira [66] we have

‖id : ⊗mε `

npm−→ `n

m

2 ‖ 1

hence factoring through ⊗mε `

npm

and using (1.14) we obtain

‖id : ⊗mε Xn −→ `n

m

2 ‖ ≺ ‖id : Xn −→ `npm‖m

(n1−1/2m

λX(n)

)m

Recall that we have

d(`nm

2 ,⊗mε Xn) = ‖id : `n

m

2 −→ ⊗mε Xn‖ ‖id : ⊗m

ε Xn −→ `nm

2 ‖ (1.31)

by proposition 1.14. Hence in passing we get from proposition 1.21 andremark 1.15 the following asymptotic description for the Banach Mazur dis-tance of ⊗m

ε Xn to the Hilbert space.

Corollary 1.22. Let X be a symmetric Banach sequence space. Then

d(`nm

2 ,⊗mε Xn)

nm/2/λX(n) X 2-convex, non trivially concave

n(m−1)/2 X 2-concave, 2m2m−1

-convex

But from here it’s only a quick step to get the asymptotic of the Banach-Mazur distance of the polynomials Pm(Xn) to the Hilbert space, which wewill take before we finish our volume ratio estimates.

Recall that a linear operator T : E −→ F between Banach spaces is called2-factorable, if it factors through some space L2(µ):

E T - FJ

JJJ

S R

L2(µ)

(1.32)

with S ∈ L(E,L2(µ)), R ∈ L(L2(µ), F ). Then γ2(T ) := inf‖R‖‖S‖ takenover all factorization operators as in (1.32) defines a norm on the spaceΓ2(E,F ) of all 2-factorable operators and makes it a Banach space. We havethat (Γ2, γ2) is a normed operator ideal (see e.g. [78, p. 95]).

From proposition 13.9 in [78, p. 103] we get

γ2(E) := γ2(idE) = d(`n2 , E)

if E is n-dimensional. Hence

d(`dn2 ,⊗m,s

εsXn) d(`n

m

2 ,⊗mε Xn) (1.33)

by theorem 1.4 since γ2 is an ideal norm (dn =(

m+n−1n−1

)is the dimension of

the space ⊗m,sεsXn). This proves


Corollary 1.23. Let X be a symmetric Banach sequence space. Then

d(`dn2 ,Pm(Xn))

n(m−1)/2 X 2-convex, 2m-concave

nm/2−1 λX(n) X 2-concave, non trivially convex

Now we have all the ingredients to give the desired volume ratio estimates.

Theorem 1.24. Let X be a symmetric Banach sequence space.

(1) If X is 2-convex with non trivial concavity or 2-concave with non trivialconvexity, then

vr(⊗mε Xn) n(m−1)/2 vr(Xn)

(2) If X is 2-convex and 2m-concave or 2-concave and non trivially convex,then

vr(⊗mπ Xn) 1

Note that in particular

vr(⊗mε Xn) d(`n

m

2 ,⊗mε Xn)

if X is 2-convex and non trivially concave or 2-concave and 2m2m−1

-convex.

Proof. We use corollary 1.20. Together with remark 1.15 and theorem 1.9we get

vr(⊗mε Xn)

nm/2/λX(n) X 2-convex, non trivially concave

n(m−1)/2 X 2-concave, non trivially convex

n(m−1)/2 vr(Xn)

In case of the projective norm by duality and proposition 1.21

‖id : `nm

2 −→ ⊗mπ Xn‖ = ‖id : ⊗m

ε (X×)n −→ `nm

2 ‖

(λX(n))m/n1/2 X 2-convex, 2m-concave

nm/2−1 λX(n) X 2-concave, non trivially convex

which together with theorem 1.16 gives the desired estimate.

Finally we are able to give an upper estimate for the volume ratio of thesymmetric injective and projective tensor product.

Remark 1.25. Let X be a symmetric Banach sequence space, 2-convex andnon trivially concave or 2-concave and non trivially convex. Then for α = εor π we have

vr(⊗m,sαsXn) ≺ vr(⊗m

αXn)


Proof. Define In := id : `dn2 −→ ⊗m,s

αsXn (with dn =

(m+n−1

n−1

)the dimension of

⊗m,sαsXn) and vn := ‖In‖−1 In. Then vn(B`dn

2) =: En is an ellipsoid contained

in the unit ball of ⊗m,sαsXn and hence

vr(⊗m,sαsXn) ≤

(volB⊗m,sαs Xn

)1/dn

(vol En)1/dn

= ‖id : `dn2 −→ ⊗m,s

αsXn‖

(volB⊗m,sαs Xn

)1/dn

(volB`dn2

)1/dn

‖id : `nm

2 −→ ⊗mαsXn‖

(volB⊗mα Xn)1/nm

(volB`nm2

)1/nm

vr(⊗mαXn)

where the first asymptotic equality follows from theorems 1.4 and 1.17 andthe second one from proposition 1.14.

Example 1.26. Let 1 ≤ p ≤ ∞.

(a) ‖id : ⊗mε `

np −→ `n

m

2 ‖

1 1 ≤ p ≤ 2m

2m−1

nm(1− 1p)− 1

2 2m2m−1

≤ p ≤ 2

nm2− 1

p 2 ≤ p ≤ ∞

(b) d(`nm

2 ,⊗mε `

np ) d(⊗m,s

2 `n2 ,⊗mεs`np )

nm( 1

p− 1

2) 1 ≤ p ≤ 2m2m−1

nm−1

22m

2m−1≤ p ≤ 2

nm2− 1

p 2 ≤ p ≤ ∞

(c) d(`dn2 ,Pm(`np )) d(⊗m,s

2 `n2 ,⊗mπs`np ) d(⊗m

2 `n2 ,⊗m

π `np )

n

m2

+ 1p−1 p ≤ 2

nm−1

2 2 ≤ p ≤ 2m

nm( 12− 1

p) p ≥ 2m

where dn =(

m+n−1n−1

)denotes the dimension of Pm(`np )

(d) vr(⊗mε `

np )

nm−1

2 1 < p ≤ 2

nm2

+ 1p 2 ≤ p <∞

(e) vr(⊗mπ `

np ) 1 for 1 < p ≤ 2m


(f) vr(Pm(`np )) ≺ vr(⊗mε `

np′)

nm2− 1

p+1 1 < p ≤ 2

nm−1

2 2 ≤ p <∞

Proof. These are applications of the results given in this section. For theremaining cases in (a), (b) and (c) by (1.31) and (1.33) it is enough toconsider 1 ≤ p ≤ pm := 2m

2m−1in (a). Factorization gives

‖id : ⊗mε `

np −→ `n

m

2 ‖ ≤ ‖id : ⊗mε `

np −→ ⊗m

ε `npm‖ ‖id : ⊗m

ε `npm−→ `n

m

2 ‖

‖id : `np −→ `npm‖m nm(1− 1

pm)− 1

2 = 1

2. Domains of convergence

It follows the main part of the thesis. First we define the domain of con-vergence for holomorphic functions on Reinhardt domains and we motivatethis definition by some examples. Then we prove lower and upper inclusionsfor the domain of convergence of the bounded holomorphic functions on abounded Reinhardt domain in a symmetric Banach sequence space. For thelower inclusion we use arguments of Bohr (see introduction, theorem 0.1) anda generalization of the results of Ryan [70] and Lempert [53] for holomorphicfunctions on multiples of the unit ball in `1 to Reinhardt domains. To obtainthe upper bound we define and estimate the arithmetic Bohr radius and useideas from the work of Dineen and Timoney [38]. Then we investigate thedomain of convergence of the space of m-homogeneous polynomials and, afterthat, of all polynomials on a Banach sequence space. We examine the specialrole of the space `1 in our theory and answer a question of Lempert. Finallywe prove a direct link between domains of convergence and unconditionalityin spaces of homogeneous polynomials.

Domains of convergence of holomorphic

functions

Let R be a Reinhardt domain in Cn. Then every holomorphic function f onR has a power series expansion which converges to f in every point of R.More precisely, for every f ∈ H(R) we can find a unique family of scalars(cα)α∈Nn

0such that

f(z) =∑α∈Nn

0

cαzα

for every z ∈ R, and the coefficients can be calculated by

cα =∂αf(0)

α!=

(1

2πi

)n ∫[|z|=r]

f(z)

zα+1dz

where r ∈ Rn>0 such that the polydisc [|z| ≤ r] is contained in R.

52 2. Domains of convergence

Now let f be a holomorphic function on a Reinhardt domain R in a Banachsequence space X. Then f has a power series expansion

∑α∈Nn

0c(n)α zα on

every finite dimensional component Rn = R ∩Xn of R with

c(n)α =

(1

2πi

)n ∫|z1|=r1

· · ·∫|zn|=rn

f(z1, . . . , zn, 0, . . .)

zα1+11 · · · zαn+1

n

dz1 · · · dzn

=

(1

2πi

)n+1 ∫|z1|=r1

· · ·∫|zn+1|=rn+1

f(z1, . . . , zn+1, 0, . . .)

zα1+11 · · · zαn+1

n zn+1

dz1 · · · dzn+1

= c(n+1)α

for α ∈ Nn0 ⊂ Nn+1

0 . Thus we can find a unique family (cα)α∈N(N)

0such that

f(z) =∑

α∈N(N)0

cαzα

for all z ∈ Rn and all n ∈ N, i.e. on spanen∩R = ∪nRn. The power series∑cαz

α is called the monomial expansion of f . We write cα(f) := cα for themonomial coefficients of f . They satisfy

cα(f) =∂αf(0)

α!=

(1

2πi

)n ∫[|z|=r]

f(z)

zα+1dz (2.1)

if α ∈ Nn0 and [|z| ≤ r] ⊂ Rn and hence are continuous linear functionals on

the Banach space H∞(R) of all bounded holomorphic functions on R withthe sup-norm ‖ · ‖R. In particular, from (2.1) we deduce

‖cα(f)zα‖R ≤ ‖f‖R (2.2)

for all α ∈ N(N)0 (Cauchy inequalities for monomials).

Where does the monomial expansion series of a holomorphic function f on aReinhardt domain R converge? Do we also have

f(z) =∑

α∈N(N)0

cα(f)zα (2.3)

for all z ∈ R as in the finite dimensional situation?

We have

f(n∑

k=1

zkek) =∑α∈Nn

0

cα(f)zα

for every n ∈ N and every z ∈ R. Thus if X is a Banach sequence space withbasis and the monomial expansion of f converges in some point z ∈ R thenit converges to f(z).

Before we give some examples, let us make an observation concerning theconnection between the Taylor series expansion and the monomial series ex-pansion of a holomorphic function on a Reinhardt domain.

Domains of convergence of holomorphic functions 53

Remark 2.1. Let f be a holomorphic function on a Reinhardt domain R inCn. Then

Pmf(0)(z) =∑|α|=m

cα(f)zα

for every z ∈ Cn. In particular, for all z ∈ R the Taylor series∑

m Pmf(0)(z)converges absolutely and

f(z) =∞∑

m=0

Pmf(0)(z)

Proof. Define the m-homogeneous polynomials

Pm(z) =∑|α|=m

cα(f)zα

on Cn. Then

∞∑m=0

|Pm(z)| ≤∞∑

m=0

∑|α|=m

|cα(f)zα| =∑α∈Nn

0

|cα(f)zα| <∞ (2.4)

for z ∈ R. On the other hand, we have

f(z) =∞∑

m=0

Pmf(0)(z)

in a zero neighbourhood U ⊂ R, where the Taylor series∑

m Pmf(0) con-verges uniformly and, in particular, absolutely on U . Hence from

∞∑m=0

Pmf(0)(z) = f(z) =∑α∈Nn

0

cα(f)zα =∞∑

m=0

Pm(z) (2.5)

for z ∈ U it follows Pm = Pmf(0) (see [19, 11.10]) and now (2.4) tells us that∑m Pmf(0) converges absolutely for every z ∈ R with

f(z) =∑α∈Nn

0

cα(f)zα =∞∑

m=0

Pmf(0)(z)

We can deduce an analogue in infinite dimensions.

Corollary 2.2. Let f be a holomorphic function on a Reinhardt domain Rin a Banach sequence space X. Then

Pmf(0)(z) =∑|α|=m

cα(f)zα


for every z ∈ spanen. In particular, for all z ∈ spanen ∩ R the Taylorseries

∑m Pmf(0)(z) converges absolutely and

f(z) =∞∑

m=0

Pmf(0)(z)

Proof. By remark 2.1 it is enough to observe that for example from theCauchy integral formulas (0.13) we can see that

Pmf |Rn(0) = Pmf(0)

on every finite dimensional component Rn. The “in particular” then followsas in the proof of remark 2.1 by the convergence of the monomial expansionon spanen ∩R.

Corollary 2.2 says nothing but

cα(Pmf(0)) = cα(f) (2.6)

for every f ∈ H(R), m ∈ N0 and α ∈ N(N)0 with |α| = m.

Now we give some examples showing that the monomial expansion of a holo-morphic function on a Reinhardt domain R in a Banach sequence space Xdoes in general not converge in every point of R.

The following examples for `p-spaces are constructed with the help of specialmatrices.

Remark 2.3. Let m ≥ 2 and A = (ars)1≤r,s≤N be a matrix with complexcoefficients such that

(a)∑N

t=1 art ast = Nδrs

(b) |ars| = 1

Let

ϕm(x1, . . . , xm) =N∑

i1,...,im=1

ai1i2 · · · aim−1 imx1(i1) · · ·xm(im)

for xj ∈ CN , j = 1, . . . ,m, be the m-linear mapping on (CN)m defined by A.Then

(1) ‖ϕm‖Lm(`N∞) ≤ N

m+12

(2) ‖ϕm‖Lm(`N2 ) ≤

N

m4 m even

Nm+1

4 m odd

Part (1) including its proof is taken from [12, p. 609].


Proof. (1) Note that

‖ϕm‖Lm(`N∞) ≤ N1/2‖ϕm‖Lm(`N

∞×···×`N∞×`N

2 )

and we prove

‖ϕm‖Lm(`N∞×···×`N

∞×`N2 ) ≤ N

m2

by induction on m. For m = 2 and x1 ∈ B`N∞

, x2 ∈ B`N2

we have

|ϕm(x1, x2)| ≤( N∑

i2=1

∣∣ N∑i1=1

ai1i2x1(i1)∣∣2)1/2

=( N∑

i2=1

N∑i1,j1=1

ai1i2aj1i2x1(i1)x1(j1))1/2

=( N∑

i1,j1=1

x1(i1)x1(j1)N∑

i2=1

ai1i2aj1i2

)1/2= N1/2

( N∑i1=1

|x1(i1)|2)1/2

≤ N

Here the first inequality is that of Cauchy-Schwarz and the last equality isassumption (a).

Now assume m ≥ 3 and let x1, . . . , xm−1 ∈ B`N∞

and xm ∈ B`N2

. Then

|ϕm(x1, . . . , xm)|

≤( N∑

im=1

∣∣ N∑i1,...,im−1=1

ai1i2 · · · aim−1imx1(i1) · · ·xm−1(im−1)∣∣2)1/2

=

( N∑i1, . . . , im−1

j1,...,jm−1=1

ai1i2aj1j2 · · · aim−2im−1ajm−2jm−1x1(i1) · · ·xm−1(jm−1)

n∑im=1

aim−1imajm−1im

)1/2

= N1/2

( N∑im−1=1

∣∣ N∑i1,...,im−2=1

ai1i2 · · · aim−2im−1x1(i1) · · ·xm−2(im−2)∣∣2

|xm−1(im−1)|2)1/2

≤ N1/2

( N∑im−1=1

∣∣ N∑i1,...,im−2=1

ai1i2 · · · aim−2im−1x1(i1) · · ·xm−2(im−2)∣∣2)1/2

≤ N1/2 ‖ϕm−1‖Lm−1(`N∞×···×`N

∞×`N2 ) ≤ N

m2


(2) Again we proceed by induction. If m = 2 and x1, x2 ∈ B`N2

we have

|ϕm(x1, x2)| ≤( N∑

i2=1

∣∣ N∑i1=1

ai1i2x1(i1)∣∣2)1/2

=( N∑

i2=1

N∑i1,j1=1

ai1i2aj1i2x1(i1)x1(j1))1/2

=( N∑

i1,j1=1

x1(i1)x1(j1)N∑

i2=1

ai1i2aj1i2

)1/2

= N1/2( N∑

i1=1

|x1(i1)|2)1/2

≤ N1/2

again with the Cauchy-Schwarz inequality and assumption (a). If m = 3 andx1, x2, x3 ∈ B`N

2then in a similar way

|ϕm(x1, x2, x3)| ≤( N∑

i3=1

∣∣ N∑i1,i2=1

ai1i2ai2i3x1(i1)x2(i2)∣∣2)1/2

=

( N∑i1, i2j1,j2

=1

ai1i2aj1j2x1(i1)x1(j1)x2(i2)x2(j2)N∑

i3=1

ai2i3aj2i3

)1/2

= N1/2

( N∑i2=1

∣∣ N∑i1=1

ai1i2x1(i1)∣∣2|x2(i2)|2

)1/2

≤ N

For the last inequality we have used assumption (b) and ‖id : `N2 −→ `N1 ‖ =N1/2.


Now let m ≥ 4 and x1, . . . , xm ∈ B`N2

. Then

|ϕm(x1, . . . , xm)|

≤( N∑

im=1

∣∣ N∑i1,...,im−1=1

ai1i2 · · · aim−1imx1(i1) · · ·xm−1(im−1)∣∣2)1/2

=

( N∑i1, . . . , im−1

j1,...,jm−1=1

ai1i2aj1j2 · · · aim−2im−1ajm−2jm−1x1(i1) · · ·xm−1(jm−1)

N∑im=1

aim−1imajm−1im

)1/2

= N1/2

( N∑im−1=1

∣∣ N∑i1,...,im−2=1

ai1i2 · · · aim−2im−1x1(i1) · · ·xm−2(im−2)∣∣2

|xm−1(im−1)|2)1/2

≤ N1/2

( N∑im−2=1

∣∣ N∑i1,...,im−3=1

ai1i2 · · · aim−3im−2x1(i1) · · ·xm−3(im−3)∣∣2)1/2

Thus ‖ϕm‖ ≤ N1/2‖ϕm−2‖ and the induction hypothesis gives the claim.

The next example is from Bohnenblust and Hille [12].

Example 2.4. Let m ∈ N and m ≥ 2. Then there is an m-homogeneouspolynomial qm ∈ Pm(`∞) such that for every ε > 0 the monomial expansionof qm does not converge in some point z ∈ `2(1− 1

m)−1+ε.

We give a proof since we use the polynomials constructed here for the proofof example 2.5.

Proof. For the construction of qm as in [12] we use several steps.

1) Choose a prime number p > m and define for n ∈ N the pn × pn-matrixMn inductively by

M1 := (e2πi rsp )1≤r,s≤p

Mn := (e2πi rsp ·Mn−1)1≤r,s≤p = (a(n)

rs )1≤r,s≤pn , n ≥ 2

i.e. Mn is the Kronecker product of M1 with Mn−1 for n ≥ 2. Then Mn

satisfies

(a)∑pn

t=1 a(n)rt a

(n)st = pnδrs


(b) |a(n)rs | = 1

(c) (a(n)rs )p = 1

The properties (b) and (c) are obvious. To understand (a) note that forn = 1 we have

p∑t=1

a(1)rt a

(1)st =

p∑t=1

e2πi tp(r−s) =

p∑t=1

ζt

with ζ a p-th root which is different from one if and only if r 6= s. If n ≥ 2we have

pn∑t=1

a(n)rt a

(n)st =

p∑t=1

a(1)r′ta

(1)s′t

pn−1∑t=1

a(n−1)r′′t a

(n−1)s′′t

with r = s if and only if r′ = s′ and r′′ = s′′.

So by remark 2.3 the m-linear mapping ϕm,n defined by the Matrix Mn

satisfies‖ϕm,n‖Lm(`pn

∞ )≤ p(m+1)n/2

2) Now let qm,n = (ϕm,n)s be the m-homogeneous polynomial associated to

ϕm,n with monomial coefficients c(n)α , α ∈ Npn

0 , |α| = m, explicitely

c(n)α =

1

α!

∑σ∈Sm

a(n)iσ(1) iσ(2)

· · · a(n)iσ(m−1) iσ(m)

(2.7)

where i1 ≤ . . . ≤ im such that |k | ik = r| = αr, r = 1, . . . , pn and Sm thegroup of permutations on 1, . . . ,m. Then

(i) ‖qm,n‖P(m`pn∞ )≤ pn(m+1)/2

(ii) η := inf|c(n)α | > 0

The first part is clear from 2), since the norm of pm,n is less or equal thanthe norm of ϕm,n. The infimum in (ii) is taken over all n and all α ∈ Npn

0

with |α| = m and we now prove that this is in fact away from zero.

Sincesupα! | α ∈ N(N)

0 , |α| = m ≤ (m!)m

we can disregard the binomial factors in (2.7) and it is enough to prove that

I = ∑

σ∈Sm

a(n)iσ(1) iσ(2)

· · · a(n)iσ(m−1) iσ(m)

| i1 ≤ . . . ≤ im

is a finite set of non zero elements. For this choose a primitive p-th root ζ,

for example ζ = e2πip . Then every element of I can be written as

p−1∑k=0

λkζk


with λk ∈ N0 andp−1∑k=0

λk = |Sm| = m! (2.8)

in particular cannot be zero, since this would imply

p−1∑k=1

(λk − λ0)ζk = 0

and henceλ0 = λ1 = . . . = λp−1 =: λ

since ζ, ζ2, . . . , ζp−1 is linearly independent over Q. But that would leadto

p · λ =

p−1∑k=0

λk = m!

which contradicts the fact that p is a prime number exceeding m. We alsoconclude from (2.8) that

I ⊂ p−1∑k=0

λkζk | (λ0, . . . , λp−1) ∈ 0, . . . ,m!p

must be finite.

3) Now we define weights

λn :=1

n2p−n(m+1)/2

andqm(z) :=

∑n

λnqm,n(zn) (2.9)

for z = (zn)n with zn = (zn(1), . . . , zn(pn)) ∈ Cpn, i.e.

z = (z1(1), . . . , z1(p), z2(1), . . . z2(p2), . . .)

Then qm is an m-homogeneous polynomial on `∞ with ‖qm‖ ≤∑

n1n2 and

monomial coefficients

cα(qm) =

λn c

(n)β α = 0× β ∈ 0p+p2+...+pn−1 × Npn

0 , |α| = m

0 otherwise(2.10)

4) Finally let ε > 0 and define

s :=2

1− 1m

+ ε >2

1− 1m

=2m

m− 1=: r


Then 0 < δ := r/s < 1 and since p > m we can choose 0 < b < 1 with

p1−δbδ > 1. Defining zn :=(

bp

)n/s

· 1 ∈ Cpnwe have

∑n

pn∑i=1

|zn(i)|s =∑

n

bn <∞

hence z := (zn)n ∈ `s but∑α

|cα(qm) zα| ≥ η∑

n

λn

∑α∈Npn

0|α|=m

|zαn |

≥ η

m!

∑n

λn

( pn∑i=1

|zn(i)|)m

=η

m!

∑n

1

n2p−n(m+1)/2

(pn

(b

p

)n/s)m

=η

m!

∑n

1

n2

[(p1−δbδ)(m−1)/2

]n=∞

In particular, this example provides us for all p > 2 and all Reinhardt do-mains R ⊂ `p with a holomorphic function f on R satisfying dom(f) 6= Rwhich can be chosen to be bounded if R is bounded.

A modification of the example of Bohnenblust and Hille shows how to con-struct “bad” functions for 4

3< p ≤ 2.

Example 2.5. For all m ≥ 2 there is an m-homogeneous polynomial qm ∈Pm(`2) such that for every ε > 0 the monomial expansion of qm does notconverge in some point z ∈ ` 4

3+ε.

Proof. For an arbitrary natural number m ≥ 2 and a prime number p > mwe construct the polynomials qm,n exactly as in example 2.4. Then

‖qm,n‖Pm(`pn

2 )≤

pmn/4 if m even

p(m+1)n/4 if m odd

by remark 2.3, (2).

So assuming that m is even we can define the m-homogeneous polynomialqm on `2 as in (2.9) but with the weights

λn =1

n2p−mn/4


such that again ‖qm‖ ≤∑

1n2 and qm has monomial coefficients as in (2.10).

Let ε > 0 and define s := 43

+ ε. Then δ := 1s< 3

4and we choose a positive

real number b < 1 such that

p1−δ− 14 bδ > 1

Defining zn :=(

bp

)n/s

1 ∈ Cpnas in example 2.4 4) then again z := (zn) ∈ `s

and∑α

|cαzα| ≥ η

m!

∑n

1

n2p−nm/4

(pn

(b

p

)n/s)m

=η

m!

∑n

1

n2(p1−δ− 1

4 bδ)mn =∞

It remains to consider the case when m is odd. Then, by assumption, m ≥3 and we know that there are qm−1 ∈ Pm−1(`2) and z ∈ ` 3

4+ε such that∑

|α|=m−1|cα(qm−1)zα| = ∞. Take ϕ := e′1 ∈ `′2, i.e. ϕ(u) = u1 for u ∈ `2.

Thenqm := ϕ · qm−1

defines an m-homogeneous polynomial on `2 with ‖qm‖ ≤ ‖qm−1‖ and mono-mial coefficients

cα(qm) =

cα−e1(qm−1) α1 6= 0

0 otherwise

Thus (we can choose z1 6= 0)∑|α|=m

|cα(qm)zα| = |z1|∑|α|=mα1 6=0

|cα−e1(qm−1)zα−e1| = |z1|

∑|α|=m−1

|cα(qm−1)zα| =∞

After these examples the following definition makes sense.

Definition 2.6. Let R be a Reinhardt domain in a Banach sequence spaceX. For a holomorphic function f on R we define

dom(f) := z ∈ R |∑

α∈N(N)0

|cα(f) zα| <∞

to be the set of all points z of R in which its monomial expansion seriesconverges and call it the domain of convergence of f .

Note that dom(f) is balanced.

Corollary 2.7. Let p > 43

and R ⊂ `p (or R ⊂ c0) be a Reinhardt domain.Then we can find a holomorphic function f on R with

dom(f) 6= R

If R is bounded then f can be chosen to be bounded.


In a later section we will describe the domain of convergence of the m-homogeneous polynomials on a Banach sequence space. Special cases ofour results will prove the existence of such polynomials as in the precedingexamples. We will even settle the case X = `p with 1 ≤ p ≤ 4/3 (see example2.54).

As we mentioned in the introduction, Bohr’s theorem 0.1 about the conver-gence of power series in infinitely many variables can be proved for boundedholomorphic functions.

Theorem 2.8. Let f ∈ H∞(B`∞). Then∑α∈N(N)

0

|cα(f)zα| <∞

for all z ∈ `2 ∩B`∞.

Proof. Let x ∈ `2 ∩ B`∞ . Choose 0 < r < 1 such that x ∈ rB`∞ and definefr ∈ H∞(1

rB`∞) by fr(z) := f(rz). Then cα(fr) = cα(f) r|α|. Let n ∈ N. The

monomials zα in Cn are orthonormal on the torus Πn = z ∈ Cn | |z| = 1:

(zα|zβ)L2(Πn) =

(1

2π

)n ∫[0,2π]n

ei(α1θ1+···αnθn) e−i(β1θ1+···βnθn)dθ

=

(1

2π

)n ∫[0,2π]n

ei(α−β|θ)dθ =n∏

k=1

1

2π

∫ 2π

0

ei(αk−βk)θdθ

=

1 if α = β

0 otherwise

Hence ∑α∈Nn

0

|cα(fr)|2 =∑α∈Nn

0

cα(fr) cα(fr)(zα|zα)L2(Πn)

=∑

α, β∈Nn0

cα(fr) cβ(fr)(zα|zβ)L2(Πn)

=(∑

α∈Nn0

cα(fr)zα∣∣∣ ∑

α∈Nn0

cα(fr)zα)

L2(Πn)

= ‖fr‖2L2(Πn) ≤ supz∈Πn

|fr(z)|2 ≤ ‖f‖2B`∞

Since n was arbitrary this implies (cα(fr))α∈N(N)0∈ `2(N(N)

0 ). Furthermore

recall that the series∑|zα| converges if and only if z ∈ `1 ∩ DN with limit∑

α∈N(N)0

|zα| = limm→∞

∑α∈Nm

0

|zα| = limm→∞

m∏n=1

∞∑k=0

|zn|k =∞∏

n=1

1

1− |zn|


By the Cauchy-Schwarz inequality it follows∑α∈N(N)

0

|cα(f)xα| ≤( ∑

α∈N(N)0

|cα(fr)|2)1/2 ( ∑

α∈N(N)0

|(r−1x)2 α|)1/2

≤ ‖f‖B`∞

(∏n

1

1− |r−1xn|2)1/2

Note that in the definition of the number S defined by Bohr (see the intro-duction, p. 11) one could also take the power series bounded on the domainBc0 = z ∈ B`∞ | zn −→ 0 instead of B`∞ = [|zi| ≤ 1] (since ‖ε‖∞ < 1 forε ∈ `q ∩ ] 0 , 1[ N) and they can be identified with the monomial expansions ofthe bounded holomorphic functions on Bc0 . More generally we have:

Remark 2.9. Let R be a Reinhardt domain in a Banach sequence space Xwith basis and cα ∈ C, α ∈ N(N)

0 . The following are equivalent:

(1)∑cαz

α is the monomial expansion of an f ∈ H∞(R).

(2)∑cαz

α is bounded on R:

supn

supz∈Rn

∣∣ ∑α∈Nn

0

cαzα∣∣ <∞

In this case‖f‖R = sup

nsupz∈Rn

∣∣ ∑α∈Nn

0

cαzα∣∣

Proof. It is easy to see that (1) implies (2): If n ∈ N and z ∈ Rn then∣∣ ∑α∈Nn

0

cαzα∣∣ = |f(z)| ≤ ‖f‖R

and hencesup

nsupz∈Rn

∣∣ ∑α∈Nn

0

cαzα∣∣ ≤ ‖f‖R <∞

On the other hand, suppose∑cαz

α is bounded on R. Then∑

α∈Nn0cαz

α

converges on Rn for every n and we can define the function

f0(z) :=∑

α∈N(N)0

cαzα

on ∪nRn which is bounded with

‖f0‖∪nRn = supn‖f0‖Rn = sup

nsupz∈Rn

∣∣ ∑α∈Nn

0

cαzα∣∣


and holomorphic on every finite dimensional section Rn of R. If for m ∈ Nand z ∈ X0 := spanen =

⋃nXn we define the m-homogeneous polynomial

Pm(z) :=∑|α|=m

cαzα

then we can write

f0(z) =∞∑

m=0

Pm(z) (2.11)

for every z ∈ ∪nRn. Now let m,n ∈ N and z ∈ Rn. Then

Pm(z) =1

2πi

∫|λ|=1

f0(λz)

λm+1dλ

(see e.g. [19, p. 175]) and hence

|Pm(z)| ≤ ‖f0‖Rn ≤ supn‖f0‖Rn <∞

which implies that Pm is bounded on ∪nRn and hence on rBX0 if r > 0 suchthat rBX ⊂ R. Thus Pm is continuous and can be extended to X0 = X with

‖Pm‖R = ‖Pm‖∪nRn ≤ supn‖f0‖Rn

Since for all 0 < r < 1 we have

∞∑m=1

‖Pm‖rR =∞∑

m=1

rm‖Pm‖R ≤supn‖f0‖Rn

1− r<∞

the series∑

m Pm converges on R and by [19, theorem 14.23], the function

f(z) =∞∑

m=0

Pm(z)

is holomorphic on R. From (2.11) we deduce that f is an extension of f0

with monomial expansion∑cαz

α. In particular, f is bounded and

‖f‖R = supn‖f0‖Rn

The notion of boundedness in (2) is a translation of Hilbert’s definition ofboundedness “in the domain |zi| ≤ 1” (i.e. on B`∞) to Reinhardt domains ininfinite dimensions. In this sense, investigating bounded holomorphic func-tions is nothing but examining bounded power series in infinitely many vari-ables.

Lower inclusions for domH∞(R) 65

Definition 2.10. Let R be a Reinhardt domain in a Banach sequence spaceX and F(R) ⊂ H(R). We call

domF(R) :=⋂

f∈F(R)

dom(f)

the domain of convergence of F(R).

Note thatdomF(R) = dom spanF(R) (2.12)

and domF(R) as an intersection of balanced sets is balanced.

The operator domH∞( · ) is increasing in R and positive homogeneous. Moreprecisely:

Remark 2.11. Let X and Y be Banach sequence spaces, X ⊂ Y and R ⊂ X,S ⊂ Y be Reinhardt domains. Let c > 0.

(1) If R ⊂ S then domH∞(R) ⊂ domH∞(S).

(2) domH∞(cR) = c domH∞(R).

Proof. (1) The natural inclusion mapping X → Y is linear and continu-ous, hence holomorphic. This implies that the restriction f |R of a mappingf ∈ H∞(S) is a bounded holomorphic function with the same monomialcoefficients as f and so

domH∞(R) ⊂ dom(f |R) ⊂ dom(f)

for all f ∈ H∞(S) which gives the claim.

(2) Only note that if f ∈ H∞(cR) then the function fc defined by fc(z) :=f(cz) is a bounded holomorphic function on R with

dom(fc) = c dom(f)

Lower inclusions for domH∞(R)

In this section, we are going to determine a large set of elements which iscontained in the domain of convergence of the bounded holomorphic functionson a given Reinhardt domain R in a Banach sequence space X. We willprove that for a sequence z = (zn) ∈ R it is enough to satisfy an additionalsummability condition to become an element of domH∞(R), namely z ∈ `1or z ∈ X · `2 (theorem 2.25). In a later section, we see that for a wide class


of Banach sequence spaces this condition turns out to be “almost” necessary(theorem 2.44).

ForX = `1 the situation is very nice. Matos proved in [60] that a holomorphicfunction f : U −→ C on an open subset U of `1 has a monomial expansionin a neighbourhood of every point z ∈ U (here cα,z(f) = ∂αf(z)

α!, see (2.1)). In

our case z = 0 this can be improved. The following result is due to Ryan [70]and Lempert [53]. Ryan investigated the `1-case (here rB`1 := `1 for r =∞)and Lempert proved the general result while investigating the ∂-equation inan open subset of a Banach space and the approximation of holomorphicfunctions on balls in Banach spaces by functions holomorphic on the wholespace (see [52, 53, 54]).

Theorem 2.12. Let 0 < r ≤ ∞. The monomial expansion of a holomor-phic function f on rB`1 converges uniformly absolutely on compacta, i.e.∑|cα(f)zα| converges uniformly on compact subsets of rB`1.

Our goal is to generalize this theorem to arbitrary Reinhardt domains (theo-rem 2.21). This implies that `1∩R ⊂ domH(R) is always true (see corollary3.3) and, in particular, gives one part of theorem 2.25. The second part isthen obtained from theorem 2.8 via a “multiplication trick” (remark 2.24).

The main work lies in the generalization of theorem 2.12. The idea is to splitK ⊂ R compact into the product of two compact sets Km ⊂ U and Km ⊂rB`1 , where U is a Reinhardt domain in a finite dimensional space Cm and U×rB`1 ⊂ R. Then we can apply the lemmata 2.13 and 2.16 (the latter resultingfrom an analysis of the proof of theorem 2.12) to prove that

∑|cα(f)zα| is

bounded on compact sets for every f ∈ H(R) (proposition 2.18) which, by aresult about equicontinuous families of holomorphic functions (lemma 2.19),implies what we want (theorem 2.21).

Lemma 2.13. Let R be a Reinhardt domain in Cm and K a compact subsetof R. Then we can find a constant c > 0 and a compact set K ⊂ R containingK such that for all countable families (fν)ν∈I of holomorphic functions fν onR we have ∑

ν∈I

∑α∈Nm

0

‖cα(fν)zα‖K ≤ c ·∥∥∑

ν∈I

|fν |∥∥

K

Proof. We can assume that K is circular and balanced. (Otherwise takeK := B`∞K ⊂ R, see the proof of remark 2.17.) Choose r > 0 and t > 1such that K := t(K + rB`m

∞) ⊂ R and suppose ‖∑

ν∈I |fν |‖K < ∞. Let

x ∈ K. Then s := t(|x| + r · 1) ∈ K. It follows [|z| ≤ s] ⊂ K since K iscircular and balanced and

cα(fν) =

(1

2πi

)m ∫|z|=s

fν(z)

zα+1dz


Thus for all z ∈ x+ rB`m∞ and all ν ∈ I we have

|cα(fν)zα| ≤ |zα|sα

(1

2π

)m ∣∣∣∣ ∫[0,2π]m

fν(s · eiθ)

(eiθ)αdθ

∣∣∣∣≤ t−|α|

(1

2π

)m ∫[0,2π]m

|fν(s · eiθ)|dθ

(where eiθ := (eiθ1 , . . . , eiθm) for θ ∈ [0, 2π]m) and using Lebesgue’s theoremwe get∑

ν∈I

∑α∈Nm

0

‖cα(fν)zα‖x+rB`m∞≤∑

α∈Nm0

t−|α|(

1

2π

)m ∫[0,2π]m

∑ν∈I

|fν(s · eiθ)|dθ

≤ (1− t−1)−m ‖∑ν∈I

|fν |‖K

Precompactness gives K ⊂ x1, . . . , xn+ rB`m∞ for some xj ∈ K and hence

the claim.

The proof of the following lemma is a skilful utilization of Stirling’s formulaas was used by Lempert in his proof of theorem 2.12 (see [53, lemma 4.1] andalso the exposition in [36, p. 304 ff.]).

Lemma 2.14. Let θ ∈ B+c0. Then there exists an N ∈ N and a zero sequence

ξ ∈ B+c0

such that

|α||α|

ααθα ≤ eN |α|!

α!ξα

for all α ∈ N(N)0 .

Here B+c0

:= Bc0 ∩ RN>0.

Proof. Let η = ‖θ‖1/2∞ and choose ` ∈ N such that θn ≤ η/e for n ≥ ` and,

with the help of Stirling’s formula, an m ∈ N with (eη)n ≤ nn/n! for all

n ≥ m. Let α ∈ N(N)0 and I = n ∈ N | n < ` , αn ≥ m . Then

α!∏i∈I

(eη)αi =∏i/∈I

αi!∏i∈I

(eη)αiαi! ≤ αα

by the choice of m and the fact that n! ≤ nn. Since nn ≤ enn! we get

|α||α|

αα≤ e|α|∏

i∈I(eη)αi· |α|!α!

=e

Pi/∈I αi

ηP

i∈I αi· |α|!α!≤ e

Pi≥` αi +` m

ηP

i<` αi· |α|!α!

hence|α||α|

ααθα ≤ e` m |α|!

α!ξα


for all α ∈ N(N)0 where ξ ∈ B+

c0is defined by

ξn =

1ηθn if n < `

e θn if n ≥ `

Proposition 2.15. Let θ ∈ B+c0. Then there exists a constant c > 0 such

that for all subsets A ⊂ N(N)0 , all families (cα)α∈A of scalars and all r > 0 we

have ∑m

supu∈rB`1

∑α∈A|α|=m

|cα(θu)α| ≤ c · supα∈A|cα|

αα

|α||α|r|α| θ α/2

Proof. Let A ⊂ N(N)0 and cα, α ∈ A be scalars. By lemma 2.14 we have

|α||α|

ααθα/2 ≤ eN |α|!

α!ξα

for some N ∈ N and ξ ∈ B+c0

and all α ∈ N(N)0 . Then∑

m

supu∈rB`1

∑α∈A|α|=m

|cα(θ u)α| ≤ eN∑m

supu∈B`1

∑α∈A|α|=m

|cα|αα

|α||α|r|α| θα/2 |α|!

α!|(ξ u)α|

≤ eN supα∈A|cα|

αα

|α||α|r|α| θα/2

∑m

supu∈B`1

∑α∈A|α|=m

|α|!α!|(ξ u)α|

By the multinomial formula∑|α|=m

|α|!α!|(ξ u)α| = ‖ξ u‖m`1 ≤ ‖ξ‖

m∞

if u ∈ B`1 and by calculating the arising geometric series the claim follows

with c = eN

1−‖ξ‖∞ .

We recall here the fact that (ξB`1)ξ∈B+c0

is a fundamental system of compact

subsets of B`1 (see [70] or [36]): For every ξ ∈ Bc0 the set ξB`1 = Γξnen iscompact in B`1 , and conversely forK ⊂ B`1 compact we choose 0 < t < t′ < 1with K ⊂ tB`1 and a strictly increasing sequence (nk) of natural numberssuch that

supz∈K

∑j>nk

|zj| ≤1− δk 2k+1

where δ = t/t′. This is possible since precompactness of K in `1 means

supz∈K

∑j≥n

|zj|n→∞−−−→ 0


Then

ξn :=

t′ 1 ≤ n ≤ n1

12k

nk < n ≤ nk+1

defines a positive zero sequence with ‖ξ‖∞ < 1 such that for all z ∈ K

∞∑n=1

ξ−1n |zn| =

n1∑n=1

ξ−1n |zn|+

∞∑k=1

nk+1∑n=nk+1

ξ−1n |zn|

=1

t′

n1∑n=1

|zn|+∞∑

k=1

2k

nk+1∑n=nk+1

|zn|

≤ δ + (1− δ)∞∑

k=1

2−k = 1

Then of course (ξB`1)ξ∈rB+c0

is a fundamental system of compact subsets of

rB`1 for 0 < r ≤ ∞ where rB`1 := `1 and rB+c0

:= c+0 = ξ ∈ c0 | ξ > 0 ifr =∞.

Lemma 2.16. Let r > 0 and K ⊂ rB`1 compact. Then there exists c > 0and a compact set K ⊂ rB`1 which contains K such that∥∥ ∑

α∈N(N)0

|cα(f)zα|∥∥

K≤ c · ‖f‖K

for all f ∈ H(rB`1).

Proof. We can assume r = 1. Since (ξB`1)ξ∈B+c0

is a fundamental system

of compact subsets in B`1 we can choose θ ∈ B+c0

with K ⊂ θB`1 . Defineξ := θ1/2. By the Cauchy inequalities (2.2) we get

|cα(f)| αα

|α||α|ξα = ‖cα(f)zα‖ξB`1

≤ ‖f‖ξB`1

for every f ∈ H(B`1) and all α, hence we obtain our result from proposition2.15 with K = ξB`1 .

The next remark allows us in the proof of proposition 2.18 to restrict ourselvesto compact sets K which are circular (u ∈ CN, v ∈ K and |u| = |v| impliesu ∈ K) and balanced.

Remark 2.17. Let R be a Reinhardt domain in `1 and K ⊂ R be compact.Then we can find a compact, circular and balanced subset K ⊂ R whichcontains K and ηB`∞ for some η ∈ `+1 .


Proof. The set K := B`∞ K ⊂ R is circular, balanced and contains K and0. Further K is closed in `1 with

limm→∞

supz∈K

∑n≥m

|zn| = limm→∞

supz∈K

∑n≥m

|zn| = 0

since K is relatively compact in `1. Thus K is compact in `1.

Now choose r > 0 such that K + rB`1 ⊂ R. Then η := (r/2n+1)n ∈ `+1 andK := K + ηB`∞ is the set that we were looking for.

Proposition 2.18. Let R be a Reinhardt domain in `1 and K a compactsubset of R. Then there is a constant c > 0 and a compact set K ⊂ Rcontaining K such that for all holomorphic functions f on R we have∥∥ ∑

α∈N(N)0

|cα(f)zα|∥∥

K≤ c · ‖f‖K

Proof. By remark 2.17 we can assume K to be circular and balanced. Wechoose t > 1 and 0 < r < 1 such that t(K + 2rB`1) ⊂ R and then m ∈ Nsatisfying

supz∈K

∑n≥m

|zn| < r

which is possible since K is relatively compact in `1. Denote πm : `1 −→ `m1the m-th projection and πm : `1 −→ `1, π

m(z) := (zn)n≥m+1. Then

Km := πm(K) ⊂ `m1

Km := πm(K) ⊂ `1

are compact subsets with K ⊂ Km ×Km and we have

Km ⊂ U := t(Km + rB`m1

)

Km ⊂ rB`1

where U is a Reinhardt domain in Cm and

U × rB`1 ⊂ t((Km + rB`m1

)× rB`1) ⊂ t(Km + 2rB`1) ⊂ R

Now we choose cm > 0 and Km ⊂ Km ⊂ U as in lemma 2.13, as well ascm > 0 and Km ⊂ Km ⊂ rB`1 as in lemma 2.16. Then for f ∈ H(R) andx ∈ U we have f(x, · ) ∈ H(rB`1) with

fβ(x) := cβ(f(x, · )) =

(1

2πi

)n ∫|y|=b

f(x, y)

yβ+1dy


if β ∈ Nn0 and b ∈ Rn

>0 such that [|y| ≤ b] ⊂ rB`n1. Thus fβ is a continuous

function on U . Now we choose a ∈ Rm>0 with [|x| ≤ a] ⊂ U . If x ∈ U and

α := (0, β) ∈ Nm0 × Nn

0 then (x, 0) ∈ U × rB`n1⊂ R and by Fubini’s theorem

fβ(x) =

(1

2πi

)m+n ∫|y|=b

∫|ζ|=a

f(ζ, y)

(ζ − x)1yβ+1dζ dy

=∂αf(x, 0)

α!

=

(1

2πi

)m+n ∫|ζ|=a

∫|y|=b

f(ζ, y)

(ζ − x)1yβ+1dy dζ

=

(1

2πi

)m ∫|ζ|=a

fβ(ζ)

(ζ − x)1dζ

Thus fβ has a local Cauchy representation and hence is holomorphic on U .For γ = (α, β) ∈ Nm+n

0 = Nm0 × Nn

0 we get

cγ(f) =∂γf(0)

γ!=∂α

α!

(∂βf(0)

β!

)=∂αfβ(0)

α!= cα(fβ)

Hence for z = (x, y) ∈ K ⊂ Km ×Km we estimate∑γ∈N(N)

0

|cγ(f)zγ| =∑

β∈N(N)0

∑α∈Nm

0

|cα(fβ)xαyβ|

≤ cm supx∈Km

∑β

|cβ(f(x, · ))yβ|

≤ cm cm sup

x∈Km

‖f(x, · )‖Km

= c ‖f‖K

with c = cm cm and

K ⊂ Km ×Km ⊂ K := Km × Km ⊂ U × rB`1 ⊂ R

In [6, proposition 37] it is proved that, in a locally convex Baire space E,a set F(Ω) ⊂ H(Ω) of holomorphic functions on an open subset Ω ⊂ E islocally bounded if it is bounded on finite dimensional compact sets and fromthis it is standard to conclude that F(Ω) is equicontinuous, if E is a Banachspace. For the sake of completeness we give a direct proof of the next lemma,including the arguments of [6] for the Banach space case.

Lemma 2.19. Let Ω be an open subset of a Banach space E and F(Ω) ⊂H(Ω) be bounded on finite dimensional compact subsets of Ω. Then F(Ω) isequicontinuous.


Proof. Let ξ ∈ Ω and V be some multiple of the open unit ball BE in Esuch that ξ + V ⊂ Ω. For every x ∈ V we have that K := ξ + Dx is a finitedimensional compact subset of Ω and by the Cauchy integral formulas (0.12)

|Pmf(ξ)(x)| ≤ sup|λ|=1

|f(ξ + λx)| ≤ supf∈F(Ω)

‖f‖K <∞

so that we can write V = ∪nVn with sets

Vn := x ∈ V | ∀m ∈ N , ∀f ∈ F(Ω) : |Pmf(ξ)(x)| ≤ n

which are closed in V . As an open subset of a Banach space, the set V is ofsecond category and by Baire’s theorem there must be an index n such that

the interior of Vn in V , i.e.V n ∩ V =

V n is non-empty. Take a ∈

V n and

choose a multiple W of BE contained in V such that a+W ⊂ Vn. Then wehave for all f ∈ F(Ω), m ∈ N and all x ∈ W

|Pmf(ξ)(x)| ≤ sup|λ|≤1

|Pmf(ξ)(λa+ x)| = sup|λ|=1

|Pmf(ξ)(λa+ x)|

= sup|λ|=1

|Pmf(ξ)(a+ λx)| ≤ supy∈W|Pmf(ξ)(a+ y)|

≤ ‖Pmf(ξ)‖Vn ≤ n

Here we have used the maximum principle for the first equality and them-homogeneity of Pmf(ξ) for the second one:

Pmf(ξ)(λa+ x) = λmPmf(ξ)(a+ λ−1x)

Thus for 0 < r < 1∞∑

m=0

‖Pmf(ξ)‖rW ≤ |f(ξ)|+ n∞∑

m=1

rm <∞

So the Taylor expansion series∑

m Pmf(ξ) of f in ξ converges uniformly onrW and hence defines a holomorphic function on rW (see [19, theorem 12.5])which, by the principle of analytic continuation ([19, theorem 12.8]), mustbe equal to f(ξ + · ) on this set, since this is true in some neighbourhoodof zero. (By the choice of W we have ξ + rW ⊂ ξ + V ⊂ U , so f(ξ + · )is holomorphic on rW .) Then we have for all f ∈ F(Ω), 0 < r < 1 and allx ∈ rW

|f(ξ + x)− f(ξ)| =∣∣ ∞∑

m=1

Pmf(ξ)∣∣ ≤ ∞∑

m=1

‖Pmf(ξ)‖rW ≤n

1− rr

r→0−−→ 0

which proves that F(Ω) is equicontinuous in ξ.

Corollary 2.20. Let Ω be an open set in a Banach space X and (fi)i∈I acountable family of functions fi ∈ H(Ω) such that

∑|fi| converges point-

wise in Ω and is bounded on compact subsets of Ω. Then∑|fi| converges

uniformly on compact subsets of Ω.


This is a stronger version of Lempert’s [53, proposition 4.2], which he proveddirectly, assuming boundedness on the whole of Ω and not being aware ofthe result [6, proposition 37] from 1977.

Proof. It is enough to prove this for I = N and∑∞

n=1|fn|. By lemma 2.19the set

N∑

j=1

εjfj | N ∈ N, |εj| = 1 ⊂ H(Ω)

is equicontinuous. Thus for z0 ∈ Ω and ε > 0 we can find δ > 0 such thatfor all z ∈ z0 + δBX ⊂ Ω and all N ∈ N

∣∣ N∑j=1

|fj(z)| −N∑

j=1

|fj(z0)|∣∣ ≤ N∑

j=1

|fj(z)− fj(z0)|

= sup|εj |=1

∣∣ N∑j=1

εj(fj(z)− fj(z0))∣∣ < ε

which implies that F(Ω) := ∑N

n=1|fn| | N ∈ N is equicontinuous in Ω.Thus for K ⊂ Ω compact F(K) = f |K | f ∈ F(Ω) is bounded andequicontinuous in C(K) and hence relatively compact by the Arzela-Ascolitheorem. It follows that (

∑Nn=1|fn|)N has a convergent subsequence and

hence is convergent in C(K), i.e.∑∞

n=1|fn| converges uniformly on K.

Now we have everything to prove our generalization of theorem 2.12.

Theorem 2.21. Let f be a holomorphic function on a Reinhardt domain Rin `1. Then the monomial expansion of f converges uniformly absolutely oncompact subsets of R.

Proof. By proposition 2.18 the series∑

α∈N(N)0|cα(f)zα| is bounded on com-

pact subsets of R and we can apply corollary 2.20.

Corollary 2.22. Let R be a Reinhardt domain in a Banach sequence spaceX. Then

`1 ∩R ⊂ domH(R)

Proof. Since the natural inclusion mapping `1 → X is continuous, the set`1 ∩ R is open and hence a Reinhardt domain in `1 which, by theorem 2.21,implies

`1 ∩R ⊂ dom(f |`1∩R) ⊂ dom(f)

for all f ∈ H(R).

The next remark allows us to apply theorem 2.8 by a simple “multiplicationtrick”.


Lemma 2.23. Let X be a Banach sequence space and c0 the space of all zerosequences. Then

X · c0 = spanen

Proof. If z = λ · µ ∈ X · c0, then

∥∥z − n∑k=1

zkek

∥∥X

= ‖πn(z)‖X ≤ ‖πn(λ)‖X ‖πn(µ)‖∞ ≤ ‖λ‖X supk>n|µk|

n→∞−−−→ 0

where πn(x) = (0. . . . , 0, xn+1, xn+2, . . .) for x ∈ X. On the other hand, ifz ∈ spanen then choose a sequence n1 < n2 < . . . < nk < . . . of naturalnumbers with ∥∥ ∑

n≥nk

znen

∥∥X<

1

k3

and then decompose z “blockwise” by zn = kzn · 1/k if nk ≤ n < nk+1.

Remark 2.24. Let X be a Banach sequence space and R a Reinhardt domainin X. Then

R ∩ (X · `2) = R · (`2 ∩B`∞)

Proof. Let z = λ · µ ∈ R ∩ (X · `2), without loss of generality z, λ, µ ≥ 0.Further we can assume that λ ∈ X0 := spanen (by lemma 2.23) andµ ∈ B`∞ (otherwise consider z = rλ · (1/r)µ with r > ‖µ‖∞).

Let a > 1 and b > 0 such that az + bBX ⊂ R. Then we can find an nsatisfying

‖∑k>n

λkek‖ < b

and hence z = u · v with

u = (az1, . . . , azn, λn+1, λn+2, . . .) ≤ az +∑k>n

λkek ∈ R

and

v = (1

a, . . . ,

1

a, µn+1, µn+2, . . .) ∈ `2 ∩B`∞

The remaining inclusion is obvious.

Finally we have reached the main theorem of this section.

Theorem 2.25. Let R be a Reinhardt domain in a Banach sequence spaceX. Then

(`1 ∪ (X · `2)) ∩R ⊂ domH∞(R)

The arithmetic Bohr radius 75

Proof. From corollary 2.22 we get `1 ∩R ⊂ domH(R) ⊂ domH∞(R).

For the remaining inclusion it is enough to prove

R · (`2 ∩B`∞) ⊂ domH∞(R)

by remark 2.24.

So let f ∈ H∞(R). For x ∈ R and z ∈ B`∞ we define fx(z) := f(xz). Thenfx is a bounded holomorphic function on B`∞ with cα(fx) = cα(f)xα and bytheorem 2.8 we have∑

α∈N(N)0

|cα(f)(xz)α| =∑

α∈N(N)0

|cα(fx)zα| <∞

for all z ∈ `2 ∩B`∞ .

The arithmetic Bohr radius

As promised at the beginning of the preceding section our aim is to provethat for a large class of Banach sequence spaces X the set (`1∪(X ·`2))∩R isan almost precise description of domH∞(R), if the Reinhardt domain R ⊂ Xis bounded (see theorem 2.44). Our main tool to prove upper inclusions fordomH∞(R) is the arithmetic Bohr radius.

The Bohr radius K(Rn) (see [11]) and the Bohr radius B(Rn) (see [3]) of aReinhardt domain Rn ⊂ Cn are defined to be the least upper bound of allpositive real numbers r such that∥∥ ∑

α∈N(N)0

|cα(f)zα|∥∥

rRn≤ ‖f‖Rn

and ∑α∈N(N)

0

‖cα(f)zα‖rRn ≤ ‖f‖Rn

respectively for all bounded holomorphic functions f on Rn. These Bohrradii have been estimated asymptotically in the dimension n. For the `np -unitballs the estimates are asymtotically correct up to a logarithmic term:

1

c

(log n/ log log n

n

)1− 1minp,2

≤ K(B`np) ≤ c

(log n

n

)1− 1minp,2

(2.13)

and1

c

(1

n

) 12+ 1

maxp,2

≤ B(B`np) ≤ c

(log n

n

) 12+ 1

maxp,2

(2.14)


(see [37, 11, 10, 23])

For 1 ≤ p ≤ 2 the lower estimate for the second Bohr radius is slightly better:

1− n

√2

3≤ B(B`n

p) (2.15)

(see [10] and [3]).

Definition 2.26. Let Rn ⊂ Cn be a Reinhardt domain, λ ≥ 1 and m ∈ N.Then we define the λ-arithmetic Bohr radius of Rn

Aλ(Rn) := sup 1

n

n∑i=1

ri | r ∈ Rn≥0 ,∀f ∈ H∞(Rn) :∑

α∈Nn0

|cα(f)|rα ≤ λ‖f‖Rn

and

Amλ (Rn) := sup 1

n

n∑i=1

ri | r ∈ Rn≥0 ,∀f ∈ Pm

∞(Rn) :∑α∈Nn

0|α|=m


where Pm∞(Rn) := Pm(Cn) ∩ H∞(Rn) is the space of all m-homogeneous

bounded holomorphic functions on Rn. We call A(Rn) := A1(Rn) the arith-metic Bohr radius of Rn.

Obviously Aλ(Rn) and Amλ (Rn) are increasing in λ and we have Aλ(Rn) ≤

Amλ (Rn) for all m.

For bounded Reinhardt domains Rn, Sn ⊂ Cn we use the notation

S(Rn, Sn) := inft > 0 | Rn ⊂ tSn

from [25]. Note that if Rn ⊂ Cn is a bounded Reinhardt domain and X andY are Banach sequence spaces then

S(Rn, BXn) = supz∈Rn

‖z‖X

S(BXn , BYn) = ‖id : Xn −→ Yn‖

Remark 2.27. Let Rn, Sn ⊂ Cn be bounded Reinhardt domains and λ ≥ 1.Then

Aλ(Rn) ≤ S(Rn, Sn)Aλ(Sn) (2.16)

In particular, Aλ(tRn) = t Aλ(Rn) for all t > 0 and Aλ is increasing, i.e.Aλ(Rn) ≤ Aλ(Sn) if Rn ⊂ Sn.


Proof. If r ∈ Rn≥0 satisfies ∑

α∈Nn0


for all f ∈ H∞(Rn) then, writing s = S(Rn, Sn) for short, we have∑α∈Nn

0

|cα(f)|( r

s+ ε

)α ≤ λ∥∥ ∑

α∈Nn0

cα(f)( z

s+ ε

)α∥∥Rn≤ λ ‖f‖Sn

for every f ∈ H∞(Sn) and all ε > 0, hence

1

n

n∑i=1

ri ≤ S(Rn, Sn)Aλ(Sn)

There is a general relationship between the Bohr radius K(Rn) and the λ-arithmetic Bohr radius Aλ(Rn) for bounded Reinhardt domains Rn.

Remark 2.28. Let Rn be a bounded Reinhardt domain in Cn and λ ≥ 1.Then

Aλ(Rn) ≥S(Rn, B`n

1)

nK(Rn)

Proof. Since Aλ(Rn) is increasing in λ it is enough to prove the inequalityfor the arithmetic Bohr radius. We have

S(Rn, B`n1) = sup

z∈Rn

‖z‖`n1

thus for 0 < ε < K(Rn) we can find an element z0 ∈ Rn such that

‖z0‖`n1≥ S(Rn, B`n

1)− ε

Let t := K(Rn) − ε, u := tz0 and r := t|z0| = |u|. Since u ∈ tRn andt < K(Rn) we have∑

α∈Nn0

|cα(f)|rα =∑α∈Nn

0

|cα(f)uα| ≤∥∥ ∑

α∈Nn0

|cα(f)zα|∥∥

tRn≤ ‖f‖Rn

and it follows

A(Rn) ≥ 1

n

n∑i=1

ri =K(Rn)− ε

n‖z0‖`n

1≥ K(Rn)− ε

n(S(Rn, B`n

1)− ε)


We first give estimates for the λ-arithmetic Bohr radius of the `np -unit balls.We can deduce a lower bound from remark 2.28. For an upper bound we usea probabilistic estimate of Boas [10] as in the case of the Bohr radii K(Rn)and B(Rn). He proved that for every 1 ≤ p ≤ ∞ there are signs εα ∈ −1, 1such that

∥∥ ∑|α|=m

εαm!

α!zα∥∥

B`np

≤√

32m log(6m)

n

12 (m!)1− 1

p if p ≤ 2

n12+( 1

2− 1

p)m(m!)

12 if p ≥ 2

(2.17)This is a symmetric version of a result contained in [59].

We use it to get the following estimate.

Proposition 2.29. Let 1 ≤ p ≤ ∞. There is a constant c > 0 such that forall m and all n

Amλ (B`n

p) ≤ c

m√λ

2m

√m logm (m!)2(1− 1

min2,p) n

(1

n

) 12+ 1

max2,p

Proof. Let m ∈ N. Take r ∈ Rn≥0 with

∑|α|=m|cα(f)| rα ≤ λ‖f‖B`n

pfor every

f ∈ Pm(Cn). Then we can choose signs εα ∈ −1, 1 such that

( n∑i=1

ri

)m

=∑|α|=m

m!

α!rα ≤ λ

∥∥ ∑|α|=m

εαm!

α!zα∥∥

B`np

≤ λcm

by the above cited result, where cm is the right side of (2.17). Taking them-th root and dividing by n we obtain

Amλ (B`n

p) ≤

m√λcmn

for every m and n.

Theorem 2.30. Let 1 ≤ p ≤ ∞. Then there is a constant c > 0 such that

1

c

(log n/ log log n)1− 1minp,2

n12+ 1

max2,p≤ Aλ(B`n

p) ≤ c λ2/ log n (log n)1− 1

minp,2

n12+ 1

max2,p

for all n ∈ N.

Proof. We get the lower bound from remark 2.28 and (2.13). For the upperbound by proposition 2.29 we only have to show that there is a constantc > 0 such that for all n we can find an m with

m√λ

2m

√m logm (m!)2(1− 1

min2,p) n ≤ c λ2/ log n (log n)1− 1minp,2 (2.18)


Since 2m√m logm is bounded in m and m

√m! ≤ m/

√2, the left hand side of

(2.18) can be estimated up to a constant independent of m and n by

m√λm1− 1

min2,p 2m√n (2.19)

Putting m = [log n] the greatest integer less or equal log n, we have log n ≤2m and hence (2.19) can be estimated from above by

λ2/ log n (log n)1− 1min2,p e

Different from the situation for all versions of Bohr radii defined for holo-morphic functions in the literature, where no exact value for any Reinhardtdomain Rn ⊂ Cn in dimension n ≥ 2 is known, we have the following result.

Theorem 2.31.

A(B`n1) =

1

3n

Proof. First we prove A(B`n1) ≤ 1

3n. For the proof of K(D) ≤ 1/3, one can

use the linear fractional transformation

fa(z) =z − a1− az

0 < a < 1 (see Boas [10]; in Bohr’s original paper [14], the function 1−z1−az

is used). This function is bounded by 1 on the unit disc by the maximumprinciple, since for all z ∈ C with |z| = 1 we have |fa(z)| = 1 and from

fa(z) = (z − a)∞∑

k=0

(az)k = −a+∞∑

k=1

(1− a2)ak−1zk

we see that∞∑

k=0

|ck(fa)zk| = a+∞∑

k=1

(1− a2)ak−1|z|k = 2a+ fa(|z|)

which exceeds one for

|z| > 1

1 + 2a

For our purposes, we use the function

fa(z) =s(z)− a

1− a s(z), z ∈ B`n

1

where again 0 < a < 1 and s(z) = z1 + . . .+ zn. Then fa is bounded by onesince s(B`n

1) = D and we have

fa(z) = −a+ (1− a2)∞∑

k=1

ak−1s(z)k = −a+ (1− a2)∞∑

k=1

ak−1∑|α|=k

k!

α!zα


from which it follows c0(fa) = −a and

cα(fa) = (1− a2)a|α|−1 |α|!α!

for α ∈ N(N)0 , α 6= 0. Hence we have again∑

α∈N(N)0

|cα(fa)zα| = 2a+ fa(|z|) = 2a+‖z‖`n

1− a

1− a‖z‖`n1

which exceeds one if and only if

‖z‖`n1>

1

1 + 2a

if and only if1

n

n∑i=1

|zi| >1

n

1

1 + 2a

This implies

A(B`n1) ≤ 1

(1 + 2a)n

for all 0 < a < 1 and hence the desired estimate.

For the remaining inequality let 0 < r < 13

= K(D) and define r :=(r, 0, . . . , 0) ∈ Rn

≥0. Let f ∈ H∞(B`n1). Identifying D with D × 0n−1 we

have f |D ∈ H∞(D) and hence

∑α∈Nn

0

|cα(f)|rα =∞∑

k=0

|c(ke1)(f)|rk =∞∑

k=0

|ck(f |D)|rk ≤ ‖f |D‖D ≤ ‖f‖B`n1

It follows1

nr =

1

n

n∑i=1

ri ≤ A(B`n1)

With the same arguments we obtain a more general result.

Theorem 2.32.

Aλ(B`n1) =

Aλ(D)

n

Proof. For “≤” let r ∈ Rn≥0 with

∑α∈Nn

0|cα(f)| rα ≤ λ‖f‖B`n

1for all

f ∈ H∞(B`n1). Then we have to prove

n∑i=1

ri ≤ Aλ(D) (2.20)


Let f ∈ H∞(D). Using the function s defined in the proof of theorem 2.31we get f s ∈ H∞(B`n

1) with cα(f s) = ck(f) k!

α!for α ∈ Nn

0 with |α| = k. Itfollows

∞∑k=0

|ck(f)|‖r‖k`n1

=∞∑

k=0

|ck(f)|∑|α|=k

k!

α!rα =

∑α∈Nn

0

|cα(f s)| rα

≤ λ‖f s‖B`n1

= λ‖f‖D

and hence (2.20).

For the remaining inequality argue in the same way as in the proof of theorem2.31.

In [16, teorema B] Bombieri proved

Aλ(D) =1

3λ− 2√

2(λ2 − 1)

for 1 ≤ λ ≤√

2 (see also [17]).

Corollary 2.33. For 1 ≤ λ ≤√

2 we have

Aλ(B`n1) =

1

(3λ− 2√

2(λ2 − 1))n

Now we give general estimates of Aλ(Rn) where Rn ⊂ Cn is an arbitrarybounded Reinhardt domain. Again the key is a probabilistic estimate, apolynomial version of Chevet’s inequality, which was proved in [26, corollary3.2].

Theorem 2.34. Let (εα)|α|=m be a family of independent standard Bernoullirandom variables on a probability space (Ω, µ) and cα, |α| = m be scalars.There exists a constant 0 < Cm ≤

√m323m−1 such that for each bounded and

circled set U ⊂ Cn we have∫Ω

supz∈U

∣∣ ∑|α|=m

εαcαzα∣∣dµ

≤√

log nCm sup|α|=m

|cα|√α!

m!

∥∥ n∑i=1

|zi|∥∥

U

∥∥( n∑i=1

|zi|2)1/2∥∥m−1

U

Thus, given a circled and bounded subset U of Cn and a family (cα)|α|=m ofscalars, there are signs ±1 such that the m-homogeneous polynomial pm(z) =∑|α|=m

± cαzα satisfies

‖pm‖U ≤√m323m−1 log n sup

|α|=m

|cα|√α!

m!

∥∥ n∑i=1

|zi|∥∥

U

∥∥( n∑i=1

|zi|2)1/2∥∥m−1

U

(2.21)


Proposition 2.35. Let Rn ⊂ Cn be a bounded Reinhardt domain and λ ≥ 1.Then

Amλ (Rn) ≤ m

√λ 2m√m3m! 23m−1n log n

S(Rn, B`n2)

nfor all m ∈ N.

Proof. Let m ∈ N. Take r ∈ Rn≥0 with

∑|α|=m|cα(f)| rα ≤ λ‖f‖Rn for every

f ∈ Pm(Cn). Then by the above cited result we can choose signs εα ∈ −1, 1such that( n∑

i=1

ri

)m

=∑|α|=m

m!

α!rα ≤ λ

∥∥ ∑|α|=m

εαm!

α!zα∥∥

Rn

≤ λ√m3 23m−1 log n sup

|α|=m

√m!

α!

∥∥ n∑i=1

|zi|∥∥

Rn

∥∥( n∑i=1

|zi|2)1/2∥∥m−1

Rn

Since ‖∑n

i=1|zi|‖Rn ≤ n1/2‖(∑n

i=1|zi|2)1/2‖Rn and ‖(∑n

i=1|zi|2)1/2‖Rn =S(Rn, B`n

2), it follows

n∑i=1

ri ≤m√λ 2m√m3m! 23m−1n log n S(Rn, B`n

2)

which yields the claim.

Theorem 2.36. Let Rn ⊂ Cn be a bounded Reinhardt domain and λ ≥ 1.Then

max 1

3n

1

s1,n

,sn,1

nK(Rn) ≤ Aλ(Rn) ≤ 17 λ2/ log n

√log n

nsn,2

wheresp,n = S(B`n

p, Rn) , sn,p = S(Rn, B`n

p)

Proof. The lower bound follows from remark 2.28 and remark 2.27 togetherwith theorem 2.31 and it remains to prove the upper bound. By proposition2.35 it is enough to show that for every n ≥ 2 there is an m ∈ N with

m√λ 2m√m3m! 23m−1n log n ≤ 17λ2/ log n

√log n

and this is done as in [24, p.186], but somewhat more carefully. If n = 2, . . . , 8then m = 2 will be ok, otherwise take m = [log n] ≥ 2. Then log n ≤ 2m,m√m! ≤ m/

√2 and using x1/x ≤ e1/e for x > 0 we obtain

m√λ 2m√m3m! 23m−1n log n =

m√λ(

m√m)3/2 (m

√m!)1/2

23/2−1/(2m) 2m√n log n

≤ λ2/ log n (e1/e)3/2√

log n 25/4 (n log n)1/ log n

≤ λ2/ log n(e1/e)5/2 e√

log n 25/4

≤ 17λ2/ log n√

log n


In particular, we have

1

3n≤ Aλ(BXn) ≤ c

√log n

n

(with c depending on λ, but not on n) for every Banach sequence space Xsince `1 ⊂ X ⊂ `∞ with continuous inclusions and the extreme cases 1

3n

and√

log nn

are attained (up to a logarithm term, see (2.13)) for X = `1 and

X = `∞.

Comparing (2.14) with the estimates in theorem 2.30, we see that for `np -unit balls the arithmetic Bohr radius is very close to the Bohr radius B(Rn)defined by Aizenberg. The reason for this can be seen from the proofs, wheresimilar estimates are used. The next result also gives evidence for this fact.

Remark 2.37. Let Rn be a logarithmically convex Reinhardt domain in Cn.Then

A(Rn) = sup 1

n

n∑i=1

ri | r ∈ Rn≥0 ∩Rn, ∀f ∈ H∞(Rn) :∑

α∈Nn0

‖cα(f)zα‖rB`n∞≤ ‖f‖Rn

= sup 1

n

n∑i=1

|zi| | z ∈ Rn, ∀f ∈ H∞(Rn) :∑α∈Nn

0

|cα(f)zα| ≤ ‖f‖Rn

Here rB`n∞ of course means the set rz | z ∈ B`n

∞, i.e. the polydisc[|z| < r] ⊂ Cn with radius r.

Proof. The second equality is obvious. Denote

M := z ∈ Cn | ∀f ∈ H∞(Rn) :∑α∈Nn

0

|cα(f)zα| ≤ ‖f‖Rn

Then

supz∈M∩Rn

1

n

n∑i=1

|zi| = supz∈M∩Rn

1

n

n∑i=1

|zi| ≤ A(Rn)

and it remains to show

A(Rn) ≤ supz∈M∩Rn

1

n

n∑i=1

|zi|

So let x ∈ M and without loss of generality x = r ∈ Rn≥0. Suppose r /∈ Rn.

Then (Rn is compact) r and Rn can be divided by disjoint neighbourhoods,thus we can find δ > 0 and t > 1 such that

B(r, δ) ∩ tRn = ∅


Since Rn ⊂ tRn every f ∈ H(tRn) defines a bounded holomorphic functionon Rn and satisfies ∑

α∈Nn0

|cα(f)|rα ≤ ‖f‖Rn

(since r ∈M), hence can be extended by

f(z) :=

f(z) z ∈ tRn∑

α∈Nn0cα(f)zα |z| < r

to a holomorphic function f on tRn ∪ [|z| < r] % tRn. But this is impossiblesince by the assumption tRn is a domain of holomorphy (see e.g. [67, theorem3.28]).

Though the asymptotic is very similar, in general, the arithmetic Bohr radiusand B(Rn) are not identical. For example, we have

A(B`n1) =

1

3n< 1− n

√2

3≤ B(B`n

1)

by theorem 2.31, Bernoulli’s inequality and (2.15).

Upper inclusions for domH∞(R)

The next lemma allows us to estimate the decreasing rearrangement of theelements of domH∞(R) (proposition 2.40). Here we use ideas from [37] werethe case of power series in infinitely many variables bounded on B`∞ wasconsidered (see the introduction).

Lemma 2.38. Let R be a bounded Reinhardt domain in a Banach sequencespace X and F(R) be a closed subspace of H∞(R). Then for every z ∈domF(R) we can find a λ > 0 such that∑

α∈N(N)0

|cα(f)zα| ≤ λ‖f‖R

for all f ∈ F(R).

Proof. Denote F = F(R) and let z ∈ domF . For n ∈ N the set

Hn := f ∈ F |∑

α∈N(N)0

|cα(f)zα| ≤ n

is closed, since cα ∈ (H∞(R))′ for α ∈ N(N)0 by Cauchy’s integration formula

(see the remarks after (2.1)). Since z ∈ domF we have F = ∪nHn and

Upper inclusions for domH∞(R) 85

Baire’s theorem gives a natural number n0 such that the interior of Hn0 isnon empty. Thus we have f0 + sBF ⊂ Hn0 for some f0 ∈ F and some s > 0.So if f ∈ BF then

cα(f) =1

s(cα(f0 + sf)− cα(f0))

for α ∈ N(N)0 implies ∑

α∈N(N)0

|cα(f)zα| ≤ 2n0

s=: λ

Recall that if R is a Reinhardt domain in a Banach sequence space X thenRn = R ∩ Xn denotes the n-th section of R, i.e. Rn = πn(R), whereπn(∑∞

k=1 zkek) =∑n

k=1 zkek is the projection from X onto Xn. Then forall n ∈ N by f f πn we can consider H∞(Rn) as a subspace of H∞(R).Thus an application of the lemma to F(R) = H∞(R) gives the following:

Corollary 2.39. Let R be a bounded Reinhardt domain in a Banach sequencespace X. Then for every z ∈ domH∞(R) we can find a λ ≥ 1 such that

1

n

n∑i=1

|zi| ≤ Aλ(Rn)

for all n ∈ N.

Proposition 2.40. Let R be a bounded Reinhardt domain in a symmetricBanach sequence space X. Then

z∗n ≺√

log n

n‖id : Xn −→ `n2‖

for every z ∈ domH∞(R).

(See page 26 for the notation ≺.)

Proof. We have R ⊂ rBX for some r > 0 which implies

domH∞(R) ⊂ r domH∞(BX)

by remark 2.11. Hence it is enough to prove the claim for R = BX .

Let z ∈ domH∞(BX). We have

Aλ(BXn) ≤ c λ2/ log n

√log n

n‖id : Xn −→ `n2‖ (2.22)


by theorem 2.36. Since the inclusion mapping X → `∞ has norm 1, factor-ization gives ‖id : Xn −→ `n2‖ ≤ n1/2 and so (2.22) implies(

Aλ(BXn))

n∈ `2+ε (2.23)

for all ε > 0.

By the symmetry of BX , domH∞(BX) is invariant under permutations, thatmeans we have z σ ∈ domH∞(BX) for every bijection σ : N −→ N. Hence( 1

n

∑ni=1|zσ(i)|) converges to zero for every permutation σ by corollary 2.39

and (2.23). This implies zn −→ 0.

Thus the decreasing rearrangement z∗ ∈ BX (X is symmetric) is merely apermutation of the sequence |z| and we obtain

z∗n ≤1

n

n∑i=1

z∗i ≤ Aλ(BXn)

for some constant λ ≥ 1, which, by (2.22), finishes the proof.

In particular, factorization through `n∞ as in the above proof gives a generalupper inclusion.

Corollary 2.41. Let R be a bounded Reinhardt domain in a symmetric Ba-nach sequence space X and ε > 0. Then

domH∞(R) ⊂ `2+ε

Definition 2.42. A Banach sequence space is called regular, if( 1

λX(n)nε

)n∈ X for every ε > 0

This condition is, for example, satisfied for Lorentz sequence spaces d(w, p)if

n∑i=1

wi nwn

so, in particular, for the `p,q-spaces (see the preliminaries) and as a specialcase for the `p-spaces, 1 ≤ p <∞, as well as for c0 and `∞.

Now let ϕ be an Orlicz function and ε > 0. Given λ > 0 choose n0 such thatnε

0 > 1/λ. Then for all n ≥ n0 and all ρ > 0 with ϕ(1/ρ) ≤ 1/n we have

ϕ( 1

ρ nε λ

)= ϕ

( 1

nε λ

1

ρ+(

1− 1

nε λ

)0)≤ 1

nε λϕ(1

ρ

)≤ 1

λ

1

n1+ε

by the convexity of ϕ. This implies

ϕ( 1

λ`ϕ(n)nε λ

)≤ 1

λ

1

n1+ε

for all n ≥ n0 and hence `ϕ is regular.

For regular Banach sequence spaces we use the following lemma.

Upper inclusions for domH∞(R) 87

Lemma 2.43. Let X be a regular and symmetric Banach sequence space,1 ≤ r, s < ∞ and ε > 0. Let z = (zn) be a zero sequence. If there is aconstant K > 0 such that for all n

z∗n ≤ K

(1

λX(n)

)1/r (1

n

)1/s

log n

then z ∈ X 1r · `s+ε.

Proof. We choose 0 < ε1 < ε and δ1, δ2 > 0 such that

1

s=δ1r

+1

s+ ε1

+ δ2

Then

z∗n ≤ K

(1

nδ1 λX(n)

)1/r (1

n

)1/(s+ε1)log n

nδ2

and hence z∗ = x ·y for some x ∈ X1/r and y ∈ `s+ε, x, y > 0. Since zn −→ 0we have z∗ = |zσ| = (|zσ(n)|)n for some permutation σ : N −→ N. Thenxr

σ−1 ∈ X and yσ−1 ∈ `s+ε by the symmetry of X and `s+ε. Choose λ ∈ CN

with |λ| = 1 such that z = λ|z|. Then z = λxσ−1 · yσ−1 ∈ X1/r · `s+ε.

We are now able to prove our upper inclusions for domH∞(R), and togetherwith the lower ones (theorem 2.25) we obtain the following result.

Theorem 2.44. Let R be a bounded Reinhardt domain in a symmetric Ba-nach sequence space X and ε > 0.

(1) If X ⊂ `2 then

`1 ∩R ⊂ domH∞(R) ⊂ `1+ε ∩R

(2) If X is 2-convex and regular then

(X · `2) ∩R ⊂ domH∞(R) ⊂ (X · `2+ε) ∩R

Proof. The lower bounds are special cases of theorem 2.25.

For the upper bounds let z ∈ domH∞(R). Then we have

z∗n ≺√

log n

n‖id : Xn −→ `n2‖

by proposition 2.40. This gives (1), and to understand (2) we use‖id : Xn −→ `n2‖ n1/2/λX(n) (see (1.14)). Then

z∗n ≤ K1

λX(n)

(1

n

)1/2

log n

By corollary 2.41 we have zn −→ 0, and hence we can apply lemma 2.43 toconclude z ∈ X · `2+ε.


From proposition 2.40 we can also deduce a more general upper inclusion fordomH∞(R), which holds for a wider class of Banach sequence spaces X butis weaker than theorem 2.44. This sharpens the estimate of corollary 2.41 byposing an additional assumption on X.

Theorem 2.45. Let X be a regular and symmetric Banach sequence space,R ⊂ X be a bounded Reinhardt domain and ε > 0. Then

domH∞(R) ⊂ X1/2 · `2+ε

Proof. Let z ∈ domH∞(R). Since

‖id : Xn −→ `n2‖ = sup‖z‖X≤1

( n∑i=1

|zi|2)1/2

≤ sup‖z‖X≤1

( n∑i=1

|zi|)1/2

=∥∥∥ n∑

i=1

e′i

∥∥∥1/2

X×= (λX×(n))1/2

(2.24)

it follows by proposition 2.40 and (1.12) that

z∗n ≤ K

(1

λX(n)

)1/2 (1

n

)1/2

log n

We have z ∈ c0 by corollary 2.41, and lemma 2.43 gives z ∈ X1/2 · `2+ε.

We apply theorem 2.44 to some well known sequence spaces.

Example 2.46.

(a) For X = `p, 1 ≤ p ≤ ∞ we have

(1) if p < 2 then

`1 ∩R ⊂ domH∞(R) ⊂ `1+ε ∩R

(2) if p ≥ 2 and 1q

= 12

+ 1p

then

`q ∩R ⊂ domH∞(R) ⊂ `q+ε ∩R

(b) Let X = `p,q, 2 ≤ q < p < ∞ and r, s ≥ 1 such that 1r

= 1p

+ 12,

1s

= 1q

+ 12. Then

`r−ε,s ∩R ⊂ domH∞(R) ⊂ `r+ε,s ∩R

(c) If ϕ is a non degenerate Orlicz function which satisfies the ∆2-conditionand there is an equivalent Orlicz function ϕ such that ϕ(

√· ) is convex.

Then(`ϕ · `2) ∩R ⊂ domH∞(R) ⊂ (`ϕ · `2+ε) ∩R

A description of domPm(X) 89

Proof. For part (a) use Holder’s inequality.

To understand (b) recall that

`t,u ⊂ `v,w

if 0 < t < v ≤ ∞ and 0 < u,w <∞. This together with Holder’s inequalitygives `r−ε,s ⊂ `p,q · `2 and `p,q · `2+δ ⊂ `r+ε,s for ε > δ > 0. Under theassumptions on p and q, the space `p,q is 2-convex (see the preliminaries, p.20) and regular (see p. 86) so that we can apply theorem 2.44.

Under the conditions of (c) the Orlicz space `ϕ = hϕ is 2-convex (preliminar-ies, p. 21) and we have already checked that every Orlicz space is regular (p.86). So again the assumptions of theorem 2.44 are satisfied.

A description of domPm(X)

Continuous polynomials on a Banach sequence space X are holomorphicfunctions on X. We now examine the domain of convergence of the space ofall continuous m-homogeneous polynomials on X.

The 1-homogeneous polynomials are the continuous linear functionals on X.The monomial expansion of a functional ϕ ∈ X ′ is

∑ϕ(en)zn and we have

ϕ(z) =∑n∈N

ϕ(en)zn

for every z ∈ X if the en’s form a basis of X (which is then unconditional).Hence

domX ′ = X

and from now on in this section, we will assume m ≥ 2.

Let us first estimate the set domPm(X) from below. We start with a the-orem of Bohnenblust and Hille [12, theorem I] which is a generalization ofLittlewood’s result [58, theorem 1] about the summability of the coefficientsof a bounded bilinear form on `∞.

Theorem 2.47. Let P be a continuous m-homogeneous polynomial on `∞.Then (∑

|α|=m

|cα(P )|2(1+ 1m

)−1

) 12(1+ 1

m)

<∞

The result is actually proved for spanen = c0. (Recall that every polyno-mial P ∈ Pm(c0) can be extended to `∞ under preservation of the norm bythe Aron-Berner extension, see [36, prop. 1.51].) An application of Holder’sinequality gives


Corollary 2.48. Every continuous m-homogeneous polynomial P on c0 sat-isfies

`2(1− 1m

)−1 ⊂ dom(P )

In general we get the following lower inclusion for the domain of convergenceof the m-homogeneous polynomials (compare with theorem 2.25).

Theorem 2.49. Let X be a Banach sequence space and m ≥ 2. Then

`1 ∪X · `2(1− 1m

)−1 ⊂ domPm(X)

Proof. We have `1 ⊂ X and the inclusion mapping is continuous. Hencethe restriction of a polynomial P ∈ Pm(X) is a continuous m-homogeneouspolynomial on `1 and, in particular, a holomorphic function on `1 which bytheorem 2.25 implies `1 ⊂ dom(P ).

On the other hand, given x ∈ X and P ∈ Pm(X) the function Px defined byPx(z) = P (xz) is a continuous m-homogeneous polynomial on c0 which bycorollary 2.48 satisfies `2(1− 1

m)−1 ⊂ dom(Px) and hence

x · `2(1− 1m

)−1 ⊂ dom(P )

for all x ∈ X.

To give upper inclusions for domPm(X) we use the following lemma whichis an analogue to proposition 2.40.

Lemma 2.50. Let X be a symmetric Banach sequence space andz ∈ domPm(X). Then

z∗n ≺2m√

log n

n‖id : Xn −→ `n1‖

1m ‖id : Xn −→ `n2‖1−

1m

If X ′ has non trivial cotype then

z∗n ≺1

n‖id : Xn −→ `n1‖

1m ‖id : Xn −→ `n2‖1−

1m

Proof. Let λ > 1, n ∈ N and r ∈ Rn≥0 with∑

α∈Nn0

|α|=m

|cα(P )|rα ≤ λ‖P‖

for every P ∈ Pm(Xn). Then by (2.21) we can find εα ∈ −1, 1 and aconstant cm depending only on m such that( n∑

i=1

ri

)m

=∑|α|=m

m!

α!rα ≤ λ

∥∥ ∑|α|=m

εαm!

α!zα∥∥

≤ cm λ√

log n ‖id : Xn −→ `n1‖ ‖id : Xn −→ `n2‖m−1

(2.25)

A description of domPm(X) 91

Thus

Amλ (BXn) ≤ c′m λ

1m

2m√

log n

n‖id : Xn −→ `n1‖

1m ‖id : Xn −→ `n2‖1−

1m (2.26)

for all λ ≥ 1. Factoring through `n∞ this, in particular, implies

Amλ (BXn) ≺

2m√

log n

n12(1− 1

m)

n→∞−−−→ 0

for all λ ≥ 1. By lemma 2.38 with F(X) = Pm(X) we can find λ > 1 suchthat ∑

|α|=m

|cα(P )zα| ≤ λ‖P‖

for all P ∈ Pm(X). It follows

1

n

n∑i=1

|zi| ≤ Amλ (BXn)

and with the same argument as in the proof of proposition 2.40 and using(2.26)

z∗n ≤ Amλ (BXn) ≺

2m√

log n

n‖id : Xn −→ `n1‖

1m ‖id : Xn −→ `n2‖1−

1m

If X ′ has finite cotype s then the log-term in (2.25) can be replaced by thecotype-s constant Cs(X

′) of X ′ (see [78, (4.3)]) which yields the second partof the claim.

Theorem 2.51. Let X be a symmetric Banach sequence space and ε > 0.Then:

(1) (a) domPm(X) ⊂ `2(1− 1m

)−1+ε

(b) X regular: domPm(X) ⊂ X12(1+ 1

m) · `2(1− 1

m)−1+ε

(c) X ′ finite cotype: For all z ∈ domPm(X):

supn

(λX(n))12(1+ 1

m) n

12(1− 1

m) z∗n <∞

(2) If X ⊂ `2:

(a) domPm(X) ⊂ `(1− 12m

)−1+ε

(b) X regular: domPm(X) ⊂ X1m · `(1− 1

m)−1+ε


supn

(λX(n))1m n1− 1

m z∗n <∞

(3) If X is 2-convex:


(b) X regular: domPm(X) ⊂ X · `(1− 1m

)−1+ε


supnλX(n)n1− 1

m z∗n <∞

Proof. By lemma 2.50 we have

z∗n ≺2m√

log n

n‖id : Xn −→ `n1‖

1m ‖id : Xn −→ `n2‖1−

1m

and we can replace the log-term by 1, if X ′ has finite cotype. Further

‖id : Xn −→ `n1‖ = sup‖z‖X≤1

n∑i=1

|zi| =∥∥∥ n∑

i=1

e′i

∥∥∥X×

= λX×(n) =n

λX(n)

(for the last equalities we use symmetry, see (1.12)), and

(1) ‖id : Xn −→ `n2‖ ≤ ‖id : Xn −→ `n1‖1/2 =

(n

λX(n)

)1/2

(2) ‖id : Xn −→ `n2‖ ≤ ‖X → `2‖ <∞ (if X ⊂ `2)

(3) ‖id : Xn −→ `n2‖ ≤ cn1/2

λX(n)(if X is 2-convex)

It follows

z∗n ≤ K

(1

λX(n)

)1/r (1

n

)1/s

log n (2.27)

with

(1) r = 2(1 + 1m

)−1 , s = r′ = 2(1− 1m

)−1

(2) r = m , s = (1− 1m

)−1

(3) r = 1 , s = 2(1− 1m

)−1

and hence the claim by lemma 2.43 in the case where X is regular. If X ′ hasfinite cotype, we can replace the log-term in (2.27) by 1 and also get whatwe want. For the remaining cases we deduce from (2.27) that

z∗n ≤ K ′(

1

n

)1/s′

log n

with

(1) s′ = 2(1− 1m

)−1

(2) s′ = (1− 12m

)−1

A conjecture 93

since (1) r ≥ 1 and (2) λX(n) ≥ c λ`2(n) = c n1/2 if X ⊂ `2. This implies (a)in both cases.

For the first part of this statement also note that

domPm(`∞) ⊂ `2(1− 1m

)−1+ε (2.28)

for all ε > 0 by lemma 2.50 and if X and Y are Banach sequence spaces withX ⊂ Y then

domPm(X) ⊂ domPm(Y )

(see also remark 2.11).

Note that if R is a bounded Reinhardt domain in a Banach sequence space Xthen every continuous m-homogeneous polynomial on X defines a boundedholomorphic function on R. Hence domH∞(R) ⊂ domPm(X) for all m andwe can deduce the upper inclusions of theorem 2.44 from theorem 2.51.

Together with theorem 2.49 we can determine domPm(X) almost preciselyin the 2-convex case.

Corollary 2.52. Let X be a 2-convex, regular and symmetric Banach se-quence space. Then

X · `2(1− 1m

)−1 ⊂ domPm(X) ⊂ X · `2(1− 1m

)−1+ε

for all ε > 0.

A conjecture

From corollary 2.22 and theorem 2.51 we get

`1 ⊂ domPm(X) ⊂ X1m · `(1− 1

m)−1+ε

for all ε > 0 if X is regular and contained in `2. We presume that in analogyto corollary 2.52 domPm(X) is nearly identical with X

1m · `(1− 1

m)−1 .

Conjecture 2.53. Let X be a symmetric Banach sequence space containedin `2. Then

X1m · `(1− 1

m)−1 ⊂ domPm(X)

For `p-spaces we know the following.

Example 2.54. We consider X = `p, 1 ≤ p ≤ ∞. Define qm by1

qm= 1

p+ 1

2(1− 1

m). Then

(1) For p =∞ :

`2(1− 1m

)−1 ⊂ domPm(`∞) ⊂ `2(1− 1m

)−1+ε


(2) For 2 ≤ p <∞ :`qm ⊂ domPm(`p) ⊂ `qm,∞

(3) For 1 ≤ p ≤ 2 :

`max1,qm ⊂ domPm(`p) ⊂ `(mp′)′,∞

Note that max1, qm = qm if and only if p ≥ 2(1 + 1m

)−1.

For X = `p, 1 < p < 2 our conjecture is

`(mp′)′ ⊂ domPm(`p) (2.29)

In the following we investigate this conjecture for the `p-spaces.

Let’s see what we actually have proved to obtain the lower bound in theorem2.49.

Let P =∑cαz

α ∈ Pm(`p). For 2 ≤ p ≤ ∞ our argument gives

supz∈B`qm

∑|α|=m

|cαzα| = supx∈B`p

supy∈B`

2(1− 1m )−1

∑|α|=m

|cα(xy)α|

≤ cm supx∈B`p

supz∈Bc0

∣∣ ∑|α|=m

cα(xz)α∣∣

≤ cm ‖P‖

(2.30)

where the constant cm from the result of Bohnenblust and Hille only dependson m and not on the polynomial P (see also [48] for an estimate of theconstant).

For p = 1 we have

supz∈B`1

∑|α|=m

|cαzα| = supz∈B`1

∑|α|=m

|cα|αα

mm

mm

αα|zα|

≤ supz∈B`1

∑|α|=m

‖cαzα‖B`1

m!

α!em |zα|

≤ em ‖P‖∑|α|=m

m!

α!|zα| = em ‖P‖ ‖z‖m`1

≤ em ‖P‖ .

(2.31)

More generally we have the following result for holomorphic functions ofbounded type from [70]. A holomorphic function on a Banach space E iscalled of bounded type if it is bounded on bounded subsets of E.

Theorem 2.55. Let cα ∈ C, α ∈ N(N)0 .

The following are equivalent:

A conjecture 95

(1) (cααα

|α||α| r|α|)α ∈ `∞(N(N)

0 ) for all r > 0.

(2)∑

α∈N(N)0cαz

α is bounded on bounded subsets of `1.

(3)∑

α∈N(N)0|cαzα| is bounded on bounded subsets of `1.

In this case f(z) :=∑

α∈N(N)0cαz

α defines a holomorphic function of bounded

type on `1 with

supz∈rB`1

∑α∈N(N)

0

|cαzα| ≤ r + ε

ε‖f‖e(r+ε)B`1

for all r > 0 and all ε > 0.

These considerations lead to the following definition.

Definition 2.56. Let X and Y be Banach sequence spaces. We defineκm(Xn, Yn) to be the best constant c > 0 such that for all P ∈ Pm(Xn)

supz∈BYn

∑|α|=m

|cα(P )zα| ≤ c ‖P‖Pm(Xn)

In this connection the following notion is useful (see [10, p. 3]).

Definition 2.57. Let R be a Reinhardt domain in a Banach sequence spaceX and f be a holomorphic mapping on R. Then

Mf :dom(f) −→ C

z ∑

α∈N(N)0|cα(f)|zα

is called the majorant function of f .

In a Banach sequence space X

supz∈BXn

∑|α|=m

|cα(P )zα| = supz∈BXn

∣∣ ∑|α|=m

|cα(P )|zα∣∣

for all P ∈ Pm(Xn), thus κm(Xn, Yn) is the best constant c > 0 satisfying

‖MP‖Pm(Yn) ≤ c ‖P‖Pm(Xn)

for all P ∈ Pm(Xn).

The next result will help us to give a reformulation of our conjecture.

Theorem 2.58. Let X and Y be Banach sequence spaces. Then the followingstatements are equivalent:

(1) Y ⊂ domPm(X)


(2) supn κm(Xn, Yn) <∞

In this case supn κm(Xn, Yn) is the best constant c > 0 such that

‖MP‖Pm(Y ) ≤ c ‖P‖Pm(X) (2.32)

holds for every P ∈ Pm(X).

Proof. First let us suppose that (1) is true. For k, n ∈ N the set

Lk,n := P ∈ Pm(X) | supz∈ 1

kBY

∑|α|=m

|cα(P )zα| ≤ n

is closed and we have Pm(X) = ∪k,nLk,n : For P ∈ Pm(X) and n ∈ N theset

Hn = z ∈ Y |∑|α|=m

|cα(P )zα| ≤ n

is closed and by the assumption we have Y = ∪nHn, hence by Baire’s theoremthere must be an n such that Hn has non empty interior. As in the proof oftheorem 2.83 we conclude that 1

kBY ⊂ Hn for some natural number k, i.e.

supz∈ 1

kBY

∑|α|=m

|cα(P )zα| ≤ n

and thus P ∈ Lk,n.

Now, again by Baire’s theorem, the interior of some Lk,n is not empty. As inthe proof of theorem 2.83 this implies the existence of an s > 0 such that

supz∈ 1

kBY

∑|α|=m

|cα(P )zα| ≤ 2n

s

or

‖MP‖Pm(Y ) ≤2nkm

s

for all P ∈ BPm(X), thus supν κm(Xν , Yν) ≤ 2nkm/s.

Now let us assume that (2) is true and define c := supn κm(Xn, Yn). LetP ∈ Pm(X) and Pn := P |spane1,...,en. Then cα(P ) = cα(Pn) for α ∈ Nn

0 andhence∑

α∈Nn0

|α|=m

|cα(P )zα| ≤ ‖MPn‖Pm(Yn) ≤ κm(Xn, Yn) ‖Pn‖Pm(Xn) ≤ c ‖P‖Pm(X)

for all n and all z ∈ BY which implies

‖MP‖Pm(Y ) ≤ c ‖P‖Pm(X)

A conjecture 97

since every finite subset A ⊂ [|α| = m] is contained in Nn0 for some n. In

particular, we have BY ⊂ domPm(X) and by m-homogeneity it follows (1).

Finally suppose c′ is a constant satisfying ‖MP‖Pm(Y ) ≤ c′ ‖P‖Pm(X) for all

P ∈ Pm(X). Take P =∑

α∈Nn0 ,|α|=m cαz

α ∈ Pm(Xn) and define P (z) :=∑α∈N(N)

0cαz

α, z ∈ X where cα := cα if α ∈ Nn0 and zero otherwise. Then

‖MP‖Pm(Yn) = ‖MP‖Pm(Y ) ≤ c′ ‖P‖Pm(X) = c′ ‖P‖Pm(Xn)

and hence c ≤ c′.

Thus our conjecture (2.29) for `p-spaces is equivalent to

supnκm(`np , `

n(mp′)′) <∞

for 1 < p < 2. To decide this we investigate the asymptotic of κm(`np , `nq ) for

general p and q. As we can see from the calculations (2.30) and (2.31) wehave supn κm(`np , `

nqm

) ≤ cm and supn κm(`n1 , `n1 ) ≤ em.

We are going to identify κm(Xn, Yn) with another constant which is easierfor us to handle.

Definition 2.59. Let E and F be Banach spaces, x1, . . . , xn ∈ E and T ∈L(E,F ). We define χ(T, (x1, . . . , xn)) to be the best constant satisfying

∥∥ n∑i=1

εiµiTxi

∥∥ ≤ c∥∥ n∑

i=1

µixi

∥∥for all µ1, . . . , µn ∈ K and ε1, . . . , εn ∈ K with |εi| ≤ 1 and call it theunconditional constant of T with respect to (x1, . . . , xn).

If x1, . . . , xn are linearly independent, E = F and T = idE is the identitymap then χ(idE, (x1, . . . , xn)) = χ((x1, . . . , xn)) is the unconditional basisconstant of (x1, . . . , xn).

Note that

χ(T, (x1, . . . , xn)) = χ(T, (xσ(1), . . . , xσ(n)))

for all permutations σ ∈ Sn. Thus it makes sense to write χ(T, (xj)j∈J) fora finite family of elements xj ∈ E and we have

χ(T, (xi)i∈I) ≤ χ(T, (xi)i∈J) (2.33)

if I ⊂ J . Further χ( · , (xj)j∈J) is obviously homogeneous and satisfies thetriangle inequality on L(E,F ).


Remark 2.60. Let E, F , G, H be Banach spaces, (xi)i∈I a finite family ofelements xi ∈ E and T ∈ L(E,F ), S ∈ L(F,G), R ∈ L(G,H):

ET−→ F

S−→ GR−→ H

Let I0 := i ∈ I | Txi 6= 0 . Then

χ(RST, (xi)i∈I) ≤ ‖R‖χ(S, (Txi)i∈I0) ‖T‖

If (xi)i∈I and (yj)j∈J are finite families of elements in E then by (xi)i∈I ⊂(yj)j∈J we mean that there is an injective mapping ϕ : I −→ J such thatxi = yϕ(i) for all i ∈ I. Then (xi)i∈I = (yj)j∈J means (xi)i∈I ⊂ (yj)j∈J and(yj)j∈J ⊂ (xi)i∈I and (xi)i∈I∪(yj)j∈J defines a family (zk)k∈K where the indexset K has a disjoint decomposition K = K1∪K2 such that (zk)k∈K1 = (xi)i∈I

and (zk)k∈K2 = (yj)j∈J .

The following two results are easy to check.

Remark 2.61. Let E and F be Banach spaces, (xi)i∈I and (yj)j∈J be finitefamilies in E and R,S ∈ L(E,F ). Then

χ(idF , (Rxi)i∈I ∪ (Syj)j∈J) ≤ (‖R‖+ ‖S‖)χ(idE, (xi)i∈I ∪ (yj)j∈J)

Lemma 2.62. Let E and F be Banach spaces, (xs)s∈S a finite family in Eand I ∈ L(E,F ) be an isomorphism. Then

a ‖I−1‖−1 ≤ χ(I, (xs)s∈S) ≤ b ‖I‖

where

a = maxχ(idE, (xs)s∈S), χ(idF , (Ixs)s∈S)

b = min χ(idE, (xs)s∈S), χ(idF , (Ixs)s∈S)

Let X and Y be Banach sequence spaces and J ∗(m,n) := α ∈ Nn | |α| =m . Then the monomials (zα)α∈J ∗(m,n) form a basis of Pm(Xn). For T ∈L(Pm(Xn),Pm(Yn)) we define

χM(T ) := χ(T, (zα)α∈J ∗(m,n))

The next remark can be seen as a generalization of [24, lemma 2.1].

Remark 2.63. Let X and Y be Banach sequence spaces. Then

κm(Xn, Yn) = χM(id : Pm(Xn) −→ Pm(Yn))

A conjecture 99

Proof. Let P =∑

|α|=m cαzα ∈ Pm(Xn). Define εα := |cα|/cα if cα 6= 0 and 1

otherwise. Then

‖MP‖Pm(Yn) =∥∥ ∑|α|=m

εαcαzα∥∥Pm(Yn)

≤ χM(id : Pm(Xn) −→ Pm(Yn))‖P‖Pm(Xn)

hence κm(Xn, Yn) ≤ χM(id). On the other hand, given scalars cα and εα with|εα| ≤ 1 we have P :=

∑|α|=m cαz

α ∈ Pm(Xn) and∥∥ ∑|α|=m

εαcαzα∥∥Pm(Yn)

≤∥∥ ∑|α|=m

|cαzα|∥∥Pm(Yn)

= ‖MP‖Pm(Yn)

≤ κm(Xn, Yn) ‖P‖Pm(Xn)

which implies the other inequality.

Now our problem is to find asymptotic estimates for

χM(id : Pm(`np ) −→ Pm(`nq ))

and we use lemma 1.1 and the identification Pm(Xn) = ⊗m,sεs

(Xn)′ for Banachsequence spaces X to shift the problem to full tensor products.

Recall that (ei)i∈M(m,n) is a basis of ⊗mKn and (σm(ej))j∈J (m,n) a basis of⊗m,sKn (see the preliminaries, page 16). In particular, if X is a Banachsequence space and e′k denote the coefficient functionals with respect to thestandard unit vectors ek (and e′j = e′j1 ⊗ · · · ⊗ e′jm

for j ∈ J (m,n)) then(σm(e′j))j∈J (m,n) is a basis of ⊗m(Xn)′ and the identification ⊗m

εs(Xn)′ =

Pm(Xn) is given by σm(e′j) = zα where j ∈ J (m,n) and α ∈ J ∗(m,n) suchthat αi is the number of elements k ∈ 1, . . . ,m with jk = i for i = 1, . . . , n.

This suggests the following definition of χM for operators between arbitrarysymmetric and full m-fold tensor products of Banach sequence spaces: Fornorms α, β on ⊗mXn and ⊗mYn and αs, βs on ⊗m,sXn and ⊗m,sYn, respec-tively and operators T ∈ L(⊗m

αXn,⊗mβ Yn) and S ∈ L(⊗m,s

αsXn,⊗m,s

βsYn) let

χM(T ) := χ(T, (ei)i∈M(m,n))

χM(S) := χ(S, (σm(ej))j∈J (m,n))

To relate χM for identities between symmetric and full tensor products wecomplete the elements σm(ej) ∈ ⊗m,sKn ⊂ ⊗mKn, j ∈ J (m,n) with elementsei, i ∈M(m,n) to a basis of ⊗mK.

Proposition 2.64. Let m,n ∈ N and Γ :=M(m,n) \ J (m,n). Then

(σm(ej))j∈J (m,n) ∪ (ej)j∈Γ

is a basis of ⊗mKn.


Proof. Let ∑j∈J (m,n)

λjσm(ej) +∑j∈Γ

λjej = 0

with λj ∈ K. Denote [j] the set of all permutations of j and |j| = |[j]|. Using[21, lemma 1] we get

∑j∈J (m,n)

λjσm(ej) +∑j∈Γ

λjej =∑

j∈J (m,n)

λj

|j|ej +

∑j∈J (m,n)

∑i∈[j]\j

(λj

|j|+ λi

)ei

and this implies λj = 0 for all j ∈ J (m,n) and then also λi = −λj/|j| = 0for all j ∈ J (m,n) and i ∈ [j] \ j.

Theorem 2.65. Let α, β be tensor norms with equivalent s-tensor norms αs

and βs, respectively. Let X and Y be Banach sequence spaces. Then

χM(id : ⊗m,sαsXn −→ ⊗m,s

βsYn) χM(id : ⊗m


Proof. Writing idn : Xn −→ Yn we have the commutative diagram

⊗m,sαsXn ⊗m,s

βsYn

⊗mαXn ⊗m

β Yn

-

-?

6

⊗m idn

ιm σm

⊗m idn

and using remark 2.60, (2.33) and remark 2.61 together with proposition 2.64we obtain

χM(⊗m,s idn) = χ(σm ⊗m idn ιm, (σm(ej))j∈J (m,n))

≤ ‖σm‖χ(⊗m idn, (ιm σm(ej))j∈J (m,n))‖ιm‖

≤ ‖σm‖‖ιm‖χ(⊗m idn, (ιm σm(ej))j∈J (m,n) ∪ (ej)j∈Γ)

≤ ‖σm‖‖ιm‖ (‖σm‖‖ιm‖ + 1)χ(⊗m idn, (ej)j∈M(m,n))

≤ cm χM(⊗m idn)

with cm ≤ (mm/m!)2((mm/m!)2 + 1).

For the reverse asymptotic relation we check the conditions of lemma 1.1.

χM satisfies (i) (see the remarks right after definition 2.59).

(ii) Given the situation

⊗mαXNn

Pi−→ ⊗mαXn

⊗m idn−−−−→ ⊗mβ Yn

Ii−→ ⊗mβ YNn

A conjecture 101

with Pi and Ii defined as in (1.1) for i ∈ M(m,Nn) and let J0 := j ∈M(m,Nn) | Pi(ej) 6= 0 . Then (1.2) shows us (Pi(ej))j∈J0 ⊂ (ej)j∈M(m,n)

and hence using remark 2.60 and (2.33) we obtain

χM(Ii ⊗m idn Pi) = χ(Ii ⊗m idn Pi, (ej)j∈M(m,Nn))

≤ ‖Ii‖‖Pi‖χ(⊗m idn, (Piej)j∈J0)

≤ χ(⊗m idn, (ej)j∈M(m,n))

= χM(⊗m idn)

where ‖Ii‖ = 1 = ‖Pi‖ because of the metric mapping property of α and β.In particular, we can take c1 = 1.

(iii) Let N1 ≤ N2. As in the proof of proposition 1.2 we define the naturalinclusion I : YN1 −→ YN2 and projection P : XN2 −→ XN1 . Then

⊗mαXN1 ⊗m

β YN1

⊗mαXN2 ⊗m

β YN2

-

-?

6

⊗m idN1

⊗mI ⊗mP

⊗m idN2

is commutative and hence

χM(⊗m idN1) = χ(⊗mP ⊗m idN2 ⊗m I, (ej)j∈M(m,N1))

≤ (‖P‖‖I‖)m χ(⊗m idN2 , (ej)j∈M(m,N2))

= χM(idN2)

by remark 2.60 and (2.33) since ⊗mI(ej) = ej. In particular, c2 = 1 isadequate.

(iv) As in the proof of proposition 1.2 we use the factorization

⊗mαXn ⊗m

β Yn

⊗m,sαsXmn+k ⊗m,s

βsYmn+k

-

-?

6

⊗m idn

I := σmI1⊗···⊗Im P := m! P1⊗···⊗Pm ιm

⊗m,s idmn+k

By (1.4) we have I(ej) = σm(ej+n(i−1)) for i = (1, . . . ,m), thus(I(ej))j∈J (m,n) ⊂ (σm(ej))j∈J (m,mn+k) and again using remark 2.60 and (2.33)we estimate

χM(⊗m idn) = χ(P ⊗m,s idmn+k I, (ej)j∈M(m,n))

≤ ‖P‖‖I‖χ(⊗m,s idmn+k, (σm(ej))j∈J (m,mn+k))

= ‖P‖‖I‖χM(⊗m,s idmn+k)


with ‖P‖‖I‖ ≤ (mm

m!)2 =: c3.

Now lemma 1.1 implies χM(⊗m idn) ≤ c χM(⊗m,s idn) for all n ≥ m withc ≤ (mm

m!)2(m+ 1)m, and this finishes the proof.

Corollary 2.66. Let X and Y be symmetric Banach sequence spaces. Then

χM(id : Pm(Xn) −→ Pm(Yn)) χM(id : ⊗mε (Xn)′ −→ ⊗m

ε (Yn)′)

With this at hand, let us first reprove the lower inclusions in example 2.54.

The next lemma mimics the “multiplication trick” used in the proof of the-orem 2.49.

Lemma 2.67. Let 1 ≤ p ≤ q ≤ r ≤ ∞ and pr, qr ∈ [1,∞] such that1pr

= 1p− 1

rand 1

qr= 1

q− 1

r. Then

χM(id : ⊗mε `

npr−→ ⊗m

ε `nqr

) ≤ χM(id : ⊗mε `

np −→ ⊗m

ε `nq )

Proof. We are simply using the fact that B`np′r

= B`nrB`n

p′and B`n

q′r= B`n

rB`n

q′

by Holder’s inequality since 1/p′r = 1/r + 1/p′ and 1/q′r = 1/r + 1/q′.

Let I :=M(m,n) and µi, εi ∈ K with |εi| ≤ 1 for i ∈ I. Then∥∥∑i∈I

εi µi ei

∥∥⊗m

ε `nqr

= supxj∈B`n

q′r

∣∣∑i∈I

εi µi x1(i1) · · ·xm(im)∣∣

= supyj∈B`n

r

supzj∈B`n

q′

∣∣∑i∈I

εi µi y1(i1) · · · ym(im) z1(i1) · · · zm(im)∣∣

= supyj∈B`n

r

∥∥∑i∈I

εi µi y1(i1) · · · ym(im)ei

∥∥⊗m

ε `nq

≤ χM(id : ⊗mε `

np −→ ⊗m

ε `nq ) sup

yj∈B`nr

∥∥∑i∈I

µi y1(i1) · · · ym(im)ei

∥∥⊗m

ε `np

= χM(id : ⊗mε `

np −→ ⊗m

ε `nq )∥∥∑

i∈I

µi ei

∥∥⊗m

ε `npr

Theorem 2.68. Let 1 ≤ p ≤ ∞. Then

χM(id : ⊗mε `

np −→ ⊗m

ε `n∞) ≤ 1

If 1 ≤ p ≤ 2(1− 1m

)−1 and q such that 1q

= 1p− 1

2(1− 1

m) then

χM(id : ⊗mε `

np −→ ⊗m

ε `nq ) ≤ (

√2)m−1

A conjecture 103

Proof. Factorization (see remark 2.60) gives

χM(id : ⊗mε `

np −→ ⊗m

ε `n∞)


n∞ −→ ⊗m

ε `n∞) ‖id : ⊗m

ε `np −→ ⊗m

ε `n∞‖

= χ(id : `nm

∞ −→ `nm

∞ , (ek)nm

k=1) ‖id : `np −→ `n∞‖m

= 1

in the general case and for p = 1, q = 2(1 + 1m

)−1 we have

χM(id : ⊗mε `

n1 −→ ⊗m

ε `nq )

≤ ‖id : `nm

q −→ ⊗mε `

nq ‖χ(id : `n

m

q −→ `nm

q , (ek)nm

k=1) ‖id : ⊗mε `

n1 −→ ⊗m

q `nq ‖

≤ ‖id : ⊗mε `

n1 −→ ⊗m

ε `nq ‖

≤ (√

2)m−1

The boundedness of ‖id : ⊗mε `

n1 −→ ⊗m

ε `nq ‖ in n is a restatement of the

Bohnenblust and Hille result (theorem 2.47), the constant was proved byKaijser [48] with the help of Khintchin’s inequality. The remaining cases1 < p < 2(1− 1

m)−1, 1

q= 1

p− 1

2(1− 1

m) can be reduced to the case p = 1 by

lemma 2.67.

Now we give estimates for general 1 ≤ p, q ≤ ∞. We start with a lemma.

Lemma 2.69. Let (ri)i∈M(m,n) be a family of independent Bernoulli variableson a probability space (Ω, µ). Then∫

Ω

∥∥ ∑i∈M(m,n)

ri(ω)ei

∥∥⊗m

ε `npdµ ≤ cm

nm(1/p−1/2)+1/2 p ≤ 2

n1/p p ≥ 2

This is proved in [21, p. 138].

Theorem 2.70. Let 1 ≤ p, q ≤ ∞. Then

χM(id : ⊗mε `

np −→ ⊗m

ε `nq ) ≺

n(m−1)/2 p ≤ q ≤ 2

n(m−1)/q 2p≤ q

nm(1/q−1/p+1/2)−1/2 q ≤ p ≤ 2

nm/q−1/p 2q≤ p

and

χM(id : ⊗mε `

np −→ ⊗m

ε `nq )

nm(1/q−1/p−1/2)−1/2 1 ≤ p ≤ 2

nm/q−1/p 2 ≤ p ≤ ∞


In particular,

χM(id : ⊗mε `

np −→ ⊗m

ε `nq )

nm(1/q−1/p−1/2)−1/2 1 ≤ p ≤ 2

nm/q−1/p 2 ≤ p ≤ ∞

if p ≥ q.

(See p. 26 for the notation ≺ and .)

Proof. In [21] it is proved that

χM(⊗mε `

nr ) χM(⊗m

ε `nr )

n(m−1)/2 r ≤ 2

n(m−1)/r r ≥ 2

hence

an := maxχM(⊗mε `

np ), χM(⊗m

ε `nq ) n(m−1)/ minp,q,2

bn := minχM(⊗mε `

np ), χM(⊗m

ε `nq ) n(m−1)/ maxp,q,2

Together with

‖id : ⊗mε `

np −→ ⊗m

ε `nq ‖ = ‖id : `np −→ `nq ‖m =

1 p ≤ q

nm(1/q−1/p) p ≥ q

‖id−1 : ⊗mε `

np −→ ⊗m

ε `nq ‖−1 = ‖id : `nq −→ `np‖−m =

nm(1/q−1/p) p ≤ q

1 p ≥ q

this implies the upper estimate and

χM(id : ⊗mε `

np −→ ⊗m

ε `nq )

nm(1/q−1/p+1/2)−1/2 p ≤ 2q

nm/q−1/p 2 ≤ p ≤ q

n(m−1)/2 q ≤ 2p

n(m−1)/q 2 ≤ q ≤ p

sincean ‖id−1‖−1 ≤ χM(id : ⊗m

ε `np −→ ⊗m

ε `nq ) ≤ bn ‖id‖

by lemma 2.62. For q ≤ p this is not optimal. We will get a better esti-mate with a more direct argument using a method of [21]. Take a family(ri)i∈M(m,n) of independent Bernoulli variables on a probability space (Ω, µ).Then for any p and q we have

nm/q =∥∥ n∑

i=1

ei

∥∥m

`nq

=∥∥ ∑

i∈M(m,n)

ei

∥∥⊗m

ε `nq

=∥∥ ∑

i∈M(m,n)

ri(ω) ri(ω) ei

∥∥⊗m

ε `nq


np −→ ⊗m

ε `nq )∥∥ ∑

i∈M(m,n)

ri(ω) ei

∥∥⊗m

ε `np

The domain of convergence of the polynomials 105

for all ω ∈ Ω, hence

nm/q ≤ χM(id)

∫Ω

∥∥ ∑i∈M(m,n)

ri(ω)ei

∥∥⊗m

ε `npdµ

which, by lemma 2.69, yields the announced lower bound.

By theorem 2.58 and lemma 2.63 as well as theorem 2.65, for the correctnessof our conjecture (2.29), it is enough to prove

supn

χM(id : ⊗mε `

np −→ ⊗m

ε `nm p) <∞

for p > 2. Our theorem 2.70 only shows

1 ≤ χM(id : ⊗mε `

np −→ ⊗m

ε `nm p) ≤ n(1−1/m)1/p

The domain of convergence of the polyno-

mials

Let X be a Banach sequence space. Then

P(X) := span⋃m

Pm(X)

is the space of all continuous polynomials on X. Every non-zero polynomialP ∈ P(X) can be written as a finite sum

P = P0 + P1 + . . .+ Pn

of uniquely determined homogeneous polynomials Pm ∈ Pm(X), m = 1, . . . , nwith Pn 6= 0 and n is called the degree of P (see [19, chapter 4]).

Our investigations so far permit us to give a description for domP(X).

Theorem 2.71. Let X be a Banach sequence space. Then

`1 ∪X · `2 ⊂ domP(X)

Proof. Since every continuous polynomial on X is holomorphic and boundedon every bounded subset of X we have

domH∞(R) ⊂ domP(X)

for every bounded Reinhardt domain R in X and theorem 2.25 implies `1 ⊂domP(X). On the other hand by (2.12)

domP(X) = dom⋃m

Pm(X) =⋂m

domPm(X) (2.34)

and theorem 2.49 implies the second part of the claim.


Using (2.34) we can deduce upper inclusions for domP(X) from the resultsof section 5.

Theorem 2.72. Let X be a symmetric Banach sequence space and ε > 0.Then

domP(X) ⊂ `2+ε

If X is regular, thendomP(X) ⊂ X

12 · `2+ε

Proof. By theorem 2.51 (1)(a) and (2.34) we have

domP(X) ⊂ `2(1− 1m

)−1+ε

for all m and this implies the first statement. The second one follows in thesame way from theorem 2.51 (1)(b): We choose m such that 2(1 − 1

m)−1 <

2 + ε/2. Then

domP(X) ⊂ X12(1+ 1

m) · `2(1− 1

m)−1+ ε

2⊂ X

12 · `2+ε

since Xr ⊂ Xs if r > s.

Finally we obtain an analogue to theorem 2.44.

Theorem 2.73. Let X be a symmetric Banach sequence space and ε > 0.

(1) If X ⊂ `2 then`1 ⊂ domP(X) ⊂ `1+ε

(2) If X is 2-convex and regular, then

X · `2 ⊂ domP(X) ⊂ X · `2+ε

Proof. We have already proved the lower inclusions (theorem 2.71). Theupper inclusion of (1) follows from theorem 2.51 (2)(b) since the Banach

sequence space X1m is contained in `∞ for all m. For the remaining inclusion

in (2) use theorem 2.51 (3)(b).

Using this result we get an improvement of theorem 2.44.

If R is a bounded Reinhardt domain in a Banach sequence space X then twocontinuous polynomials on X are equal if they are equal on R and we canconsider P(X) as a subset of H∞(R).

Corollary 2.74. Let R be a bounded Reinhardt domain in a symmetric Ba-nach sequence space X and F(R) ⊂ H(R) be a subset of holomorphic func-tions on R with P(X) ⊂ F(R) ⊂ H∞(R). Let ε > 0.

The exceptional role of `1 107

(1) If X ⊂ `2 then

`1 ∩R ⊂ domF(R) ⊂ `1+ε ∩R

(2) If X is 2-convex and regular, then

(X · `2) ∩R ⊂ domF(R) ⊂ (X · `2+ε) ∩R

Proof.

domH∞(R) ⊂ domF(R) ⊂ domP(X)

If X is symmetric and R is bounded then under the assumptions of theo-rem 2.44 on X we see that domH∞(R) and R ∩ domP(X) have the samedescription. So these sets are identical “up to an ε”.

Problem 2.75.

domH∞(R)?= R ∩ domP(X)

The exceptional role of `1

In [53] Lempert treated the problem of solving the ∂-equation in an opensubset of a Banach space and using monomial expansions (see theorem 2.12)he was able to prove a result for (0, 1)-forms on balls rB`1 in `1. He asked:“Why `1 of all Banach spaces”? His investigations naturally led him to con-sider monomial expansions of holomorphic functions as were investigated byRyan in [70], also for specifically the space `1. This space seems to play aspecial role in infinite dimensional holomorphy. Lempert gave part of an an-swer to his question by explaining how his approach depends on the structureof `1. He pointed out that, as in finite dimensions the geometric series is animportant tool, he based his arguments on the convergence of an appropriateseries in infinitely many variables, basically

∑α∈N(N)

0

|α||α|

αα|zα| (2.35)

This series converges if and only if the series∑α∈N(N)

0

|α|!α!|zα| (2.36)

converges. To understand this, note that we have |α|!/α! ≤ |α||α|/αα andfrom the proof of lemma 2.14 (which uses the arguments of Lempert) we can


see that for every z ∈ B`1 we can find a w ∈ B`1 and a constant c > 0 suchthat

|α||α|

αα|zα| ≤ c

|α|!α!|wα|

for every α ∈ N(N)0 . Thus (2.35) converges if and only if z ∈ B`1 since this is

true for (2.36). In this case we have

∑α∈N(N)

0

|α|!α!|zα| =

∞∑n=1

∑|α|=n

n!

α!|zα| =

∞∑n=1

‖z‖n`1 =1

1− ‖z‖`1

Lempert also remarked that the series (2.35) dominates∑|zα| which con-

verges if and only if z ∈ `1 ∩ DN with limit∏

n 1/(1− |zn|).

Note that the series∑

α∈N(N)0

|α||α|αα zα defines a holomorphic function on B`1

since the convergence of (2.35) is, as Lempert proved, even uniform on com-pact subsets of B`1 . More generally Lempert gave a necessary and sufficientcondition on the coefficients cα such that f(z) =

∑cαz

α defines a holomor-phic function on rB`1 , extending the work of Ryan. Our formulation is aslight extension of that result.

Theorem 2.76. Let cα ∈ C, α ∈ N(N)0 and r > 0. The following conditions

are equivalent:

(1) (cααα

|α||α| ξα) ∈ `∞(N(N)

0 ) for all ξ ∈ rB+c0.

(2) (cααα

|α||α| ξα) ∈ c0(N(N)


(3)∑cαz

α is bounded on compact subsets of rB`1.

(4)∑|cαzα| is bounded on compact subsets of rB`1.

(5)∑cαz

α is compactly convergent on rB`1.

(6)∑|cαzα| is compactly convergent on rB`1.

In this case f(z) :=∑

α∈N(N)0cαz

α, z ∈ rB`1 defines a holomorphic function

with normally convergent Taylor series expansion∑

m Pmf(0).

By c0(N(N)0 ) we mean the space of all families (λα)

α∈N(N)0

which can be ar-

bitrarily small (in absolute value) outside finite subsets of N(N)0 , i.e. for all

ε > 0 we can find a finite subset A ⊂ N(N)0 such that

supα/∈A

|λα| < ε


Proof. Obviously (6) y (4) y (1) and (6) y (5). If∑cαz

α converges uni-formly on compact subsets of rB`1 then f(z) :=

∑cαz

α defines a continuousmapping on rB`1 , thus (5) y (3). The mapping f is even holomorphic: Letx ∈ rB`1 , y ∈ `1 and choose ε > 0 such that K := x+ εDy ⊂ rB`1 . Then Kis a compact subset of rB`1 and this implies that

∑cα(x + λy)α converges

uniformly in λ on εD. Hence λ f(x + λy) is holomorphic on εD. Thus fis G-holomorphic.

We now prove (3) y (1). Let ξ ∈ rB+c0

and α ∈ N(N)0 , say α ∈ Nn. Denote

cβ,ξ := cβξβ for β ∈ N(N)

0 . Then

gn(z) :=∑β∈Nn

0

cβ,ξzβ

defines a bounded holomorphic function on B`n1

and using the Cauchy in-equalities (2.2) we get

|cα|αα

|α||α|ξα = ‖cα,ξz

α‖B`n1≤ ‖gn‖B`n

1=∥∥ ∑

β∈N(N)0

cβzβ∥∥

ξB`n1

≤∥∥ ∑

β∈N(N)0

cβzβ∥∥

ξB`1

(1) y (2) follows from B+c0

= B+c0·B+

c0and the fact that ξα −→ 0 for ξ ∈ B+

c0:

Let ε > 0. Choose n0 such that ξn < ε for n > n0. Since ξi < 1 we can findk0 with rk

i < ε for k > k0 and i = 1, . . . , n0. Then ξα < ε outside the finite

set A = [0, k0]n0 × 0N ⊂ N(N)

0 .

Finally (2) y (6) is a consequence of proposition 2.15 which also yields that∑m

∥∥ ∑|α|=m

|cαzα|∥∥

K<∞

for all compact subsets K ⊂ rB`1 . This implies the last part of the claim.

Note that the equivalence of (4) and (6) is true for power series in arbitraryReinhardt domains R ⊂ X by corollary 2.20.

In the proof of theorem 2.76 we have deduced the normal convergence of theTaylor series from proposition 2.15. But this is always true for holomorphicfunctions if the Taylor series converges pointwise on a balanced, open set.

Remark 2.77. Let f be a holomorphic function on an open subset U of aBanach space E and

f(ξ + x) =∞∑

m=0

Pmf(ξ)(x)


the Taylor expansion of f in ξ ∈ U convergent for all x in an open, balancedsubset B ⊂ E with ξ + B ⊂ U . Then the Taylor series converges normallyin B, i.e. we have

∞∑m=0

‖Pmf(ξ)‖K <∞

for all compact subsets K ⊂ B.

Proof. Let K ⊂ B be compact. We can assume that K is balanced (seeremark 2.17). Since B is balanced and open we have

K ⊂ B =⋃t>1

1

tB

and by compactness of K there is a t > 1 such that tK ⊂ B. By the Cauchyinequalities (0.13) and m-homogeneity we have for all m ∈ N

‖Pmf(ξ)‖K = t−m‖Pmf(ξ)‖tK ≤ t−m‖f‖ξ+tK

and we conclude

∞∑m=0

‖Pmf(ξ)‖K ≤ |f(ξ)|+ ‖f‖ξ+tK

t− 1

Note that the monomial coefficients of a holomorphic function f on rB`1

satisfy condition (1) of theorem 2.76 by the Cauchy inequalities (2.2): Forξ ∈ rB+

c0the function fξ defined by fξ(z) := f(ξz), z ∈ B`1 is holomorphic

on B`1 with monomial coefficients cα(fξ) = cα(f)ξα and since

‖zα‖B`1=

αα

|α||α|(2.37)

it follows

|cα(f)| αα

|α||α|ξα = ‖cα(fξ)z

α‖B`1≤ ‖fξ‖B`1

= ‖f‖ξB`1<∞

The last expression is finite since ξB`1 is a compact subset of rB`1 .

Thus a holomorphic function on rB`1 has a compactly convergent monomialexpansion series but the convergence is in general not normal, i.e.

∑‖cαzα‖K

does in general not converge for every compact subset K ⊂ rB`1 : Definecen := 1 and cα := 0 for α /∈ en | n ∈ N. Then

supα|cα|

αα

|α||α|ξα = sup ξn <∞


for all ξ ∈ B+c0

, but ∑α∈N(N)

0

|cα|αα

|α||α|ξα =

∑n

ξn =∞

for ξn = 1/(2n).

The following characterization is immediate from the fact that (ξB`1)ξ∈rB+c0

is a fundamental system of compact subsets of rB`1 .

Remark 2.78. Let cα ∈ C, α ∈ N(N)0 and r > 0. The following are equiva-

lent:

(1)∑cαz

α converges normally on rB`1.

(2) (cααα

|α||α| ξα) ∈ `1(N(N)


So the situation for holomorphic functions on `1 is nice, but not as nice asin finite dimensions.

Now we give an answer to Lempert’s question by proving that `1 is nearlythe only Banach sequence space X in which the situation domH∞(R) = Rfor some bounded Reinhardt domain R ⊂ X is possible.

Theorem 2.79. Let X be a symmetric Banach sequence space. If X con-tains a bounded Reinhardt domain R such that domH∞(R) is absorbant, inparticular, if domH∞(R) = R, then

`1 ⊂ X ⊂ `1+ε

for all ε > 0.

Proof. By corollary 2.41 we have domH∞(R) ⊂ `2+ε and thus X ⊂ `2+ε forall ε > 0 since by the assumption for all z ∈ X there is a λ > 0 such thatλz ∈ domH∞(R). But now, since the inclusion X ⊂ `2+ε is automaticallycontinuous, factorization through `n2+ε gives

‖id : Xn −→ `n2‖ ≤ n1/2−1/(2+ε)

for all ε > 0. Thus the estimate in proposition 2.40 implies that domH∞(R)and hence X is contained in `1+ε for all ε > 0.

The assumption of absorbance in the preceding theorem is not so muchweaker than domH∞(R) = R. In fact, it is equivalent to say that domH∞(R)contains a zero neighbourhood (see theorem 2.83).

This raises the following question.


Problem 2.80. Can we take ε = 0 in theorem 2.79 ?

In the previous section we asked if domH∞(R) = R ∩ domP(X) since theyhave the same description under our assumptions on the underlying Banachsequence space X. What if we assume domP(X) to be absorbant? We firstconsider homogeneous polynomials.

Proposition 2.81. Let X be a symmetric Banach sequence space such thatdomPm(X) is absorbant. Then

`1 ⊂ X ⊂ `(1− 1m

)−1+ε

for all ε > 0.

Proof. Theorem 2.51 (1)(a) and the absorbance of domPm(X) implyX ⊂ `2(1− 1

m)−1+ε for all ε > 0. Let s > 2 and z ∈ X. Without loss of

generality we can assume that z ∈ domPm(X). Then we use the estimatein lemma 2.50 and factorize both id : Xn −→ `n1 and id : Xn −→ `n2 through`ns(1− 1

m)−1 to get

z∗n ≺2m√

log n

nn(1− 1

s(1− 1

m)) 1

m n(12− 1

s(1− 1

m))(1− 1

m)

=2m√

log n

n(12

+ 1s)(1− 1

m)

This implies z ∈ `(1− 1m

)−1 ts+ε for all ε > 0, where ts = (12

+ 1s)−1 ↓ 1 for s ↓ 2

and hence the result.

This immediately implies the following result.

Theorem 2.82. Let R be a Reinhardt domain in a symmetric Banach se-quence space X and P(X) ⊂ F(R) ⊂ H(R). If domF(R) is absorbant then

`1 ⊂ X ⊂ `1+ε

for all ε > 0.

Proof. We have

domF(R) ⊂ domP(X) ⊂ domPm(X)

implying that domPm(X) is absorbant and hence by proposition 2.81

`1 ⊂ X ⊂ `(1− 1m

)−1+ε

holds for all m and every ε > 0.

Domains of convergence, Bohr radii and unconditionality 113

In particular, theorem 2.82 is true for F(R) = P(X). However, in thiscase, we know that we cannot take ε = 0. In [28] an example of a Lorentzsequence space X = d(w, 1) different from `1 is given such that for everym the monomials with respect to the standard basis vectors en form anunconditional basic sequence of Pm(X). In other words:

supnχM(Pm(Xn)) = sup

nχM(idPm(Xn)) <∞

This implies X ⊂ domPm(X) by theorem 2.58 and remark 2.63.

Domains of convergence, Bohr radii and un-

conditionality

Our main result in this section gives a good insight into the connection be-tween domains of convergence, Bohr radii and unconditionality. Additionallywe will have a short look at the border case X = `∞ .

Theorem 2.83. Let X and Y be Banach sequence spaces, R ⊂ X be abounded Reinhardt domain and F(R) a closed subspace of H∞(R) which con-tains the space P(X) of all polynomials on X. The following are equivalent:

(1) tBY ⊂ domF(R) for some t > 0.

(2) domF(R) absorbs Y , i.e. for every z ∈ Y there is a t > 0 such thattz ∈ domF(R).

(3) There is an 0 < r < 1 and a constant λ > 0 such that

supz∈rBY

∑α∈N(N)

0

|cα(f)zα| ≤ λ‖f‖R (2.38)

for all f ∈ F(R).

(4) There is a constant c > 0 such that for all m

supnχM(id : Pm(Xn) −→ Pm(Yn)) ≤ cm

Before we start with the proof note that, for example, condition (2) impliesthat Y ⊂ X. Condition (3) says that there exists some sort of infinite di-mensional Bohr radius (for λ = 1, X = Y , R = BX and F(R) = H∞(R) thiswould be exactly the existence of the Bohr radius in the infinite dimensionalReinhardt domain BX). Condition (4) is a stronger version of condition (2)given in theorem 2.58 for fixed m (see also remark 2.63).


Proof. Of course (1) implies (2). We now prove that (2) implies (3). We willuse a Baire argument. For natural numbers m and n the sets

Fm,n = f ∈ F | supz∈ 1

mBY

∑α∈N(N)

0

|cα(f)zα| ≤ n

are closed in F := F(R) since the cα are continuous linear functionals onH∞(R) by the Cauchy integration formulas (2.1) and we have F = ∪m,nFm,n:

Let f ∈ F . Since domF absorbs Y by the assumption, we have Y =∪m,nHm,n with

Hm,n = z ∈ Y |∑

α∈N(N)0

|cα(f)( zm

)α

| ≤ n

and these sets are closed in Y : If (zν) is a sequence in Hm,n converging

to z ∈ Y then it converges coordinatewise. A finite subset A ⊂ N(N)0 is

contained in Nk0 for some k and hence zα

ν = zν(1)α1 · · · zν(k)αk converges tozα for every α ∈ A which implies that the finite sum

∑α∈A|cα

(zm

)α| as the

limit of∑

α∈A|cα(

zν

m

)α| is bounded by n. Thus z ∈ Hm,n.

Then by Baire’s theorem, the interior of some Hm,n must be non-empty, i.e.we can find z ∈ Hm,n and k ∈ N such that z + 1

kBY ⊂ Hm,n and this even

implies 1kB ⊂ Hm,n: Let u ∈ 1

kBY . Define θ = (θn) by θn := zn/|zn| if zn 6= 0

and zero otherwise. Then θ|u| ∈ 1kBY since BY is a Reinhardt domain and

for

w := θ(|z|+ |u|) = z + θ|u| ∈ z +1

kBY

we have |u| ≤ |w|, hence u ∈ Hm,n. This proves f ∈ Fkm,n.

Then again Baire’s theorem says that we can find m,n ∈ N such thatint(Fm,n) 6= ∅, which means f0 + sBF ⊂ Fm,n for some f0 ∈ Fm,n and s > 0.So if f ∈ BF then

cα(f) =1

s(cα(f0 + sf)− cα(f0))

for α ∈ N(N)0 implies

supz∈ 1

mBY

∑α∈N(N)

0

|cα(f)zα| ≤ 2n

s=: λ

which is (3) with r = 1/m.

If (3) is true then, in particular, for all m and all P ∈ Pm(X)

supz∈BY

∑|α|=m

|cα(P )zα| ≤ (r−1λ)m‖P‖R ≤ (r−1λs)m‖P‖


if R ⊂ sBX (without loss of generality λ ≥ 1). By theorem 2.58 and remark2.63 this is the same as (4) where c = λs/r.

Finally if we choose t > 0 with BX ⊂ tR then (4), as just remarked, implies

supz∈BY

∑|α|=m

|cα(P )zα| ≤ cm ‖P‖ ≤ (ct)m ‖P‖R

for all m ∈ N and all P ∈ Pm(X).

So if we take 0 < r < 1/(ct) then for all z ∈ rBY and all f ∈ F∑|α|=m

|cα(f)zα| = rm∑|α|=m

|cα(Pmf(0))(r−1z)α| ≤ (rct)m ‖Pmf(0)‖R

≤ (rct)m ‖f‖R

by (2.6) and the Cauchy inequalities (0.13) and hence

∑α∈N(N)

0

|cα(f)zα| = |c0(f)|+∞∑

m=1

∑|α|=m

|cα(f)zα|

≤ |c0(f)|+ ‖f‖R∞∑

m=1

(rct)m <∞

With the help of theorem 2.83 we can get an improvement of theorem 2.79.

For the following proposition we use arguments from [27, lemma 4.5].

Proposition 2.84. Let X, Y and F(R) be as in theorem 2.83 and satisfyingone of the conditions (1)–(4). Then there is a constant c > 0 such that forall m and n

∥∥ n∑i=1

e′i∥∥

Y ′ ≤ cm1/2(log n)1/(2m)∥∥ n∑

i=1

e′i∥∥(m+1)/(2m)

X′ (2.39)

In particular, if X = Y then for all ε > 0 there is a constant c > 0 such thatfor all n ∥∥ n∑

i=1

e′i∥∥

X′ ≤ c (log log n)1+ε (2.40)

Proof. We use (2.21),

‖id : Xn −→ `n1‖ = supz∈BXn

n∑i=1

|zi| =∥∥ n∑

i=1

e′i∥∥

X′


and (2.24) to estimate

∥∥ n∑i=1

e′i∥∥m

Y ′ =∥∥( n∑

i=1

e′i

)m∥∥Y ′

=∥∥ ∑

j∈J (m,n)

|j|e′j∥∥Pm(Yn)

=∥∥ ∑

α∈Nn0

|α|=m

m!

α!zα∥∥Pm(Yn)

≤√m!χM(id : Pm(Xn) −→ Pm(Yn))

∥∥ ∑α∈Nn

0|α|=m

±√m!

α!zα∥∥Pm(Xn)

≤√m!m323m−1 log nκm(Xn, Yn) ‖id : Xn −→ `n1‖‖id : Xn −→ `n2‖m−1

≤ c√m!m323m−1 log nκm(Xn, Yn)

∥∥ n∑i=1

e′i∥∥(m+1)/2

X′

This gives the first claim sincem! ≤ mm and χM(id : Pm(Xn) −→ Pm(Yn)) ≤cm by the assumption with a constant c independent of m and n. If X = Ythen it follows ∥∥ n∑

i=1

e′i∥∥

X′ ≤ c mm

m−1 (log n)1

m−1

So if ε > 0 we choose m = [log log n] such that mm−1

< 1 + ε. Then

∥∥ n∑i=1

e′i∥∥

X′ ≤ c (log log n)1+ε (log n)1

log log n−2 = c (log log n)1+ε elog log n

log log n−2

≤ c′ (log log n)1+ε

Lemma 2.85. Let X be a symmetric Banach sequence space satisfying thatfor all ε > 0 there is a constant c > 0 such that for all n

λX(n) ≥ c n1−ε

Then X ⊂ `p for all p > 1.

Proof. Let p > 1 and choose ε > 0 with p(1 − ε) > 1. Then for x ∈ X wehave

n1−ε x∗n ≤ cε λX(n)x∗n = cε∥∥ n∑

i=1

x∗nei

∥∥ ≤ cε∥∥ n∑

i=1

x∗i ei

∥∥ ≤ cε ‖x∗‖ = cε ‖x‖

Hence x∗n ≤ cε ‖x‖ 1/n1−ε, which implies that x∗ and by symmetry also x isan element of `p.

We can now improve theorem 2.79.


Theorem 2.86. Let X be a symmetric Banach sequence space. If there isa bounded Reinhardt domain R ⊂ X and a closed subspace F(R) ⊂ H∞(R)containing P(X) such that domF(R) is absorbant, then

λX(n) n

(log log n)1+ε

for all ε > 0. In particular X ⊂ `1+ε for all ε > 0.

Proof. This is nothing but inequality (2.40) if we take into account that (1.12)and λX×(n) =

∥∥∑ni=1 e

′i

∥∥X′ holds. For the last part apply lemma 2.85.

Now let R be a bounded Reinhardt domain in a symmetric Banach sequencespace X. From the preceding section we know that if domH∞(R) is ab-sorbant then X ⊂ `1+ε for all ε > 0. In view of theorem 2.44 and example2.46 (a) a natural question is the following: If domH∞(R) absorbs `2−ε forall ε > 0, how close is X to `∞ ?

Proposition 2.87. Let R be a bounded Reinhardt domain in a symmetricBanach sequence space X. Then the following are equivalent:

(1) domH∞(R) absorbs `2−ε for all ε > 0.

(2) `p ⊂ X for all 1 ≤ p <∞.

(3) `2−ε ⊂ X and `2−ε ∩R ⊂ domH∞(R) for all ε > 0.

(4) For every ε > 0 there is a λ > 0 such that λB`2−ε ⊂ domH∞(R).

Proof. We start with (1) implies (2): Suppose (1) is true. Let ε > 0 andp > 2 such that 1

2−ε+ 1

p= 1. Then by proposition 2.84

n1p =

∥∥ n∑i=1

ei

∥∥`p

=∥∥ n∑

i=1

e′i∥∥

(`2−ε)′≤ cm1/2(log n)1/(2m)

∥∥ n∑i=1

e′i∥∥(m+1)/(2m)

X′

and thus

λX×(n) ≥ 1

c

(1

log n

)m+1 (1

m

) mm+1

nm

m+12p

Now choose m = [log n] and n large enough such that mm+1

2p≥ 1 − ε

2and

nε2/ log n ≥ 1. Then

λX×(n) ≥ 1

ce

1

log nn1− ε

2 ≥ e

cn1−ε

Hence, given p <∞, by lemma 2.85 it follows X× ⊂ `p′ with p′ > 1 the dualexponent such that 1

p+ 1

p′= 1. Then by duality (recall (Xn)′ = (X×)n from

chapter 1)

‖id : `np −→ Xn‖ = ‖id : (X×)n −→ `np′‖ ≤ ‖X× → `p′‖


for all n, and thus `p ⊂ X.

Now suppose that (2) holds and we are going to prove (3). Let 0 < ε < 1 andpε > 0 such that 1

2−ε= 1

2+ 1

pε. Using the fact that set inclusion of Banach

sequence spaces implies continuity of the inclusion mapping we get from theassumption that `pε ∩R is a Reinhardt domain in `pε , hence by example 2.46(a) and remark 2.11 we obtain

`2−ε ∩ (`pε ∩R) ⊂ domH∞(`pε ∩R) ⊂ domH∞(R)

Taking into account that pε must be greater than 2, this implies (3).

If (3) is true and ε > 0 then `2−ε ⊂ X and the inclusion mapping is contin-uous. We choose t > 0 such that tBX ⊂ R and then λ := t/‖`2−ε → X‖satisfies λB`2−ε ⊂ `2−ε ∩R ⊂ domH∞(R).

Finally (4) implies (1) since λB`2−ε is absorbant in `2−ε, if λ > 0.

Corollary 2.88. Let X be a symmetric Banach sequence space.If domH∞(BX) is absorbant, then for every ε > 0 there is a λ > 0 such that

λB`2−ε ⊂ domH∞(BX×)

Proof. By theorem 2.79 we haveX ⊂ `1+ε for all ε > 0 which implies `p ⊂ X×

for 1 ≤ p < ∞, since (`q)× = `q′ for 1 ≤ q < ∞. Apply proposition 2.87 to

get the conclusion.

3. Tensor and s-tensor orders

Let E and F be finite dimensional vector spaces with bases (e1, . . . , em) and(f1, . . . , fn), respectively. Then the elements

ei ⊗ fj ,1 ≤ i ≤ m

1 ≤ j ≤ n

form a basis of the tensor product E ⊗ F .

For infinite dimensional Banach spaces E and F with (Schauder) bases (en)n

and (fn)n we have to order the elements (ei⊗ fj)i,j to get a (Schauder) basisfor, say, the projective tensor product E⊗πF . In other words: N2 has to beordered in a certain way. This was done by Gelbaum and de Lamadrid in[41]:

(1, 1) → (1, 2) (1, 3) · · ·↓ ↓

(2, 1) ← (2, 2) (2, 3) · · ·↓

(3, 1) ← (3, 2) ← (3, 3) · · ·...

......

(3.1)

Let us write i1 · · · in for (i1, . . . , in). Then the order is

1112 22 21

13 23 33 32 3114 24 34 44 43 42 41

...

(3.2)

It is proved in [41] that the “tensor product (ei⊗fj)i,j of the bases (en)n and(fn)n” with this order is a basis for E⊗πF , and this result was extended in asimple way to π-tensor products ⊗m

π,i=1Ei of m Banach spaces for arbitrarym in [69].

120 3. Tensor and s-tensor orders

One may also ask the same question for symmetric tensor products: Given Ea Banach space with basis (en) and an s-tensor norm α, do the symmetrizedtensors

σm(ej) =1

m!

∑σ∈Sm

ejσ(1)⊗ · · · ⊗ ejσ(m)

, j ∈ J (m) (3.3)

in some order form a basis for ⊗m,sα E ? (Recall J (m) = j ∈ Nm | j1 ≤

. . . ≤ jm.) This is of particular interest since certain symmetric tensorproducts can be identified with spaces of homogeneous polynomials. Forexample ⊗m,s

εsE ′ = Pm

appr(E) is the space of approximable polynomials in theway that the functional σm(e′j) = e′j1 ⊗ · · · ⊗ e

′jm

, j ∈ J (m) is identified withthe monomial zα =

∏∞i=1(e

′i)

αi , αi = |k | jk = i|.

Thus if the coefficient functionals (e′n) with respect to the basis (en) of Eform a basis of the dual E ′, then the problem to find an order which makes(σm(e′j))j∈J (m) a basis is the same as to order the multiindices α ∈ N(N)

0 suchthat the monomials zα form a basis for Pm

appr(E). The latter was solved in[34].

In this chapter we investigate in a systematic way orders ν on Nm suchthat, given any tensor norm α and Banach spaces Ei with basis (eij)j∈N, theelements

⊗mi=1eiji

, j ∈ Nm

ordered by ν form a basis for ⊗mα,i=1Ei, and in the same way we study orders

on J (m) such that the elements (3.3) obtained from the basis (en) of aBanach space E form a basis for the symmetric tensor product ⊗m,s

α E, α anys-tensor norm.

The key is decompositions in intervals. We first define decomposition num-bers for orders on Nm and J (m) and prove some important properties relatedto them. Using these numbers we then give strong necessary conditions ofpurely combinatorial nature which permit to check whether a given order is atensor order (s-tensor order) and to generate examples in an easy way. Thenwe give three examples. The third one solves a problem raised in [69] (whichwas again posed in [44]). The next step is to relate the results to spacesof homogeneous polynomials and finally we prove that the monomial expan-sion series of any polynomial on a Banach space with basis always convergesconditionally pointwise on the whole space.

Very recently a preprint [44] was brought to our attention that deals with thesame kind of problem we consider here, though not in a systematic way aswe do it. The main aim of the paper [44] is to construct one example of whatwe call an s-tensor order here. They managed to do this going through fulltensor products. The order they got is the same as was used in [34] for thespecial case of approximable polynomials. It is maybe the most natural one,a fact that becomes even more apparent in the tensor product formulation

Orders on Nm 121

(i.e. on J (m)) that we give in example 1. In our systematic approach theproof reduces to a few lines with better constants than in [44].

This chapter was worked out without knowledge of [44], but we have addedsome remarks concerning that paper in the right places.

Orders on Nm

During this chapter k, l,m, n are always natural numbers. For vectors ofnatural numbers we use the letters i and j.

In this chapter we write Am for Nm or J (m). By an order on Am we mean abijective mapping ν : N −→ Am. Important will be minimal decompositionsof subsets of Nm in intervals I ⊂ Nm with respect to the natural partial order(i ≤ j iff in ≤ jn for n = 1, . . . ,m), i.e.

I = [a, b] = i ∈ Nm | a ≤ i ≤ b = [a1, b1]× . . .× [am, bm]

where a, b ∈ Nm, a ≤ b or I = ∅. If not stated otherwise, “≤” stands for thenatural partial order.

Definition 3.1. For A ⊂ Nm and r ⊂ Nm × Nm a partial order we define

ρr(A) := minn | A =n⋃

i=1

[ai, bi]r

and

ρr,d(A) := minn | A =n⋃

i=1

[ai, bi]r disjoint union

We only use this definition with r being the natural partial order or r beinga total order. In the first case we leave the r away, writing simply ρ and ρd

and in the second case note that ρr,d = ρr.

A union of intervals in Nm can always be written as a union of disjointintervals whose number can be controlled.

Remark 3.2. Let I1, . . . , In be intervals in Nm. Then there is k ≤ (3m)n−1

and there are disjoint intervals J1, . . . , Jk ⊂ Nm such that

n⋃i=1

Ii =k⋃

i=1

Ji

Proof. This is proved by induction on n. The case n = 1 is obvious.


For the induction step first note that if I and J are intervals in Nm then J∩Ican be written as a union of at most 2m disjoint intervals. This follows easilywith induction since

I = I1 × Nm−1 ∪ I1 × (I2 × · · · × Im)

and the intersection of two intervals in Nm is coordinatewise.

Now let the statement be true for some n ∈ N. Take intervals I1, . . . , In+1 ⊂Nm. Then by induction hypothesis we can find k′ ≤ (3m)n−1 and disjointintervals J1, . . . , Jk′ ⊂ Nm such that ∪n

i=1Ii = ∪k′i=1Ji. It follows that

n+1⋃i=1

Ii = In+1 ∪k′⋃

i=1

(Ji ∩ In+1)

is a union of k ≤ 2mk′ + 1 ≤ (3m)n disjoint intervals.

Corollary 3.3. For all A ⊂ Nm

ρ(A) ≤ ρd(A) ≤ (3m)ρ(A)−1

Notation 3.4. For a ∈ Nm, b ∈ Nn and i ∈ J (n,m+ 1) we define

a×i b :=

(a1, . . . , ai1−1, b1, ai1 , . . . , ai2−1, b2, . . . , ain−1 , . . . , ain−1, bn, ain , . . . , am)

(if ik−1 < ik = . . . = il < il+1 read it as a×ib = (. . . , aik−1, bk, bk+1, . . . , bl, ail , . . .))and then for A ⊂ Nm and B ⊂ Nn

A×i B := a×i b | a ∈ A, b ∈ B

and

Am+n× B :=

⋃i∈J (n,m+1)

A×i B

The following is immediate from the definitions.

Remark 3.5.

(1) For A ⊂ Nm, B ⊂ Nn and C ⊂ Nk we have

(Am+n× B)

m+n+k× C = A

m+n+k× (B

n+k× C)

(2) If A ⊂ Nm and B ⊂ N then

Sm+n(A×Bn) = (SmA)m+n× Bn

Orders on Nm 123

Here Sm means the group of permutations on the first m natural numbers.

Note that if I ⊂ Nm and J ⊂ Nn are intervals and i ∈ J (n,m+ 1) then

I ×i J = I1 × · · · × Ii1−1 × J1×Ii1 × · · · × Ii2−1 × J2 × · · ·

· · · ×Iin−1 × · · · × Iin−1 × Jn × Iin × · · · × Imis an interval in Nm+n. In particular if k ≤ l then

l⋃r=k

Im+1× r = I

m+1× [k, l] (3.4)

can be written as the union of m+ 1 intervals in Nm+1.

Remark 3.6. Let I ⊂ Nm and J ⊂ Nn be intervals such that Ii ∩ Jj = ∅ forall i, j. Then

ρd(Im+n× J) ≤

(m+ n

n

)Proof. The assumption Ik∩Jl = ∅ for all k, l assures that the intervals I×iJ ,i ∈ J (n,m+ 1) are pairwise disjoint. Thus

ρd(Im+n× J) ≤ |J (n,m+ 1)| =

(m+ n

n

)

Notation 3.7. Let ν be an order on Am and r be a partial order on Nm.Then we define

|ν|r :=

supn ρr(ν[1, n]) if Am = Nm

supn ρr(Smν[1, n]) if Am = J (m)

and

|ν|r,d :=

supn ρr,d(ν[1, n]) if Am = Nm

supn ρr,d(Smν[1, n]) if Am = J (m)

These values may of course be infinity. Again we write |ν| and |ν|d for rbeing the natural partial order on Nm. If r is a total order then | · |r,d = | · |r.

Note that by corollary 3.3 |ν| is finite if and only if |ν|d is and in this case

|ν| ≤ |ν|d ≤ (3m)|ν|−1 (3.5)

This definition is motivated by the characterization of sum-preserving per-mutations π : N −→ N in the case m = 1 (then |π| = |π|d). It seems that thefollowing result for X = K first appeared in [1]. The charactarization is verynatural and was rediscovered in [65, lemma 1]. For completeness we give aproof combining arguments from [65] and a nice idea of V. Ogranovich from1985, which can be found in the book [47, p. 141].


Remark 3.8. Let X be a Banach space and π : N −→ N a permutation.Then the following are equivalent:

(1) For every x ∈ XN the series∑xπ(n) converges whenever

∑xn does.

(2) |π| <∞.

In this case π does not change the sum of any convergent series in X.

Proof. Suppose that (1) is true. The space

E = x ∈ XN |∑xn convergent in X

with norm

‖x‖ = supn

∥∥ n∑i=1

xi

∥∥X

is a Banach space. By the assumption we can define the linear operator

Tπ : E −→ E , x xπ = (xπ(n))

which has closed graph and hence is bounded. We prove ‖Tπ‖ ≥ |π|. Forarbitrary n we have

π[1, n] =b⋃

i=1

[ni,mi] (3.6)

with natural numbers n1 ≤ m1 < n2 ≤ m2 < . . . < nb ≤ mb and b =ρ(π[1, n]). Choose e ∈ X with ‖e‖ = 1 and define x ∈ E by xmi

:= e,xmi+1 := −e for i = 1, . . . , b and xk := 0 otherwise. Then ‖x‖ = 1 and hence

‖Tπ‖ ≥ ‖Tπx‖ ≥∥∥ n∑

i=1

xπ(i)

∥∥X

=∥∥ ∑

i∈π[1,n]

xi

∥∥X

=∥∥ b∑

i=1

mi∑k=ni

xk

∥∥X

= ‖b e‖X = b

This implies ‖Tπ‖ ≥ |π| and hence (2).

Now let us assume that |π| <∞. Let∑xn be a convergent series in X with

limit x. Given ε > 0 we choose n0 such that for all n ≥ n0∥∥ n∑i=1

xi − x∥∥ < ε

Then we have for all n ≥ maxπ−1[1, n0] that [1, n0] ⊂ π[1, n] and hence wecan write π[1, n] as a union of intervals as in (3.6) with n1 = 1 and m1 ≥ n0.It follows∥∥ n∑

i=1

xπ(i) − x∥∥ ≤ ∥∥ m1∑

i=1

xπ(i) − x∥∥+

b∑i=2

∥∥ mi∑k=ni

xk

∥∥ ≤ (2|π| − 1)ε

Orders on Nm 125

The first characterization of convergence-preserving permutations was givenby Kronrod [50, 51] and at the same time by Levi [55]. They both used thesame condition, which can be easily seen to be equivalent to the criteriongiven above (see for example [71]).

Remark 3.9. Let ν be an order on Nm. Then |ν|d ≥ m.

Proof. Let Mm := [1m, 2m] = i ∈ Nm | ‖i‖∞ ≤ 2 (see (3.12) for thenotation). Then we have

|Mm ∩ I| ∈ 0, 1, 2, 4, . . . , 2m (3.7)

for every interval I in Nm. This is obvious for m = 1 and if it is true forsome m define Mm+1,k := i ∈ Mm+1 | im+1 = k for k = 1, 2. Then, givenan interval I ⊂ Nm+1,

|Mm+1 ∩ I| = |Mm+1,k ∩ I| = |Mm ∩ I| ∈ 0, 1, 2, 4, . . . , 2m

for some k ∈ 1, 2 or

|Mm+1 ∩ I| = 2|Mm+1,k ∩ I| = 2|Mm ∩ I| ∈ 0, 2, 4, . . . , 2m+1

where k ∈ 1, 2 = k, k′ is chosen such that |Mm+1,k ∩ I| ≥ |Mm+1,k′ ∩ I|,since i, j ∈ I with im+1 = 1, jm+1 = 2 implies (i, 2), (j, 1) ∈ I. Here we haveused notation (3.11), see page 134.

¿From (3.7) it follows that if I1, . . . , In are pairwise disjoint intervals with

|Mm ∩n⋃

i=1

Ii| = 2m − 1

then it must be n ≥ m. (Check by induction that ifa1, . . . , an ∈ 1, 2, 4, . . . , 2m−1 with

∑ni=1 ai = 2m − 1 then there are nat-

ural numbers 1 = n1 < n2 < . . . < nm = n+ 1 such that

nk+1−1∑i=nk

ai = 2k−1

for k = 1, . . . ,m.)

We choose n with|ν[1, n] ∩Mm| = 2m − 1

to finish the proof.

If ν1 and ν2 are orders on Am, i.e. bijective mappings N −→ Am, then ν−12 ν1

is a permutation on N.


Notation 3.10. Let ν1 and ν2 be orders on Am. We define

ν1 ≺ ν2 :⇐⇒ |ν−12 ν1| <∞

and we write ν1 ν2 if ν1 ≺ ν2 and ν2 ≺ ν1.

Note that the relation ≺ is reflexive and transitive.

Definition 3.11. We say that two orders ν1 and ν2 on Am are equivalent ifν1 ν2, and we call them essentially different if neither ν1 ≺ ν2 nor ν2 ≺ ν1

is true.

It follows from the theory of sum-preserving permutations that an order ν1

which can be transformed into an order ν2 by a nice, i.e. sum-preserving,permutation need not be equivalent to ν2. In fact, given an order ν1, we canalways construct an order ν2 with

ν2 ≺ ν1 6≺ ν2

This is due to the fact that the sum-preserving permutations on N do forma semigroup, but not a group (see [55, p. 582] or [65, theorem 1]). Thuswe can take π : N −→ N bijective with |π| < ∞ and |π−1| = ∞ and defineν2 := ν1π.

We give a criterion that helps to decide if a permutation is not sum-preserving.

Lemma 3.12. Let π : N −→ N be bijective and (nk)k be a strictly in-creasing sequence of natural numbers such that the sequences (π(nk))k and(π(nk+1)− π(nk))k are strictly increasing.

Then there is k0 such that

ρ(π[1, nk]) ≥ k − supi ∈ N | π(ni) ≤ nk

for all k ≥ k0.

Proof. First observe that

ak := supi ∈ N | π(ni)− π(n1) ≤ nk ≤ supi ∈ N | π(ni) ≤ nk+ 1

for k large enough since (π(nk+1)− π(nk))k is strictly increasing.

Let k ∈ N and s = sk := ρ(π[1, nk]). If s ≥ k there is nothing more toprove. Otherwise we choose intervals Ii ⊂ N with π[1, nk] =

⋃si=1 Ii and

max Ii < min Ii+1 for i = 1, . . . , s − 1. Using that (π(nr))r is increasing wecan find 0 = r0 < r1 < . . . < rs = k such that

π(nri−1+1), . . . , π(nri) ∈ Ii

Orders on Nm 127

for i = 1, . . . , s. Then

|Ii| ≥ π(nri)− π(nri−1+1) ≥ π(nri−(i−1))− π(nri−1−(i−1)+1)

by the fact that (π(nk+1)− π(nk))k is increasing and hence

nk = |π[1, nk]| =s∑

i=1

|Ii| ≥ π(nrs−(s−1))− π(n1)

Looking at the definition of ak we find k − (s− 1) = rs − (s− 1) ≤ ak or

sk = s ≥ k − ak + 1

Corollary 3.13. Let ν1, ν2 be orders on Am. Suppose there is a subset I ⊂ Nand ϕ : N −→ N bijective with

supn∈I

n− ϕ(n) =∞

as well as a sequence (in) in Am such that ν−11 (in) ≤ ν−1

2 (iϕ(n)) for infinitelymany n ∈ I. Then

ν1 6≺ ν2

Proof. Define nk := ν−11 (ik) and π := ν−1

2 ν1. Then by assumption we haveπ(nϕ(k)) ≥ nk and hence ak := supi ∈ N | π(ni) ≤ nk ≤ ϕ(k) or k − ak ≥k−ϕ(k) for infinitely many k ∈ I. It follows from the assumption on ϕ thatthe sequence (k− ak)k must be unbounded and we can apply lemma 3.12 toget |π| =∞.

If ν is an order on Nm then there is a unique injective mapping ι : N → Nwith the following two properties:

(a) ι is increasing

(b) ν(ι(N)) = J (m)

We define ν|s := ν ι : N −→ J (m) and call it the restriction of ν to J (m).This is the natural order on J (m) obtained from ν by cutting out all theelements which are not contained in J (m), i.e. we have

[i, j]ν|s = [i, j]ν ∩ J (m)

for all i, j ∈ J (m) with i ≤ν|s j. If we are working with a restriction we willoften write [i, j]s and i ≤s j whenever it is clear from the context where theorder is coming from.


Theorem 3.14. Let ν be an order on J (m) with ‖ν‖ < ∞, where‖ · ‖ = | · | or | · |d.

Then there is an order ω on Nm with

(1) ‖ω‖ ≤ ‖ν‖+m!

(2) ν = ω|s

We can choose ω such that |ν|ω = 1.

Proof. Let (nk) be the unique sequence of natural numbers such that

|Smν(k)| = nk − nk−1 (3.8)

(n0 = 1) and then choose ω such that

ω([nk−1, nk − 1]) = Smν(k) (3.9)

Then

Smν[1, k] =k⋃

r=1

Smν(r) =k⋃

r=1

ω[nr−1, nr − 1] = ω[1, nk − 1]

for all k gives |ν|ω = 1.

Further given n ∈ N there is a unique k with nk−1 ≤ n < nk and thus

ρ(ω[1, n]) ≤ ρ(ω[1, nk−1 − 1]) + nk − nk−1 = ρ(Smν[1, k − 1]) + |Smν(k)|

≤ |ν|+m!

with ρ = ρ or ρd, implying ‖ω‖ ≤ ‖ν‖+m!.

For restrictions of orders ν on Nm the value |ν|s|ν will be of particular im-portance. There is a general relation between |ν|s| and |ν|s|ν which can beexpressed using the following notation (see also [47, p. 141] for the casem = 1).

Notation 3.15. Let r be a partial order on Nm and ρ = ρr or ρr,d. For anorder ν on Am we define

ρ(ν) :=

supk≤l ρ(ν[k, l]) if Am = Nm

supk≤l ρ(Smν[k, l]) if Am = J (m)

Here decompositions are taken over all intervals of N instead of initial ones,and we have |ν| ≤ ρ(ν) and |ν|d ≤ ρd(ν). Note that ρω,d = ρω if r = ω is atotal order.

Orders on Nm 129

For a permutation π : N −→ N the value |π| is finite if and only if ρ(π) is.In fact, using the proof-technique of remark 3.8 it is not difficult to see that

|π| ≤ ρ(π) ≤ 4|π| (3.10)

We don’t know if the finiteness of |ν| also implies ρ(ν) <∞ when m ≥ 2.

The following relations are evident from the definitions.

Remark 3.16. Let ν be an order on J (m) and ω an order on Nm. Then

(a) |ν| ≤ |ν|ω ρ(ω) and |ν|d ≤ |ν|ω ρd(ω)

(b) ρ(ν) ≤ ρω(ν) ρ(ω) and ρd(ν) ≤ ρω(ν) ρd(ω)

Remark 3.17. Let ν be an order on Am and π be a permutation on N. Then

(a) |νπ| ≤ ρ(ν)|π| and |νπ|d ≤ ρd(ν)|π|(b) ρ(νπ) ≤ ρ(ν)ρ(π) and ρd(νπ) ≤ ρd(ν)ρ(π)

In particular it follows: If ν1 ≺ ν2 and ρ(ν2) <∞ then

|ν1| ≤ ρ(ν2)|ν−12 ν1|

is finite.

We can also prove an extension result in the same way as above.

Theorem 3.18. Let ν be an order on J (m) with ρ(ν) < ∞, whereρ = ρ or ρd.

Then there is an order ω on Nm with

(1) ρ(ω) ≤ ρ(ν) + 2m!

(2) ν = ω|s

We can choose ω such that |ν|ω = 1.

Remark 3.19. Let ν be an order on Nm. Then

ρν(ν|s) ≤ 4(|ν|s|ν +m!)

Proof. Let ω be the extension of ν|s defined in the proof of theorem 3.14 andπ := ν−1ω. Then

|ν|s|ν = supkρ(π[1, nk]) , ρν(ν|s) = sup

l≤kρ(π[nl, nk])

andρ(π[1, n]) ≤ ρ(π[1, nk]) +m!

if nk ≤ n < nk+1 (here (nk) is the sequence of natural numbers connectedwith the extension ω, see the proof of theorem 3.14). Using (3.10) it follows

ρν(ν|s) ≤ ρ(π) ≤ 4|π| ≤ 4(|ν|s|ν +m!)


Orders for tensor products

Definition 3.20.

(1) An order ν on Nm is called tensor order if whenever we take Banachspaces Ei with basis (eij)j∈N, i = 1, . . . ,m and a tensor norm α of orderm then (eν(n))n∈N is a basis for ⊗m

α,i=1Ei, where we denote ej := ⊗mi=1eiji

for j ∈ Nm.

(2) An order ν on J (m) is called s-tensor order if for every Banach spaceE with basis (en) and every s-tensor norm α (of order m) the sequence(σm(eν(n)))n∈N is a basis of ⊗m,s

α E.

If ν1 and ν2 are orders on Am with ν1 ≺ ν2 and ν2 is a tensor order (s-tensororder) then by definition of the relation ≺ and remark 3.8 it is immediatethat also ν1 is a tensor order (s-tensor order). In particular, if ν1 and ν2 areequivalent, then ν1 is a tensor order (s-tensor order) if and only if ν2 is.

Recall from (3.5) that |ν| is finite if and only if |ν|d is.

Theorem 3.21. Let ν be an order on Nm.

If |ν| <∞ then ν is a tensor order.

Moreover if E1, . . . , Em are Banach spaces with basis (eij)j∈N, i = 1, . . . ,mand α is a tensor norm of order m then we have

b(eν(n)) ≤ |ν|d 2m b(e1j) · · · b(emj)

for the basis constants.

Proof. Let e′ij be the coefficient functionals and

Pin =n∑

j=1

e′ij ⊗ eij

the n-th projection of the basis (eij)j∈N in Ei. Denote e′j := ⊗mi=1e

′iji∈

(⊗mα,i=1Ei)

′ for j ∈ Nm. Then

e′i(ej) = δij

for i, j ∈ Nm, i.e. ((eν(n)), (e′ν(n)))n∈N is a biorthogonal system in ⊗m

α,i=1Ei.

Hence it is enough to show that span ej | j ∈ Nm is dense in ⊗mα,i=1Ei and

that the projections

Pn =n∑

k=1

e′ν(k) ⊗ eν(k) , n ∈ N

Orders for tensor products 131

are uniformly bounded (see e.g. [56]).

To see the first, take z ∈ ⊗mi=1Ei. By linearity we can assume z = x1⊗· · ·⊗xm.

Then

zn := P1n x1 ⊗ · · · ⊗ Pmn xm

=n∑

j1,...,jm=1

〈x1, e′1j1〉 · · · 〈xm, e

′mjm〉e1j1 ⊗ · · · ⊗ em jm ∈ span ej | j ∈ Nm

converges to z because the tensor product mapping is continuous relative toany tensor norm.

For the second claim the main observation is that by the m-linearity of thetensor product mapping an element e′j ⊗ ej can be written as the tensorproduct

(e′1j1⊗ e1j1)⊗ . . .⊗ (e′mjm

⊗ emjm)

of one dimensional projections (check with elementary tensors).

Let n ∈ N. By the assumption we have a decomposition ν[1, n] = ∪ri=1[ai, bi]

in r ≤ |ν|d disjoint intervals, hence

Pn =n∑

k=1

e′ν(k) ⊗ eν(k) =r∑

i=1

∑j∈[ai,bi]

e′j ⊗ ej

=r∑

i=1

bi(1)∑j1=ai(1)

· · ·bi(m)∑

jm=ai(m)

(e′1j1⊗ e1j1)⊗ . . .⊗ (e′mjm

⊗ emjm)

=r∑

i=1

( bi(1)∑j=ai(1)

e′1j ⊗ e1j

)⊗ · · · ⊗

( bi(m)∑j=ai(m)

e′mj ⊗ emj

)

=r∑

i=1

(P1bi(1) − P1ai(1)−1)⊗ · · · ⊗ (Pmbi(m) − Pmai(m)−1)

By the metric mapping property of α this implies

‖Pn‖ ≤ |ν|d 2m b(e1j) · · · b(emj)

with b(eij) = supn‖Pin‖ the basis constant of (eij)j∈N, i = 1, . . . ,m.

Theorem 3.22. Let ν be an order on J (m).

If |ν| <∞ then ν is an s-tensor order.

Moreover if E is a Banach space with basis (en) and α an s-tensor norm oforder m then we have

b(σm(eν(n))) ≤ |ν|d(mm

m!

)2(2 b(en))m

for the basis constants.


Proof. We have that spanei | i ∈ Nm is dense in ⊗mαE and so is

spanσm(ej) | j ∈ J (m) = σm(spanei | i ∈ Nm)

in ⊗m,sαsE = σm(⊗m

αE) (and hence in ⊗m,sαsE).

For j ∈ J (m) let µm(j) be the number of all permutations σ ∈ Sm withσj = j and σm(e′j) := 1

m!

∑σ∈Sm

e′jσ(1)⊗ · · · ⊗ e′jσ(m)

. Then we have for

i, j ∈ J (m)

〈σm(ei), σm(e′j)〉 =1

m!

∑σ∈Sm

〈eiσ(1), e′j1〉 · · · 〈eiσ(m)

, e′jm〉

=

µm(i)/m! i = j

0 otherwise

Note that |j| := m!/µm(j) is the number of elements i ∈ Nm with σi = j forsome σ ∈ Sm (see also [21], lemma 4.1). It follows that(σm(ej), |j|σm(e′j))j∈J (m) is a biorthogonal system and it remains to showthat the associated coordinate projections Ps,n are uniformly bounded. Wehave

Ps,n =n∑

k=1

|ν|s(k)|σm(e′ν|s(k))⊗ σm(eν|s(k)) =n∑

k=1

∑j∈Smν|s(k)

σm(e′j)⊗ σm(ej)

=∑

j∈Smν|s[1,n]

σm(e′j)⊗ σm(ej) = σm ( ∑

j∈Smν|s[1,n]

e′j ⊗ ej

) ιm

and arguing in the same way as in the proof of theorem 3.21 we obtain∑j∈Smν|s[1,n]

e′j ⊗ ej =r∑

i=1

∑j∈[ai,bi]

e′j ⊗ ej

=r∑

i=1

(Pbi(1) − Pai(1)−1)⊗ · · · ⊗ (Pbi(m) − Pai(m)−1)

with r ≤ |ν|d, where

Pn =n∑

j=1

e′j ⊗ ej

denotes the n-th projection of the basis (ej)j∈N in E. Hence

‖Ps,n‖ ≤ |ν|d ‖σm‖ ‖ιm‖ (2 supn‖Pn‖)m

Recall that a basis in a Banach space is called shrinking if the coordinatefunctionals form a basis of the dual space.

Orders for tensor products 133

Theorem 3.23. Let E be a Banach space with basis (en)n∈N and ν be anorder on Nm. Let αs be an s-tensor norm of order m and α an equivalenttensor norm such that (eν(n))n∈N is a basis for ⊗m

αE.

If |ν|s|ν < ∞ then (σm(eν|s(n)))n∈N is a basis for ⊗m,sαsE, which is shrinking

whenever (eν(n))n∈N is. Moreover

b(σm(eν|s(n))) ≤ |ν|s|ν 2b(eν(n)) ‖σm‖‖ιm‖

where ιm : ⊗m,sαsE −→ ⊗m

αE is the natural injection and σm : ⊗mαE −→

⊗m,sαsE the natural projection.

For an order ν on Nm satisfying |ν|s|ν < ∞ we will say that it has therestriction property.

Proof. We have seen in the proof of theorem 3.22 that spanσm(ej) = ⊗m,sαsE

and we only have to show the finiteness of the basis constant.

For the n-th projection we have

Ps,n =n∑

k=1

|ν|s(k)|σm(e′ν|s(k))⊗ σm(eν|s(k)) =n∑

k=1

∑j∈Smν|s(k)


=∑

j∈Smν|s[1,n]

σm(e′j)⊗ σm(ej) =t∑

i=1

∑j∈ν[ni,mi]


with n1 ≤ m1 < n2 ≤ m2 < . . . < nt ≤ mt and t ≤ |ν|s|, and it follows

Ps,n = σm ( t∑

i=1

∑j∈ν[ni,mi]

e′j ⊗ ej

) ιm = σm

( t∑i=1

Pmi− Pni−1

) ιm

with Pk =∑k

l=1 e′ν(l) ⊗ eν(l) the k-th basis projection of the basis (eν(l))l∈N in

⊗mαE, P0 := 0. We conclude

‖Ps,n‖ ≤ ‖σm‖|ν|s|ν 2b(eν(n)) ‖ιm‖

Finally, if the basis (eν(n))n∈N is even shrinking, then, given x′ ∈ (⊗m,sαsE)′

and ε > 0, we have y′ := x′ σm ∈ (⊗mαE)′ and hence we can find n0 such

that‖y′ (id−Pn)‖ < ε

for all n ≥ n0. Choose k0 such that

Smν|s[1, k0] ⊃ ν[1, n0]

Then for all n ≥ k0 an optimal decomposition

Smν|s[1, n] =t⋃

i=1

ν[ni,mi]


with 1 = n1 ≤ m1 < n2 ≤ m2 < . . . < nt ≤ mt and t = ρν(Smν|s[1, n])satisfies m1 ≥ n0. Hence from the representation

Pn,s = σm (Pm1 +

t∑i=2

Pmi− Pni−1

) ιm

it follows

x′ (ids−Ps,n) = x′ σm (id−Pm1 − (

t∑i=2

Pmi− Pni−1)

) ιm

=(y′ (id−Pm1)−

t∑i=2

y′ (Pmi− Pni−1)

) ιm

and thus‖x′ (ids−Ps,n)‖ < ‖ιm‖ (2|ν|s|ν − 1)ε

Corollary 3.24. Let ν be a tensor order on Nm.

If |ν|s|ν <∞ then the restriction ν|s is an s-tensor order.

In this case: Given a Banach space E with basis (en)n∈N and an s-tensornorm αs of order m with equivalent tensor norm α, then

b(σm(eν|s(n))) ≤ |ν|s|ν 2b(eν(n)) ‖σm‖‖ιm‖

and if the basis (eν(n))n∈N of ⊗mαE is shrinking, then also (σm(eν|s(n)))n∈N is

a shrinking basis of ⊗m,sαsE.

Recall from remark 3.16 that

|ν|s| ≤ |ν|s|ν ρ(ν)

Thus given a tensor order ν with ρ(ν) <∞ then |ν|s|ν <∞ implies |ν|s| <∞,but the other direction does not hold. In fact we can find a tensor order ν(with ρ(ν) < ∞) whose restriction ν|s is an s-tensor order (with |ν|s| < ∞)but which fails to have the restriction property (see example 3 below).

Examples

The following notation will be convenient in this section.

Let i ∈ Nm. Then by i we mean cutting off the last element:

i := (i1, . . . , im−1) (3.11)

Examples 135

This can of course be extended to subsets A ⊂ Nm by A = a | a ∈ A.

Our examples will be orders νm on Am = Nm or J (m) defined for everym. ν1 will always be the natural order on N. To compare elements withrespect to these orders we write i ≤νm j or i ≤ν j or simply i ≤ j if thenecessary informations are clear from the context. In this case we don’tmake a difference between an element j ∈ Nm and its position ν−1

m (j), whichallows to compare vectors of arbitrary length. For example j ≤ j meansν−1

m−1(j) ≤ ν−1m (j) or we can write jm ≤ j, i.e. jm ≤ ν−1

m (j) or νm(jm) ≤ j ifyou want. This will be very convenient in example 3.

Writing vectors i = (i1, . . . , im) ∈ Nm we leave brackets and commas away asoften as possible as is done in (3.2), and we use the notation km for the vectorof length m with all entries equal to the natural number k. For example

1nkm−n = (1, . . . , 1︸︷︷︸n

, k, . . . , k︸︷︷︸m−n

) ∈ Nm (3.12)

Example 1

Perhaps the most natural order on Am is the following order defined on J (m).

Definition 3.25. For i, j ∈ J (m):

i ≤ j :⇐⇒(1) im < jm

or (2) im = jm and i ≤ j

Read it as an inductive definition: for m = 1 we start with the natural orderon N and if m ≥ 2 then i ≤ j in (2) refers to the order on J (m− 1).

This can be considered as an extension of the Gelbaum and Gil de Lamadridorder (3.2) restricted to J (2).

Remark 3.26. The order νm on J (m) defined in 3.25 is an s-tensor orderwith

|νm|d ≤ 2m

Proof. Let us write [·, ··]s for the intervals corresponding to the given order.Then for any m and any j ∈ J (m+ 1), j 6= 1m+1

[1m+1, j]s = ([1m+1, (jm+1 − 1)m+1] ∩ J (m+ 1)) ∪ [1m, j]s × jm+1

and hence applying remark 3.5 (2)

Sm+1[1m+1, j]s = [1m+1, (jm+1 − 1)m+1] ∪ (Sm[1m, j]s)m+1× jm+1 (3.13)

(This already implies that νm is an s-tensor order with |νm| ≤ (m+ 1)! .)


Now fix m. Given j ∈ J (m), j 6= 1m we define

In := [1m−n, (jm−n − 1)m−n]

for n = 0, . . . ,m0 := maxr ∈ N0 | jm−r > 1 and

Im0+1 :=

[1m−m0−1, 1m−m0−1] if m0 < m− 1

∅ if m0 = m− 1

Then by (3.13) and remark 3.5 (1) we obtain

Sm[1m, j]s = I0 ∪m0+1⋃n=1

Inm× (jm−n+1, . . . , jm) (3.14)

The sets on the right side are pairwise disjoint. This is due to the fact that

a vector from Inm× (jm−n+1, . . . , jm) has no component equal to jm−n.

Using remark 3.6 it follows (ρd(∅) := 0)

ρd(Sm[1m, j]s) ≤ 1 +

m0+1∑n=1

ρd(Inm× (jm−n+1, . . . , jm)) ≤

m∑n=0

(m

n

)= 2m

Thus if E is a Banach space with normalized basis (en) and α an s-tensornorm then the basis constant Cm of the basis (σm(eν(n))) in ⊗m,s

α E can beestimated by

4m(mm

m!

)2(theorem 3.22). This is better than [44] where they obtain an estimate worsethan

Cm ≤m−1∑n=0

(m

n

)Cn

But then supmm√Cm need not be finite.

Using the equality

Sm+1[i, j]s = (Sm[i, (im+1)m]s)m+1× im+1

∪ [1m, (jm+1 − 1)m]m+1× [im+1 + 1, jm+1 − 1]

∪ (Sm[1m, j]s)m+1× jm+1

we can get ρ(νm) ≤ m!3m as a rough estimate, but this can be considerablyimproved.

Examples 137

Remark 3.27. The order νm on J (m) defined in 3.25 satisfies

ρd(νm) ≤ (4m +m!) 2m (3m− 1)

Proof. Let ωm be the order 3.28 defined in example 2. Then ωm|s = νm. Weuse remarks 3.16, 3.19, corollary 3.30 and theorem 3.31 to estimate

ρd(νm) ≤ ρωm(νm) ρd(ωm) ≤ 4(|ωm|s|ωm +m!)m

2(3m− 1)

≤ (4m +m!) 2m (3m− 1)

By theorems 3.14 and 3.18 we can extend the order 3.25 to a tensor order νm

with |νm|d ≤ 2m +m! and ρd(νm) ≤ (4m +m!) 2m (3m− 1) + 2m!. The orderνm could be constructed in the way that every element of J (m) is followedby all its permutations in some order (see the proof of theorem 3.14). In thecase m = 3 this could be for example:

111

112 121 211

122 212 221

222

113 131 311

123 213 132 231 312 321

223 232 322

133 313 331

233 323 332

333

114 141 411

124 214 142 241 412 421

224 242 422

134 314 143 341 413 431

...

(3.15)

This shows that the extension theorems 3.14 and 3.18 actually are very strongresults (try a direct proof!). In fact, this is the order considered in [44],but there it was used the additional assumption that the tensor norms aresymmetric.

Example 2

For this example we use the following notation: For i ∈ Nm we define


µ(i) := maxk | ik = ‖i‖∞ i

:= (i1, . . . , iµ(i)−1, iµ(i)+1, . . . , im)

So if j ∈ J (m) then j = j.

Definition 3.28. For i, j ∈ Nm we define

i ≤ j :⇐⇒(1) ‖i‖∞ < ‖j‖∞

or (2) ‖i‖∞ = ‖j‖∞and (a) µ(i) < µ(j)

or (b) µ(i) = µ(j) and i ≤ j

In the case m = 2 we get

1121 12 22

31 32 13 23 3341 42 43 14 24 34 44

...

(3.16)

which is not far away from the order of Gelbaum and Gil de Lamadrid. Ituses the same principle of successively “walking down the inner sides of theupper left hand blocks”.

Let’s have a short look at the situation m = 3.

Examples 139

111

211

121221

112212 122 222

311321 312 322

131231 132 232

331 332

113213 123 223

313 323 133 233 333

411421 412 422

431 432 413 423 433

141241 142 242

341 342 143 243 343441 442 443

114214 124 224

314 324 134 234 334414 424 434 144 244 344 444

...

(3.17)

Note that the elements of J (3) are occuring in the right half of every (3k−2)-th pyramid.

It is not hard to see that any extension, say ωm, of the order given in example1, obtained in such a way that every element j ∈ J (m) is followed by allits permutations (as for example in (3.15)) is not equivalent to the order νm


defined in 3.28. More precisely

ωm ≺ νm 6≺ ωm

for m ≥ 2.

The order νm is optimal with respect to the decomposition number |νm|d.

Theorem 3.29. The order νm on Nm defined in 3.28 is a tensor order with

|νm|d = m

Proof. By remark 3.9 it remains to prove that |νm|d ≤ m.

For m, k ∈ N define

Im(k) := j ∈ Nm | ‖j‖∞ ≤ k

andImr (k) := j ∈ Nm | ‖j‖∞ = k , µ(j) = r

for r = 1, . . . ,m. By the definition of the order Im(k), Im1 (k+1), . . . , Im

m (k+1)are consecutive νm-intervals, and we have

Im(k) = [1m, km] = [1m, km]νm

andImr (k + 1) = [(1r−1, k + 1, 1m−r), ((k + 1)r, km−r)]νm

as well as

Im(k) ∪r⋃

i=1

Imi (k + 1) = [1m, ((k + 1)r, km−r)] = [1m, ((k + 1)r, km−r)]νm

for all k,m ∈ N and r = 0, . . . ,m.

Further define for a νm-interval I = νm[k, n]

ρ∗(I) := supk≤l≤n

ρd(νm[k, l])

Then|νm|d = sup

kρ∗(Im(k))

and it will be enough to show that

ρ∗(Im(k)) ≤ m Am(k)

holds for all natural numbers m and k. We prove by induction that

∀ m : Am(k) A(k)

Examples 141

is true for all k. We start with Im(1) = 1m and hence

ρ∗(Im(1)) = 1 ≤ m

for all m. Now suppose that A(k) holds for some k ∈ N. Then we have toprove that Am(k + 1) is true for all m ∈ N. This is done by induction.

Since ν1 is the natural order on N we get

ρ∗(I1(k + 1)) = ρ∗([1, k + 1]) = 1

Let m ∈ N with Am(k + 1). We have

Im+1(k + 1) = Im+1(k) ∪m+1⋃r=1

Im+1r (k + 1)

with ρ∗(Im+1(k)) ≤ m+ 1 by the induction hypothesis A(k). Further

Im+11 (k + 1)

= j ∈ Nm | ‖j‖∞ ≤ k = Im(k)

(here A

for a subset A ⊂ Nm+1 of course means the set j | j ∈ A) and

Im+1r (k + 1)

= Im(k) ∪ j ∈ Nm | ‖j‖∞ = k + 1 , µ(j) ≤ r − 1

= Im(k) ∪r−1⋃i=1

Imi (k + 1)

= [1m, ((k + 1)r−1, km−r+1)]νm

for r = 2, . . . ,m+ 1 are initial νm-intervals contained in Im(k + 1).

Now observe that for a subset A ⊂ Nm+1 with |µ(A)| = 1 and ‖j‖∞ = l forsome l ∈ N and all j ∈ A the equality

ρd(A) = ρd(A)

holds. This implies

ρ∗(Im+1r (k + 1)) = ρ∗(Im+1

r (k + 1)) ≤ ρ∗(Im(k + 1)) ≤ m

for r = 1, . . . ,m+ 1 by the induction hypothesis Am(k + 1).

Thus for all η = 0, . . . ,m+ 1 we have

Im+1(k + 1) = Im+1(k) ∪η⋃

r=1

Im+1r (k + 1) = [1m+1, ((k + 1)η, km+1−η)]

withρ∗(Im+1(k)) ≤ m+ 1


andρ∗(Im+1

r (k + 1)) ≤ m

for r = 1, . . . ,m+ 1. But then it must be

ρ∗(Im+1(k + 1)) ≤ m+ 1

Corollary 3.30. For the order νm on Nm defined in 3.28 we have

ρd(νm) ≤ m

2(3m− 1)

Proof. Let i, j ∈ Nm with i ≤νm j and m0 = µ(i), m′0 = µ(j), k = ‖i‖∞,

k′ = ‖j‖∞. Then using the notation of the preceding proof we have

[i, j]νm = [i, (km0 , (k − 1)m−m0)]νm

∪m⋃

r=m0+1

Imr (k) ∪

k′−1⋃l=k+1

m⋃r=1

Imr (l) ∪

m′0−1⋃

r=1

Imr (k′)

∪ [(1m′0−1, k

′, 1m−m′0), j]νm

(3.18)

For any natural numbers 1 < u < u′ and 1 ≤ r ≤ m we have

Imr (u′) = [1, u′]r−1 × u′ × [1, u′ − 1]m−r

= [1, u′]r−1 × u′ × [1, u− 1]m−r

∪m⋃

n=r+1

[1, u′]r−1 × u′ × [1, u′ − 1]n−r−1 × [u, u′ − 1]× [1, u− 1]m−n

¿From this it can be deduced that

k′−1⋃l=k+1

m⋃r=1

Imr (l) ∪

m′0−1⋃

r=1

Imr (k′)

=

m′0−1⋃

n=1

[1, k′]n−1 × [k + 1, k′]× [1, k]m−n

∪m⋃

n=m′0

[1, k′]m′0−1 × [1, k′ − 1]n−m′

0 × [k + 1, k′ − 1]× [1, k]m−n

Further we have

[i, (km0 , (k − 1)m−m0)]νm = [i, (km0−1, (k − 1)m−m0)]νm−1 ×m0 k

Examples 143

and[(1m′

0−1, k′, 1m−m′

0), j]νm = [1m−1, j

]νm−1 ×m′

0k′

Using all these informations and theorem 3.29 we obtain from (3.18) that

ρd([i, j]νm) ≤ ρd(νm−1) + 3m− 2

and hence

ρd(νm) ≤m∑

n=1

(3n− 2) =m

2(3m− 1)

We already know from example 1 that the restriction νm|s of the order definedin 3.28 is an s-tensor order. We will now prove that it even has the restrictionproperty.

Theorem 3.31. The order νm on Nm defined in 3.28 satisfies

|νm|s|νm ≤ 4m

Proof. Let j ∈ J (m), j 6= 1m.

¿From (3.14) in the proof of remark 3.26 (and the sets In and m0 defined asthere) we have

Sm[1m, j]s = I0 ∪m0+1⋃n=1

Inm× (jm−n+1, . . . , jm)

Since I0 = [1m, (jm − 1)m] = [1m, (jm − 1)m]νm and

Inm× (jm−n+1, . . . , jm) =

⋃i∈J (n,m)

[1m−n ×i bn, (jm−n − 1)m−n ×i bn]νm

with bn = (jm−n+1, . . . , jm) for n = 1, . . . ,m0 (and

Im0+1

m× (jm−m0 , . . . , jm) =

⋃i∈J (m0+1,m)

1m−m0−1 ×i (jm−m0 , . . . , jm)

if m0 < m− 1) it follows

ρνm(Sm[1m, j]s) ≤ 1 +

m0+1∑n=1

|J (n,m)| = 1 +

m0+1∑n=1

(m+ n− 1

n

)

≤m∑

n=0

(2m

n

)= 4m


Example 3

We have seen that the orders of example 1 and 2 are related by ≺. Wewill now give an example of an order that, together with its restriction,is essentially different from all examples up to now and fails to have therestriction property.

In [69] an extension of the order (3.2), say ν2, of Gelbaum and Gil deLamadrid to Nm for arbitrary m is given, which can be described as follows:For m ≥ 2 the mapping νm+1 satisfies

νm+1(n) =(νm(ν2(n)1) , ν2(n)2

), n ∈ N (3.19)

We write down some elements in the case m = 3.

111112 122 121

113 123 223 222 221114 124 224 214 213 212 211

115 125 225 215 135 134 133 132 131...

(3.20)

Thus if (jk) are the elements of Nm ordered by νm and we order (jk, n)k,n∈N byν2 then we obtain the same sequence as if we order Nm+1 by νm+1. It is thenimmediate from the associativity (⊗m

π,i=1Ei)⊗πEm+1 = ⊗m+1π,i=1Ei of the π-

tensor product that this order works for π-tensor products of arbitrary lengthand moreover for every m-fold α-tensor product, whenever α is associative.

However there are tensor norms which are not associative. Actually most ofthe examples we get by extending well known 2-fold tensor norms (see e.g.[22]) are not associative.

We describe the order above in a way that proves to be useful in a moment.

Definition 3.32. For i, j ∈ Nm we define

i ≤ j :⇐⇒(1) j ≤ jm

and (a) im < jm and i < jm

or (b) im = jm and i ≤ j

or (2) j > jm

and (a) im < jm and i < j

or (b) jm ≤ im ≤ j and i ≤ j

For the order ν2 on N2 we have |ν2| = |ν2|d = 2 (see (3.2)).

Examples 145

Theorem 3.33. The order νm on Nm defined in 3.32 is a tensor order with

|νm| ≤ 2 · 6m−2

andρ(νm) ≤ 6m−1

for m ≥ 2.

Proof. For m ≥ 2 we can write

νm+1[1, n] =t⋃

i=1

νm[ni,mi]× [ki, li] =t⋃

i=1

ti⋃ri=1

[aij, bij]× [ki, li]

with t ≤ |ν2| = 2 and ti ≤ ρ(νm), where ni ≤ mi, ki ≤ li are natural numbersand aij ≤ bij are elements of Nm (use (3.19)). This implies |νm+1| ≤ 2 ρ(νm).The same argument gives

ρ(νm+1) ≤ ρ(ν2)ρ(νm)

for m ≥ 2, hence by induction ρ(νm) ≤ ρ(ν2)m−1 and |νm| ≤ 2 ρ(ν2)

m−2 forall m ≥ 2 and we now prove ρ(ν2) ≤ 6 to finish.

Have a look at (3.2) to see what’s going on.

For the moment let us write the elements of N2 as column vectors. If i, j ∈N2 with i ≤ν2 j then there are unique natural numbers k ≤ n such thati ∈[(

1k

),(

k1

)]ν2

and j ∈[(

1n

),(

n1

)]ν2

.

If k = n then

[i, j]ν2 =

[(

i1n

),(

j1n

)]if(

1n

)≤2 i, j ≤2

(nn

)[(

nj2

),(

ni1

)]if(

nn

)≤2 i, j ≤2

(n1

)[(

i1n

),(

nn

)]∪[(

nj2

),(

nn

)]if i ≤2

(nn

)≤2 j

If k < n then

[i, j]ν2 =[i,(

k1

)]ν2∪⋃n−1

r=k+1

[(1r

),(

r1

)]ν2∪[(

1n

), j]ν2

By the case that we have just considered, the first and the last ν2-interval inthis decomposition can be represented by a union of maximal two intervalsin N2 (with the natural partial order). So it will be enough to prove that forall natural numbers k ≤ n⋃n

r=k

[(1r

),(

r1

)]ν2

=[(

1k

),(

nn

)]∪[(

k1

),(

nk−1

)](3.21)

We fix k and proceed by induction on n. For any natural number n we have[(1n

),(

n1

)]ν2

=[(

1n

),(

nn

)]∪[(

n1

),(

nn−1

)]


(see (3.2)). Suppose now that (3.21) is true for some n ≥ k. We have[(1k

),(

n+1n+1

)]=[(

1k

),(

nn

)]∪[(

n+1k

),(

n+1n

)]∪[(

1n+1

),(

n+1n+1

)]and [(

k1

),(

n+1k−1

)]=[(

k1

),(

nk−1

)]∪[(

n+11

),(

n+1k−1

)]Hence [(

1k

),(

n+1n+1

)]∪[(

k1

),(

n+1k−1

)]=[(

1k

),(

nn

)]∪[(

k1

),(

nk−1

)]∪[(

1n+1

),(

n+1n+1

)]∪[(

n+11

),(

n+1n

)]=⋃n

r=k

[(1r

),(

r1

)]ν2∪[(

1n+1

),(

n+11

)]ν2

=⋃n+1

r=k

[(1r

),(

r1

)]ν2

Remark 3.34. The order νm on Nm defined in 3.32 does not have the re-striction property for m ≥ 3.

Proof. Let p := (νm|s)−1. Then p(1m−1, k) = 1 + (k − 1)2 and p(k, 1m−1) =k2(m−1). It follows

p(n+ 1, 1m−1)− p(n, 1m−1) ≥ n2m−3 ≥ n3 ≥ m! p(1m−1, k)

> |Sm[1m, (1m−1, k)]νm|s |

for all n ≥(m!(1 + (k − 1)2)

)1/3=: ck. Thus if k − n ≥ ck and

Sm[1m, (1m−1, k)]νm|s =⋃r

i=1[ai, bi]νm with ai, bi ∈ Nm, then the elements(k − l + 1, 1m−1) ∈ Sm[1m, (1m−1, k)]νm|s must be in different intervals forl = 1, . . . , n+ 1 since the distance of two different elements’ position exceedsthe number of elements of the set Sm[1m, (1m−1, k)]νm|s . This implies

ρν(Smνm|s[1, 1 + (k − 1)2]) ≥ [k − ck] + 1 ≥ k − ckk→∞−−−→∞

where [k − ck] means the biggest integer not exceeding k − ck.

Remark 3.35. The order νm on Nm defined in 3.32 and its restriction νm|son J (m) are essentially different to the orders given in example 1 and 2 ifm ≥ 3.

Proof. Let ω1,m be any extension to Nm of the order 3.25 defined in example1 such that each element j ∈ J (m) is followed by all its permutations (seefor example (3.15) in the case m = 3). Further let ω2,m be the order 3.28 onNm given in example 2.

Then ω1|s = ω2|s is the order defined in 3.25 of example 1 and it is not hardto check that

Examples 147

ω−11 (1m−1k) = (k − 1)m + 1

ω−11 (km) = km

ω−12 (1m−1k) = km − km−1 + 1

ω−12 (km) = km

(ω2|s)−1(km) =∑k

i=1(ω2|s)−1(im−1)

(ω2|s)−1(1m−1k) = (ω2|s)−1((k − 1)m) + 1

Thus for ω = ωi, i = 1, 2 we have

ω−1(1m−1k) ∼ ω−1(km) ∼ km

(ω|s)−1(1m−1k) ∼ (ω|s)−1(km) ∼ km

asymptotically in k. On the other hand

ν−1(1m−1k) = (k − 1)2 + 1

ν−1(km) =∑ν−1(km−1)

i=1 (2i−1)−(k−1) = (ν−1(km−1))2−(k−1) ∼ k2m−1

and for the restriction define

M(km) := [1m, km]ν|s

M(km) := [1m, km]ν \ [1m, km]ν|s

and A(km) := |M(km)|. Then (ν|s)−1(km) can be written as

(ν|s)−1(km) = ν−1(km)− A(km)

and since

A(km) = A(km−1) ν−1(km−1) +

∑j∈M(km−1)\km−1

(‖j‖∞ − 1)

≤ A(km−1) ν−1(km−1) + (ν|s)−1(km−1) · ν−1(km−2)

it follows

(ν|s)−1(km) ≥ (ν−1(km−1)− ν−1(km−2)) (ν|s)−1(km−1)− (k − 1)

k2m−2

(ν|s)−1(km−1) k2m−1

We have (ν|s)−1(km) ≤ ν−1(km) and hence

(ν|s)−1(km) ∼ k2m−1

In a similar way we obtain (ν|s)−1(1m−1k) ∼ k2.

Apply corollary 3.13 to see that ωi,m 6≺ νm 6≺ ωi,m and ωi,m|s 6≺ νm|s 6≺ ωi,m|sfor i = 1, 2 and m ≥ 3.


We will now prove that the restriction of νm is an s-tensor order (theorem3.56). This needs some preparations. The following results should bringsome light into the structure of the order, as far as we need it for the proofof theorem 3.56. It might be helpful to read this proof first.

In the following ≤ is always with respect to the order νm and ≤s with respectto the restriction νm|s. Writing i ≤s j requires i, j ∈ J (m) for some m.

Notation 3.36. For i ∈ Nm+1 let

m(i) := max i , im+1

Lemma 3.37. Let i, j ∈ Nm+1. Then

(a) if i ≤ j then m(i) ≤ m(j)

(b) if m(i) < m(j) then i < j

(c) m(i) = r if and only if (1m, r) ≤ i < (1m, r + 1)

(d) m(j) ≤ j andif j 6= 1m+1 then j < j,if j 6= 1m+1, 1m2 then jm+1 < j

(e) if jm = jm+1 and j 6= 1m+1, 1m−122 then jm+1 < j

If i, j ∈ J (m+ 1) then

(f) if i < j and im = im+1 then m(i) < m(j)

(g) if jm = jm+1 and m(i) ≤ m(j) then i ≤ j

(h) if im = im+1 and jm = jm+1 then i < j implies i < j

(i) if i′, j′ ∈ J (m) with i′ < j′ then (i′, i′m) < (j′, j′m)

Proof. (a) If j ≤ jm+1 then i ≤ j implies

im+1 < jm+1 and i < jm+1 (3.22)

orim+1 = jm+1 and i ≤ j (3.23)

In the first case we have m(i) < jm+1 = m(j) and in the second case im+1 ≤jm+1 and i ≤ j ≤ jm+1, hence m(i) ≤ jm+1 = m(j).

If j > jm+1 then i ≤ j implies

im+1 < jm+1 and i < j (3.24)

orjm+1 ≤ im+1 ≤ j and i ≤ j (3.25)

Thus in any case im+1 ≤ j and i ≤ j and hence m(i) ≤ j = m(j).

Examples 149

(b) If j ≤ jm+1 then im+1, i ≤ m(i) < m(j) = jm+1 and hence by definition3.32 (1) (a) it follows i ≤ j. Since im+1 < jm+1 it cannot be i = j.

If j > jm+1 then i ≤ m(i) < m(j) = j. Thus if im+1 < jm+1 then i ≤ j bydefinition 3.32 (2)(a) and if jm+1 ≤ im+1 then jm+1 ≤ im+1 ≤ m(i) < m(j) =j and hence i ≤ j by definition 3.32 (2)(b). In any case it cannot be i = jsince i < j.

(c) If m(i) = r then m(i) < r + 1 = m(1m, r + 1) and hence i < (1m, r + 1)by (b). For the other inequality if i ≤ im+1 then r = im+1 and 1m ≤ i whichimplies (1m, r) ≤ i by definition 3.32 (1)(b). If i > im+1 then im+1 ≤ r = iand 1m ≤ i imply (1m, r) ≤ i by definition 3.32 (2)(b).

Now let r ∈ N with (1m, r) ≤ i < (1m, r + 1). Then from (a) we obtainr = m(1m, r) ≤ m(i) ≤ m(1m, r + 1) = r + 1, and m(i) = r + 1 would imply(1m, r + 1) ≤ i by what we have just proved, a contradiction.

(d) From (c) we have j ≥ (1m,m(j)) = (m(j) − 1)2 + 1 ≥ m(j), where thelast inequality is strict, if m(j) ≥ 3.

If j 6= 1m+1 then m(j) 6= 1 and m(j) = 2 implies j = 1m2 or j = 1m−122,hence j = 1m = 1 < 2 = 1m2 = j or j = 1m−12 = 2 < 3 = 1m−122 = j.

If j 6= 1m+1 and j 6= 1m2 then again m(j) 6= 1 and m(j) = 2 impliesj = 1m−122 and thus jm+1 = 2 < 3 = j.

(e) If jm = jm+1 and j 6= 1m+1, 1m−122 then using (d) we obtain j > jm =jm+1.

(f) Applying (e) we get m(i) = i.

If j ≤ jm+1 then i ≤ j implies (3.22) or (3.23). In the first case we havem(i) = i < jm+1 = m(j). In the second case it must be i < j (since i 6= j)and then m(i) = i < j ≤ m(j).

If j > jm+1 then i ≤ j implies (3.24) or (3.25). Suppose that m(i) = m(j).Then i = m(i) = m(j) = j (again (e) for the last equality) shows that wemust be in the second case and hence im = jm ≤ jm+1 ≤ im+1 = im givesim+1 = jm+1 and thus i = j, a contradiction.

(g) If m(i) < m(j) then i < j by (b).

Let m(i) = m(j).

If j = 1m+1 then m(i) = 1, hence i = 1m+1 ≤ j.

If j = 1m−122 then m(i) ≤ m(j) = 2 and it follows i = 1m+1 or i = 1m2 ori = 1m−122. In any case we have i ≤ j.

Now let j 6= 1m+1, 1m−122. Then (e) gives j > jm+1.

If i ≤ im+1 then im+1 = m(i) = m(j) = j. Thus we have jm+1 ≤ im+1 = jand i ≤ m(i) ≤ m(j) = j which implies i ≤ j by definition 3.32 (2)(b).


If i > im+1 then (using (e)) i = m(i) = m(j) = j and hence jm+1 = jm =im ≤ im+1 < i = j, implying i ≤ j by definition 3.32 (2)(b).

(h) Using (e) and (f) we get i = m(i) < m(j) = j.

(i) By (e) we have m(i′, i′m) = i′ < j′ = m(j′, j′m) which by (b) implies(i′, i′m) < (j′, j′m).

Remark 3.38. Let i ∈ Nm and j = mini < j′ | j′ ∈ J (m).

Thenim ≤ jm + 1

andjm + 1 ≤ i

if i 6= 1m, 1m−12, 1m−222.

Proof. Suppose jm +2 ≤ im. Then we must be in the case (2)(b) of definition3.32. Thus if we define j′ ∈ Nm by j′m := jm + 1 and j′n := jn for n < mthen j′m < im ≤ j = j′ and i ≤ j = j′, implying i < j′ (definition 3.32 (2)(b);i = j′ cannot be since j′m < im). Further we have jm < j′m < j and j′ = jwhich implies j′ < j, and hence i < j′ < j with j′ ∈ J (m), since j ∈ J (m),j′n = jn for n < m and jm ≤ j′m. By the definition of j this is a contradiction.Thus it must be im ≤ jm + 1.

Let i 6= 1m, 1m−12, 1m−222. By lemma 3.37(c) we have

(1m−1,m(i)) ≤ i < (1m−1,m(i) + 1)

Hence j ≤ (1m−1,m(i) + 1) by the way j is defined. Thus if m(i) ≥ 3 thenusing lemma 3.37(a) it follows

jm ≤ m(j) ≤ m(1m−1,m(i) + 1) = m(i) + 1 < (m(i)− 1)2 + 1

= (1m−1,m(i)) ≤ i

If i = 1m−221 then j = 1m3 and jm + 1 = 4 = i.

Notation 3.39. For j ∈ J (m) we define j(0) := 1m and

j(n) := maxi ≤s j | in = . . . = im

Remark 3.40. Let j ∈ J (m+ 1) with j 6= j(m). Then

j(m) = maxi ∈ J (m) | i < m(j)

Proof. Since j 6= j(m) it must be j(m) < j and hence by lemma 3.37 (f)

it follows j(m) = m(j(m)) < m(j). For the remaining inequality let i ∈J (m) with i < m(j). Then also im ≤ i < m(j) (lemma 3.37(d)) and hencem(i, im) < m(j) which implies i′ := (i, im) < j by lemma 3.37(b) and thusi′ ≤ j(m) by the definition of j(m), since i′m = i′m+1. Then applying lemma

3.37(a) and (e) we get i = m(i′) ≤ m(j(m)) = j(m).

Examples 151

Lemma 3.41.

(a) If j ∈ J (m+ 1) with jm = jm+1 then j(n) = (j)(n) for all n ≤ m.

(b) If j ∈ J (m+ 1) then j(n) = (j(k))(n) for n ≤ k ≤ m.

(c) If j ∈ J (m) and n ≤ m then j(n)m ≤ jm.

Proof. (a) If j = 1m+1 then j(n) = 1m = (j)(n).

If j = 1m−122 then j(n) = j and hence j(n) = j = (j)(n) for n = m and

j(n) = 1m+1, j = 1m−12, thus j(n) = 1m = (j)(n) for n < m.

If j 6= 1m+1, 1m−122 then j > jm+1 by lemma 3.37(e). Thus j(n) ≤ j implies

j(n) ≤ j and it follows j(n) ≤ (j)(n) =: i. Since j(n)m = j

(n)m+1 and jm = jm+1

we obtain j(n) ≤ (i, im) ≤ j by lemma 3.37(i) from which we conclude j(n) =

(i, im) (by the definition of j(n)) and hence j(n) = i = (j)(n).

(b) j(n) ≤ j(k) =: j′ ≤ j implies j(n) = (j′)(n) ≤ j′ with j′m = j′m+1 (since

k ≤ m) and hence j(n) = (j′)(n) = (j(k))(n) by (a).

(c) This is proved by induction. For m = 1 we have j(0)1 = 1 ≤ j = j1.

Now suppose (c) is true for an m ∈ N. Let j ∈ J (m+ 1) and n ≤ m+ 1 andlet us assume that j(n) < j. Then it must be n ≤ m.

We first consider the case n = m. We have j(m) < j, in particular jm < jm+1.

If j ≤ jm+1 then j(m)m+1 ≤ m(j(m)) ≤ m(j) = jm+1 (lemma 3.37(a),(e)).

If j > jm+1 then we verify

j = minj(m) < j′ | j′ ∈ J (m) =: j∗

to apply remark 3.38: j(m)m+1 = j

(m)m ≤ jm + 1 ≤ jm+1. ¿From j

(m)m = j

(m)m+1

and j(m) < j we deduce j(m) = m(j(m)) < m(j) = j by lemma 3.37(e),(f)and hence j ≥ j∗. The supposition j > j∗ implies j∗m ≤ j∗ < j (lemma3.37(d)), hence m(j∗, j∗m) < j = m(j) from which it follows (j∗, j∗m) ≤ j bylemma 3.37(b) and thus (j∗, j∗m) ≤ j(m) by the definition of j(m). But lemma

3.37(i) gives j(m) < (j∗, j∗m) since j(m) < j∗ (by the definition of j∗). This isa contradiction. Hence it must be j = j∗, and this proves the case n = m.

Now let n < m. Define j′ := j(m). Then using (b) (with k = m) we get

j(n) = (j′)(n) ≤ j′ ∈ J (m), hence by induction hypothesis j(n)m+1 = j

(n)m =

(j′)(n)m ≤ j′m = j

(m)m . It follows j

(n)m+1 ≤ j

(m)m = j

(m)m+1 ≤ jm+1 by what we’ve

just proved.

Notation 3.42. For j ∈ J (m) and n = 0, . . . ,m+ 1 we define

tn(j) :=

j(n)m if n ≤ m

m(j) + 1 if n = m+ 1


and for n = 1, . . . ,m+ 1

tn,1(j) :=

minim | j(n−1) <s i ≤s j

(n) if j(n−1) < j(n) and n < m

tn(j) else

Remark 3.43. Let j ∈ J (m). Then

tn(j) ≤ tn+1,1(j) ≤ tn+1(j)

for n = 0, . . . ,m. In particular: For every r ≤ m(j) there is a unique n ≤ mwith tn(j) ≤ r < tn+1(j).

Proof. First we prove tn(j) ≤ tn+1(j). We have tm(j) = jm ≤ m(j) + 1 =tm+1(j) and for n < m from j(n) ≤ j(n+1) (by definition) it follows j(n) =

maxi ≤s j(n+1) | in = . . . = im, hence tn(j) = j

(n)m ≤ j

(n+1)m = tn+1(j) by

lemma 3.41 (c).

Further we have

tn+1,1(j) =

minim | j(n) <s i ≤s j

(n+1) if j(n) < j(n+1) and n+ 1 < m

tn+1(j) else

≤

j(n+1)m = tn+1(j) if j(n) < j(n+1) and n+ 1 < m

tn+1(j) else

= tn+1(j)

Finally for tn(j) ≤ tn+1,1(j) we only need to check the case where j(n) < j(n+1)

and n + 1 < m since otherwise we have tn+1(j) = tn+1,1(j) by the definitionof tn+1,1(j) (and we have already seen that tn(j) ≤ tn+1(j)). In this caselet j(n) <s i ≤s j

(n+1) with im = mini′m | j(n) <s i′ ≤s j

(n+1) = tn+1,1(j).Then j(n) = maxi′ ≤s i | i′n = . . . = i′m = i(n), and lemma 3.41(c) gives

tn(j) = j(n) = i(n)m ≤ im = tn+1,1(j).

Remark 3.44. Let n+ 1 ≤ m and j ∈ J (m+ 1) with j(n) < j(n+1). Then

tn+1,1(j) = minim+1 | j(n) <s i ≤s j(n+1) , in+1 = . . . = im+1

Proof. We use induction.

If m = 1 then it must be n = 0 and we have j(0) = 11 and j(1) = kk for somek ≥ 2. It follows

t1,1(j) = mini2 | 11 <s i ≤2 kk = 2 = mini2 | 11 <s i ≤s kk , i1 = i2

since 11 <s i = 22 ≤s kk.

Now let m ≥ 2 and suppose the statement holds for m− 1.

Examples 153

Let n+ 1 ≤ m and j ∈ J (m+ 1) with j(n) < j(n+1). We have to prove that

tn+1,1(j) = minim+1 | j(n) <s i ≤s j(n+1) , in+1 = . . . = im+1 =: an(j)

holds. Obviously “ ≤ ” is true. For the remaining inequality let j(n) <s i ≤s

j(n+1) with im+1 = tn+1,1(j). Then ik < ik+1 = . . . = im+1 for some k ≥ n bythe definition of j(n).

If k = n then tn+1,1(j) = im+1 ≥ an(j).

If n < k < m then by lemma 3.41(b) we can assume without loss of generalitythat jm = jm+1 (otherwise take j(m) instead of j). Using lemma 3.41(a) andlemma 3.37(h) we get (j)(n) <s i ≤s (j)(n+1), and since n+ 1 < k+ 1 ≤ m byinduction hypothesis there must be an element (j)(n) <s i

∗ ≤s (j)(n+1) withi∗n+1 = . . . = i∗m ≤ im. Then lemma 3.37(i) gives j(n) <s i

′ := (i∗, i∗m) ≤s

j(n+1) with i′n+1 = . . . = i′m+1 = i∗m ≤ im = im+1 and hence tn+1,1(j) =im+1 ≥ an(j).

If k = m then it will be enough to find an element (j)(n) <s i∗ ≤s (j)(n+1)

with i∗m ≤ im+1, because in this case we obtain j(n) <s j∗ := (i∗, i∗m) ≤s j

(n+1)

by lemma 3.37(i), with j∗k′ < j∗k′+1 = . . . = j∗m+1 for some n ≤ k′ < m, andthen im+1 ≥ i∗m = j∗m+1 ≥ an(j) by what we’ve already proved.

To find i∗ first note that j(n) < m(i) ≤ j(n+1) by lemma 3.37(f) and (a).

Thus if i ≥ im+1 then j(n) < i ≤ j(n+1) and i∗ := i satisfies i∗m = im ≤ im+1.

If i < im+1 then we define r := m(i) = im+1.

If νm(r) ∈ J (m) then j(n) < i∗ := νm(r) ≤ j(n+1) with i∗m ≤ i∗ = νm(r) =im+1 (lemma 3.37(d)).

If νm(r) /∈ J (m) then j(n) < νm(r) < j(n+1) implies that

i∗ := minνm(r) < j′ | j′ ∈ J (m) satisfies j(n) < i∗ ≤ j(n+1) and i∗m <νm(r) = im+1 by remark 3.38 (since νm(r) < i∗, νm(r) /∈ J (m) it must bei∗ 6= 1m, 1m−12, 1m−222).

Corollary 3.45. Let n+ 1 ≤ m and j ∈ J (m+ 1) with j(n) < j(n+1). Then

tn+1,1(j) = minim | j(n) <s i ≤s j(n+1)

Proof. To see “≤” take j(n) <s i ≤s j(n+1). Then j(n) <s i′ := (i, im) ≤s

j(n+1) by lemma 3.37(i). It follows tn+1,1(j) ≤ i′m+1 = im.

Now we prove “≥”. Let j(n) <s i ≤s j(n+1) with im+1 = tn+1,1(j). Byremark 3.44 we can assume that im = im+1. Using lemma 3.37(h) it follows

j(n) <s i ≤s j(n+1) and thus

im+1 = im ≥ mini′m | j(n) <s i′ ≤s j(n+1)


Lemma 3.46. Let i, j ∈ J (m + 1) with im+1 < k ≤ jm+1 and i < j suchthat i′m < i′m+1 for all i ≤ i′ < j. Then i < 1mk.

Proof. Suppose i > im+1. Then m(i) ≤ m(j) by lemma 3.37(a), and itmust be m(i) < m(j), since otherwise j > jm+1 implies i = j and hencejm+1 ≤ im+1 by definition 3.32(2)(b), and j ≤ jm+1 implies i < jm+1 bydefinition 3.32(1)(a) and then jm+1 = m(j) = m(i) = i < jm+1.

But m(i) < m(j) is impossible too, since then m(i, im) = m(i) < m(j),implying i ≤ (i, im) =: i′ < j by lemma 3.37(g) and (b), with i′m = im = i′m+1,which contradicts the assumptions.

Thus it must be i ≤ im+1. Then i < (1m,m(i) + 1) ≤ (1m, k) by lemma3.37(c) and since m(i) + 1 = im+1 + 1 ≤ k.

Lemma 3.47. Let n+ 1 ≤ m, j ∈ J (m+ 1), tn+1,1(j) < tn+1(j) and j(n) <s

i ≤s j(n+1) with in+1 = . . . = im+1 = l < k ≤ tn+1(j).

Then there is an element i <s i′ ≤s j

(n+1) with i′n+1 = . . . = i′m+1 = k.

Proof. We prove by induction. For m = 1 it must be n = 0. Then t1,1(j) =

min i2 | 11 <s i ≤s j(1) < t1(j) = j

(1)2 implies j(1) = rr with r ≥ 3. We

have 11 <s i ≤s j(1) with i1 = i2 = l < k ≤ r, and hence i = ll < kk =: i′ ≤

rr = j(1).

Now let m ≥ 2 and suppose our statement is true for m− 1.

Let n+ 1 ≤ m, j ∈ J (m+ 1) with tn+1,1(j) < tn+1(j) and j(n) <s i ≤s j(n+1)

such that in+1 = . . . = im+1 = l < k ≤ tn+1(j).

If n + 1 ≤ m − 1 then by lemma 3.41(b) we can assume without loss ofgenerality that jm = jm+1 (otherwise take j(m) instead of j). An applicationof lemma 3.41(a), lemma 3.37(h) and corollary 3.45 gives (j)(n) <s i ≤s

(j)(n+1) withtn+1,1(j) = tn+1,1(j) < tn+1(j) = tn+1(j)

and in+1 = . . . = im = l < k ≤ tn+1(j) = tn+1(j). Hence the induction

hypothesis assures the existence of an element i <s i∗ ≤s j(n+1) with i∗n+1 =

. . . = i∗m = k, and lemma 3.37(i) implies i <s i′ := (i∗, i∗m) ≤s j

(n+1) withi′n+1 = . . . = i′m+1 = k.

If n + 1 = m then we define i′ := (1m−1, k, k). With k ≤ tm(j) = j(m)m+1 =

j(m)m ≤ m(j(m)) and lemma 3.37(c) we have

(1m−1, k) ≤ (1m−1,m(j(m))) ≤ j(m)

and hence i′ = (1m−1, k, k) ≤ j(m) by lemma 3.37(i).

Further i ≤ j(m) implies i ≤ j(m) by lemma 3.37(h) and we have im = im+1 =

l < tm(j) = j(m)m+1 = j

(m)m , hence i < j(m) with im = l < k ≤ j

(m)m . It cannot

Examples 155

be i ≤s i′′ <s j(m) with i′′m−1 = i′′m since in this case i ≤ i∗ := (i′′, i′′m) < j(m)

with i∗m−1 = i∗m = i∗m+1 by lemma 3.37(i) which implies i∗ ≤ j(m−1) < i ≤ i∗.

Thus we can apply lemma 3.46 and obtain i < (1m−1, k) and hence i <(1m−1, k, k) = i′ again using lemma 3.37(i).

Remark 3.48. Let n+1 ≤ m and j ∈ J (m+1) such that tn+1,1(j) < tn+1(j).Then for all tn+1,1(j) ≤ k ≤ tn+1(j) there is an element j(n) <s i ≤s j

(n+1)

with in+1 = . . . = im+1 = k.

Moreover

max i ≤s j(n+1) | im+1 = k = max i ≤s j

(n+1) | in+1 = . . . = im+1 = k

Proof. In view of lemma 3.47 it is enough to prove the case k := tn+1,1(j).Again we use induction. If m = 1 then the assumptions are n = 0, j(0) =11 < j(1) = rr (hence r ≥ 2) and t1,1(j) = mini2 | 11 <s i ≤s rr = 2 < r.Then i := 22 satisfies 11 <s i ≤s rr and i1 = i2 = t1,1(j), and we havemaxi ≤s rr | i2 = 2 = 22 = maxi ≤s rr | i1 = i2 = 2.

Let m ≥ 2 and assume that the statement (with k = tn+1,1(j)) is true form− 1.

Let n + 1 ≤ m and j ∈ J (m + 1) with tn+1,1(j) < tn+1(j). Let j(n) <s i ≤s

j(n+1) with im+1 = k (such an element exists by the definition of tn+1,1(j)).Then there is a n ≤ k′ ≤ m with ik′ < ik′+1 = . . . = im+1.

Suppose k′ = m. Then im < im+1. We have j(n) < m(i) ≤ j(n+1) by lemma3.37(f), (a) and (e).

If i ≥ im+1 then j(n) < i ≤ j(n+1) which implies j(n) <s (i, im) =: i′ ≤s j(n+1)

(lemma 3.37(i)) with i′m+1 = im < im+1 = k = tn+1,1(j), in contradiction tothe definition of tn+1,1(j).

If i < im+1 = m(i) =: r then arguing as above we get j(n) <s (i∗, i∗m) =: i′ ≤s

j(n+1) withi∗ := minνm(r) ≤ j′ | j′ ∈ J (m)

It follows i∗ = νm(r) = 1m−12 since otherwise i′m+1 = i∗m < νm(r) = r = im+1

(use remark 3.38, if νm(r) /∈ J (m) and otherwise lemma 3.37(d)), again acontradiction. But then by lemma 3.37(d) we have i ≤ m(i) = r = i∗ = 2 and

hence i = 1m2, j(n) = 1m+1. Since 2 = im+1 = tn+1,1(j) < tn+1(j) = j(n+1)m+1 it

must be j(n+1) ≥ 1m−133 and hence j(n) ≥ 1m−2222, a contradiction.

Thus it must be k′ < m.

If k′ = n then we are done.

If n < k′ < m then n + 1 ≤ m− 1 and j ∈ J (m) such that (without loss ofgenerality jm = jm+1)

tn+1,1(j) = tn+1,1(j) = k < tn+1(j) = tn+1(j)


and (j)(n) <s i ≤s (j)(n+1) (see the proof of lemma 3.47) with im = k. Thus

by induction hypothesis there must be an element j(n) <s i ≤s i′ ≤s j(n+1)

with i′n+1 = . . . = i′m = k, and by lemma 3.37(i) we have j(n) <s i ≤s i′′ :=

(i′, i′m) ≤s j(n+1) with i′′(n+1) = . . . i′′(m+1) = k.

In particular it follows

maxi ≤s j(n+1) | im+1 = k ≤ maxi ≤s j

(n+1) | in+1 = . . . = im+1 = k

and the remaining inequality is obvious.

In particular, if n < m and j ∈ J (m) with tn,1(j) < tn(j) = j(n)m then there

is an i ≤s j(n) with in = . . . = im = j

(n)m − 1.

Notation 3.49. For j ∈ J (m) and n = 1, . . . ,m+ 1 we denote

j(n,2) :=

maxi ≤s j

(n) | in = . . . = im = j(n)m − 1 if n < m, tn,1(j) < tn(j)

j if n = m+ 1

j(n) else

Remark 3.50. Let j ∈ J (m). Then

j(n) ≤ j(n+1,2) ≤ j(n+1)

for n = 0, . . . ,m. More precisely:

If n+ 1 < m and tn+1,1(j) < tn+1(j) then

j(n) < j(n+1,2) < j(n+1)

and in all other cases we have j(n) ≤ j(n+1,2) = j(n+1).

Proof. If n+ 1 < m and tn+1,1(j) < tn+1(j) then it follows from remark 3.48or remark 3.44 that

j(n) < maxi ≤s j(n+1) | in+1 = . . . = im = j(n+1)

m − 1 = j(n+1,2)

and then obviously j(n+1,2) < j(n+1). In the remaining cases we have j(n) ≤j(n+1) = j(n+1,2) by definition.

Lemma 3.51. Let n + 1 ≤ m and j ∈ J (m + 1) with tn+1,1(j) < tn+1(j).Then

(a) j(n+1,2) = maxi ≤s j(n+1) | im+1 = j

(n+1)m+1 − 1

(b) j(n+1,2) =(j(k))(n+1,2)

for n+ 1 ≤ k ≤ m.

Examples 157

Proof. (a) follows from remark 3.48.

(b) Using lemma 3.41(b), corollary 3.45 and (a) we see that

tn+1,1(j(k)) = tn+1,1(j) < tn+1(j) = tn+1(j) = tn+1(j(k))

and (j(k))(n+1,2)

= maxi ≤s j(n+1) | im = j(n+1)m − 1

We have j(n+1,2) ≤ j(n+1) (remark 3.50) which implies j(n+1,2) ≤ j(k) by

lemma 3.37(h). Since j(n+1,2)m = j

(n+1)m − 1 it follows j(n+1,2) ≤

(j(k))(n+1,2)

.

For the other inequality take i ≤s j(n+1) with im = j(n+1)m − 1. Then i′ :=

(i, im) ≤ j(n+1) by lemma 3.37(i) with i′m+1 = j(n+1)m − 1 = j

(n+1)m+1 − 1. With

(a) it follows i′ ≤ j(n+1,2) and then i = i′ ≤ j(n+1,2) by lemma 3.37(h).

Lemma 3.52. Let i ∈ J (m) and k ∈ N. Then km < i implies k < im.

Proof. We use induction. For m = 1 we have k < i = i1. Now let thestatement be true for some m ∈ N and take i ∈ J (m + 1) and k ∈ N withkm+1 < i. By lemma 3.37(f) we have m(km+1) < m(i). Thus if i ≤ im+1 thenk ≤ m(km+1) < m(i) = im+1 and if i > im+1 then km = m(km+1) < m(i) = iwhich implies k < im ≤ im+1 by induction hypothesis.

Notation 3.53. For i ∈ J (m) and r < im we denote

ϕr(i) := (i1, . . . , ik, r, ik+1, . . . , im−1) (∈ J (m))

with k = max(0 ∪ n | in ≤ r).

Remark 3.54. Let i ∈ J (m) and r, k ∈ N with 1 ≤ r < k and r < im.

Then i ≤ km implies ϕr(i) ≤ km.

Proof. Again we proceed by induction.

For m = 1 we have i ≤ k and r < i, hence ϕr(i) = r < i ≤ k.

Let m ∈ N and the statement be true for m. Take i ∈ J (m + 1), i ≤ km+1

and r, k ∈ N with 1 ≤ r < k and r < im+1. Then m(i, r) ≤ m(i) ≤ m(km+1)(lemma 3.37(a)) and hence ϕr(i) = (i, r) ≤ km+1 by lemma 3.37(g) if r ≥ im.

If r < im we have ϕr(i) ≤ km = m(km+1) by induction hypothesis and alsoim ≤ im+1 ≤ m(i) ≤ m(km+1). Thus m(ϕr(i), im) ≤ m(km+1), and againapplying lemma 3.37(g) we conclude ϕ(i) = (ϕr(i), im) ≤ km+1.

Proposition 3.55. Let i, j ∈ J (m), r < im and n, n′ ≤ m such that tn(j) ≤r < tn+1(j) and tn′(j) ≤ im < tn′+1(j).

If (1) i ≤ j(n) and tn(j) ≤ r < tn+1,1(j)


or (2) i ≤ j(n+1,2) and tn+1,1(j) ≤ r < tn+1(j)

then

(a) tn′(j) ≤ im < tn′+1,1(j) implies ϕr(i) ≤ j(n′)

(b) tn′+1,1(j) ≤ im < tn′+1(j) implies ϕr(i) ≤ j(n′+1,2)

Proof. First note that by remark 3.50 the numbers n and n′ exist since inany case we have i ≤ j and hence r < im ≤ m(i) ≤ m(j), and it must ben ≤ n′ since r ≤ im.

We proceed by induction. For m = 1 we have t0(j) = 1, t1(j) = t1,1(j) =t2,1(j) = j, t2(j) = j + 1 and j(0) = 1, j(1) = j(1,2) = j(2,2) = j. We showthat neither (1) nor (2) is satisfied, i.e. there is nothing to prove and thestatement is true. If n = 0 it would be (1) 1 = t0(j) ≤ r < i ≤ j(0) = 1or (2) j = t1,1(j) ≤ r < i ≤ j(1,2) = j, and if n = 1 we would have (1)j = t1(j) ≤ r < t2,1(j) = j or (2) j = t2,1(j) ≤ r < t2(j) = j + 1 and hencej = r < i ≤ j, a contradiction in any case.

Now suppose the statement is true for some m ∈ N and let i, j ∈ J (m+ 1),r < im+1 and n, n′ ≤ m+1 such that tn(j) ≤ r < tn+1(j) and tn′(j) ≤ im+1 <tn′+1(j).

(1) i ≤ j(n) and tn(j) ≤ r < tn+1,1(j):

(a) If tn′(j) ≤ im < tn′+1,1(j), then we have to prove ϕr(i) ≤ j(n′).

First suppose r ≥ im. Then ϕr(i) = (i, r).

If n = m + 1 then also n′ = m + 1 and hence tm+1(j) ≤ r < im+1 < tm+2(j)and j(n′) = j. Since jm+1 = tm+1(j) ≤ r < im+1 it follows j > jm+1

and then i ≤ j implies m(i) ≤ m(j) = j (lemma 3.37(a)). Thus we havejm+1 ≤ r < im+1 ≤ j and i ≤ j which shows that (i, r) ≤ j (definition3.32(2)(b)).

If n ≤ m then r < im+1 and i ≤ j(n) imply m(i, r) ≤ m(i) ≤ m(j(n)) bylemma 3.37(a) and applying lemma 3.37(g) and remark 3.50 (n ≤ n′) we get(i, r) ≤ j(n) ≤ j(n′).

Now suppose that r < im. Then we have ϕr(i) = (ϕr(i), im).

If n = m+ 1 then again jm+1 = tm+1(j) ≤ r < im+1 implies j > jm+1 and inparticular i ≤ m(i) ≤ m(j) = j. From

tm(j) = jm ≤ jm+1 ≤ r < im ≤ m(i) ≤ m(j) < tm+1(j) = tm+1,1(j)

it follows ϕr(i) ≤ j by induction hypothesis, and thus (ϕr(i), im) ≤ j, sincejm+1 ≤ r < im ≤ m(j) ≤ j (lemma 3.37(d), definition 3.32(2)(b)).

Examples 159

If n = m then i ≤ j(m) implies i ≤ m(i) ≤ m(j(m)) = j(m) =: j′ and it follows

tm(j′) = j′m = j(m)m = j

(m)m+1 = tm(j) ≤ r < im

≤ m(i) ≤ m(j′) < tm+1(j′) = tm+1,1(j

′)

so that from the induction hypothesis we get ϕr(i) ≤ j′. We also have im ≤m(j′) ≤ j′ and hence m(ϕr(i), im) ≤ j′ = j(m) = m(j(m)). An application oflemma 3.37(g) and remark 3.50 (m = n ≤ n′) gives (ϕr(i), im) ≤ j(m) ≤ j(n′).

If n < m define j′ := j(m). Then from lemma 3.41(b) and corollary 3.45 weget tn(j′) = tn(j) and tn+1,1(j

′) = tn+1,1(j), hence tn(j′) ≤ r < tn+1,1(j′).

Further i ≤ j(n) implies i ≤ j(n) = (j′)(n) (again lemma 3.41(b)). By remark3.43 there is a unique n′′ ≤ m with tn′′(j

′) ≤ im < tn′′+1(j′), and it must be

n ≤ n′′ ≤ n′ since r < im and tn′′(j) = tn′′(j′) ≤ im ≤ im+1.

Now if tn′′(j′) ≤ im < tn′′+1,1(j

′) then ϕr(i) ≤ (j′)(n′′) = j(n′′) = m(j(n′′)) byinduction hypothesis, lemma 3.41(b) and lemma 3.37(e). Also im ≤ m(i) ≤m(j(n)) ≤ m(j(n′′)) by lemma 3.37(a) and remark 3.50, hence using lemma3.37(g) and remark 3.50 we conclude (ϕr(i), im) ≤ j(n′′) ≤ j(n′).

If tn′′+1,1(j′) ≤ im < tn′′+1(j

′) then it must be n′′ < m (since by definitiontm+1,1(j

′) = tm+1(j′)) and n′′ < n′ (using lemma 3.41(b) and corollary 3.45

we have tn′′+1,1(j) = tn′′+1,1(j′) ≤ im). By induction hypothesis and remark

3.50 ϕr(i) ≤ (j′)(n′′+1,2) ≤ (j′)(n′′+1) = j(n′′+1) = m(j(n′′+1)) together withim ≤ m(i) ≤ m(j(n)) ≤ m(j(n′′+1)) gives ϕr(i) = (ϕr(i), im) ≤ j(n′′+1) ≤ j(n′)

(again lemma 3.37(g) and remark 3.50).

(b) If tn′+1,1(j) ≤ im+1 < tn′+1(j) then it must be n′ < m, since by definitiontn′+1,1(j) = tn′+1(j) for n′ ≥ m.

We have to prove that ϕr(i) ≤ j(n′+1,2).

If r ≥ im then as in case (a) m(i, r) ≤ m(i) ≤ m(j(n)) implies (i, r) ≤ j(n) ≤j(n′) ≤ j(n′+1,2).

If r < im then again as in case (a) for j′ := j(m) we have tn(j′) = tn(j) ≤r < tn+1,1(j) = tn+1,1(j

′), i ≤ j(n) = (j′)(n) and tn′′(j′) ≤ im < tn′′+1(j

′) withn ≤ n′′ ≤ n′.

If tn′′(j′) ≤ im < tn′′+1,1(j

′) then exactly as in case (a) it follows ϕr(i) =(ϕr(i), im) ≤ j(n′′) ≤ j(n′) ≤ j(n′+1,2).

If tn′′+1,1(j′) ≤ im < tn′′+1(j

′) then ϕr(i) ≤ (j′)(n′′+1,2) = j(n′′+1,2) = m(j(n′′+1,2))by induction hypothesis, im ≤ m(i) ≤ m(j(n)) ≤ m(j(n′′+1,2)), hence ϕr(i) =(ϕr(i), im) ≤ j(n′′+1,2) ≤ j(n′+1,2) by lemma 3.37(g).

(2) i ≤ j(n+1,2) and tn+1,1(j) ≤ r < tn+1(j):

In this case n < m since tn+1,1(j) = tn+1(j) for n ≥ m.


(a) If tn′(j) ≤ im+1 < tn′+1,1(j) then n < n′ since otherwise tn′+1,1(j) =tn+1,1(j) ≤ r < im+1 < tn′+1,1(j).

We have to prove that ϕr(i) ≤ j(n′).

If r ≥ im then i ≤ j(n+1,2) and r < im+1 imply m(i, r) ≤ m(i) ≤ m(j(n+1,2))and since n+ 1 ≤ m and n+ 1 ≤ n′ it follows ϕr(i) = (i, r) ≤ j(n+1,2) ≤ j(n′)

by lemma 3.37(g) and remark 3.50.

If r < im then we define j′ := j(m). Since n < m we have tn+1,1(j′) = tn+1,1(j)

and tn+1(j′) = tn+1(j), thus tn+1,1(j

′) ≤ r < tn+1(j′). Further i ≤ j(n+1,2)

implies i ≤ j(n+1,2) = (j′)(n+1,2) and there is a unique n ≤ n′′ ≤ n′ withtn′′(j

′) ≤ im < tn′′+1(j′) (see case (1)(a)).

If tn′′(j′) ≤ im < tn′′+1,1(j

′) and n′ ≤ m then by induction hypothesis ϕr(i) ≤(j′)(n′′) ≤ (j′)(n′) = j(n′) = m(j(n′)) and im ≤ m(i) ≤ m(j(n+1,2)) ≤ m(j(n′))(since n < n′) and hence ϕr(i) = (ϕr(i), im) ≤ j(n′) by lemma 3.37(g).

If n′ = m + 1 then the induction hypothesis gives ϕr(i) ≤ (j′)(n′′) ≤ j′ =

j(m) = m(j(m)) and we have im ≤ m(i) ≤ m(j(n+1,2)) ≤ m(j(m)) (sincen < m). Again an application of lemma 3.37(g) gives ϕr(i) = (ϕr(i), im) ≤j(m) ≤ j = j(n′).

If tn′′+1,1(j′) ≤ im < tn′′+1(j

′) then it must be n′′ < m and n′′ < n′.

Then ϕr(i) ≤ (j′)(n′′+1,2) = j(n′′+1,2) = m(j(n′′+1,2)) by induction hypothesisand also im ≤ m(i) ≤ m(j(n+1,2)) ≤ m(j(n′′+1,2)). From lemma 3.37(g) weobtain ϕr(i) = (ϕr(i), im) ≤ j(n′′+1,2) ≤ j(n′) since n′′ < n′.

(b) If tn′+1,1(j) ≤ im+1 < tn′+1(j) then n′ < m.

We have to prove that ϕr(i) ≤ j(n′+1,2) is true.

If r ≥ im then m(i, r) ≤ m(i) ≤ m(j(n+1,2)) ≤ m(j(n′+1,2)) implies ϕr(i) =(i, r) ≤ j(n′+1,2) since n′ < m.

If r < im then as in the corresponding case (1)(b) for j′ := j(m) we have

tn+1,1(j′) = tn+1,1(j) ≤ r < tn+1(j) = tn+1(j

′), i ≤ j(n+1,2) = (j′)(n+1,2) andtn′′(j

′) ≤ im < tn′′+1(j′) for some n ≤ n′′ ≤ n′.

In any case the induction hypothesis implies ϕr(i) ≤ (j′)(n′′+1,2) ≤ (j′)(n′+1,2) =

j(n′+1,2) = m(j(n′+1,2)) (since n′ < m) and also im ≤ m(i) ≤ m(j(n+1,2)) ≤m(j(n′+1,2)), hence ϕr(i) = (ϕr(i), im) ≤ j(n′+1,2) by lemma 3.37(g).

Theorem 3.56. The restriction νm|s on J (m) of the order νm on Nm definedin 3.32 is an s-tensor order with

|νm|s| ≤ 2m(m!)2

Proof. We use induction. For m = 1 we have the natural order on N, thus|ν1|s| = 1.

Examples 161

Now suppose |νm|s| ≤ 2m(m!)2 for some m ∈ N and take j ∈ J (m+ 1). It isenough to prove that

Sm+1[1m+1, j]s

=m+1⋃n=1

tn+1,1−1⋃r=tn

(Sm[1m, j(n)]s)m+1× r ∪

tn+1−1⋃r=tn+1,1

(Sm[1m, j(n+1,2)]s)m+1× r

with natural numbers tn ≤ tn+1,1 ≤ tn+1, n = 1, . . .m+ 1. Then we have

Sm[1m, j(n)]s =

ϕ1(n)⋃i=1

I(n)i , Sm[1m, j(n+1,2)]s =

ϕ2(n)⋃i=1

J(n)i

with (natural order) intervals I(n)i and J

(n)i in Nm and ϕi ≤ |νm|s| for i = 1, 2

and hence by (3.4) we can find intervals I(n)ij and J

(n)ij ⊂ Nm+1 such that

Sm+1[1m+1, j]s

=m+1⋃n=1

ϕ1(n)⋃i=1

I(n)i

m+1× [tn, tn+1,1 − 1] ∪

ϕ2(n)⋃i=1

J(n)i

m+1× [tn+1,1, tn+1 − 1]

=m+1⋃n=1

ϕ1(n)⋃i=1

m+1⋃j=1

I(n)ij ∪

ϕ2(n)⋃i=1

m+1⋃j=1

J(n)ij

This is a union of (m+1)2 (ϕ1(n)+ϕ2(n)) ≤ 2 (m+1)2 |νm|s| intervals, hence

ρ(Sm+1[1m+1, j]s) ≤ 2 (m+ 1)2 |νm|s| ≤ 2m+1((m+ 1)!)2

by induction hypothesis.

To prove the formula for Sm+1[1m+1, j]s we use several steps.

First define

Ar :=

(i, r) ∈ J (m+ 1) | i < m(j) r < jm+1

(i, r) ∈ J (m+ 1) | i < j jm+1 ≤ r ≤ m(j)

Then

(1) [1m+1, j]s =

m(j)⋃r=1

Ar


This follows straight from definition 3.32: If j ≤ jm+1 then

[1m+1, j]s = (i, r) ∈ J (m+ 1) | r < jm+1 and i < jm+1

∪ (i, r) ∈ J (m+ 1) | r = jm+1 and i ≤ j

=

jm+1−1⋃r=1

(i, r) ∈ J (m+ 1) | i < jm+1 = m(j)

∪m(j)⋃

r=jm+1

(i, r) ∈ J (m+ 1) | i ≤ j

and if j > jm+1 then

[1m+1, j]s = (i, r) ∈ J (m+ 1) | r < jm+1 and i < j

∪ (i, r) ∈ J (m+ 1) | jm+1 ≤ r ≤ j and i ≤ j

=

jm+1−1⋃r=1

(i, r) ∈ J (m+ 1) | i < j = m(j)

∪m(j)⋃

r=jm+1

(i, r) ∈ J (m+ 1) | i ≤ j

Next we define

Bn :=

tn+1,1(j)−1⋃r=tn(j)

(i, r) ∈ J (m+ 1) | i ≤ j(n)

∪tn+1(j)−1⋃

r=tn+1,1(j)

(i, r) ∈ J (m+ 1) | i ≤ j(n+1,2)

for n = 0, . . . ,m+ 1, and we claim

(2)

m(j)⋃r=1

Ar =m+1⋃n=0

Bn

We have

Bm+1 =

m(j)⋃r=tm+1(j)

(i, r) ∈ J (m+ 1) | i ≤ j =

m(j)⋃r=jm+1

Ar

and it remains to prove that⋃tm+1(j)−1

r=1 Ar =⋃m

n=0Bn.

“⊂”: Let (i, r) ∈ J (m + 1) with i < m(j) = tm+2(j) − 1 and r < jm+1 =tm+1(j). Then by remark 3.43 there is a unique n ≤ m such that tn(j) ≤ r <tn+1(j), and we claim that (i, r) ∈ Bn.

Examples 163

If tn(j) ≤ r < tn+1,1(j) then we have to prove i ≤ j(n).

Since i < m(j) it follows i < m(j(m)) = j(m) if j = j(m) and i < j(m) if

j 6= j(m) by remark 3.40. Thus in any case i < j(m) =: j′ ∈ J (m) and byremark 3.50 (and lemma 3.41(b)) there is a unique 0 ≤ k ≤ m − 1 with

j(k) = (j′)(k) <s i ≤s (j′)(k+1) = j(k+1).

Suppose i > j(n). Then it must be k ≥ n and hence

im ≥ mini′m | j(k) <s i′ ≤s j(k+1) = tk+1,1(j) ≥ tn+1,1(j)

(corollary 3.45 and remark 3.43). But (i, r) ∈ J (m+ 1) and the choice of nimply im ≤ r < tn+1,1(j), a contradiction.

If tn+1,1(j) ≤ r < tn+1(j) then it must be n < m. We have to prove that

i ≤ j(n+1,2).

Since r < tn+2,1(j) with n+ 1 ≤ m we can argue exactly as above to see that

i ≤ j(n+1). Further we have j(n+1)m+1 − 1 < j

(n+1)m+1 ≤ m(j(n+1)) = j(n+1), hence

m(i, j(n+1)m+1 − 1) ≤ j(n+1), implying i′ := (i, j

(n+1)m+1 − 1) ≤ j(n+1) by lemma

3.37(g) and i′ ∈ J (m+ 1) since im ≤ r < tn+1(j) = j(n+1)m+1 . It follows

i′ ≤ maxi′′ ≤s j(n+1) | i′′m+1 = j(n+1) − 1 = j(n+1,2)

(lemma 3.51(a)) and thus i = i′ ≤ m(i′) ≤ m(j(n+1,2)) = j(n+1,2).

“⊃”: Let (i, r) ∈ Bn for some 0 ≤ n ≤ m. We have to prove that r < jm+1

and i < m(j).

If i ≤ j(n) and tn(j) ≤ r < tn+1,1(j) then r < tn+1,1(j) ≤ tm+1(j) = jm+1

and i ≤ j(n) ≤ j(m) < m(j) if j(m) 6= j by remark 3.43. If j(m) = j thenit must be n < m since tn(j) < tn+1(j), hence j(n) < j(n+1) which implies

i ≤ j(n) = m(j(n)) < m(j(n+1)) ≤ m(j) (lemma 3.37(f)).

If i ≤ j(n+1,2) and tn+1,1(j) ≤ r < tn+1(j) then n < m, in particular r <tm(j) ≤ tm+1(j) = jm+1. Further tn+1.1(j) < tn+1(j) implies j(n+1,2) <

j(n+1) (n + 1 < m + 1, remark 3.50) from which we deduce i ≤ j(n+1,2) =m(j(n+1,2)) < m(j(n+1)) ≤ m(j) again using lemma 3.37(f).

Defining

Cn :=

tn+1,1(j)−1⋃r=tn(j)

[1m, j(n)]s × r ∪tn+1(j)−1⋃

r=tn+1,1(j)

[1m, j(n+1,2)]s × r

for n = 0, . . . ,m+ 1 we now claim

(3) Sm+1

(m+1⋃n=0

Bn

)= Sm+1

(m+1⋃n=0

Cn

)


Obviously we have Bn ⊂ Cn and thus “⊂” holds.

To prove “⊃” it will be enough to show that for every n ≤ m + 1 and(i, r) ∈ Cn with r < im we have (ϕr(i), im) ∈ Bn′ for some n′ ≤ m+ 1.

So let n ≤ m+ 1 and (i, r) ∈ Cn with r < im. Then tn(j) ≤ r < tn+1(j) and

i ≤ j(n) or i ≤ j(n+1,2). For j′ = j(n) or j(n+1,2) we have j′ ≤ j and hencej′ ≤ m(j′) ≤ m(j), implying im ≤ i ≤ m(j) (lemma 3.37(a), (d)). Thus byremark 3.43 there is n′ ≤ m + 1 such that tn′(j) ≤ im < tn′+1(j). We claimthat (ϕr(i), im) ∈ Bn′ , i.e. we have to prove

(a) tn′(j) ≤ im < tn′+1,1(j) implies ϕr(i) ≤ j(n′)

(b) tn′+1,1(j) ≤ im < tn′+1(j) implies ϕr(i) ≤ j(n′+1,2)

If n′ < m then with j′ := j(m) we have

tk(j′) = tk(j) for k = n, n+ 1, n′, n′ + 1 (3.26)

as well as

tk+1,1(j′) = tk+1,1(j)

(j′)(k) = j(k), (j′)(k+1,2) = j(k+1,2)for k = n, n′ (3.27)

by lemma 3.41(b) and corollary 3.45. It follows tn(j′) ≤ r < tn+1(j′) and

tn′(j′) ≤ im < tn′+1(j), and since (i, r) ∈ Cn, r < im we have

i ≤ j(n) = (j′)(n) and tn(j′) ≤ r < tn+1,1(j′)

or i ≤ j(n+1,2) = (j′)(n+1,2) and tn+1,1(j′) ≤ r < tn+1(j

′)(3.28)

An application of proposition 3.55 gives

ϕr(i) ≤

(j′)(n′) = j(n′) if tn′(j′) ≤ im < tn′+1,1(j

′)

(j′)(n′+1,2) = j(n′+1,2) if tn′+1,1(j′) ≤ im < tn′+1(j

′)

which by (3.26) and (3.27) is what we want.

If n′ = m then tn′+1,1(j) = tn′+1(j) and there is only the one case (a) tm(j) ≤im < tm+1(j), i.e. we have to prove that ϕr(i) ≤ j(m).

If n < m then again with j′ := j(m) we have tn(j′) ≤ r < tn+1(j′) and (3.28).

In particular im ≤ m(i) ≤ m(j′) and thus there is a unique n ≤ n′′ ≤ m withtn′′(j

′) ≤ im < tn′′+1(j′) by remark 3.43. If n′′ < m then by proposition 3.55

we have ϕr(i) ≤ (j′)(n+1,2) ≤ j′ and if n′′ = m = n′ then proposition 3.55

gives ϕr(i) ≤ j′. Thus in any case ϕr(i) ≤ j′ = j(m).

If n = m then we have tm(j) ≤ r < im < tm+1(j). Since i ≤ j(m) =: j′

it follows im ≤ m(i) ≤ m(j′) < tm+1(j′) and hence tm(j′) = tm(j) ≤ r <

Examples 165

im < tm+1(j′) such that again we can apply proposition 3.55 and obtain

ϕr(i) ≤ j′ = j(m).

If n′ = m + 1 then again tn′+1,1(j) = tn′+1(j) and there is only case (a)tm+1(j) ≤ im < tm+2(j). We have to prove ϕr(i) ≤ j.

If n ≤ m then in the same way as for the case n′ = m we see that ϕr(i) ≤ j(m).

Then ϕr(i) ≤ j(m) = m(j(m)) ≤ m(j) = j. The last equality follows from

lemma 3.37(e) if j(m) = j. If j(m) 6= j we have i ≤ j(m) or n = m and

i ≤ j(m+1,2) = j. In the first case we can apply remark 3.40 to see thatjm+1 = tm+1(j) ≤ im ≤ i < m(j). In the second case, note that it must beim ≤ m(j), since im < tm+2(j) = m(j) + 1 by assumption. Thus we havejm+1 = tm+1(j) ≤ im < m(j) or m(j) = im ≤ i ≤ j.

If n = m + 1 = n′ then i ≤ j and we have to prove that ϕr(i) ≤ j. Thisfollows from

tm(j) = jm ≤ jm+1 = tm+1(j) ≤ r < im ≤ m(i) ≤ m(j) < tm+1(j)

by proposition 3.55.

Now from (1), (2) and (3) we obtain Sm+1[1m+1, j]s = Sm+1

(⋃m+1n=0 Cn

)and

using remark 3.5 (2) it follows

(4) Sm+1[1m+1, j]s =

m+1⋃n=0

tn+1,1(j)−1⋃r=tn(j)

(Sm[1m, j(n)]s)m+1× r ∪

tn+1(j)−1⋃r=tn+1,1(j)

(Sm[1m, j(n+1,2)]s)m+1× r

This formula can be improved. We claim that

(5) Sm+1[1m+1, j]s =

m+1⋃n=1

tn+1,1−1⋃r=tn

(Sm[1m, j(n)]s)m+1× r ∪

tn+1−1⋃r=tn+1,1

(Sm[1m, j(n+1,2)]s)m+1× r

holds with t1 := 1, tn := tn(j) and tn,1 := tn,1(j) for n = 2, . . . ,m+ 2.

To understand this first note that

B0 = B′0 :=

t1(j)−1⋃r=1

(i, r) ∈ J (m+ 1) | i ≤ j(1)

If j(1) < 2m+1 then j(1) = 1m+1 = j(1,2) = j(0), t1,1(j) = 1 = t1(j) and thusB0 = ∅ = B′

0.

If j(1) ≥ 2m+1 then t1,1(j) = 2 ≤ t1(j), and j(0) ≤ j(1,2) ≤ j(1) implying

j(0) ≤ j(1,2) ≤ j(1) gives B0 ⊂ B′0.


For the other inclusion take (i, r) ∈ J (m+1) with i ≤ j(1) and 1 ≤ r < t1(j).If r = 1 then (i, r) = 1m+1 ∈ B0.

If r > 1 then 2 ≤ r < t1(j) and it is enough to prove i ≤ j(1,2). If j(1) = 2m+1

then j(1,2) = j(1) and we are done. If j(1) > 2m+1 then j(1) = km+1 for somek ≥ 3 and j(1,2) = (k − 1)m+1, and 2 = t1,1(j) ≤ r ≤ t1(j) − 1 = k − 1.

Suppose i > j(1,2) = (k − 1)m. Then im > k − 1 by lemma 3.52 and thus

r ≥ im > k − 1, a contradiction. Hence it must be i ≤ j(1,2).

Then using (2) it follows Sm+1[1m+1, j]s = Sm+1(B′0∪⋃m+1

n=1 Bn) and we claimthat

Sm+1

(B′

0 ∪m+1⋃n=1

Bn

)= Sm+1

(C ′

0 ∪m+1⋃n=1

Cn

)where C ′

0 :=⋃t1(j)−1

r=1 [1m, j(1)]s × r. By (3) it is enough to prove

C ′0 ⊂ Sm+1

(B′

0 ∪m+1⋃n=1

Bn

)To understand this observe that

im ≤ i ≤ j(1) = m(j(1)) ≤ m(j) < tm+2(j)

implies M := maxim | i ≤ j(1) < tm+2(j) and thus

M⋃r=1

(i, r) ∈ J (m+ 1) | i ≤s j(1) ⊂ B′0 ∪

m+1⋃n=1

Bn

Hence it is enough to prove that if i ≤ j(1) and 1 ≤ r < t1(j) with r < imthen (ϕr(i), im) ∈ Sm+1(

⋃Mr=1(i, r) ∈ J (m + 1) | i ≤s j(1)), which is the

case if and only if ϕr(i) ≤ j(1). This is true, since we can apply remark 3.54

with k := t1(j) = j(1)m+1.

We have

C ′0 ∪ C1

=

t1(j)−1⋃r=1

[1m, j(1)]s × r ∪t2,1(j)−1⋃r=t1(j)

[1m, j(1)]s × r ∪t2(j)−1⋃

r=t2,1(j)

[1m, j(2,2)]s × r

=

t2,1−1⋃r=1

[1m, j(1)]s × r ∪t2−1⋃

r=t2,1

[1m, j(2,2)]s × r

and then

Sm+1[1m+1, j]s = Sm+1

(C ′

0 ∪m+1⋃n=1

Cn

)= Sm+1

(m+1⋃n=1

tn+1,1−1⋃r=tn

[1m, j(n)]s × r ∪tn+1−1⋃

r=tn+1,1

[1m, j(n+1,2)]s × r)

Bases for symmetric tensor products and spaces of polynomials 167

which finally gives (5) and the theorem is proved.

Bases for symmetric tensor products and

spaces of polynomials

The space Pm(E) of m-homogeneous polynomials on a Banach space E withbasis can be identified with the dual space of the symmetric projective tensorproduct (⊗m,s

πsE)′ in such a way that the monomial

zα =∞∏i=1

(e′i)αi , α ∈ N(N)

0 , |α| =∞∑i=1

αi = m

with respect to the basis (en) on E (with e′n the coefficient functionals) cor-responds to the functional σm(e′j) = e′j1 ⊗ · · · ⊗ e′jm

with j ∈ J (m) andαi = |k | jk = i|.

Thus ordering J (m) means ordering the monomials of order m. For examplethe s-tensor order given in example 1 can be expressed as follows:

For α ∈ N(N)0 define n(α) := maxk | αk 6= 0 and α by

αn :=

αn − 1 n = n(α)

αn otherwise

Then for multiindices α, β of order m we have

α ≤ β ⇐⇒(1) n(α) < n(β)

or (2) n(α) = n(β) and α ≤ β

(3.29)

Theorem 3.57. Let E be a Banach space with shrinking basis (en). Then theassociated monomials (zα)|α|=m ordered by an s-tensor order form a basis forPm

appr(E) and Pmnuc(E). In particular: If we additionally assume that E ′ has

the Radon-Nikodym property, then the monomials form a basis for Pmint(E).

For the approximable polynomials with the order (3.29) this is due to Dimantand Dineen [34, proposition 10]. See also [36], proposition 4.4 (and 2.8).

Proof. We have ⊗m,sεsE ′ = Pm

appr(E) (see [36], p. 104) and ⊗m,sπsE ′ = Pm

nuc(E)isometrically since E ′ has the approximation property. The nuclear poly-nomials on E can be identified with the integral ones if E ′ has the Radon-Nikodym property (see [39], 4.3 and 4.5 for the last two statements).


Next we prove that if the Banach space E has not only a shrinking basis butalso the Dunford-Pettis property, the monomials with respect to the basiseven form a basis for the whole space Pm(E) of all m-homogeneous polyno-mials (with norm ‖p‖ = supz∈BE

|p(z)|), as Ryan claimed in his thesis [69].This is based on his ideas. Nevertheless we cannot work with an inductiveargument as he suggested, and some multilinear generalizations have to beinvestigated.

Recall that a Banach space E is said to have the Dunford-Pettis propertyif every weakly compact operator T from E into some Banach space (theimage of the unit ball is relatively weakly compact) is fully complete, i.e. ittransforms weakly convergent into norm convergent sequences. A reflexiveinfinite dimensional Banach space cannot have the Dunford-Pettis property.

The Dunford-Pettis property has simple characterizations (see e.g. [32]):

Remark 3.58. Let E be a Banach space. The following are equivalent:

(1) E has the Dunford-Pettis property.

(2) If (xn) is a weak Cauchy sequence in E and (x′n) is a weak zero sequencein E ′ then (〈xn, x

′n〉) is a zero sequence.

(3) If (xn) is a weak zero sequence in E and (x′n) is a weak Cauchy sequencein E ′ then (〈xn, x

′n〉) is a zero sequence.

(4) If (xn) is a weak Cauchy sequence in E and (x′n) a weak Cauchy se-quence in E ′ then (〈xn, x

′n〉) is a Cauchy sequence.

Lemma 3.59. Let E1, . . . , Em be Banach spaces with Dunford-Pettis prop-erty, F be a Banach space and (xi

k)k∈N weak Cauchy sequences in Ei fori = 1, . . . ,m. If φ ∈ Lm(E1, . . . , Em;F ) then

(φ(x1

k, . . . , xmk ))

k∈N is a weakCauchy sequence in F .

Proof. First we take F = K and prove by induction. The case m = 1 is clear,since φ ∈ E ′

1 is uniformly weakly continuous.

Now let m ∈ N and assume that every continuous m-linear functional on∏mi=1Ei maps σ(Ei, E

′i)-Cauchy sequences to a Cauchy sequence.

For φ ∈ Lm+1(E1, . . . , Em+1) we then define the m-linear and continuousmapping

ψ :

∏mi=1Ei −→ E ′

m+1

x φ(x, · )

Then we have x′′ ψ ∈ Lm(E1, . . . , Em) for every x′′ ∈ E ′′m+1, hence by the

induction hypothesis((x′′ ψ)(x1

k, . . . , xmk ))

kis Cauchy for every x′′ ∈ E ′′

m+1,

i.e.(ψ(x1

k, . . . , xmk ))

kis weak Cauchy in E ′

m+1. Since (xm+1k )k is weak Cauchy

in Em+1 the sequence(φ(x1

k, . . . , xm+1k )

)k

is Cauchy in Em+1 by the Dunford-Pettis property of Em+1.

Bases for symmetric tensor products and spaces of polynomials 169

For F arbitrary we just apply the above to y′ φ for every functional y′ onF , so

(φ(x1

k, . . . , xmk ))

kis weak Cauchy in F .

Corollary 3.60. Let E1, . . . , Em be Banach spaces with the Dunford-Pettisproperty, ϕ ∈ Lm(E1, . . . , Em) and (xi

k)k weak Cauchy sequences, i = 1, . . . ,m,one of which converges weakly to zero. Then

(ϕ(x1

k, . . . , xmk ))

kis a zero se-

quence.

Proof. The case m = 1 is obvious, so let m ≥ 2. We can assume that (xmk )k

is a weak zero sequence. The mapping

φ :

∏m−1i=1 Ei −→ E ′

m

x ϕ(x, · )

is (m−1)-linear and continuous, so lemma 3.59 assures that(φ(x1

k, . . . , xm−1k )

)k

is weak Cauchy in E ′m. Hence the Dunford-Pettis property of Em gives the

claim.

With corollary 3.60 at hand the following multilinear extension of [69, propo-sition 5.1] can be proved in the same way as it is done there.

Proposition 3.61. Let Ei be a Banach space with shrinking bases (eij)j∈N,i = 1, . . . ,m and Dunford-Pettis property and ν be an order on Nm such that(eν(n))n∈N is a basis for ⊗m

π,i=1Ei and ν[1, km] = [1, k]m holds for every k ∈ N.Then (eν(n))n∈N is shrinking.

See definition 3.20 for the notation eν(n).

Proof. We define X := ⊗mπ,i=1Ei, I := idX and let Pn be the n-th coordinate

projection of the basis (eν(k))k∈N.

Suppose the basis is not shrinking. Taking into account the fact that (I −Pn)X ⊂ (I −Pl)X for n ≥ l, we can choose a ϕ ∈ X ′ = Lm(E1, . . . , Em) anda δ > 0 such that ‖ϕ|(I−Pn)X‖ ≥ δ for every n.

With Ii = idEi, Pi,k the k-th basis projection of Ei and

Ri,k := P1,k ⊗ · · · ⊗ P(i−1),k ⊗ (Ii − Pi,k)⊗ Ii+1 ⊗ · · · ⊗ Im

for i = 1, . . . ,m we obtain

I − Pkm = ⊗mi=1Ii −⊗m

i=1Pi,k =m∑

i=1

Ri,k

and hence δ ≤∑m

i=1‖ϕ|Ri,kX‖ for every k. So there must be an index i0 ∈1, . . . ,m such that

‖ϕ|Ri0,kX‖ ≥δ

m


for infinitely many k. Note that Ri0,kX = ⊗mπ,i=1Fi with

Fi =

Pi,kEi i = 1, . . . , i0 − 1

(Ii − Pi,k)Ei i = i0

Ei i = i0 + 1, . . . ,m

Then there are sequences (xik)k in Fi ∩BEi

such that

|ϕ(x1k, . . . , x

mk )| ≥ δ

2m

for all k. Since E ′i is separable, (BEi

, σ(Ei, E′i)) is a precompact metric space,

and we can assume that (xik)k is σ(Ei, E

′i)-Cauchy for i = 1, . . . ,m. We have

xi0k ∈ (Ii0 − Pi0,k)Ei0 and hence

|〈xi0k , x

′〉| = |〈xi0k , x

′ (Ii0 − Pi0,k)〉| ≤ ‖x′ (Ii0 − Pi0,k)‖ −→ 0

for every functional x′ ∈ E ′i0

, since the basis (ei0j)j of Ei0 is shrinking.

But then (xi0k )k is a weak zero sequence in Ei0 and corollary 3.60 gives

ϕ(x1k, . . . , x

mk ) −→ 0, a contradiction.

Theorem 3.62. Let E be a Banach space with shrinking basis and Dunford-Pettis property. Then the associated monomials of degree m with the order(3.29) form a basis for Pm(E).

Proof. Let (en) be a shrinking basis of E. The order (3.29) of example 1 isthe restriction ν|s of the order ν given in example 2. By proposition 3.61 itfollows that (eν(n))n∈N is a shrinking basis of ⊗m

π E and hence theorem 3.31together with corollary 3.24 imply that (σm(eν|s(n)))n∈N is a shrinking basisof ⊗m,s

πsE.

Pointwise convergence

Theorem 3.63. Let E be a Banach space with basis (en) and ν be an orderon J (m) such that (σm(eν(n)))n∈N is a basis for ⊗m,s

πsE. Then the monomials

with respect to the basis (en) in order coming from ν form a basis of Pm(E)in the topology of pointwise convergence.

Proof. The coefficient functionals x′n in a Banach space X with basis (xn)build a weak*-basis of the dual space X ′, since for ϕ ∈ X ′ andx =

∑∞n=1 x

′n(x)xn ∈ X we have

ϕ(x) =∞∑

n=1

ϕ(xn)x′n(x)

Pointwise convergence 171

By assumption the family (σm(ej))j∈J (m) ordered by ν gives a basis for ⊗m,sπsE

with coefficient functionals |j|σm(e′j), where |j| is the number of elementsi ∈ Nm with σi = j for some permutation σ (see the proof of theorem 3.22).So if ϕ is the linear functional on ⊗m,s

πsE corresponding to a polynomial

P ∈ Pm(E) and z ∈ E we have

P (z) = ϕ(σm(⊗mz)) =∑

j∈J (m)

ϕ(σm(ej)) |j|σm(e′j)(σm(⊗mz)) (3.30)

(summation in order ν) and we prove that the series on the right side is themonomial expansion of P in z.

The elements ei = ei1 ⊗ · · · ⊗ eim , i ∈ Nm (ordered by some tensor order)form a basis for ⊗m

π E with coefficient functionals e′i = e′i1 ⊗ · · · ⊗ e′im . We

havee′i(⊗mz) = e′i1(z) · · · e′im(z) = zi1 · · · zim

so the element ⊗mz in ⊗mπ E has the basis representation

⊗mz =∑i∈Nm

zi1 · · · zimei

where summation is carried out in some tensor order . Then

σm(⊗mz) =∑i∈Nm

zi1 · · · zimσm(ei)

and applying the coefficients functionals |j|σm(e′j) of the basis (σm(ej))j∈J (m)

of ⊗m,sπsE we get

|j|σm(e′j)(σm(⊗mz)) = |j| zj1 · · · zjm (3.31)

since

〈|j|σm(e′j), σm(ei)〉 =

1 ∃ permutation σ : σi = j

0 otherwise

The equation∑|α|=m

cα(P )uα = P (u) = ϕ(σm(⊗mu)) =∑

j∈J (m)

|j|uj1 · · ·ujmϕ(σm(ej))

for u ∈ spanen implies

cα(P ) = |j|ϕ(σm(ej))

if αi = |k | jk = i|, and this together with (3.31) and (3.30) gives

P (z) =∑

j∈J (m)

|j|zj1 · · · zjmϕ(σm(ej)) =∑|α|=m

cα(P )zα

where both series are summed up in the order given by ν.


Now we extend the result to the space PN(E) of all polynomials of degreeless or equal than N .

Let us define J (0) = (), where the “empty sequence” () is some elementnot contained in N(N). For i ∈ N we define i := (). By an order on J (0) wemean the mapping ν0 : 1 −→ J (0).

Given orders νm on J (m) for m = 0, . . . , N , by ⊕Nm=0νm we fix some order

ν on⋃N

m=0 J (m) (i.e. ν : N −→⋃N

m=0 J (m) bijective) whose restriction toJ (m) is νm for all m:

ν|J (m) = νm , m = 0, . . . , N

This simply means that there are ϕm : N −→ N strictly increasing, m =1, . . . , N and ϕ0 : 1 −→ N such that

(a) ν ϕm = νm for m = 0, . . . , N

(b) Imϕi ∩ Imϕj = ∅ for i 6= j

i.e. the elements of J (m) appear in (ν(n))n in the order νm.

Such an order can be obtained as an extension of example 1 simply by definingthe “empty sequence” () to be the smallest element: () ≤ j for all j ∈ N(N).

Definition 3.64. For i, j ∈⋃N

m=1 J (m):

i ≤ j :⇐⇒(1) im < jm

or (2) im = jm and i ≤ j

Pointwise convergence 173

We write down the first elements.

()

1 11 111 · · · 1N

2 12 112 · · · 1N−12

22 122 · · · 1N−222

222 · · · 1N−3222

. . ....

2N

3 13 113 · · · 1N−13 33 133 · · · 1N−233 3N

23 123 · · · 1N−223 233 · · · 1N−3233

223 · · · 1N−3223. . .

... · · ·. . .

... 2N−233

2N−13

...

In this way we get an order on⋃N

m=0 J (m) whose restriction to J (m) is ans-tensor order for m = 1, . . . , N .

The set⋃N

m=0 J (m) corresponds to the set α ∈ N(N)0 | |α| ≤ N in the way

described at the beginning of the last section, where the empty sequence ismapped on the zero sequence. Defining the zero sequence to be the smallestelement, the order on α ∈ N(N)

0 | |α| ≤ N can be described with (3.29).

Let us identify the sets J (m) and [|α| = m] in the usual way. Then the nextresult is an immediate consequence of theorem 3.63.

Corollary 3.65. Let N ∈ N and νm be an s-tensor order on J (m) form = 1, . . . , N . Then for every Banach space E with basis (en) the monomialswith respect to the basis ordered by ⊕N

m=0νm form a basis of PN(E) in thetopology of pointwise convergence.

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Curriculum Vitae

Name Christopher PrengelBorn 28.10.1975 in Hanover, GermanyNationality german

Education

1995 School-leaving

1995–96 Civilian service

1996 Preparatory course for music studies at theconservatory in Hanover

1997–2002 Student at the University of Oldenburg

1997 Studies of music and mathematics to become ateacher (Gymnasiallehramt)

1998–2002 Studies of mathematics

2002 Diploma in mathematics

2002–05 Candidate for a doctor of science at the Universityof Oldenburg

10/04 Research visit at the University in Valencia

Employment

1999–2004 Tutor at the University of Oldenburg

2002–04 Teacher for preparatory courses in mathematicsat the technical college (Fachhochschule)Oldenburg/Ostfriesland/Wilhelmshaven in Oldenburgand Bremen

2003–05 GradFoG Stipendium of the Land Niedersachsen(State of Lower Saxony)

Publications

2005 Survey Harald Bohr meets Stefan Banach(published by the London Mathematical Society),with A. Defant

2006 Domains of convergence of monomial expansionsof holomorphic functions (preprint),with A. Defant and M. Maestre

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