Domains of Convergence in Infinite Dimensional Holomorphy

Domains of convergence in infinite dimensional holomorphy

Von der Fakultät fürMathematik und Naturwissenschaften der Carl von Ossietzky UniversitätOldenburg zur Erlangung des Grades und Titels eines Doktors der Naturwissenschaften (Dr. rer. nat.) angenommene Dissertation von

Christopher Prengel geboren am 28.10.1975 in Hannover 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 domain m R in a Banach sequence space X, mainly for H∞(R), P (X) and P(X), we give precise descriptions of the domain of convergence dom F(R), i.e. the set of all points of R where the monomial expansion

X ∂αf(0) zα α! (N) α∈N0 of every function f ∈ F(R) is (unconditionally) convergent. The results are obtained by an interplay of complex analysis and local Banach space theory, improving work of Ryan and Lempert about monomial expansions of holomorphic functions on `1 and using methods from the theory of multidimensional Bohr radii. The problem of conditional convergence is solved for polynomials. It is closely connected with bases for full and symmetric tensor products and convergence preserving permutations. Zusammenfassung

Fürgewisse Teilmengen F(R) holomorpher Funktionen auf einem Reinhardt- m gebiet R in einem Banachfolgenraum X, vor allem für H∞(R), P (X) und P(X), geben wir eine genaue Beschreibung des Konvergenzbereichs dom F(R), d.h. der Menge aller Punkte aus R, in denen die Monomial- entwicklung X ∂αf(0) zα α! (N) α∈N0 einer jeden Funktion f ∈ F(R) (unbedingt) konvergiert. Die Resultate werden durch ein Zusammenspiel von komplexer Analysis und lokaler Banachraumtheorie erzielt, setzen Arbeiten von Ryan und Lempert über Monomialentwicklungen holomorpher Funktionen auf `1 fort und bedie- nen sich der Theorie mehrdimensionaler Bohrradien. Das Problem bedingter Konvergenz wird fürPolynome gelöst. Es ist eng verknüpft mit Basen fürvolle und symmetrische Tensorprodukte und kon- vergenzerhaltenden Permutationen. I thank my supervisor Andreas Defant for his guidance during the preparation of this thesis. Gracias a Manuel Maestre por las discusiónes útilesy por su invitacióna Va- lencia. Tambiéna todas las personas que hac´ıan que mi estancia in Valencia fuera muy agradable, especialmente Pilar y su familia, Pablo y Marta, y por supuesto Manolo. Thank you Melissa for correcting my English.

Contents

Introduction 9

Preliminaries 15

1. Norm estimates of identities between symmetric tensor products of Banach sequence 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 dom H∞(R) ...... 65 The arithmetic Bohr radius ...... 75

Upper inclusions for dom H∞(R) ...... 84 A description of dom Pm(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 series expansion X α f(z) = cα(f)z n α∈N0 ∂αf(0) with coefficients cα(f) = α! that converges to f in every point of R. In infinite dimensions, if f is holomorphic on a Reinhardt domain R in a ∂αf(0) Banach sequence space X and we define cα(f) := α! then

X α f(z) = cα(f)z

(N) α∈N0 holds for finite sequences z = (z1, . . . , zn, 0,...) ∈ R and we ask: Where on R does this monomial expansion series converge? We call the set dom(f) of all those points the domain of convergence of f. In this thesis we give precise descriptions of the domain of convergence \ dom F(R) = dom(f) f∈F(R) of certain subsets F(R) ⊂ H(R) of holomorphic functions. Historically the problem that we consider here is connected with the definition of holomorphy in infinite dimensions. Holomorphy in finite dimensions can be characterized by convergence of monomial expansion series in the neighbourhood of every point and it was this approach that Hilbert tried at the beginning of the 20th century. In [46] he proposed a theory of analytic functions in infinitely many variables based on representations by power series

X α cαz (0.1)

(N) α∈N0 (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 Fr´echet and Gˆateaux that expansions in power series of homogeneous polynomials are more suitable, which formally can be obtained from (0.1) by building blocks

X α X X α cαz = cαz

(N) m |α|=m α∈N0

This theory was continued by students of Fréchet. The current form for functions between normed spaces can be found in papers of Graves [43] and Taylor [75, 76] from the mid 1930’s, where the equivalence of Fréchet differentiability, Gâteaux differentiability plus continuity, and representation by a power series of homogeneous polynomials in the neighbourhood of every point is proved. A natural environment where it is possible to work with monomial expansions are fully nuclear locally convex spaces with basis, as was shown in 1978 by Boland and Dineen [15], see [35], [37] and [36, sec. 3.3, 4.5] for more results in this direction. Matos [60] characterized those holomorphic functions on an open subset of a Banach space with unconditional basis having monomial series representations in the neighbourhood of every point to be functions of a certain holomorphy type (see also [61]). He proved that in `1 these are all holomorphic functions. 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 is more restrictive than he actually thought. He gives us a criterion to decide 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- P α ing sense: Every finite dimensional section n cαz of (0.1) (putting α∈N0 the variables zn+1, zn+2, zn+3,... equal to zero and identifying (α1, . . . , αn) = (N) (α1, . . . , αn, 0, 0,...) ∈ N0 ) is convergent for |z1| ≤ ε1,..., |zn| ≤ εn and

X α sup sup cαz < ∞ (0.3) n∈N |zi|≤εi n α∈N0

It can be seen from an example of Bohnenblust and Hille [12] that this is not 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 (see remark 2.9), that `1 is almost the only Banach sequence space where the criterion 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 holomorphic function 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 the beginning of the 20th century. He studied Dirichlet series

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

6 uniform convergence -

absolute convergence - -

T = sup a − u P an ns u a He attacked this problem using the nowadays so called Bohr method (see [13]). Using the fundamental theorem of arithmetic he established a one to one correspondence between Dirichlet series and power series in inﬁnitely many variables, given by

X an X α c z , c = a α (0.5) ns ! α α p n (N) α∈N0 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 P α P α |cα| ε < ∞ for every power series cαz bounded in Hilbert sense on |zn| ≤ 1, i.e. with X α sup sup cα z < ∞ (0.6) n |zi|≤1 n α∈N0 and every ε ∈ `q ∩ ] 0 , 1[ N and used the prime number theorem to prove

1 T = (0.7) S

Thus Bohr shifted the problem to the setting of bounded power series in inﬁnitely many variables and he was able to prove S ≥ 2 ([13, Satz III]): 12 Introduction

P α Theorem 0.1. Let cαz be a power series bounded in the domain [|zi| ≤ 1] and ε ∈ `2 ∩ ] 0 , 1[ N. Then

X α |cα| ε < ∞

(N) α∈N0

Bohr did not know if 2 was the exact value of S, not even if S < ∞ (i.e. if T > 0, in other words: if there is a Dirichlet series (0.4) whose absissa a of absolute convergence and u of uniform convergence do not coinside). Um dies Problem zu erledigen, ist ein tieferes Eindringen in die Theorie der Potenzreihen unendlich vieler Variabeln n¨otig,als es mir in §3 gelungen ist.1 Bohr hoped to solve this problem by studying the relation between the absolute value of a power series and the sum of the absolute values of the individual terms. He went down to dimension one and asked (see [14]):

P n Given 0 < r < 1, does there always exist a power series cnz convergent and bounded by one on the unit disc such that

∞ X n |cn| r > 1 n=1 is true ? This question found a satisfying answer.

Theorem 0.2 (Bohr’s power series theorem). P n Let cnz be a power series in C with

∞ X n cnz ≤ 1 (0.8) n=1 for all |z| < 1. Then ∞ X n |cnz | ≤ 1 n=1 for all |z| < 1/3 and the value 1/3 is optimal, i.e. for every r > 1/3 there is P n P∞ n a power series cnz satisfying (0.8) and n=1|cn|r > 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 by constructing for every δ > 0 an ε ∈ `4+δ ∩ ] 0 , 1[ N and a 2-homogeneous P α P α polynomial |α|=2 cαz bounded on [|zi| ≤ 1] with |α|=2|cα|ε = ∞. Then after some time Bohnenblust and Hille [12] proved in 1931 that 2 is in fact the exact value of S. They showed that the upper bound 4 of Toeplitz is best possible 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 polynomials for arbitrary m to get the result (see example 2.4). In 1989 Dineen and Timoney [37] studied the existence of absolute monomial bases for spaces of holomorphic functions on locally convex spaces and saw a possibility to solve their basis problem with the help of a polydisc version of Bohr’s theorem. In [38] they used this polydisc result to reprove the estimate S ≤ 2 in the way that Bohr originally might have thought of, showing that if P α P α ε ∈] 0 , 1[ N is such that |cα|ε < ∞ for all power series cαz bounded on [|zi| ≤ 1], then ε ∈ `2+δ for all δ > 0 (a sort of converse of theorem 0.1). In order to prove their multidimensional extension of theorem 0.2 they used a probabilistic method from [59] which goes back to the mid 70’s and of course was not accessible to Bohr. The initial step for our theory is to see that the considerations above about bounded power series in inﬁnitely many variables also work for bounded holomorphic functions on B`∞ . The power series in the deﬁnition of S can be replaced by these functions:

S = sup{r > 0 | `r ∩ B`∞ ⊂ dom H∞(B`∞ )} and the results above can be summarized in a very simple way as

`2 ∩ B`∞ ⊂ dom H∞(B`∞ ) ⊂ `2+ε ∩ B`∞ (0.9) for every ε > 0. Here the lower inclusion is a restatement of Bohr’s theorem 0.2 and the upper one reﬂects the result of Dineen and Timoney. The examples of Bohnenblust and Hille give

`2+ε ∩ B`∞ 6⊂ dom H∞(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 function f on some multiple rB`1 of the unit ball in `1, again underlining the fact that this Banach space plays a special role in infinite dimensional complex analysis. In 1997 a systematic study of multidimensional extensions of Bohr’s power series theorem started with the work [11]. Several so called Bohr radii where defined and estimated, for example in [3, 2, 4, 5, 10, 11, 24, 25, 62]. Aside from techniques of complex analysis a keypoint in these estimates is often the use of probabilistic methods. On the other hand questions related to the one dimensional Bohr phenomenon were investigated, as in [7, 16, 17]. The tools developed in these investigations provide important ideas for our theory and made it possible to describe the domain of convergence dom H∞(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 the domain of convergence of the spaces Pm(X) of m-homogeneous polynomials and P(X) of all polynomials. This is done in chapter 2. Our results arise from an interplay of complex analysis and local Banach space theory. There is a close connection with unconditionality in spaces of m-homogeneous polynomials, and during the investigation of the domains of convergence in chapter 2 the problem occurs to give estimates for the unconditional constant of identity mappings between those spaces which is solved going through full tensor products. That this can be done is justified in chapter 1 with a general theorem, providing a helpful tool to give estimates for many important invariants of local Banach space theory, such as type or cotype constants as well as the `-norm, of identities between symmtric tensor products working in the full tensor product, which in general is easier to handle. As an application, we give in chapter 1 asymptotically correct m volume estimates for unit balls of m-homogeneous polynomials P (Xn) and of the volume ratio for the m-fold injective and projective tensor product m ⊗α Xn, α = ε or π. The convergence of the monomial expansion series that we consider in chapter 2 is the convergence of the net of finite sums or unconditional convergence. In chapter 3 we investigate the problem of conditional convergence. This is solved for polynomials. Here the task is to order the monomials in an appropriate way, and we overcome this problem with a systematic study of orders that produce bases for tensor and s-tensor products of Banach spaces with basis. Preliminaries

Tensor products and m-homogeneous polynomials

m Let E1,...,Em and F be vector spaces over K. We write L (E1,...,Em; F ) Qm for the vector space of all m-linear mappings T : i=1 Ei −→ F and then m m L (E,F ) if E1 = ... = Em = E and Ls (E,F ) for the subspace of symmetric m m m mappings. If F = K we simple write L (E1,...,Em), L (E) and Ls (E).

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

kT k = sup kT (x1, . . . , xm)k kxik≤1

Linearization of m-linear mappings produces the (algebraic) tensor product m ⊗i=1Ei, i.e. we have

m m L (E1,...,Em; F ) = L(⊗i=1Ei,F ) for all vector spaces F via T LT , LT (x1 ⊗ · · · ⊗ xm) = T (x1, . . . , xm). m m If E = E1 = ... = Em we write ⊗ E = ⊗i=1Ei for the m-fold tensor product of E and ⊗m,sE for the symmetric m-fold tensor product of E which can be obtained by linearization of symmetric m-linear maps. ⊗m,sE is a m m,s m complemented subspace of ⊗ E and we denote ιm : ⊗ E −→ ⊗ E the natural injection and

⊗mE −→ ⊗mE 1 X σm : x ⊗ · · · ⊗ x x ⊗ · · · ⊗ x 1 m m! σ(1) σ(m) σ∈Sm 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, . . . , n}m and its subset J (m, n) = {j ∈ M(m, n) | j1 ≤ ... ≤ jm} as well as

m J (m) = {j ∈ N | j1 ≤ ... ≤ jm} 16 Preliminaries

Further the notation xi := xi1 ⊗ · · · ⊗ xim for elements x1, . . . , xn of a vector space 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,s M(m, n)} and {σm(xj) | j ∈ J (m, n)} are bases of ⊗ E and ⊗ E, respec- m m m,s m+n−1 tively, in particular dim ⊗ E = n and dim ⊗ E = n−1 . m The projective tensor norm π on the tensor product ⊗i=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 ˜ m m L(⊗π,i=1Ei,F ) = L (E1,...,Em; F ) isometrically for all normed spaces F . The injective tensor norm ε is the m m 0 0 norm induced by the embedding ⊗i=1Ei ,→ (⊗π,i=1Ei) . 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)

m m m k⊗i=1Ti : ⊗α,i=1Ei −→ ⊗α,i=1Fik = kT1k · · · kTmk If α is reasonable but does not necessarily satisfy the metric mapping prop- m erty, we speak of a tensor product norm (of order m). We write ⊗α,i=1Ei for m ˜ m the space ⊗i=1Ei equipped with the norm α and ⊗α,i=1Ei for its completion. An m-homogeneous polynomial P : E −→ F between vector spaces E and F 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 to be symmetric, and in this way the space P m(E,F ) of all m-homogeneous m polynomials P : E −→ F can be identiﬁed with Ls (E,F ). If E and F are normed spaces the subspace of continuous m-homogeneous polynomials is denoted by Pm(E,F ), equipped with norm

kP k = sup kP (x)k kxk≤1

In the same way as above the projective s-tensor norm πs is characterized by ˜ m m the fact that L(⊗πs E,F ) = P (E,F ) isometrically for all normed spaces E m,s and F and the injective s-tensor norm εs on ⊗ E is the norm induced by m,s m,s 0 0 ⊗ E,→ (⊗πs E ) . Then an s-tensor norm αs of order m assigns to each m,s m,s normed space E a norm αs = αs( · ; ⊗ E) on ⊗ E such that εs ≤ αs ≤ πs (αs is reasonable) and the metric mapping property is satisﬁed:

m,s m,s m,s m k⊗ T : ⊗αs E −→ ⊗αs F k = kT k for all normed spaces E and F and all operators T : E −→ F , where ⊗m,sT = m σm ◦ ⊗ T ◦ 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 to symmetric tensor products gives an s-tensor norm.

Now let α be a tensor norm and αs an s-tensor norm, both of order m. We say that α and αs are equivalent – notation α ∼ αs – if there is a constant m,s m cm > 0 such that the natural injection ιm : ⊗αs E −→ ⊗α E and projection m m,s σm : ⊗α E −→ ⊗αs E have norm smaller than cm for all normed spaces E. By the norm extension theorem of Floret [40], given an s-tensor norm αs of order m, there is always an equivalent tensor norm α, and the constant cm can be chosen to be smaller than mm/m!.

Now let 1 ≤ p ≤ ∞. The natural norm ∆p of order m on the p-integrable functions assigns to each m-tuple (Lp(µ1),...,Lp(µm)) of Lp-spaces (over m m measure spaces (Ωi, µi)) the norm ∆p = ∆p( · ; ⊗i=1Lp(µi)) on ⊗i=1Lp(µi) coming from the natural inclusion

m ⊗i=1Lp(µi) ,→ Lp(µ1 ⊗ · · · ⊗ µm)

˜ ˜ m where the element f1 ⊗ · · · ⊗ fm ∈ ⊗i=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 the m-fold tensor product of discreet L -spaces. We write ⊗m`n := ⊗m `n and ⊗m,s`n := ⊗m,s `n. p p p ∆p p p p ∆p,s p So

X X p1/p λiei m n = |λi| ⊗p `p i∈M(m,n) i∈M(m,n) X X p1/p λjej m,s n = |λj| ⊗p `p j∈J (m,n) j∈J (m,n)

m m n n m,s n dn m+n−1 i.e. ⊗p `p = `p and ⊗p `p = `p isometrically, where dn = n−1 is the m,s n dimension of ⊗p `p .

Note that ∆p and ∆p,s are reasonable on the class of Banach spaces where they 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 as m-homogeneous polynomials see [22], [39], [40] and [36].

Banach sequence spaces

By a Banach sequence space we mean a Banach space X ⊂ CN of sequences in C such that `1 ⊂ X ⊂ `∞, and satisfying that |x| ≤ |y| implies x ∈ X and kxk ≤ kyk whenever x ∈ CN, y ∈ X. In particular, the canonical unit vectors en = (δnk)k belong to X. 18 Preliminaries

Defined in this way a Banach sequence space X in particular is a complex Köthefunction space in the sense of [57] (the complexification of the real N Banach sequence space XR = {x ∈ X | x ∈ R }) 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-unconditional basic sequence and in particular every inﬁnite dimensional complex Banach space with 1-unconditional Schauder basis satisfying the maximality condition 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 set inclusion X ⊂ Y between Banach sequence spaces X and Y already implies that the inclusion mapping X,→ Y is continuous.

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

∗ xn := inf{ sup |xi| | I ⊂ N , |I| < n} i∈N\I If x is a zero sequence this is nothing else than the sequence |x| rearranged in some non-increasing order. For a Banach sequence space X with basis being symmetric it is equivalent to say that any permutation of an element of X remains in X with equal norm or equivalently the basis (en) of X is symmetric and ∞ ∞ X X anen = aneπ(n) n=1 n=1 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 sequence space X always think of functions N −→ C. For example |x| = (|xn|)n, x > 0 p p means xn > 0 for all n, x = (xn)n for x ≥ 0 and p > 0, xy is the sequence with n-th coordinate xnyn for x, y ∈ X and so on. Let 0 < p ≤ ∞. A Banach sequence space X is called p-convex if the p-convexity constant of X

(p) n n M (X) := sup kid : `p (X) −→ Mp (X)k n is ﬁnite and p-concave if its p-concavity constant

n n M(p)(X) := sup kid : Mp (X) −→ `p (X)k n Preliminaries 19

n n n is ﬁnite, where Mp (X) and `p (X) denote X with the quasi norm (norm, if p ≥ 1)  n  X p1/p  |xi| X p < ∞ kxk = i=1 k max |xi|kX p = ∞  1≤i≤n

n and kxk = k(kx1kX ,..., kxnkX )k`p , respectively. For 1 ≤ p ≤ ∞, p-convexity is decreasing and p-concavity increasing in p ([57, 1.d.5]). For p-convexity this is also true if p < 1 (see [20, lemma 4]). Every Banach sequence space is 1-convex and ∞-concave, we say trivially convex and trivially concave, with

(p) M (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 K¨othedual of X is deﬁned to be the space

× N X = {y ∈ C | xy ∈ `1}

It is again a Banach sequence space with norm kyk × = sup kxyk X kxkX ≤1 `1 (this number is ﬁnite for every y ∈ X× by the closed graph theorem). The K¨othedual X× can be regarded as a subspace of X0 via

∞ X λ [x λnxn] n=1

× 0 and we have X = X if and only if X = span{en}. Further X ⊂ Y implies Y × ⊂ 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 r 1/r X := {x ∈ CN | |x| ∈ X} and the product X · Y := {xy | x ∈ X, y ∈ Y } of X and Y . Pn The fundamental function of X is defined by λX (n) = k i=1 eik, and we denote n-th section Xn = span{e1, . . . , en} of X to be the space generated by the first n standard unit vectors. We mention two types of examples for Banach sequence spaces from the literature. Let 1 = w1 ≥ w2 ≥ ... ≥ wn ≥ ... ≥ 0 be a sequence of weights P∞ with n=1 wn = ∞ and 1 ≤ p < ∞. Then ∞ N X p d(w, p) := {x ∈ C | sup wn|xπ(n)| < ∞} π n=1 20 Preliminaries together with the norm ∞ 1/p X p kxk = sup wn|xπ(n)| π n=1 is a Banach space which is called Lorentz sequence space. Here the sup is 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 this basis is 1-unconditional. The norm can be expressed as ∞ 1/p X ∗ p kxk = wn|xn| n=1 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 sequence space in our sense. For more general information on Lorentz sequence spaces see [56, 4.e].

Taking all weights equal to one we get the usual `p-spaces. More general if q −1 we deﬁne wn := n p 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 interpolation theory (see e.g. [8, chapter IV] or [57, 2.b.8]). The space `p,q is r-convex if 0 < 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 and limt→∞ ϕ(t) = ∞. ϕ is called a degenerate Orlicz function if ϕ(t) = 0 for some t > 0. The Orlicz function generates a sequence space ∞ X |xn| ` = {x ∈ N | ∃ : ϕ < ∞} ϕ C ρ>0 ρ n=1 which is a symmetric Banach sequence space with norm ∞ X |xn| kxk = inf{ρ > 0 | ϕ ≤ 1} ρ n=1 (see e.g. [68, theorem 10]). The unit vectors form a normalized 1-unconditional basis of the subspace ∞ X |xn| h := {x ∈ ` | ∀ ρ > 0 : ϕ < ∞} ϕ ϕ ρ n=1

(see [56, prop. 4.a.2]) which is equal to `ϕ if and only if the Orlicz function ϕ satisﬁes the ∆2-property ϕ(2t) lim sup < ∞ t→∞ ϕ(t) Preliminaries 21

In this case and under the additional assumption that ϕ is non-degenerate the following conditions are equivalent (see [49], proposition 14 and lemma 5):

(1) `ϕ is p-convex. (2) ϕ is equivalent to an Orlicz functionϕ ˜ which satisﬁes

ϕ˜(λx) ≥ λpϕ˜(x)

for all x ≥ 0 and λ ≥ 1. √ (3) ϕ is equivalent to an Orlicz functionϕ ˜ such thatϕ ˜( p · ) is convex.

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

n n 1/p X X p xi ≤ c kxik i=1 i=1

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

Here equivalence of ϕ andϕ ˜ means that there exists t0 > 0 and a constant c > 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.A function f : U −→ F is called Fréchetdifferentiable if for all ξ ∈ U we can find a continuous linear mapping A : E −→ F such that

kf(x) − f(ξ) − A(x − ξ)k lim = 0 x→ξ kx − ξk where f 0(ξ) := Df(ξ) := A is called the Fréchet derivative of f in ξ. For m ≥ 2 the function f is m times Fréchet differentiable if f (m−1) : U −→ L(E, Lm−2(E,F )) = Lm−1(E,F ) is Fréchet differentiable. By Schwarz’ the- (m) m m orem we have f ∈ Ls (E,F ). We denote by C (U, F ) the vector space of all m-times Fréchet differentiable mappings f : U −→ F and Cm(U) := Cm(U, C). The function f : U −→ F is Gâteaux holomorphic or G-holomorphic if for all ξ ∈ U and all x ∈ E the function λ f(ξ + λx) is holomorphic on U(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 0 and every finite dimensional subspace E0 ⊂ E the function ϕ ◦ f is holomorphic on U ∩ E0. Let HG(U, F ) denote the space of all G-holomorphic mappings f : U −→ F and HG(U) := HG(U, C). The function f : U −→ F is holomorphic iff it has a Taylor series expansion at each point ξ of U: For all ξ ∈ U we can find a unique sequence of continuous m m-homogeneous polynomials Pmf(ξ) ∈ P (E,F ) and a radius r > 0 such that ∞ X f(ξ + x) = Pmf(ξ)(x) (0.11) m=0 or ∞ X m f(ξ + x) = am(f)x m=1 uniformly on rBE, where am(f) ∈ Ls(E,F ) are the symmetric m-linear mappings corresponding to the m-homogeneous polynomials such that m ∞ Pmf(ξ)(x) = am(f)x := am(f)(x, . . . , x). In this case f ∈ C (ξ + rBE,F ) (m) and am(f) = f (ξ)/m!. We denote H(U, F ) the vector space of all holomorphic mappings f : U −→ F and H(U) := H(U, C). A function f : U −→ F is holomorphic if and only if it is Fréchet differentiable if 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 we have for all x ∈ B 1 Z f(ξ + λx) Pmf(ξ)(x) = m+1 dλ , m = 0, 1, 2,... (0.12) 2πi |λ|=1 λ (Cauchy integral formulas), in particular

kPmf(ξ)kB ≤ kfkξ+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 we have f ∈ H(U) if and only if f is continuous and f is holomorphic on every ﬁnite dimensional component of U.

We denote H∞(U) the Banach space of all bounded holomorphic functions on U with norm kfkU = supz∈U |f(z)|. By a Reinhardt domain R ⊂ Cn we mean an open set which is n-circular n and balanced, i.e. it satisﬁes λR ⊂ R for all λ ∈ B`∞ . 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 assumption 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 an open, circular and balanced subset of X, in other words satisfying λR ⊂ R for all λ ∈ B`∞ . Note that if R ⊂ X is a Reinhardt domain then the n-th n projection Rn := R ∩ Xn of R is a Reinhardt domain in C . An important example for a Reinhardt domain in a Banach sequence space X is the open unit ball BX , whose n-th projection BXn is even a logarithmically convex Reinhardt domain in Cn.

1. Norm estimates of identities between symmetric tensor products of Banach sequence spaces

An active research area of Banach space theory in recent years is the local theory which investigates the structure of infinite dimensional Banach spaces by relating it to the structure of their finite dimensional subspaces. It has its roots in the work of Grothendieck in the 1950s. An important method is to derive properties of a Banach space from the asymptotical behaviour of isometric invariants of its finite dimensional subspaces when the dimension grows to infinity. For example in [28] an asymptotically correct estimate in m n n of the unconditional basis constant of the space P (`p ) of m-homogeneous n m polynomials on `p is used to prove that P (E) cannot have an unconditional basis for every infinite dimensional Banach space E with unconditional basis. Other important invariants of the local theory are for example type and cotype constants or the Gordon-Lewis constant. These constants are defined for general operators between Banach spaces and we will see that it can be useful to know their asymptotic behaviour (in the dimension n) for identity mappings id : ⊗m,sX −→ ⊗m,sY αs n βs n between spaces of symmetric tensor products, where X and Y are Banach sequence spaces. For example in chapter 2 we are confronted with the problem m m of giving the asymptotic for χM (id : P (Xn) −→ P (Yn)), a generalization of the unconditional basis constant of the monomials (see definition 2.59), m m,s 0 which by the identification P (Xn) = ⊗εs Xn for Banach sequence spaces X is a problem of the above type. In general it is easier to calculate those estimates for identities between full tensor products, since there is a well developed theory for the α-tensor product E⊗˜ αF of two Banach spaces E and F which in many cases can be easily ˜ m extended to m-fold tensor products ⊗α,i=1Ei (see e.g. Floret [39]). For ex- ˜ m ample in the case of α = ε or π the α-tensor product ⊗α,i=1Ei is associative, which makes it possible to proceed by induction and to deduce an estimate m m for id : ⊗α Xn −→ ⊗β 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 for some constant c > 0 independent of n and we write an bn if an ≺ bn and bn ≺ an. In this chapter we are going to give general estimates which permit in many cases to work in the full tensor product. In particular we prove that A(id : ⊗m,sX −→ ⊗m,sY ) A(id : ⊗mX −→ ⊗mY ) αs n βs n α n β n if α ∼ αs, β ∼ βs and A is an ideal norm. Here the asymptotic is with respect to n (m is always ﬁxed in this chapter). This generalizes results from [21] m,s n m n m,s and [29]. There it is proved that gl(⊗εs `p ) gl(⊗ε `p ) and C2(⊗εs Xn) m C2(⊗ε 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-

m totic descriptions for the volume of the unit ball BP (Xn) and the volume ratio m vr(⊗α Xn), α = ε or π for a wide class of symmetric Banach sequence spaces X.

Identities between symmetric and full tensor products

As in [21] and [29] for our estimates we use the fact that for α = ε, and in fact for an arbitrary tensor norm α with equivalent s-tensor norm αs, the m,s m symmtric tensor product ⊗αs Xn is a complemented subspace of ⊗α Xn and with a certain construction the full α-tensor product is also complemented m,s in the symmetric one (more precisely in ⊗αs Xnm). First we deﬁne the algebraic mappings which are involved in the complement construction (see also [29, section 3] or [39, 1.10]).

For N, n ∈ N, k = 0,...,N and i = 1,...,N we deﬁne the mappings n n n Nn+k X X Ii : K −→ K , λj ej λj en(i−1)+j j=1 j=1 Nn+k n (1.1) Nn+k n X X Pi : K −→ K , λj ej λn(i−1)+j ej j=1 j=1

m and Ii := Ii1 ⊗ · · · ⊗ Iim , Pi := Pi1 ⊗ · · · ⊗ Pim for i ∈ {1,...,N} . Then X X Ii λjej = λj−n(i−1)ej j∈{1,...,n}m n(i−1)+1≤j≤ni (1.2) X X Pi λjej = λn(i−1)+jej j∈{1,...,Nn+k}m j∈{1,...,n}m Identities between symmetric and full tensor products 27

(where 1 = (1,..., 1) ∈ Nm if we write i + 1 with i ∈ Nm). m m,s n m n Further let idn, ⊗ idn and ⊗ idn denote the identity map on K , ⊗ K m,s n and ⊗ K , respectively. Note that Ii is an injection, Pi a projection and N X idNn = Ii ◦ idn ◦Pi i=1 (1.3) m X m ⊗ idNn = Ii ◦ ⊗ idn ◦Pi i∈{1,...,N}m

m n m Recall that (ej)j∈{1,...,n} is a basis of ⊗ K and (σm(ej))1≤j1≤...≤jm≤n a basis of ⊗m,sKn. From (1.2) we get X X σm ◦ Ii λjej = λj−n(i−1)σm(ej) (1.4) j∈{1,...,n}m n(i−1)+1≤j≤ni and X X m!Pi ◦ ιm λjσm(ej) = λn(i−1)+jej (1.5) j∈{1,...Nn+k}m j∈{1,...,n}m since ( 1 m! ej−n(i−1) n(i − 1) + 1 ≤ j ≤ ni Pi(σm(ej)) = 0 otherwise In particular

m n m,s mn+k I := σm ◦ I1 ⊗ · · · ⊗ Im : ⊗ K −→ ⊗ K (1.6) is an injection,

m,s mn+k m n P := m! P1 ⊗ · · · ⊗ Pm ◦ ιm : ⊗ K −→ ⊗ K (1.7) m m n m n a projection and PI = ⊗ idn is the identity on ⊗ K , i.e. ⊗ K is a complemented subspace of ⊗m,sKmn+k and the diagram

m ⊗ idn ⊗mKn - ⊗mKn 6

I1⊗···⊗Im m! P1⊗···⊗Pm ? ⊗mKmn+k ⊗mKmn+k (1.8) 6 σm ιm

? m,s ⊗ idmn+k ⊗m,sKmn+k - ⊗m,sKmn+k commutes.

Note that the norms α, β and αs, βs in the following can be ∆p respectively m m,s ∆p,s. In this case for example ⊗α Xn or ⊗αs Xn requires X = `p. 28 1. Norm estimates of identities between symmetric tensor products

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

[ m m f : L(⊗α Xn, ⊗β Yn) −→ R≥0 n [ g : L(⊗m,sX , ⊗m,sY ) −→ αs n βs n R≥0 n be functions and ci > 0, i = 1, 2, 3 be constants such that for all n

m m m,s m,s (i) f|L(⊗ Xn,⊗ Yn) and g|L(⊗ X ,⊗ Y ) satisfy the triangle inequality. α β αs n βs n (ii) For all N ∈ N and all i ∈ {1,...,N}m

m m m f(Ii ◦ ⊗ idn ◦Pi : ⊗α XNn −→ ⊗β YNn) m m m ≤ c1 f(⊗ idn : ⊗α Xn −→ ⊗β Yn)

(iii) For all M ≤ N

m m m m m m f(⊗ idM : ⊗α XM −→ ⊗β YM ) ≤ c2 f(⊗ idN : ⊗α XN −→ ⊗β YN )

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

m m m f(⊗ idn : ⊗α Xn −→ ⊗β Yn) ≤ c g(⊗m,s id : ⊗m,sX −→ ⊗m,sY ) 3 mn+k αs mn+k βs mn+k

Then

f(⊗m id : ⊗mX −→ ⊗mY ) ≤ c g(⊗m,s id : ⊗m,sX −→ ⊗m,sY ) n α n β n n αs n βs n

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

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

m m m f(⊗ idNn) ≤ c1 N f(⊗ idn) for all N. Together with (iii) and (iv) this implies

m m m m f(⊗ idmn+k) ≤ c2 f(⊗ id(m+1)n) ≤ c1 c2 (m + 1) f(⊗ idn) m m,s ≤ c1 c2 c3 (m + 1) g(⊗ idmn+k) for k = 0, . . . , m. Proposition 1.2. Let (A, A) be a normed operator ideal, α, β tensor product norms 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-tensor norm βs.

(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 : ⊗mX −→ ⊗mY ) ≤ c A(id : ⊗m,sX −→ ⊗m,sY ) α n β n αs n βs n

(mm)2 m m for all n ≥ m where (a) c ≤ m! (m + 1) , (b) c ≤ (m(m + 1)) , (c) (m(m+1))m m c ≤ m! and (d) c ≤ (m + 1) .

Proof. We are going to check (i) – (iv) of lemma 1.1 for A = f = g. A is m m an ideal norm, hence satisﬁes (i) and for i ∈ {1,...,N} and Ii : ⊗β Yn −→ m m m ⊗β YNn and Pi : ⊗α XNn −→ ⊗α Xn (as deﬁned right before (1.2)) we have

m m A(Ii ◦ ⊗ idn ◦Pi) ≤ kIikkPikA(⊗ idn)

We claim kIik = 1 = kPik. Since X and Y are symmetric, the mapping Il : Yn −→ YNn is an isometry and Pl : XNn −→ Xn a norm-1-projection for l = 1,...,N (see (1.1)). Thus if α and β are tensor norms then the claim follows by the metric mapping property. On the other hand if α = ∆p and m n m Nn β = ∆q then a glance at (1.2) assures that Ii : ⊗q `q −→ ⊗q `q is even an m Nn m n isometry and Pi : ⊗p `p −→ ⊗p `p 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 the natural projection and injection, respectively. Then

⊗m id m M - m ⊗α XM ⊗β YM 6 ⊗mI ⊗mP ? ⊗m id m N - m ⊗α XN ⊗β YN commutes, and again if α and β are tensor norms then the vertical mappings have norm 1 by the metric mapping property. For the other cases just note m m that (⊗ I)(ej) = ej and (⊗ P )(ej) = ej if j1, . . . , jm ≤ M and zero other- m m M m N wise which implies that ⊗ I : ⊗q `q −→ ⊗q `q is the natural injection and m m N m M ⊗ P : ⊗p `p −→ ⊗p `p the natural projection. Hence

m m m m m A(⊗ idM ) ≤ k⊗ Ikk⊗ P kA(⊗ idN ) = A(⊗ idN ) and we can take c2 = 1. 30 1. Norm estimates of identities between symmetric tensor products

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

⊗m id m n - m ⊗α Xn ⊗β Yn 6 IP ? ⊗m,s id ⊗m,sX mn+k - ⊗m,sY αs mn+k βs mn+k commutes by (1.8) and we calculate the norms of I and P to get the constant c3. 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 are tensor norms with equivalent s-tensor norms αs and βs, respectively, then 1 m we have kIk and m! kP k bounded by m /m! since this is a bound for kσmk (mm)2 and kιmk. Thus we can take c3 = m! in case (a), c3 = 1 in case (d) and m c3 = m 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 in proposition 1.2 satisfying one of the four conditions (a)–(d) and X and Y symmetric Banach sequence spaces. Then A(id : ⊗m,sX −→ ⊗m,sY ) ≤ c A(id : ⊗mX −→ ⊗mY ) αs n βs n α n β n

mm 2 mm mm for all n ≥ m where (a) c ≤ m! , (b) c ≤ m! , (c) c ≤ m! , (d) c ≤ 1.

Proof. Note that the mapping

m,s n m n ⊗p `p −→ ⊗p `p ιm,p : P P j∈J (m,n) λjσm(ej) j∈J (m,n) λjej m n m,s is an isometry and σm,q : ⊗q `q −→ ⊗q `q deﬁned by ( σm(ej) j ∈ J (m, n) σm,q(ej) := 0 otherwise is a norm-1-projection. All the diagrams

⊗m id ⊗m id ⊗m,sX n - ⊗m,sY ⊗m,s`n n - ⊗m,sY αs n βs n p p βs n 6 6 (a) ιm σm (b) ιm,p σm ? ? ⊗m id ⊗m id m n - m m n n - m ⊗α Xn ⊗β Yn ⊗p `p ⊗β Yn

⊗m id ⊗m id m,s n - m,s n m,s n n - m,s n ⊗αs Xn ⊗q `q ⊗p `p ⊗q `q 6 6 (c) ιm σm,q (d) ιm,p σm,q ? ? ⊗m id ⊗m id m n - m n m n n - m n ⊗α Xn ⊗q `q ⊗p `p ⊗q `q Volume estimates for unit balls of spaces of m-homogeneous polynomials 31

are commutative which gives the claim since kιmk and kσmk are bounded by mm/m!.

We have proved the following theorem.

Theorem 1.4. Let (A, A) be a normed operator ideal and α, αs, β, βs be tensor product norms satisfying one of the conditions (a)–(d) of proposition 1.2. Let X and Y be symmetric Banach sequence spaces. Then

A(id : ⊗m,sX −→ ⊗m,sY ) A(id : ⊗mX −→ ⊗mY ) αs n βs n α n β n

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 volume n ratio of the tensor product of two `p -spaces is estimated, a result which was originally obtained by Sch¨utt[73].

By the volume vol B of a Borel subsets B ⊂ RN we mean the Lebesgue m m n n m,s n dn measure of B and we identify ⊗ R = R and ⊗ R = R , dn = m,s n m+n−1 dim ⊗ R = n−1 by

m n nm m,s n d ⊗ R −→ R ⊗ R −→ R n P , P i∈M(m,n) λiei (λi)i∈M(m,n) j∈J (m,n) λjσm(ej) (λj)j∈J (m,n) when we write vol B for a subset B ⊂ ⊗mRn or B ⊂ ⊗m,sRn. It is well known that 1/n −1/p n (vol B`p ) n (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

 m (λX (n))  X 2-convex, non trivially concave  n(m+1)/2 1/dn  m vol BP (Xn) λ (n)  X X 2-concave, non trivially convex  n

m+n−1 m where dn = n−1 stands for the dimension of P (Xn). 32 1. Norm estimates of identities between symmetric tensor products

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

 −(m( 1 − 1 )+ 1 ) n 2 p 2 2 ≤ p < ∞ 1/dn  vol BPm(`n) p −(1− 1 ) n p 1 < p ≤ 2

Before we start working on the proof of the theorem let us collect some known facts about Banach sequence spaces that we will use in the following. Let X be a Banach sequence space. Then also the K¨othe dual X× is a Banach × 0 sequence space and (X )n = (Xn) isometrically. For 1 ≤ p ≤ ∞ we have that X is p-convex if and only if X× is p0-concave and in this case

(p) × M (X) = M(p0)(X ) (1.10) where p0 is the conjugate exponent satisfying 1/p + 1/p0 = 1 ([57, 1.d.4]). Our interest will be the case p = 2, so we mention that in particular

(2) 0 M (Xn) = M(2)((Xn) ) (1.11)

(2) (2) Note that M (Xn) ≤ M (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 < ∞  1 X p-convex n  kid : `p −→ Xnk (1.13) −1/p n λX (n) X p-concave and dually

 1/p −1 n (λX (n)) X p-convex n  kid : Xn −→ `p k (1.14) 1 X p-concave

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

1/2 n n M(2)(Xn) kid : Xn −→ `2 k (1.15) λX (n) whenever X is 2-convex with ﬁnite concavity and

λ (n) M(2)(X ) X kid : `n −→ X k (1.16) n n1/2 2 n 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 projective tensor product

n Let γn be the standard Gaussian probability measure on R with density Pn 2 −1/2 i=1 xi n/2 n e /(2π) and let E be a Banach space. Then L(`2 ,E) is a closed n subspace of L2(R , γn,E) and the induced norm is called the `-norm on n L(`2 ,E), denoted by `. In other words, if g1, . . . , gn are standard Gaussian random variables on a probability space (Ω, Σ, µ), i.e. they√ are independent −x2/2 and all have the standard normal distribution γ1 = e / 2π dx, then Z n X 2 1/2 `(u) = gi(ω)uei Eµ(dω) Ω i=1 n for u ∈ L(`2 ,E), and by the rotational invariance of γn the standard unit n vectors e1, . . . , en can be replaced by any orthonormal basis of `2 (see [33, p. 239] for a proof of this fact). The `-norm can be extended to operators u : E −→ F between arbitrary Banach spaces by

n `(u) = sup{`(uv) | n ∈ N , v ∈ L(`2 ,E) , kvk ≤ 1} and all operators u with `(u) < ∞ together with ` form the normed operator ideal of Gaussian operators (see [78, p. 81]). Remark 1.6. There is a constant c > 0 such that for all ﬁnite dimensional Banach spaces E = (RN , k · k) −1 N −1 1/N −1 N 0 c `(id : `2 −→ E) ≤ (vol BE) ≤ c N `(id : `2 −→ E )

Proof. Let E = (RN , k · k) be a Banach space. Then we have 1/N 0 ! vol T B 0 E ≤ N −1/2 `(T ) (1.17) vol B N `2

N for all linear operators T : `2 −→ E (see [64, remark 3.14]). Since −1 −1/2 1/n −1/2 c n ≤ (vol B n ) ≤ c n (1.18) `2 for some constant c independent of n this implies the upper estimate:

1/N 1/N −1/2 N 0 −1 N 0 (vol BE) ≤ vol(B N ) N `(id : ` −→ E ) ≤ c N `(id : ` −→ E ) `2 2 2 On the other hand the converse of Santalo’s inequality by Bourgain and Milman gives a constant c0 > 0 such that for all n and all convex symmetric and bounded subsets B ⊂ Rn with non-empty interior ◦ 1/n vol B vol B 0 2 ≥ c (vol B n ) `2 34 1. Norm estimates of identities between symmetric tensor products

N (see [64, corollary 7.2]). Thus using again (1.17) with T = id : `2 −→ E and (1.18) we obtain

1/N 1/N vol B vol B 0 c0 ≤ E E vol B N vol B N `2 `2 −1/N −1/2 N 1/N ≤ (vol B N ) N `(id : ` −→ E) (vol BE) `2 2 N 1/N ≤ c `(id : `2 −→ E) (vol BE)

n Corollary 1.7. Let (En)n be a sequence of Banach spaces (R , k · kn) such that n n 0 `(id : `2 −→ En) `(id : `2 −→ En) ≺ n Then

1/n n −1 −1 n 0 (vol BEn ) `(id : `2 −→ En) n `(id : `2 −→ En)

m m,s 0 m 0 m,s Now we have P (Xn) = ⊗εs (Xn) and (P (Xn)) = ⊗πs Xn and an estimate of the form

m,s n m,s m,s n m,s 0 `(id : ⊗2 `2 −→ ⊗πs Xn) `(id : ⊗2 `2 −→ ⊗εs (Xn) ) ≺ dn (1.19) would provide us with an asymptotical description of the desired volume in m terms of the `-norm. Since dn n and

m,s n m,s m n m `(id : ⊗2 `2 −→ ⊗αs Yn) `(id : ⊗2 `2 −→ ⊗α Yn) for α ∈ {ε, π} and any Banach sequence space Y by theorem 1.4 it will be enough to work in the full tensor product. So what we are going to do is to m n m give asymptotically correct estimates for `(id : ⊗2 `2 −→ ⊗α Xn), α = ε or π in terms of the fundamental function λX (n). This will imply (1.19) and 1/d m n hence also the estimates for (vol BP (Xn)) . We ﬁrst treat the case α = ε.

Lemma 1.8. There is a constant cm > 0 such that for all n ∈ N, all Banach n spaces E and all linear operators u : `2 −→ E

m−1 m m n m m−1 kuk `(u) ≤ `(⊗ u : ⊗2 `2 −→ ⊗ε E) ≤ cm kuk `(u)

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

n n max{kuk `(v) , kvk `(u)} ≤ `(u ⊗ v : `2 ⊗2 `2 −→ E ⊗ε F ) ≤ c (kuk `(v) + kvk `(u)) `-norm estimates for the injective and projective tensor product 35

n for all n ∈ N, Banach spaces E, F and operators u : `2 −→ E and n v : `2 −→ F (see also [18, p. 243]). n For the lower bound let x = x1 ∈ `2 with norm 1 and complete to an n orthonormal basis x1, . . . , xn of `2 . Let gij, i = 1, . . . , m, j = 1, . . . , n be a family of standard Gaussian random variables. Then by the contraction principle (see [33, p. 231])

m n 1/2 Z X X 2 `(u ⊗ v) = gij(ω)uxi ⊗ vxj µ(dω) E⊗ F Ω i=1 j=1 ε

m 1/2 Z X 2 ≥ kvx1k gi1(ω)uxi µ(dω) E Ω i=1 = kvxk `(u) implying kvk `(u) ≤ `(u ⊗ v) and with the same argument we get kuk `(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

n `(id : `2 −→ Xn) λX (n) (1.20) (see [29, p. 294]). This is due to Kahane’s inequality and the fact that in this case Gaussian and Rademacher averages can be estimated by each other with universal constants (see [78, p. 15 and 16]).

Theorem 1.9. Let X be a symmetric Banach sequence space. Then  X 2-convex, λ (n)  X non trivially concave `(id : ⊗m`n −→ ⊗mX ) 2 2 ε n (λ (n))m  X X 2-concave  n(m−1)/2

Proof. By lemma 1.8 we have

m n m n m−1 n `(id : ⊗2 `2 −→ ⊗ε Xn) kid : `2 −→ Xnk `(id : `2 −→ 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 the following. 36 1. Norm estimates of identities between symmetric tensor products

n Denote rn = sgn(sin 2 π( · )) the n-th Rademacher function and for a Banach n n Pn space E let R2 (E) be E with the semi-norm kxk = k i=1 rixikL2([0,1],E) and n n n W = E with norm kxk = sup k(ϕ(x )) k n . Now let E and F be p ϕ∈BE0 i i=1 `p Banach spaces and u : E −→ F be a linear operator. For n ∈ N denote n n n n n u : E −→ F the mapping u (x) = (u(xi))i=1. Recall that u has type p/cotype p/is p-summing if the type p constant

n n n Tp(u) = sup ku : `p (E) −→ R2 (F )k n the cotype p constant

n n n Cp(u) = sup ku : R2 (E) −→ `p (F )k n and the p-summing constant

n n n πp(u) = sup ku : Wp (E) −→ `p (F )k n 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 Banach space is said to have type p/cotype p/to be p-summing if the identity opertor idE 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 is defined: Given a Banach operator ideal (A, A), a Banach space E is said to 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 constant c > 0 such that for all Banach spaces E and every operator u : `2 −→ E

0 `(u) ≤ c T2(E)π2(u ) (1.21) holds (see [78, p. 83 and theorem 25.1]).

m First we give an asymptotic description of the type 2 constant of ⊗π Xn which is dual to that in [29]. There it is shown that

m (m−1)/2 C2(⊗ε Xn) n 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 only if it is 2-convex and non trivially concave (see [33, 16.20]) and it has cotype 2 if and only if it is 2-concave ([57, 1.d.6]). From [57, 1.d.6] and Kahane’s inequality it follows that

(2) T2(Xn) M (Xn) (1.23) `-norm estimates for the injective and projective tensor product 37 and C2(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 and non trivially concave or 2-concave and non trivially convex. Then

m (m−1)/2 T2(⊗π Xn) n T2(Xn)

Proof. The proof is based on duality. For the lower bound we use (1.22) and the fact that 0 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

(2) 0 0 T2(Xn) M (Xn) M(2)((Xn) ) C2((Xn) ) and together with (1.22) we obtain

(m−1)/2 m 0 m 0 m 0 n T2(Xn) C2(⊗ε (Xn) ) = C2((⊗π Xn) ) ≤ T2((⊗π Xn) ) by the duality of ε and π. Now we prove the upper estimate. An inequality reverse to (1.25) is in general not true (see for example [22, p. 105]), and we are going to factorize m through suitable spaces [Xn] and [Xn]m which permit an appropriate reverse estimate. First assume that X is 2-convex and non trivially concave. As in [29] we deﬁne inductively the space [X]m of all functions f : Nm −→ R satisfying m−1 m−1 m−1 f(i, ·) ∈ X for all i ∈ N and (kf(i, ·)kX )i∈N ∈ [X] with norm

m m−1 m−1 kfk[X] = k(kf(i, ·)kX )i∈N k[X] which satisﬁes all conditions of a Banach sequence space (choose an order of Nm) 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: By induction it is easy to check that

m m M(p)([X] ) ≤ M(p)(X) is true).

m m For the subspace [Xn] of all functions f ∈ [X] which are zero outside m m 0 × m {1, . . . , n} we have ([Xn] ) = [(X )n] isometrically and from [29, p. 304] we get × m × m C2([(X )n] ) ≺ M(2)((X )n) 38 1. Norm estimates of identities between symmetric tensor products hence together with (1.23), (1.11) and (1.24)

m (2) m m 0 T2([Xn] ) M ([Xn] ) M(2)(([Xn] ) ) (1.26) × m × m (2) m C2([(X )n] ) ≺ M(2)((X )n) M (Xn)

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

α(x1 ⊗ · · · ⊗ xm) = kx1k · · · kxmk

m m and hence α ≤ π. Thus factorizing the identity id : ⊗π Xn −→ ⊗π Xn m through [Xn] we obtain

m m m m m m T2(⊗π Xn) ≤ kid : ⊗π Xn −→ [Xn] k T2([Xn] ) kid : [Xn] −→ ⊗π Xnk m m ≺ kid : [Xn] −→ ⊗π Xnk Now in [29, lemma 5.5] it is proved that

m m m−1 m−1 (m−1)/2 kid : ⊗ε Yn −→ [Yn] k ≤ KG M(2)(Yn) n for every Banach sequence space Y , where KG denotes the Grothendieck constant. Hence by duality

m m m × × m (m−1)/2 kid : [Xn] −→ ⊗π Xnk = kid : ⊗ε (X )n −→ [(X )n] k ≺ n

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

Now let X be 2-concave and non trivially convex. Then we use the space [X]m m m−1 of functions f : N −→ R satisfying f(·, i) ∈ X and (kf(·, i)kX )i∈N ∈ [X]m−1 with norm

m−1 kfk[X]m = k(kf(·, i)kX )i∈N k[X]m−1

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

m M(2)([X]m) ≤ M(2)(X)

0 × As above we have ([Xn]m) = [(X )n]m isometrically for the subspace [Xn]m m of all functions f ∈ [X]m which are zero outside {1, . . . , n} and from [29, p. 307] we get × × m C2([(X )n]m) ≺ M(2)((X )n) Again using (1.23), (1.11) and (1.24) we obtain

(2) 0 T2([Xn]m) M ([Xn]m) M(2)(([Xn]m) ) × × m C2([(X )n]m) ≺ M(2)((X )n) `-norm estimates for the injective and projective tensor product 39 and with [29, lemma 6.2] it follows

1/2 m−1 × n T2([Xn]m) ≺ M(2)((X )n) (1.27) λX× (n) Here we use that X is 2-concave with non trivial convexity. Again we have m ⊗ Xn = [Xn]m naturally by x1 ⊗ · · · ⊗ xm f with f(i) = x1(i1) ··· xm(im) m and the norm β induced on ⊗ Xn satisﬁes

β(x1 ⊗ · · · ⊗ xm) = kx1k · · · kxmk implying β ≤ π. Hence factorizing through [Xn]m gives

m m T2(⊗π Xn) ≤ T2([Xn]m) kid : [Xn]m −→ ⊗π Xnk and since m m−1 kid : ⊗ε Yn −→ [Yn]mk ≤ λY (n) for any Banach sequence space Y by [29, lemma 6.4], using (1.27), duality, (1.11) and (1.23) we conclude

1/2 m−1 m × n m−1 (m−1)/2 T2(⊗π Xn) ≺ M(2)((X )n) λX× (n) n T2(Xn) λX× (n)

In view of (1.21) our next aim is to calculate the 2-summing norm m nm π2(id : ⊗ε Xn −→ `2 ). The following deﬁnition we have taken from [30, p. 10]. By O(n) we denote the orthogonal group.

Deﬁnition 1.11. An n-dimensional Banach space E = (Rn, k · k) is said to have 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 a multiple of the identity.

If E = (Rn, k · k) has enough symmetries in O(n) then n −1 E = kid : ` −→ Ek B n (1.28) E 2 `2 for the ellipsoid EE of maximal volume in BE (see e.g. [30, p. 11]) and in particular n n n d(`2 ,E) = kid : `2 −→ Ek kid : E −→ `2 k (1.29) since in this case the Banach Mazur distance to the Hilbert space equals n kid : E −→ Emaxk where Emax = R with the norm generated by EE (see [9, lemma 4.6]). Another consequence of (1.28) is the following lemma which can be found in [30, lemma 3]. 40 1. Norm estimates of identities between symmetric tensor products

n n Lemma 1.12. Let E = (R , k · k1) and F = (R , k · k2) be n-dimensional Banach spaces with enough symmetries in O(n). Then n 1/2 kid : `2 −→ F k π2(id : E −→ F ) = n n kid : `2 −→ Ek We will prove an analogue for tensor products (proposition 1.14) which is already included in [74]. This is done for the convenience of the reader and also with a diﬀerent proof, which is shorter and for us seems to be more transparent. Let S(Rn) be the group of bijections generated by n n n X n Mε : R −→ R , x εkxkek , ε ∈ {−1, 1} k=1 n n n X Tσ : R −→ R , x xσ(k)ek , σ ∈ Πn k=1 where Πn are the permutations on {1, . . . , n}. Then m n m n n S(⊗ R ) = ⊗ S(R ) = {S1 ⊗ · · · ⊗ Sm | Si ∈ S(R ) } is a group of bijections on ⊗mRn. If X is a symmetric Banach sequence n m space then all S ∈ S(R ) are isometries on Xn. A norm α on ⊗ Xn is called symmetrically invariant if all symmetries S ∈ S(⊗mRn) are isometries on m m ⊗α Xn. Every m-fold tensor norm α is symmetrically invariant on ⊗ Xn: m n m For all S = S1 ⊗ · · · ⊗ Sm ∈ S(⊗ R ) we have, writing E = ⊗α Xn for short m 0 0 0 Y kS : E −→ E k = kS : E −→ Ek = kSi : Xn −→ Xnk = 1 i=1 by the metric mapping property of α and hence also 0 −1 −1 0 k(S ) kL(E0) = k(S ) kL(E0) = 1 0 This implies S BE0 = BE0 , thus 0 0 0 kSzkE = sup |hz , Szi| = sup |hS z , zi| = kzkE 0 0 z ∈BE0 z ∈BE0

A proof of the following result can be found in [29]. We use a diﬀerent argument from [42, p. 35] which was given there for m = 2 and can be directly applied to the general case. Lemma 1.13. Let X be a symmetric Banach sequence space and α be a m m nm symmetrically invariant norm on ⊗ Xn. Then ⊗α Xn = (R , k · kα) has enough symmetries in O(nm), the identiﬁcation being given by

m nm ⊗ Xn −→ R I : P λi ei (λi) −1 and k · kα = α (I ( · )). `-norm estimates for the injective and projective tensor product 41

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

−1 −1 −1 Tσ(ei) = eσ (i1) ⊗ · · · ⊗ eσ (im) = eσ (i) for i ∈ {1, . . . , n}m. Thus we have X X S λiei = ε1(i1) ··· εm(im) λσ(i) ei i∈M(m,n) i∈M(m,n)

m n which implies that S is an isometry on ⊗2 `2 and by the assumption on α, m S is also an isometry on ⊗α Xn. Now assume that the linear operator u : ⊗mRn −→ ⊗mRn commutes with each element of S(⊗mRn). Then in particular

0 0 0 hei, ueji = hei,MεuMεeji = hei, ueji ε1(i1) ··· εm(im) ε1(j1) ··· εm(jm) and 0 0 0 0 0 hei, ueji = hei,TσuTσ−1 eji = hTσei, uTσ−1 eji = heσ(i), ueσ(j)i n m for all i, j ∈ M(m, n), ε = (ε1, . . . , εm) ∈ ({−1, 1} ) and products σ = 0 0 0 σ1 × · · · × σm of permutations σi ∈ Πn, where ei = ei1 ⊗ · · · ⊗ eim , i ∈ m n M(m, n) are the coeﬃcient functionals of the basis ei, i ∈ M(m, n) of ⊗ R , 0 0 generated from the coeﬃcient functionals e1, . . . , en of e1, . . . , en. This implies 0 0 0 hei, ueji = 0 for i 6= j and c := he1, ue1i = hei, ueii for all i ∈ M(m, n) since m n we can choose σ1, . . . , σm ∈ Πn independently. Thus u = c id⊗ R and

−1 m n G := {ISI | S ∈ S(⊗ R ) } is the requested subgroup of O(nm).

Lemma 1.13 in particular proves that (1.28), (1.29) and lemma 1.12 hold for certain tensor products. The ﬁrst two statements of the next proposition will be used in section 4. Proposition 1.14. Let X and Y be symmetric Banach sequence spaces and m m α, β be symmetrically invarivant norms on ⊗ Xn and ⊗ Yn, respectively. Then

nm m −1 (1) E⊗mX = kid : ` −→ ⊗ Xnk B nm α n 2 α `2 nm m nm m m nm (2) d(`2 , ⊗α Xn) = kid : `2 −→ ⊗α Xnk kid : ⊗α Xn −→ `2 k nm m m m m/2 kid : `2 −→ ⊗β Ynk (3) π2(id : ⊗α Xn −→ ⊗β Yn) = n nm m kid : `2 −→ ⊗α Xnk 42 1. Norm estimates of identities between symmetric tensor products

nm m It is not hard to estimate the norm of the identity id : `2 −→ ⊗ε Xn. Remark 1.15. Let X be a symmetric Banach sequence space. Then  1 X 2-convex nm m  kid : `2 −→ ⊗ε Xnk 1/2 m (λX (n)/n ) X 2-concave

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

nm m m n m n m kid : `2 −→ ⊗ε Xnk ≤ kid : ⊗ε `2 −→ ⊗ε Xnk = kid : `2 −→ Xnk m m m = sup kxkXn = sup k⊗ xk⊗ε Xn kxk n ≤1 kxk n ≤1 `2 `2 m n m m ≤ kid : ⊗ ` −→ ⊗ X k sup k⊗ xk m n 2 2 ε n ⊗2 `2 kxk n ≤1 `2 nm m = kid : `2 −→ ⊗ε Xnk

nm m n m hence kid : `2 −→ ⊗ε Xnk = kid : `2 −→ Xnk and a glance at (1.16) ﬁnishes the proof.

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

Theorem 1.16. Let X be a symmetric Banach sequence space. Then  X 2-convex, n(m−1)/2 (λ (n))m  X non trivially concave m n m  `(id : ⊗2 `2 −→ ⊗π Xn) X 2-concave, nm−1 λ (n)  X non trivially convex

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

nm m m m × nm `(id : `2 −→ ⊗π Xn) ≤ T2(⊗π Xn) π2(id : ⊗ε (X )n −→ `2 )

(m−1)/2 m/2 nm m × −1 n T2(Xn) n kid : `2 −→ ⊗ε (X )nk which implies the upper estimate, since  1 X 2-convex, non trivially concave T2(Xn) 1/2 λX (n)/n X 2-concave, non trivially convex by (1.23) and (1.16) and

 1/2 m (λX (n)/n ) X 2-convex nm m × −1  kid : `2 −→ ⊗ε (X )nk 1 X 2-concave `-norm estimates for the injective and projective tensor product 43 by remark 1.15,(1.12) and the remark following it. Then with the help of theorem 1.9 we get

nm m nm m 0 m `(id : `2 −→ ⊗π Xn) `(id : `2 −→ ⊗ε (Xn) ) ≺ n 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

nm m nm m 0 m `(id : `2 −→ ⊗α Xn) `(id : `2 −→ (⊗α Xn) ) n (1.30) for α = ε or π if X is 2-convex with non trivial concavity or 2-concave with non trivial convexity, hence

m 1/n nm m −1 m vol B⊗α Xn `(id : `2 −→ ⊗α Xn) by corollary 1.7 and together with the remarks that follow it we have proved the following result.

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

m 1/dn 1/n vol B m,s vol B m ⊗αs Xn ⊗α Xn

 X 2-convex, λ (n)−1  X non trivially concave   α = ε  X 2-concave,  (m−1)/2 m n /(λX (n))  non trivially convex   (m−1)/2 m −1 X 2-convex, (n (λX (n)) )  non trivially concave  α = π  X 2-concave, (nm−1 λ (n))−1  X non trivially convex

This in particular implies theorem 1.5.

Example 1.18. Let 1 < p < ∞.

( m 1 − 1 − 1 m ( 2 p ) 2 1/dn 1/n n p ≤ 2 (a) vol B m,s n vol B m n 1 ⊗εs `p ⊗ε `p − n p p ≥ 2

( 1− 1 −m m p 1/dn 1/n n p ≤ 2 (b) vol B m,s n vol B m n 1 1 1 ⊗πs `p ⊗π `p − + m n 2 ( 2 p ) p ≥ 2 44 1. Norm estimates of identities between symmetric tensor products

m,s n m,s n m n m n (c) `(id : ⊗2 `2 −→ ⊗εs `p ) `(id : ⊗2 `2 −→ ⊗ε `p )

( m 1 − 1 + 1 n ( p 2 ) 2 p ≤ 2 1 n p p ≥ 2

m,s n m,s n m n m n (d) `(id : ⊗2 `2 −→ ⊗πs `p ) `(id : ⊗2 `2 −→ ⊗π `p ) ( m−1+ 1 n p p ≤ 2 1 + 1 m− 1 n( 2 p ) 2 p ≥ 2

( m 1 2 + p −1 m n (m−1)/2 n n p ≤ 2 (e) T2(⊗π `p ) n T2(`p ) m−1 n 2 p ≥ 2

Volume ratio of m-fold tensor products

The calculations of the last section permit to give asymptotically correct estimates for the volume ratio of the m-fold injective and projective tensor product.

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

vol B 1/n vr(E) = E vol EE where EE is the ellipsoid of maximal volume in BE. See [64], [78] or [33] for more information.

n Remark 1.19. Let (En)n be a sequence of Banach spaces (R , k · kn) with enough symmetries in O(n) such that

n n 0 `(id : `2 −→ En) `(id : `2 −→ En) ≺ n

Then n 1/2 kid : `2 −→ Enk vr(En) n n `(id : `2 −→ En)

Proof. We have

n 1/2 n 1/n 1/2 kid : `2 −→ Enk vr(En) = n kid : `2 −→ Enk (vol BEn ) n n `(id : `2 −→ 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 nm m m m/2 kid : `2 −→ ⊗α Xnk vr(⊗α Xn) n nm m `(id : `2 −→ ⊗α Xn) for α ∈ {ε, π}.

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

 1/2 n λX (n) X 2-convex, non trivially concave vr(Xn) 1 X 2-concave, non trivially convex

m Thus to express vr(⊗α Xn) asymptotically in terms of the fundamental function λX (n) it remains to prove an appropriate estimate for nm m kid : `2 −→ ⊗π Xnk, 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

 nm/2  X 2-convex, non trivially concave λX (n) m nm  kid : ⊗ε Xn −→ `2 k  nm−1/2  2m  m X 2-concave, 2m−1 -convex (λX (n))

Proof. By (1.22) and factorization we have

(m−1)/2 m n C2(Xn) = C2(⊗ε Xn) m nm nm nm m ≤ kid : ⊗ε Xn −→ `2 k C2(`2 ) kid : `2 −→ ⊗ε Xnk m nm nm m = kid : ⊗ε Xn −→ `2 k kid : `2 −→ ⊗ε Xnk

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

nm If X is 2-convex then for all λ ∈ `2

1/2 1/2 X X 2 1/2 X 2 kλk`nm = |λi| ≤ n max |λi| 2 i1 i1 (i2,...,im) (i2,...,im)

(m−1)/2 n n ≤ n kid : Xn −→ `2 k max k(λ(i1,...,im))im=1kXn i1,...,im−1

(m−1)/2 n X ≤ n kid : Xn −→ ` k λi ei 2 m ⊗ε Xn which implies the upper bound in case of ﬁnite concavity (for m = 2 and X = `p this argument was used in [72, p. 126]). 46 1. Norm estimates of identities between symmetric tensor products

2m Now let pm := 2m−1 and suppose X to be 2-concave and pm-convex. Then by the multilinear version of a Hardy-Littlewood theorem [45] by Praciano- Pereira [66] we have m n nm kid : ⊗ε `pm −→ `2 k 1 m n hence factoring through ⊗ε `pm and using (1.14) we obtain 1−1/2m m m nm n m n kid : ⊗ε Xn −→ `2 k ≺ kid : Xn −→ `pm k λX (n)

Recall that we have nm m nm m m nm d(`2 , ⊗ε Xn) = kid : `2 −→ ⊗ε Xnk kid : ⊗ε Xn −→ `2 k (1.31) by proposition 1.14. Hence in passing we get from proposition 1.21 and remark 1.15 the following asymptotic description for the Banach Mazur dis- m tance of ⊗ε Xn to the Hilbert space. Corollary 1.22. Let X be a symmetric Banach sequence space. Then

 m/2 n /λX (n) X 2-convex, non trivially concave nm m  d(`2 , ⊗ε Xn) (m−1)/2 2m n X 2-concave, 2m−1 -convex

But from here it’s only a quick step to get the asymptotic of the Banach- m Mazur distance of the polynomials P (Xn) to the Hilbert space, which we will take before we ﬁnish our volume ratio estimates. Recall that a linear operator T : E −→ F between Banach spaces is called 2-factorable, if it factors through some space L2(µ):

E T - F J SRJ (1.32) JJ^ L2(µ) with S ∈ L(E,L2(µ)), R ∈ L(L2(µ),F ). Then γ2(T ) := infkRkkSk taken over all factorization operators as in (1.32) deﬁnes a norm on the space Γ2(E,F ) of all 2-factorable operators and makes it a Banach space. We have that (Γ2, γ2) is a normed operator ideal (see e.g. [78, p. 95]). From proposition 13.9 in [78, p. 103] we get n γ2(E) := γ2(idE) = d(`2 ,E) if E is n-dimensional. Hence

m dn m,s n m d(`2 , ⊗εs Xn) d(`2 , ⊗ε Xn) (1.33) m+n−1 by theorem 1.4 since γ2 is an ideal norm (dn = n−1 is the dimension of m,s the space ⊗εs Xn). This proves Volume ratio of m-fold tensor products 47

Corollary 1.23. Let X be a symmetric Banach sequence space. Then  n(m−1)/2 X 2-convex, 2m-concave dn m d(`2 , P (Xn)) m/2−1 n 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 trivial convexity, then m (m−1)/2 vr(⊗ε Xn) n vr(Xn) (2) If X is 2-convex and 2m-concave or 2-concave and non trivially convex, then m vr(⊗π Xn) 1

Note that in particular

m nm m vr(⊗ε Xn) d(`2 , ⊗ε Xn) 2m if X is 2-convex and non trivially concave or 2-concave and 2m−1 -convex.

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

 m/2 n /λX (n) X 2-convex, non trivially concave m  vr(⊗ε Xn) n(m−1)/2 X 2-concave, non trivially convex

(m−1)/2 n vr(Xn) In case of the projective norm by duality and proposition 1.21

nm m m × nm kid : `2 −→ ⊗π Xnk = kid : ⊗ε (X )n −→ `2 k

 m 1/2 (λX (n)) /n X 2-convex, 2m-concave m/2−1 n 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 the symmetric injective and projective tensor product. Remark 1.25. Let X be a symmetric Banach sequence space, 2-convex and non trivially concave or 2-concave and non trivially convex. Then for α = ε or π we have m,s m vr(⊗αs Xn) ≺ vr(⊗α Xn) 48 1. Norm estimates of identities between symmetric tensor products

Proof. Deﬁne I := id : `dn −→ ⊗m,sX (with d = m+n−1 the dimension of n 2 αs n n n−1 m,s −1 ⊗ Xn) and vn := kInk In. Then vn(B dn ) =: En is an ellipsoid contained αs `2 m,s in the unit ball of ⊗αs Xn and hence

1/d (vol B m,s ) n vr(⊗m,sX ) ≤ ⊗αs Xn αs n 1/d (vol En) n

1/d (vol B m,s ) n dn m,s ⊗αs Xn = kid : ` −→ ⊗ Xnk 2 αs 1/dn (vol B dn ) `2 1/nm m (vol B m ) kid : `n −→ ⊗m X k ⊗α Xn 2 αs n 1/nm (vol B nm ) `2 m vr(⊗α Xn) where the ﬁrst asymptotic equality follows from theorems 1.4 and 1.17 and the second one from proposition 1.14. Example 1.26. Let 1 ≤ p ≤ ∞.

 2m 1 1 ≤ p ≤  2m−1 m n nm  m 1− 1 − 1 (a) kid : ⊗ ` −→ ` k ( p ) 2 2m ε p 2 n 2m−1 ≤ p ≤ 2   m − 1 n 2 p 2 ≤ p ≤ ∞

 m( 1 − 1 ) 2m n p 2 1 ≤ p ≤  2m−1 m  n m n m,s n m n m−1 2m (b) d(` , ⊗ ` ) d(⊗ ` , ⊗ ` ) 2 2 ε p 2 2 εs p n 2m−1 ≤ p ≤ 2   m − 1 n 2 p 2 ≤ p ≤ ∞

dn m n m,s n m n m n m n (c) d(`2 , P (`p )) d(⊗2 `2 , ⊗πs `p ) d(⊗2 `2 , ⊗π `p )

 m + 1 −1 n 2 p p ≤ 2   m−1 n 2 2 ≤ p ≤ 2m   m 1 − 1 n ( 2 p ) p ≥ 2m

m+n−1 m n where dn = n−1 denotes the dimension of P (`p )

 m−1 n 2 1 < p ≤ 2 (d) vr(⊗m`n) ε p m + 1 n 2 p 2 ≤ p < ∞

m n (e) vr(⊗π `p ) 1 for 1 < p ≤ 2m Volume ratio of m-fold tensor products 49

 m − 1 +1 n 2 p 1 < p ≤ 2 m n m n  (f) vr(P (` )) ≺ vr(⊗ ` 0 ) p ε p m−1 n 2 2 ≤ p < ∞

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

m n nm m n m n m n nm kid : ⊗ε `p −→ `2 k ≤ kid : ⊗ε `p −→ ⊗ε `pm k kid : ⊗ε `pm −→ `2 k

n n m m(1− 1 )− 1 pm 2 kid : `p −→ `pm k n = 1

2. Domains of convergence

It follows the main part of the thesis. First we define the domain of convergence for holomorphic functions on Reinhardt domains and we motivate this definition by some examples. Then we prove lower and upper inclusions for the domain of convergence of the bounded holomorphic functions on a bounded Reinhardt domain in a symmetric Banach sequence space. For the lower inclusion we use arguments of Bohr (see introduction, theorem 0.1) and a generalization of the results of Ryan [70] and Lempert [53] for holomorphic functions on multiples of the unit ball in `1 to Reinhardt domains. To obtain the upper bound we define and estimate the arithmetic Bohr radius and use ideas from the work of Dineen and Timoney [38]. Then we investigate the domain of convergence of the space of m-homogeneous polynomials and, after that, of all polynomials on a Banach sequence space. We examine the special role of the space `1 in our theory and answer a question of Lempert. Finally we prove a direct link between domains of convergence and unconditionality in spaces of homogeneous polynomials.

Domains of convergence of holomorphic functions

Let R be a Reinhardt domain in Cn. Then every holomorphic function f on R 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 (c ) n such that α α∈N0 X α f(z) = cαz n α∈N0 for every z ∈ R, and the coeﬃcients can be calculated by

∂αf(0) 1 n Z f(z) cα = = α+1 dz α! 2πi [|z|=r] z

n where r ∈ R>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 Banach P (n) α sequence space X. Then f has a power series expansion n cα z on α∈N0 every ﬁnite dimensional component Rn = R ∩ Xn of R with n Z Z (n) 1 f(z1, . . . , zn, 0,...) cα = ··· dz1 ··· dzn α1+1 αn+1 2πi |z1|=r1 |zn|=rn z1 ··· zn n+1 Z Z 1 f(z1, . . . , zn+1, 0,...) = ··· dz1 ··· dzn+1 α1+1 αn+1 2πi |z1|=r1 |zn+1|=rn+1 z1 ··· zn zn+1 (n+1) = cα

n n+1 for α ∈ N0 ⊂ N0 . Thus we can ﬁnd a unique family (cα) (N) such that α∈N0

X α f(z) = cαz

(N) α∈N0 for all z ∈ Rn and all n ∈ N, i.e. on span{en} ∩ R = ∪nRn. The power series P α cαz is called the monomial expansion of f. We write cα(f) := cα for the monomial coeﬃcients of f. They satisfy ∂αf(0) 1 n Z f(z) cα(f) = = α+1 dz (2.1) α! 2πi [|z|=r] z

n if α ∈ N0 and [|z| ≤ r] ⊂ Rn and hence are continuous linear functionals on the Banach space H∞(R) of all bounded holomorphic functions on R with the sup-norm k · kR. In particular, from (2.1) we deduce

α kcα(f)z kR ≤ kfkR (2.2)

(N) for all α ∈ N0 (Cauchy inequalities for monomials). Where does the monomial expansion series of a holomorphic function f on a Reinhardt domain R converge? Do we also have

X α f(z) = cα(f)z (2.3)

(N) α∈N0 for all z ∈ R as in the ﬁnite dimensional situation? We have n X X α f( zkek) = cα(f)z n k=1 α∈N0 for every n ∈ N and every z ∈ R. Thus if X is a Banach sequence space with basis and the monomial expansion of f converges in some point z ∈ R then it converges to f(z). Before we give some examples, let us make an observation concerning the connection between the Taylor series expansion and the monomial series expansion 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 in Cn. Then X α Pmf(0)(z) = cα(f)z |α|=m n P for every z ∈ C . In particular, for all z ∈ R the Taylor series m Pmf(0)(z) converges absolutely and

∞ X f(z) = Pmf(0)(z) m=0

Proof. Deﬁne the m-homogeneous polynomials

X α Pm(z) = cα(f)z |α|=m on Cn. Then ∞ ∞ X X X α X α |Pm(z)| ≤ |cα(f)z | = |cα(f)z | < ∞ (2.4) n m=0 m=0 |α|=m α∈N0 for z ∈ R. On the other hand, we have

∞ X f(z) = Pmf(0)(z) m=0 P in a zero neighbourhood U ⊂ R, where the Taylor series m Pmf(0) converges uniformly and, in particular, absolutely on U. Hence from

∞ ∞ X X α X Pmf(0)(z) = f(z) = cα(f)z = Pm(z) (2.5) n m=0 α∈N0 m=0 for z ∈ U it follows Pm = Pmf(0) (see [19, 11.10]) and now (2.4) tells us that P m Pmf(0) converges absolutely for every z ∈ R with

∞ X α X f(z) = cα(f)z = Pmf(0)(z) n α∈N0 m=0

We can deduce an analogue in inﬁnite dimensions. Corollary 2.2. Let f be a holomorphic function on a Reinhardt domain R in a Banach sequence space X. Then

X α Pmf(0)(z) = cα(f)z |α|=m 54 2. Domains of convergence

for every z ∈ span{en}. In particular, for all z ∈ span{en} ∩ R the Taylor P series m Pmf(0)(z) converges absolutely and

∞ X f(z) = Pmf(0)(z) m=0

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

Pmf|Rn (0) = Pmf(0) on every ﬁnite dimensional component Rn. The “in particular” then follows as in the proof of remark 2.1 by the convergence of the monomial expansion on span{en} ∩ R.

Corollary 2.2 says nothing but

cα(Pmf(0)) = cα(f) (2.6)

(N) for every f ∈ H(R), m ∈ N0 and α ∈ N0 with |α| = m. Now we give some examples showing that the monomial expansion of a holomorphic function on a Reinhardt domain R in a Banach sequence space X does in general not converge in every point of R.

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

Remark 2.3. Let m ≥ 2 and A = (ars)1≤r,s≤N be a matrix with complex coeﬃcients such that PN (a) t=1 art ast = Nδrs

(b) |ars| = 1 Let N X ϕm(x1, . . . , xm) = ai1i2 ··· aim−1 im x1(i1) ··· xm(im)

i1,...,im=1 N N m for xj ∈ C , j = 1, . . . , m, be the m-linear mapping on (C ) deﬁned by A. Then

m+1 (1) kϕ k m N ≤ N 2 m L (`∞) ( m N 4 m even (2) kϕmkLm(`N ) ≤ m+1 2 N 4 m odd

Part (1) including its proof is taken from [12, p. 609]. Domains of convergence of holomorphic functions 55

Proof. (1) Note that

1/2 kϕmk m N ≤ N kϕmk m N N N L (`∞) L (`∞×···×`∞×`2 ) and we prove m kϕmk m N N N ≤ N 2 L (`∞×···×`∞×`2 ) by induction on m. For m = 2 and x1 ∈ B N , x2 ∈ B N we have `∞ `2

N N N N X X 21/2 X X 1/2 |ϕm(x1, x2)| ≤ ai1i2 x1(i1) = ai1i2 aj1i2 x1(i1)x1(j1) i2=1 i1=1 i2=1 i1,j1=1 N N N X X 1/2 1/2X 21/2 = x1(i1)x1(j1) ai1i2 aj1i2 = N |x1(i1)|

i1,j1=1 i2=1 i1=1 ≤ N

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

Now assume m ≥ 3 and let x1, . . . , xm−1 ∈ B N and xm ∈ B N . Then `∞ `2

|ϕm(x1, . . . , xm)|

N N 1/2 X X 2 ≤ ai1i2 ··· aim−1im x1(i1) ··· xm−1(im−1) im=1 i1,...,im−1=1

N X = ai1i2 aj1j2 ··· aim−2im−1 ajm−2jm−1 x1(i1) ··· xm−1(jm−1) i1, . . . , im−1=1 j1,...,jm−1 n 1/2 X aim−1im ajm−1im im=1 N N 1/2 X X 2 = N ai1i2 ··· aim−2im−1 x1(i1) ··· xm−2(im−2) im−1=1 i1,...,im−2=1 1/2 2 |xm−1(im−1)|

N N 1/2 1/2 X X 2 ≤ N ai1i2 ··· aim−2im−1 x1(i1) ··· xm−2(im−2) im−1=1 i1,...,im−2=1

1/2 m ≤ N kϕm−1k m−1 N N N ≤ N 2 L (`∞×···×`∞×`2 ) 56 2. Domains of convergence

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

N N X X 21/2 |ϕm(x1, x2)| ≤ ai1i2 x1(i1) i2=1 i1=1 N N X X 1/2 = ai1i2 aj1i2 x1(i1)x1(j1)

i2=1 i1,j1=1 N N X X 1/2 = x1(i1)x1(j1) ai1i2 aj1i2

i1,j1=1 i2=1 N 1/2X 21/2 = N |x1(i1)|

i1=1 ≤ N 1/2 again with the Cauchy-Schwarz inequality and assumption (a). If m = 3 and x1, x2, x3 ∈ B N then in a similar way `2

N N 1/2 X X 2 |ϕm(x1, x2, x3)| ≤ ai1i2 ai2i3 x1(i1)x2(i2) i3=1 i1,i2=1 N N 1/2 X X = ai1i2 aj1j2 x1(i1)x1(j1)x2(i2)x2(j2) ai2i3 aj2i3 i1, i2 =1 i3=1 j1,j2 N N 1/2 1/2 X X 2 2 = N ai1i2 x1(i1) |x2(i2)| i2=1 i1=1 ≤ N

N N For the last inequality we have used assumption (b) and kid : `2 −→ `1 k = N 1/2. Domains of convergence of holomorphic functions 57

Now let m ≥ 4 and x1, . . . , xm ∈ B N . Then `2

|ϕm(x1, . . . , xm)|

N N 1/2 X X 2 ≤ ai1i2 ··· aim−1im x1(i1) ··· xm−1(im−1) im=1 i1,...,im−1=1

N X = ai1i2 aj1j2 ··· aim−2im−1 ajm−2jm−1 x1(i1) ··· xm−1(jm−1) i1, . . . , im−1=1 j1,...,jm−1 N 1/2 X aim−1im ajm−1im im=1 N N 1/2 X X 2 = N ai1i2 ··· aim−2im−1 x1(i1) ··· xm−2(im−2) im−1=1 i1,...,im−2=1 1/2 2 |xm−1(im−1)|

N N 1/2 1/2 X X 2 ≤ N ai1i2 ··· aim−3im−2 x1(i1) ··· xm−3(im−3) im−2=1 i1,...,im−3=1

1/2 Thus kϕmk ≤ N kϕm−2k 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-homogeneous m polynomial qm ∈ P (`∞) such that for every ε > 0 the monomial expansion of qm does not converge in some point z ∈ ` 1 −1 . 2(1− m ) +ε

We give a proof since we use the polynomials constructed here for the proof of 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-matrix Mn inductively by

2πi rs M1 := (e p )1≤r,s≤p rs 2πi p (n) Mn := (e · Mn−1)1≤r,s≤p = (ars )1≤r,s≤pn , n ≥ 2 i.e. Mn is the Kronecker product of M1 with Mn−1 for n ≥ 2. Then Mn satisﬁes

Ppn (n) (n) n (a) t=1 art ast = p δrs 58 2. Domains of convergence

(n) (b) |ars | = 1

(n) p (c) (ars ) = 1 The properties (b) and (c) are obvious. To understand (a) note that for n = 1 we have p p p (1) (1) t X X 2πi p (r−s) X t art ast = e = ζ t=1 t=1 t=1 with ζ a p-th root which is diﬀerent from one if and only if r 6= s. If n ≥ 2 we have pn p pn−1 X (n) (n) X (1) (1) X (n−1) (n−1) art ast = ar0t as0t ar00t as00t t=1 t=1 t=1 with r = s if and only if r0 = s0 and r00 = s00.

So by remark 2.3 the m-linear mapping ϕm,n deﬁned by the Matrix Mn satisﬁes (m+1)n/2 n kϕm,nk m p ≤ p L (`∞ )

2) Now let qm,n = (ϕm,n)s be the m-homogeneous polynomial associated to (n) pn ϕm,n with monomial coeﬃcients cα , α ∈ N0 , |α| = m, explicitely

1 X (n) (n) c(n) = a ··· a (2.7) α α! iσ(1) iσ(2) iσ(m−1) iσ(m) σ∈Sm

n where i1 ≤ ... ≤ im such that |{k | ik = r}| = αr, r = 1, . . . , p and Sm the group of permutations on {1, . . . , m}. Then

n n(m+1)/2 (i) kqm,nk m p ≤ p P( `∞ ) (n) (ii) η := inf|cα | > 0

The ﬁrst part is clear from 2), since the norm of pm,n is less or equal than pn the norm of ϕm,n. The inﬁmum in (ii) is taken over all n and all α ∈ N0 with |α| = m and we now prove that this is in fact away from zero. Since (N) m sup{α! | α ∈ N0 , |α| = m} ≤ (m!) we can disregard the binomial factors in (2.7) and it is enough to prove that

X (n) (n) I = { a ··· a | i ≤ ... ≤ i } iσ(1) iσ(2) iσ(m−1) iσ(m) 1 m σ∈Sm is a ﬁnite set of non zero elements. For this choose a primitive p-th root ζ, 2πi for example ζ = e p . Then every element of I can be written as

p−1 X k λkζ k=0 Domains of convergence of holomorphic functions 59 with λk ∈ N0 and p−1 X λk = |Sm| = m! (2.8) k=0 in particular cannot be zero, since this would imply

p−1 X k (λk − λ0)ζ = 0 k=1 and hence λ0 = λ1 = ... = λp−1 =: λ since {ζ, ζ2, . . . , ζp−1} is linearly independent over Q. But that would lead to p−1 X p · λ = λk = m! k=0 which contradicts the fact that p is a prime number exceeding m. We also conclude from (2.8) that

p−1 X k p I ⊂ { λkζ | (λ0, . . . , λp−1) ∈ {0, . . . , m!} } k=0 must be ﬁnite. 3) Now we deﬁne weights 1 λ := p−n(m+1)/2 n n2 and X qm(z) := λnqm,n(zn) (2.9) n n pn for z = (zn)n with zn = (zn(1), . . . , zn(p )) ∈ C , i.e.

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

P 1 Then qm is an m-homogeneous polynomial on `∞ with kqmk ≤ n n2 and monomial coeﬃcients

( (n) p+p2+...+pn−1 pn λn cβ α = 0 × β ∈ {0} × N0 , |α| = m cα(qm) = (2.10) 0 otherwise

4) Finally let ε > 0 and deﬁne 2 2 2m s := 1 + ε > 1 = =: r 1 − m 1 − m m − 1 60 2. Domains of convergence

Then 0 < δ := r/s < 1 and since p > m we can choose 0 < b < 1 with n/s 1−δ δ b pn p b > 1. Deﬁning zn := p · 1 ∈ C we have

pn X X s X n |zn(i)| = b < ∞ n i=1 n hence z := (zn)n ∈ `s but

X α X X α |cα(qm) z | ≥ η λn |zn | α n pn α∈N0 |α|=m pn η X X m ≥ λ |z (i)| m! n n n i=1 n/s m η X 1 b = p−n(m+1)/2 pn m! n2 p n η X 1 n = (p1−δbδ)(m−1)/2 = ∞ m! n2 n

In particular, this example provides us for all p > 2 and all Reinhardt domains R ⊂ `p with a holomorphic function f on R satisfying dom(f) 6= R which can be chosen to be bounded if R is bounded. A modiﬁcation of the example of Bohnenblust and Hille shows how to con- 4 struct “bad” functions for 3 < p ≤ 2.

Example 2.5. For all m ≥ 2 there is an m-homogeneous polynomial qm ∈ m P (`2) such that for every ε > 0 the monomial expansion of qm does not converge in some point z ∈ ` 4 . 3 +ε

Proof. For an arbitrary natural number m ≥ 2 and a prime number p > m we construct the polynomials qm,n exactly as in example 2.4. Then ( pmn/4 if m even n kqm,nkPm(`p ) ≤ 2 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 polynomial qm on `2 as in (2.9) but with the weights 1 λ = p−mn/4 n n2 Domains of convergence of holomorphic functions 61

P 1 such that again kqmk ≤ n2 and qm has monomial coeﬃcients as in (2.10). 4 1 3 Let ε > 0 and deﬁne s := 3 + ε. Then δ := s < 4 and we choose a positive real number b < 1 such that

1−δ− 1 δ p 4 b > 1

n/s b pn Deﬁning zn := p 1 ∈ C as in example 2.4 4) then again z := (zn) ∈ `s and n/sm X α η X 1 −nm/4 n b η X 1 1−δ− 1 δ mn |c z | ≥ p p = (p 4 b ) = ∞ α m! n2 p m! n2 α n n

It remains to consider the case when m is odd. Then, by assumption, m ≥ m−1 3 and we know that there are qm−1 ∈ P (`2) and z ∈ ` 3 such that 4 +ε P α 0 0 |α|=m−1|cα(qm−1)z | = ∞. Take ϕ := e1 ∈ `2, i.e. ϕ(u) = u1 for u ∈ `2. Then qm := ϕ · qm−1 deﬁnes an m-homogeneous polynomial on `2 with kqmk ≤ kqm−1k and monomial coeﬃcients ( cα−e1 (qm−1) α1 6= 0 cα(qm) = 0 otherwise

Thus (we can choose z1 6= 0)

X α X α−e1 X α |cα(qm)z | = |z1| |cα−e1 (qm−1)z | = |z1| |cα(qm−1)z | = ∞ |α|=m |α|=m |α|=m−1 α16=0

After these examples the following definition makes sense. Definition 2.6. Let R be a Reinhardt domain in a Banach sequence space X. For a holomorphic function f on R we define

X α dom(f) := {z ∈ R | |cα(f) z | < ∞}

(N) α∈N0 to be the set of all points z of R in which its monomial expansion series converges and call it the domain of convergence of f.

Note that dom(f) is balanced. 4 Corollary 2.7. Let p > 3 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. 62 2. Domains of convergence

In a later section we will describe the domain of convergence of the m- homogeneous polynomials on a Banach sequence space. Special cases of our results will prove the existence of such polynomials as in the preceding examples. We will even settle the case X = `p with 1 ≤ p ≤ 4/3 (see example 2.54). As we mentioned in the introduction, Bohr’s theorem 0.1 about the convergence of power series in inﬁnitely many variables can be proved for bounded holomorphic functions.

Theorem 2.8. Let f ∈ H∞(B`∞ ). Then

X α |cα(f)z | < ∞

(N) α∈N0 for all z ∈ `2 ∩ B`∞ .

Proof. Let x ∈ `2 ∩ B`∞ . Choose 0 < r < 1 such that x ∈ rB`∞ and deﬁne 1 |α| fr ∈ H∞( r B`∞ ) by fr(z) := f(rz). Then cα(fr) = cα(f) r . Let n ∈ N. The α n n monomials z in C are orthonormal on the torus Πn = {z ∈ C | |z| = 1}: 1 n Z α β i(α1θ1+···αnθn) −i(β1θ1+···βnθn) (z |z )L2(Πn) = e e dθ 2π [0,2π]n

n n 1 Z Y 1 Z 2π = ei(α−β|θ)dθ = ei(αk−βk)θdθ 2π n 2π [0,2π] k=1 0 ( 1 if α = β = 0 otherwise Hence X 2 X α α |cα(fr)| = cα(fr) cα(fr)(z |z )L2(Πn) n n α∈N0 α∈N0

X α β = cα(fr) cβ(fr)(z |z )L2(Πn) n α, β∈N0 X α X α = cα(fr)z cα(fr)z n n L2(Πn) α∈N0 α∈N0 2 2 2 = kfrk ≤ sup |fr(z)| ≤ kfk L2(Πn) B`∞ z∈Πn

(N) Since n was arbitrary this implies (cα(fr)) (N) ∈ `2(N0 ). Furthermore α∈N0 P α recall that the series |z | converges if and only if z ∈ `1 ∩ DN with limit m ∞ ∞ X α X α Y X k Y 1 |z | = lim |z | = lim |zn| = m→∞ m→∞ 1 − |z | ( ) m n N α∈N0 n=1 k=0 n=1 α∈N0 Domains of convergence of holomorphic functions 63

By the Cauchy-Schwarz inequality it follows

(N) (N) (N) α∈N0 α∈N0 α∈N0 Y 1 1/2 ≤ kfkB `∞ 1 − |r−1x |2 n n

Note that in the deﬁnition of the number S deﬁned by Bohr (see the introduction, p. 11) one could also take the power series bounded on the domain

Bc0 = {z ∈ B`∞ | zn −→ 0} instead of B`∞ = [|zi| ≤ 1] (since kεk∞ < 1 for ε ∈ `q ∩ ] 0 , 1[ N) and they can be identiﬁed with the monomial expansions of the bounded holomorphic functions on Bc0 . More generally we have: Remark 2.9. Let R be a Reinhardt domain in a Banach sequence space X (N) with basis and cα ∈ C, α ∈ N0 . The following are equivalent: P α (1) cαz is the monomial expansion of an f ∈ H∞(R). P α (2) cαz is bounded on R:

X α sup sup cαz < ∞ n z∈Rn n α∈N0

In this case X α kfkR = sup sup cαz n z∈Rn n α∈N0

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

X α cαz = |f(z)| ≤ kfkR n α∈N0 and hence X α sup sup cαz ≤ kfkR < ∞ n z∈Rn n α∈N0 P α P α On the other hand, suppose cαz is bounded on R. Then n cαz α∈N0 converges on Rn for every n and we can deﬁne the function

X α f0(z) := cαz

(N) α∈N0 on ∪nRn which is bounded with

X α kf0k∪nRn = supkf0kRn = sup sup cαz n n z∈Rn n α∈N0 64 2. Domains of convergence and holomorphic on every ﬁnite dimensional section Rn of R. If for m ∈ N S and z ∈ X0 := span{en} = n Xn we deﬁne the m-homogeneous polynomial

X α Pm(z) := cαz |α|=m then we can write ∞ X f0(z) = Pm(z) (2.11) m=0 for every z ∈ ∪nRn. Now let m, n ∈ N and z ∈ Rn. Then Z 1 f0(λz) Pm(z) = m+1 dλ 2πi |λ|=1 λ

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

|Pm(z)| ≤ kf0kRn ≤ supkf0kRn < ∞ n

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

kPmkR = kPmk∪nRn ≤ supkf0kRn n Since for all 0 < r < 1 we have

∞ ∞ X X sup kf0kR kP k = rmkP k ≤ n n < ∞ m rR m R 1 − r m=1 m=1 P the series m Pm converges on R and by [19, theorem 14.23], the function

∞ X f(z) = Pm(z) m=0 is holomorphic on R. From (2.11) we deduce that f is an extension of f0 P α with monomial expansion cαz . In particular, f is bounded and

kfkR = supkf0kRn n

The notion of boundedness in (2) is a translation of Hilbert’s definition of boundedness “in the domain |zi| ≤ 1” (i.e. on B`∞ ) to Reinhardt domains in infinite dimensions. In this sense, investigating bounded holomorphic functions is nothing but examining bounded power series in infinitely many variables. Lower inclusions for dom H∞(R) 65

Deﬁnition 2.10. Let R be a Reinhardt domain in a Banach sequence space X and F(R) ⊂ H(R). We call \ dom F(R) := dom(f) f∈F(R) the domain of convergence of F(R).

Note that dom F(R) = dom span F(R) (2.12) and dom F(R) as an intersection of balanced sets is balanced.

The operator dom H∞( · ) is increasing in R and positive homogeneous. More precisely:

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 dom H∞(R) ⊂ dom H∞(S).

(2) dom H∞(cR) = c dom H∞(R).

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

dom H∞(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 dom H∞(R)

In this section, we are going to determine a large set of elements which is contained in the domain of convergence of the bounded holomorphic functions on a given Reinhardt domain R in a Banach sequence space X. We will prove that for a sequence z = (zn) ∈ R it is enough to satisfy an additional summability condition to become an element of dom H∞(R), namely z ∈ `1 or z ∈ X · `2 (theorem 2.25). In a later section, we see that for a wide class 66 2. Domains of convergence of Banach sequence spaces this condition turns out to be “almost” necessary (theorem 2.44).

For X = `1 the situation is very nice. Matos proved in [60] that a holomorphic function f : U −→ C on an open subset U of `1 has a monomial expansion ∂αf(z) in a neighbourhood of every point z ∈ U (here cα,z(f) = 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 in an open subset of a Banach space and the approximation of holomorphic functions on balls in Banach spaces by functions holomorphic on the whole space (see [52, 53, 54]).

Theorem 2.12. Let 0 < r ≤ ∞. The monomial expansion of a holomorphic function f on rB`1 converges uniformly absolutely on compacta, i.e. P α |cα(f)z | converges uniformly on compact subsets of rB`1 .

Our goal is to generalize this theorem to arbitrary Reinhardt domains (theorem 2.21). This implies that `1 ∩ R ⊂ dom H(R) is always true (see corollary 3.3) and, in particular, gives one part of theorem 2.25. The second part is then 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 split m K ⊂ R compact into the product of two compact sets Km ⊂ U and K ⊂ m rB`1 , where U is a Reinhardt domain in a ﬁnite dimensional space C and U× rB`1 ⊂ R. Then we can apply the lemmata 2.13 and 2.16 (the latter resulting P α from an analysis of the proof of theorem 2.12) to prove that |cα(f)z | is bounded on compact sets for every f ∈ H(R) (proposition 2.18) which, by a result 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 subset of R. Then we can ﬁnd a constant c > 0 and a compact set K˜ ⊂ R containing K such that for all countable families (fν)ν∈I of holomorphic functions fν on R we have X X α X kcα(fν)z kK ≤ c · |fν| K˜ m ν∈I α∈N0 ν∈I

Proof. We can assume that K is circular and balanced. (Otherwise take ˆ K := B`∞ K ⊂ R, see the proof of remark 2.17.) Choose r > 0 and t > 1 ˜ m P such that K := t(K + rB`∞ ) ⊂ R and suppose k ν∈I |fν|kK˜ < ∞. Let x ∈ K. Then s := t(|x| + r · 1) ∈ K˜ . It follows [|z| ≤ s] ⊂ K˜ since K˜ is circular and balanced and m Z 1 fν(z) cα(fν) = α+1 dz 2πi |z|=s z Lower inclusions for dom H∞(R) 67

(where eiθ := (eiθ1 , . . . , eiθm ) for θ ∈ [0, 2π]m) and using Lebesgue’s theorem we get m Z X X α X −|α| 1 X iθ kcα(fν)z kx+rB m ≤ t |fν(s · e )|dθ `∞ m m 2π [0,2π]m ν∈I α∈N0 α∈N0 ν∈I

−1 −m X ≤ (1 − t ) k |fν|kK˜ ν∈I

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

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

+ Lemma 2.14. Let θ ∈ Bc0 . Then there exists an N ∈ N and a zero sequence + ξ ∈ Bc0 such that |α||α| |α|! θα ≤ eN ξα αα α! (N) for all α ∈ N0 .

+ N Here Bc0 := Bc0 ∩ R>0.

1/2 Proof. Let η = kθk∞ 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) n ≥ m. Let α ∈ N0 and I = { n ∈ N | n < ` , αn ≥ m }. Then

Y αi Y Y αi α α! (eη) = αi! (eη) αi! ≤ α i∈I i/∈I i∈I by the choice of m and the fact that n! ≤ nn. Since nn ≤ enn! we get

|α| |α| P α P α +` m |α| e |α|! e i∈ /I i |α|! e i≥` i |α|! ≤ · = P · ≤ P · α Q αi αi αi α i∈I (eη) α! η i∈I α! η i<` α! hence |α||α| |α|! θα ≤ e` m ξα αα α! 68 2. Domains of convergence

(N) + for all α ∈ N0 where ξ ∈ Bc0 is deﬁned by

 1  η θn if n < ` ξn = e θn if n ≥ `

+ Proposition 2.15. Let θ ∈ Bc0 . Then there exists a constant c > 0 such (N) that for all subsets A ⊂ N0 , all families (cα)α∈A of scalars and all r > 0 we have α X X α α |α| α/2 sup |cα(θu) | ≤ c · sup |cα| |α| r θ α∈A |α| m u∈rB`1 α∈A |α|=m

(N) Proof. Let A ⊂ N0 and cα, α ∈ A be scalars. By lemma 2.14 we have |α||α| |α|! θα/2 ≤ eN ξα αα α!

+ (N) for some N ∈ N and ξ ∈ Bc0 and all α ∈ N0 . Then X X X X αα |α|! sup |c (θ u)α| ≤ eN sup |c | r|α| θα/2 |(ξ u)α| α α |α||α| α! m u∈rB`1 α∈A m u∈B`1 α∈A |α|=m |α|=m α N α |α| α/2 X X |α|! α ≤ e sup |cα| |α| r θ sup |(ξ u) | α∈A |α| α! m u∈B`1 α∈A |α|=m By the multinomial formula X |α|! |(ξ u)α| = kξ ukm ≤ kξkm α! `1 ∞ |α|=m

if u ∈ B`1 and by calculating the arising geometric series the claim follows with c = eN . 1−kξk∞

+ We recall here the fact that (ξB`1 ) is a fundamental system of compact ξ∈Bc0 subsets of B`1 (see [70] or [36]): For every ξ ∈ Bc0 the set ξB`1 = Γ{ξnen} is 0 compact in B`1 , and conversely for K ⊂ B`1 compact we choose 0 < t < t < 1 with K ⊂ tB`1 and a strictly increasing sequence (nk) of natural numbers such that X 1 − δ sup |zj| ≤ k+1 z∈K k 2 j>nk 0 where δ = t/t . This is possible since precompactness of K in `1 means

X n→∞ sup |zj| −−−→ 0 z∈K j≥n Lower inclusions for dom H∞(R) 69

Then ( 0 t 1 ≤ n ≤ n1 ξn := 1 2k nk < n ≤ nk+1 deﬁnes a positive zero sequence with kξk∞ < 1 such that for all z ∈ K

∞ n1 ∞ nk+1 X −1 X −1 X X −1 ξn |zn| = ξn |zn| + ξn |zn| n=1 n=1 k=1 n=nk+1 n ∞ nk+1 1 X1 X X = |z | + 2k |z | t0 n n n=1 k=1 n=nk+1 ∞ X ≤ δ + (1 − δ) 2−k = 1 k=1

+ Then of course (ξB`1 ) is a fundamental system of compact subsets of ξ∈rBc0 + + rB`1 for 0 < r ≤ ∞ where rB`1 := `1 and rBc0 := c0 = {ξ ∈ c0 | ξ > 0} if r = ∞.

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

X α |cα(f)z | K ≤ c · kfkK˜ (N) α∈N0

for all f ∈ H(rB`1 ).

+ Proof. We can assume r = 1. Since (ξB`1 ) is a fundamental system ξ∈Bc0 + of compact subsets in B`1 we can choose θ ∈ Bc0 with K ⊂ θB`1 . Deﬁne ξ := θ1/2. By the Cauchy inequalities (2.2) we get

α α α α |cα(f)| ξ = kcα(f)z kξB ≤ kfkξB |α||α| `1 `1

for every f ∈ H(B`1 ) and all α, hence we obtain our result from proposition ˜ 2.15 with K = ξB`1 .

The next remark allows us in the proof of proposition 2.18 to restrict ourselves to compact sets K which are circular (u ∈ CN, v ∈ K and |u| = |v| implies u ∈ 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 which + contains K and ηB`∞ for some η ∈ `1 . 70 2. Domains of convergence

ˆ Proof. The set K := B`∞ K ⊂ R is circular, balanced and contains K and ˆ 0. Further K is closed in `1 with X X lim sup |zn| = lim sup |zn| = 0 m→∞ ˆ m→∞ z∈K z∈K n≥m n≥m

ˆ since K is relatively compact in `1. Thus K is compact in `1. ˆ n+1 + Now choose r > 0 such that K + rB`1 ⊂ R. Then η := (r/2 )n ∈ `1 and ˜ ˆ K := K + ηB`∞ is the set that we were looking for.

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

X α |cα(f)z | K ≤ c · kfkK˜ (N) α∈N0

Proof. By remark 2.17 we can assume K to be circular and balanced. We choose t > 1 and 0 < r < 1 such that t(K + 2rB`1 ) ⊂ R and then m ∈ N satisfying X sup |zn| < r z∈K n≥m

m which is possible since K is relatively compact in `1. Denote πm : `1 −→ `1 m m the m-th projection and π : `1 −→ `1, π (z) := (zn)n≥m+1. Then

m Km := πm(K) ⊂ `1 m m K := π (K) ⊂ `1

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

K ⊂ U := t(K + rB m ) m m `1 m K ⊂ rB`1 where U is a Reinhardt domain in Cm and

U × rB ⊂ t((K + rB m ) × rB ) ⊂ t(K + 2rB ) ⊂ R `1 m `1 `1 m `1

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

1 n Z f(x, y) fβ(x) := cβ(f(x, · )) = β+1 dy 2πi |y|=b y Lower inclusions for dom H∞(R) 71

n n if β ∈ and b ∈ such that [|y| ≤ b] ⊂ rB n . Thus f is a continuous N0 R>0 `1 β m function on U. Now we choose a ∈ R>0 with [|x| ≤ a] ⊂ U. If x ∈ U and m n α := (0, β) ∈ × then (x, 0) ∈ U × rB n ⊂ R and by Fubini’s theorem N0 N0 `1 1 m+n Z Z f(ζ, y) fβ(x) = 1 β+1 dζ dy 2πi |y|=b |ζ|=a (ζ − x) y ∂αf(x, 0) = α! 1 m+n Z Z f(ζ, y) = 1 β+1 dy dζ 2πi |ζ|=a |y|=b (ζ − x) y m Z 1 fβ(ζ) = 1 dζ 2πi |ζ|=a (ζ − x)

Thus fβ has a local Cauchy representation and hence is holomorphic on U. m+n m n For γ = (α, β) ∈ N0 = N0 × N0 we get ∂γf(0) ∂α ∂βf(0) ∂αf (0) c (f) = = = β = c (f ) γ γ! α! β! α! α β

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

= c kfkK˜

m with c = cm c and m ˜ ˜ ˜ m K ⊂ Km × K ⊂ K := Km × K ⊂ 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 is locally bounded if it is bounded on ﬁnite dimensional compact sets and from this it is standard to conclude that F(Ω) is equicontinuous, if E is a Banach space. 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) is equicontinuous. 72 2. Domains of convergence

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

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 of second 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 we have for all f ∈ F(Ω), m ∈ N and all x ∈ W

|Pmf(ξ)(x)| ≤ sup|Pmf(ξ)(λa + x)| = sup|Pmf(ξ)(λa + x)| |λ|≤1 |λ|=1

= sup|Pmf(ξ)(a + λx)| ≤ sup|Pmf(ξ)(a + y)| |λ|=1 y∈W

≤ kPmf(ξ)kVn ≤ n Here we have used the maximum principle for the ﬁrst equality and the m-homogeneity of Pmf(ξ) for the second one:

m −1 Pmf(ξ)(λa + x) = λ Pmf(ξ)(a + λ x) Thus for 0 < r < 1 ∞ ∞ X X m kPmf(ξ)krW ≤ |f(ξ)| + n r < ∞ m=0 m=1 P So the Taylor expansion series m Pmf(ξ) of f in ξ converges uniformly on rW and hence deﬁnes a holomorphic function on rW (see [19, theorem 12.5]) which, by the principle of analytic continuation ([19, theorem 12.8]), must be equal to f(ξ + · ) on this set, since this is true in some neighbourhood of 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 all x ∈ rW ∞ ∞ X X n r→0 |f(ξ + x) − f(ξ)| = Pmf(ξ) ≤ kPmf(ξ)krW ≤ r −−→ 0 1 − r m=1 m=1 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 a P countable family of functions fi ∈ H(Ω) such that |fi| converges point- P wise in Ω and is bounded on compact subsets of Ω. Then |fi| converges uniformly on compact subsets of Ω. Lower inclusions for dom H∞(R) 73

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

P∞ Proof. It is enough to prove this for I = N and n=1|fn|. By lemma 2.19 the set N X { εjfj | N ∈ N, |εj| = 1} ⊂ H(Ω) j=1 is equicontinuous. Thus for z0 ∈ Ω and ε > 0 we can ﬁnd δ > 0 such that for all z ∈ z0 + δBX ⊂ Ω and all N ∈ N

N N N X X X |fj(z)| − |fj(z0)| ≤ |fj(z) − fj(z0)| j=1 j=1 j=1 N X = sup εj(fj(z) − fj(z0)) < ε |εj |=1 j=1

PN which implies that F(Ω) := { n=1|fn| | N ∈ N} is equicontinuous in Ω. Thus for K ⊂ Ω compact F(K) = {f|K | f ∈ F(Ω)} is bounded and equicontinuous in C(K) and hence relatively compact by the Arzela-Ascoli PN theorem. It follows that ( n=1|fn|)N has a convergent subsequence and P∞ 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 R in `1. Then the monomial expansion of f converges uniformly absolutely on compact subsets of R.

P α Proof. By proposition 2.18 the series (N) |cα(f)z | is bounded on com- α∈N0 pact subsets of R and we can apply corollary 2.20. Corollary 2.22. Let R be a Reinhardt domain in a Banach sequence space X. Then `1 ∩ R ⊂ dom H(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 “multiplication trick”. 74 2. Domains of convergence

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

X · c0 = span{en}

Proof. If z = λ · µ ∈ X · c0, then

n X n n n n→∞ z − zkek X = kπ (z)kX ≤ kπ (λ)kX kπ (µ)k∞ ≤ kλkX sup|µk| −−−→ 0 k>n k=1

n where π (x) = (0...., 0, xn+1, xn+2,...) for x ∈ X. On the other hand, if z ∈ span{en} then choose a sequence n1 < n2 < . . . < nk < . . . of natural numbers with X 1 znen < X k3 n≥nk 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 domain in 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 := span{en} (by lemma 2.23) and

µ ∈ B`∞ (otherwise consider z = rλ · (1/r)µ with r > kµk∞).

Let a > 1 and b > 0 such that az + bBX ⊂ R. Then we can ﬁnd an n satisfying X k λkekk < b k>n and hence z = u · v with X u = (az1, . . . , azn, λn+1, λn+2,...) ≤ az + λkek ∈ R k>n and 1 1 v = ( ,..., , µ , µ ,...) ∈ ` ∩ B a a n+1 n+2 2 `∞

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 space X. Then

(`1 ∪ (X · `2)) ∩ R ⊂ dom H∞(R) The arithmetic Bohr radius 75

Proof. From corollary 2.22 we get `1 ∩ R ⊂ dom H(R) ⊂ dom H∞(R). For the remaining inclusion it is enough to prove

R · (`2 ∩ B`∞ ) ⊂ dom H∞(R) by remark 2.24.

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

X α X α |cα(f)(xz) | = |cα(fx)z | < ∞

(N) (N) α∈N0 α∈N0

for all z ∈ `2 ∩ B`∞ .

The arithmetic Bohr radius

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

The Bohr radius K(Rn) (see [11]) and the Bohr radius B(Rn) (see [3]) of a n Reinhardt domain Rn ⊂ C are deﬁned to be the least upper bound of all positive real numbers r such that

X α |cα(f)z | ≤ kfkR rRn n (N) α∈N0 and X α kcα(f)z krRn ≤ kfkRn

(N) α∈N0 respectively for all bounded holomorphic functions f on Rn. These Bohr n radii have been estimated asymptotically in the dimension n. For the `p -unit balls the estimates are asymtotically correct up to a logarithmic term:

1− 1 1− 1 1 log n/ log log n min{p,2} log n min{p,2} ≤ K(B n ) ≤ c (2.13) c n `p n and 1 + 1 1 + 1 1 1 2 max{p,2} log n 2 max{p,2} ≤ B(B n ) ≤ c (2.14) c n `p n 76 2. Domains of convergence

(see [37, 11, 10, 23]) For 1 ≤ p ≤ 2 the lower estimate for the second Bohr radius is slightly better: r n 2 1 − ≤ B(B n ) (2.15) 3 `p (see [10] and [3]).

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

n 1 X A (R ) := sup{ r | r ∈ n , ∀f ∈ H (R ): λ n n i R≥0 ∞ n i=1 X α |cα(f)|r ≤ λkfkRn } n α∈N0 and

n 1 X Am(R ) := sup{ r | r ∈ n , ∀f ∈ Pm(R ): λ n n i R≥0 ∞ n i=1 X α |cα(f)|r ≤ λkfkRn } n α∈N0 |α|=m

m m n where P∞(Rn) := P (C ) ∩ H∞(Rn) is the space of all m-homogeneous bounded holomorphic functions on Rn. We call A(Rn) := A1(Rn) the arithmetic Bohr radius of Rn.

m Obviously Aλ(Rn) and Aλ (Rn) are increasing in λ and we have Aλ(Rn) ≤ m Aλ (Rn) for all m. n For bounded Reinhardt domains Rn,Sn ⊂ C we use the notation

S(Rn,Sn) := inf{t > 0 | Rn ⊂ tSn}

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

S(Rn,BXn ) = sup kzkX z∈Rn

S(BXn ,BYn ) = kid : Xn −→ Ynk

n Remark 2.27. Let Rn,Sn ⊂ C 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. The arithmetic Bohr radius 77

n Proof. If r ∈ R≥0 satisﬁes

X α |cα(f)|r ≤ λkfkRn n α∈N0 for all f ∈ H∞(Rn) then, writing s = S(Rn,Sn) for short, we have

X r α X z α |cα(f)| ≤ λ cα(f) ≤ λ kfkS Rn n n s + ε n s + ε α∈N0 α∈N0 for every f ∈ H∞(Sn) and all ε > 0, hence

n 1 P ri ≤ S(Rn,Sn)Aλ(Sn) n i=1

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

n Remark 2.28. Let Rn be a bounded Reinhardt domain in C and λ ≥ 1. Then S(Rn,B`n ) A (R ) ≥ 1 K(R ) λ n n n

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

S(R ,B n ) = sup kzk n n `1 `1 z∈Rn thus for 0 < ε < K(Rn) we can ﬁnd an element z0 ∈ Rn such that

kz k n ≥ S(R ,B n ) − ε 0 `1 n `1

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

X α X α X α |cα(f)|r = |cα(f)u | ≤ |cα(f)z | ≤ kfkR tRn n n n n α∈N0 α∈N0 α∈N0 and it follows

n 1 X K(Rn) − ε K(Rn) − ε A(Rn) ≥ ri = kz0k`n ≥ (S(Rn,B`n ) − ε) n n 1 n 1 i=1 78 2. Domains of convergence

n We ﬁrst give estimates for the λ-arithmetic Bohr radius of the `p -unit balls. We can deduce a lower bound from remark 2.28. For an upper bound we use a 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, 1} such that

( 1 1 1− p X m! α p n 2 (m!) if p ≤ 2 ε z ≤ 32m log(6m) 1 1 1 α B n +( − )m 1 α! `p n 2 2 p (m!) 2 if p ≥ 2 |α|=m (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 for all m and all n

1 + 1 √ q 2 max{2,p} m m 2m 2(1− 1 ) 1 A (B n ) ≤ c λ m log m (m!) min{2,p} n λ `p n

n P α Proof. Let m ∈ . Take r ∈ with |cα(f)| r ≤ λkfkB n for every N R≥0 |α|=m `p m n f ∈ P (C ). Then we can choose signs εα ∈ {−1, 1} such that

n X m X m! X m! r = rα ≤ λ ε zα ≤ λc i α B n m α! α! `p i=1 |α|=m |α|=m by the above cited result, where cm is the right side of (2.17). Taking the m-th root and dividing by n we obtain √ m m λcm A (B n ) ≤ λ `p n for every m and n.

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

1 1 1− min{p,2} 1− min{p,2} 1 (log n/ log log n) 2/ log n (log n) ≤ Aλ(B`n ) ≤ c λ 1 + 1 p 1 + 1 c n 2 max{2,p} n 2 max{2,p} for all n ∈ N.

Proof. We get the lower bound from remark 2.28 and (2.13). For the upper bound by proposition 2.29 we only have to show that there is a constant c > 0 such that for all n we can ﬁnd an m with √ q m 2m 2 1− 1 1− 1 λ m log m (m!) ( min{2,p} ) n ≤ c λ2/ log n (log n) min{p,2} (2.18) The arithmetic Bohr radius 79 √ √ √ Since 2m m log m 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 1− 1 √ λ m min{2,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

1− 1 λ2/ log n (log n) min{2,p} e

Diﬀerent from the situation for all versions of Bohr radii deﬁned for holomorphic functions in the literature, where no exact value for any Reinhardt n domain Rn ⊂ C in dimension n ≥ 2 is known, we have the following result. Theorem 2.31. 1 A(B`n ) = 1 3n

1 Proof. First we prove A(B n ) ≤ . For the proof of K( ) ≤ 1/3, one can `1 3n D use the linear fractional transformation z − a f (z) = a 1 − az 1−z 0 < a < 1 (see Boas [10]; in Bohr’s original paper [14], the function 1−az is used). This function is bounded by 1 on the unit disc by the maximum principle, since for all z ∈ C with |z| = 1 we have |fa(z)| = 1 and from ∞ ∞ X k X 2 k−1 k fa(z) = (z − a) (az) = −a + (1 − a )a z k=0 k=1 we see that ∞ ∞ X k X 2 k−1 k |ck(fa)z | = a + (1 − a )a |z| = 2a + fa(|z|) k=0 k=1 which exceeds one for 1 |z| > 1 + 2a For our purposes, we use the function s(z) − a fa(z) = , z ∈ B`n 1 − a s(z) 1 where again 0 < a < 1 and s(z) = z1 + ... + zn. Then fa is bounded by one since s(B n ) = and we have `1 D ∞ ∞ X X X k! f (z) = −a + (1 − a2) ak−1s(z)k = −a + (1 − a2) ak−1 zα a α! k=1 k=1 |α|=k 80 2. Domains of convergence

from which it follows c0(fa) = −a and |α|! c (f ) = (1 − a2)a|α|−1 α a α!

(N) for α ∈ N0 , α 6= 0. Hence we have again

kzk n − a X α `1 |cα(fa)z | = 2a + fa(|z|) = 2a + 1 − akzk`n (N) 1 α∈N0 which exceeds one if and only if 1 kzk`n > 1 1 + 2a if and only if n 1 X 1 1 |z | > n i n 1 + 2a i=1 This implies 1 A(B`n ) ≤ 1 (1 + 2a)n for all 0 < a < 1 and hence the desired estimate.

1 For the remaining inequality let 0 < r < 3 = K(D) and deﬁner ˜ := n n−1 (r, 0,..., 0) ∈ . Let f ∈ H (B n ). Identifying with × {0} we R≥0 ∞ `1 D D have f|D ∈ H∞(D) and hence ∞ ∞ X α X k X k |cα(f)|r˜ = |c(ke1)(f)|r = |ck(f| )|r ≤ kf| k ≤ kfkB n D D D `1 n α∈N0 k=0 k=0 It follows n 1 1 X r = r˜i ≤ A(B`n ) n n 1 i=1

With the same arguments we obtain a more general result. Theorem 2.32. Aλ(D) Aλ(B`n ) = 1 n

n P α Proof. For “≤” let r ∈ ≥0 with n |cα(f)| r ≤ λkfkB n for all R α∈N0 `1 f ∈ H (B n ). Then we have to prove ∞ `1

n X ri ≤ Aλ(D) (2.20) i=1 The arithmetic Bohr radius 81

Let f ∈ H∞(D). Using the function s deﬁned in the proof of theorem 2.31 k! n we get f ◦ s ∈ H (B n ) with c (f ◦ s) = c (f) for α ∈ with |α| = k. It ∞ `1 α k α! N0 follows ∞ ∞ X k X X k! α X α |ck(f)|krk n = |ck(f)| r = |cα(f ◦ s)| r `1 α! n k=0 k=0 |α|=k α∈N0

≤ λkf ◦ skB n = λkfk `1 D and hence (2.20). For the remaining inequality argue in the same way as in the proof of theorem 2.31.

In [16, teorema B] Bombieri proved 1 Aλ(D) = 3λ − 2p2(λ2 − 1) √ for 1 ≤ λ ≤ 2 (see also [17]). √ Corollary 2.33. For 1 ≤ λ ≤ 2 we have 1 Aλ(B`n ) = 1 (3λ − 2p2(λ2 − 1)) n

n Now we give general estimates of Aλ(Rn) where Rn ⊂ C is an arbitrary bounded Reinhardt domain. Again the key is a probabilistic estimate, a polynomial version of Chevet’s inequality, which was proved in [26, corollary 3.2].

Theorem 2.34. Let (εα)|α|=m be a family of independent standard Bernoulli random variables on a probability√ space (Ω, µ) and cα, |α| = m be scalars. 3 3m−1 There exists a constant 0 < Cm ≤ m 2 such that for each bounded and circled set U ⊂ Cn we have Z X α sup εαcαz dµ z∈U Ω |α|=m

( r ) n n p α! P P 21/2 m−1 ≤ log n Cm sup |cα| |zi| U |zi| U |α|=m m! i=1 i=1

n Thus, given a circled and bounded subset U of C and a family (cα)|α|=m of scalars, there are signs ±1 such that the m-homogeneous polynomial pm(z) = P α ± cαz satisﬁes |α|=m

r n n p 3 3m−1 α! P P 21/2 m−1 kpmkU ≤ m 2 log n sup |cα| |zi| U |zi| U |α|=m m! i=1 i=1 (2.21) 82 2. Domains of convergence

n Proposition 2.35. Let Rn ⊂ C be a bounded Reinhardt domain and λ ≥ 1. Then √ n m 2pm S(Rn,B` ) Am(R ) ≤ λ m3 m! 23m−1n log n 2 λ n n for all m ∈ N.

n P α Proof. Let m ∈ N. Take r ∈ R≥0 with |α|=m|cα(f)| r ≤ λkfkRn for every m n f ∈ P (C ). Then by the above cited result we can choose signs εα ∈ {−1, 1} such that n m X X m! α X m! α ri = r ≤ λ εα z α! α! Rn i=1 |α|=m |α|=m r n n p 3 3m−1 m! P P 21/2 m−1 ≤ λ m 2 log n sup |zi| |zi| Rn Rn |α|=m α! i=1 i=1

Pn 1/2 Pn 2 1/2 Pn 2 1/2 Since k i=1|zi|kRn ≤ n k( i=1|zi| ) kRn and k( i=1|zi| ) kRn = S(R ,B n ), it follows n `2 n √ X m 2pm 3 3m−1 r ≤ λ m m! 2 n log n S(R ,B n ) i n `2 i=1 which yields the claim. n Theorem 2.36. Let Rn ⊂ C be a bounded Reinhardt domain and λ ≥ 1. Then √ 1 1 sn,1 2/ log n log n max{ , K(Rn)} ≤ Aλ(Rn) ≤ 17 λ sn,2 3n s1,n n n where

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

Proof. The lower bound follows from remark 2.28 and remark 2.27 together with theorem 2.31 and it remains to prove the upper bound. By proposition 2.35 it is enough to show that for every n ≥ 2 there is an m ∈ N with √ m 2pm λ m3 m! 23m−1n log n ≤ 17 λ2/ log n plog n and this is done as in [24, p.186], but somewhat more carefully. If n = 2,..., 8 then√ 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 2pm m 3/2 m 1/2 2pm λ m3 m! 23m−1n log n = λ m m m! 23/2−1/(2m) n log n ≤ λ2/ log n (e1/e)3/2 plog n 25/4 (n log n)1/ log n ≤ λ2/ log n(e1/e)5/2 eplog n 25/4 ≤ 17 λ2/ log n plog n The arithmetic Bohr radius 83

In particular, we have r 1 log n ≤ A (B ) ≤ c 3n λ Xn n (with c depending on λ, but not on n) for every Banach sequence space X 1 since `1 ⊂ X ⊂ `∞ with continuous inclusions and the extreme cases 3n q log n and n are attained (up to a logarithm term, see (2.13)) for X = `1 and X = `∞. n Comparing (2.14) with the estimates in theorem 2.30, we see that for `p - 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, where similar estimates are used. The next result also gives evidence for this fact.

n Remark 2.37. Let Rn be a logarithmically convex Reinhardt domain in C . Then n 1 X A(R ) = sup{ r | r ∈ n ∩ R , ∀f ∈ H (R ): n n i R≥0 n ∞ n i=1 X α kcα(f)z krB n ≤ kfkR } `∞ n n α∈N0 n 1 X X α = sup{ |zi| | z ∈ Rn, ∀f ∈ H∞(Rn): |cα(f)z | ≤ kfkRn } n n i=1 α∈N0

n n Here rB`∞ of course means the set {rz | z ∈ B`∞ }, i.e. the polydisc [|z| < r] ⊂ Cn with radius r.

Proof. The second equality is obvious. Denote

n X α M := {z ∈ C | ∀f ∈ H∞(Rn): |cα(f)z | ≤ kfkRn } n α∈N0 Then n n 1 X 1 X sup |zi| = sup |zi| ≤ A(Rn) n z∈M∩R n z∈M∩Rn i=1 n i=1 and it remains to show n 1 X A(R ) ≤ sup |z | n n i z∈M∩Rn i=1

n So let x ∈ M and without loss of generality x = r ∈ R≥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 = ∅ 84 2. Domains of convergence

Since Rn ⊂ tRn every f ∈ H(tRn) deﬁnes a bounded holomorphic function on Rn and satisﬁes X α |cα(f)|r ≤ kfkRn n α∈N0 (since r ∈ M), hence can be extended by ( f(z) z ∈ tR f˜(z) := n P α n cα(f)z |z| < r α∈N0 ˜ to a holomorphic function f on tRn ∪ [|z| < r] % tRn. But this is impossible since by the assumption tRn is a domain of holomorphy (see e.g. [67, theorem 3.28]).

Though the asymptotic is very similar, in general, the arithmetic Bohr radius and B(Rn) are not identical. For example, we have r 1 n 2 A(B`n ) = < 1 − ≤ B(B`n ) 1 3n 3 1 by theorem 2.31, Bernoulli’s inequality and (2.15).

Upper inclusions for dom H∞(R)

The next lemma allows us to estimate the decreasing rearrangement of the elements of dom H∞(R) (proposition 2.40). Here we use ideas from [37] were the case of power series in inﬁnitely many variables bounded on B`∞ was considered (see the introduction). Lemma 2.38. Let R be a bounded Reinhardt domain in a Banach sequence space X and F(R) be a closed subspace of H∞(R). Then for every z ∈ dom F(R) we can ﬁnd a λ > 0 such that

X α |cα(f)z | ≤ λkfkR

(N) α∈N0 for all f ∈ F(R).

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

X α Hn := {f ∈ F | |cα(f)z | ≤ n}

(N) α∈N0

0 (N) is closed, since cα ∈ (H∞(R)) for α ∈ N0 by Cauchy’s integration formula (see the remarks after (2.1)). Since z ∈ dom F we have F = ∪nHn and Upper inclusions for dom H∞(R) 85

Baire’s theorem gives a natural number n0 such that the interior of Hn0 is non empty. Thus we have f0 + sBF ⊂ Hn0 for some f0 ∈ F and some s > 0. So if f ∈ BF then 1 c (f) = (c (f + sf) − c (f )) α s α 0 α 0

(N) for α ∈ N0 implies X 2n0 |c (f)zα| ≤ =: λ α s (N) α∈N0

Recall that if R is a Reinhardt domain in a Banach sequence space X then Rn = R ∩ Xn denotes the n-th section of R, i.e. Rn = πn(R), where P∞ Pn πn( k=1 zkek) = k=1 zkek is the projection from X onto Xn. Then for all 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 sequence space X. Then for every z ∈ dom H∞(R) we can ﬁnd a λ ≥ 1 such that

n 1 X |z | ≤ A (R ) n i λ n i=1 for all n ∈ N. Proposition 2.40. Let R be a bounded Reinhardt domain in a symmetric Banach sequence space X. Then √ log n z∗ ≺ kid : X −→ `nk n n n 2 for every z ∈ dom H∞(R).

(See page 26 for the notation ≺.)

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

dom H∞(R) ⊂ r dom H∞(BX ) by remark 2.11. Hence it is enough to prove the claim for R = BX .

Let z ∈ dom H∞(BX ). We have √ log n A (B ) ≤ c λ2/ log n kid : X −→ `nk (2.22) λ Xn n n 2 86 2. Domains of convergence

by theorem 2.36. Since the inclusion mapping X,→ `∞ has norm 1, factor- n 1/2 ization gives kid : Xn −→ `2 k ≤ n and so (2.22) implies Aλ(BXn ) n ∈ `2+ε (2.23) for all ε > 0.

By the symmetry of BX , dom H∞(BX ) is invariant under permutations, that means we have z ◦ σ ∈ dom H∞(BX ) for every bijection σ : N −→ N. Hence 1 Pn ( n i=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 a permutation of the sequence |z| and we obtain n 1 X z∗ ≤ z∗ ≤ A (B ) n n i λ Xn i=1 for some constant λ ≥ 1, which, by (2.22), ﬁnishes the proof.

n In particular, factorization through `∞ as in the above proof gives a general upper inclusion. Corollary 2.41. Let R be a bounded Reinhardt domain in a symmetric Ba- nach sequence space X and ε > 0. Then

dom H∞(R) ⊂ `2+ε Deﬁnition 2.42. A Banach sequence space is called regular, if 1 ε ∈ X for every ε > 0 λX (n) n n

This condition is, for example, satisﬁed for Lorentz sequence spaces d(w, p) if n X wi n wn i=1 so, in particular, for the `p,q-spaces (see the preliminaries) and as a special case 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 that ε n0 > 1/λ. Then for all n ≥ n0 and all ρ > 0 with ϕ(1/ρ) ≤ 1/n we have 1 1 1 1 1 1 1 1 ϕ = ϕ + 1 − 0 ≤ ϕ ≤ ρ nε λ nε λ ρ nε λ nε λ ρ λ n1+ε by the convexity of ϕ. This implies 1 1 1 ϕ ε ≤ 1+ε λ`ϕ (n) n λ λ n for all n ≥ n0 and hence `ϕ is regular. For regular Banach sequence spaces we use the following lemma. Upper inclusions for dom H∞(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 a constant K > 0 such that for all n 1/r 1/s ∗ 1 1 zn ≤ K log n λX (n) n

1 then z ∈ X r · `s+ε.

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

1 δ1 1 = + + δ2 s r s + ε1 Then 1/r 1/(s+ε1) ∗ 1 1 log n zn ≤ K δ δ n 1 λX (n) n n 2 ∗ 1/r and hence z = x · y for some x ∈ X and y ∈ `s+ε, x, y > 0. Since zn −→ 0 ∗ we have z = |zσ| = (|zσ(n)|)n for some permutation σ : N −→ N. Then r N xσ−1 ∈ X and yσ−1 ∈ `s+ε by the symmetry of X and `s+ε. Choose λ ∈ C 1/r with |λ| = 1 such that z = λ|z|. Then z = λxσ−1 · yσ−1 ∈ X · `s+ε.

We are now able to prove our upper inclusions for dom H∞(R), and together with 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 ⊂ dom H∞(R) ⊂ `1+ε ∩ R

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

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

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

For the upper bounds let z ∈ dom H∞(R). Then we have √ log n z∗ ≺ kid : X −→ `nk n n n 2 by proposition 2.40. This gives (1), and to understand (2) we use n 1/2 kid : Xn −→ `2 k n /λX (n) (see (1.14)). Then 1/2 ∗ 1 1 zn ≤ K log n λX (n) n

By corollary 2.41 we have zn −→ 0, and hence we can apply lemma 2.43 to conclude z ∈ X · `2+ε. 88 2. Domains of convergence

From proposition 2.40 we can also deduce a more general upper inclusion for dom H∞(R), which holds for a wider class of Banach sequence spaces X but is weaker than theorem 2.44. This sharpens the estimate of corollary 2.41 by posing 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

1/2 dom H∞(R) ⊂ X · `2+ε

Proof. Let z ∈ dom H∞(R). Since n 1/2 n 1/2 n X 2 X kid : Xn −→ `2 k = sup |zi| ≤ sup |zi| kzkX ≤1 i=1 kzkX ≤1 i=1 (2.24) n 1/2 X 0 1/2 = ei = (λX× (n)) X× i=1 it follows by proposition 2.40 and (1.12) that

1/2 1/2 ∗ 1 1 zn ≤ K log n λX (n) n

1/2 We have z ∈ c0 by corollary 2.41, and lemma 2.43 gives z ∈ X · `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 ⊂ dom H∞(R) ⊂ `1+ε ∩ R

1 1 1 (2) if p ≥ 2 and q = 2 + p then

`q ∩ R ⊂ dom H∞(R) ⊂ `q+ε ∩ R

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

`r−ε,s ∩ R ⊂ dom H∞(R) ⊂ `r+ε,s ∩ R

(c) If ϕ is a non degenerate Orlicz function which satisﬁes the√ ∆2-condition and there is an equivalent Orlicz functionϕ ˜ such thatϕ ˜( · ) is convex. Then (`ϕ · `2) ∩ R ⊂ dom H∞(R) ⊂ (`ϕ · `2+ε) ∩ R A description of dom Pm(X) 89

Proof. For part (a) use H¨older’sinequality. To understand (b) recall that

`t,u ⊂ `v,w if 0 < t < v ≤ ∞ and 0 < u, w < ∞. This together with H¨older’sinequality gives `r−ε,s ⊂ `p,q · `2 and `p,q · `2+δ ⊂ `r+ε,s for ε > δ > 0. Under the assumptions 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 (preliminaries, 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 dom Pm(X)

Continuous polynomials on a Banach sequence space X are holomorphic functions on X. We now examine the domain of convergence of the space of all continuous m-homogeneous polynomials on X. The 1-homogeneous polynomials are the continuous linear functionals on X. 0 P The monomial expansion of a functional ϕ ∈ X is ϕ(en)zn and we have X ϕ(z) = ϕ(en)zn n∈N for every z ∈ X if the en’s form a basis of X (which is then unconditional). Hence dom X0 = X and from now on in this section, we will assume m ≥ 2. Let us ﬁrst estimate the set dom Pm(X) from below. We start with a theorem of Bohnenblust and Hille [12, theorem I] which is a generalization of Littlewood’s result [58, theorem 1] about the summability of the coeﬃcients of a bounded bilinear form on `∞.

Theorem 2.47. Let P be a continuous m-homogeneous polynomial on `∞. Then 1 1 2 (1+ m ) X 2(1+ 1 )−1 |cα(P )| m < ∞ |α|=m

The result is actually proved for span{en} = c0. (Recall that every polyno- m mial P ∈ P (c0) can be extended to `∞ under preservation of the norm by the Aron-Berner extension, see [36, prop. 1.51].) An application of H¨older’s inequality gives 90 2. Domains of convergence

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

` 1 −1 ⊂ dom(P ) 2(1− m ) In general we get the following lower inclusion for the domain of convergence of the m-homogeneous polynomials (compare with theorem 2.25). Theorem 2.49. Let X be a Banach sequence space and m ≥ 2. Then

m `1 ∪ X · ` 1 −1 ⊂ dom P (X) 2(1− m )

Proof. We have `1 ⊂ X and the inclusion mapping is continuous. Hence the restriction of a polynomial P ∈ Pm(X) is a continuous m-homogeneous polynomial on `1 and, in particular, a holomorphic function on `1 which by theorem 2.25 implies `1 ⊂ dom(P ). m On the other hand, given x ∈ X and P ∈ P (X) the function Px deﬁned by Px(z) = P (xz) is a continuous m-homogeneous polynomial on c0 which by corollary 2.48 satisﬁes ` 1 −1 ⊂ dom(Px) and hence 2(1− m )

x · ` 1 −1 ⊂ dom(P ) 2(1− m ) for all x ∈ X.

To give upper inclusions for dom Pm(X) we use the following lemma which is an analogue to proposition 2.40. Lemma 2.50. Let X be a symmetric Banach sequence space and z ∈ dom Pm(X). Then √ 2m ∗ log n n 1 n 1− 1 z ≺ kid : X −→ ` k m kid : X −→ ` k m n n n 1 n 2

If X0 has non trivial cotype then

∗ 1 n 1 n 1− 1 z ≺ kid : X −→ ` k m kid : X −→ ` k m n n n 1 n 2

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

X α |cα(P )|r ≤ λkP k n α∈N0 |α|=m

m for every P ∈ P (Xn). Then by (2.21) we can ﬁnd εα ∈ {−1, 1} and a constant cm depending only on m such that n m X X m! α X m! α ri = r ≤ λ εα z α! α! i=1 |α|=m |α|=m (2.25)

p n n m−1 ≤ cm λ log n kid : Xn −→ `1 k kid : Xn −→ `2 k A description of dom Pm(X) 91

Thus √ 2m m 0 1 log n n 1 n 1− 1 A (B ) ≤ c λ m kid : X −→ ` k m kid : X −→ ` k m (2.26) λ Xn m n n 1 n 2 n for all λ ≥ 1. Factoring through `∞ this, in particular, implies √ 2m m log n n→∞ A (BX ) ≺ −−−→ 0 λ n 1 (1− 1 ) n 2 m for all λ ≥ 1. By lemma 2.38 with F(X) = Pm(X) we can ﬁnd λ > 1 such that X α |cα(P )z | ≤ λkP k |α|=m for all P ∈ Pm(X). It follows

n 1 X |z | ≤ Am(B ) n i λ Xn i=1 and with the same argument as in the proof of proposition 2.40 and using (2.26) √ 2m ∗ m log n n 1 n 1− 1 z ≤ A (B ) ≺ kid : X −→ ` k m kid : X −→ ` k m n λ Xn n n 1 n 2 If X0 has ﬁnite cotype s then the log-term in (2.25) can be replaced by the 0 0 cotype-s constant Cs(X ) of X (see [78, (4.3)]) which yields the second part of the claim. Theorem 2.51. Let X be a symmetric Banach sequence space and ε > 0. Then:

m (1) (a) dom P (X) ⊂ ` 1 −1 2(1− m ) +ε m 1 (1+ 1 ) (b) X regular: dom P (X) ⊂ X 2 m · ` 1 −1 2(1− m ) +ε (c) X0 ﬁnite cotype: For all z ∈ dom Pm(X):

1 1 1 1 2 (1+ m ) 2 (1− m ) ∗ sup(λX (n)) n zn < ∞ n (2) If X ⊂ `2:

m (a) dom P (X) ⊂ ` 1 −1 (1− 2m ) +ε m 1 (b) X regular: dom P (X) ⊂ X m · ` 1 −1 (1− m ) +ε (c) X0 ﬁnite cotype: For all z ∈ dom Pm(X):

1 1 m 1− m ∗ sup(λX (n)) n zn < ∞ n (3) If X is 2-convex: 92 2. Domains of convergence

m (b) X regular: dom P (X) ⊂ X · ` 1 −1 (1− m ) +ε (c) X0 ﬁnite cotype: For all z ∈ dom Pm(X):

1 1− m ∗ sup λX (n) n zn < ∞ n

Proof. By lemma 2.50 we have √ 2m ∗ log n n 1 n 1− 1 z ≺ kid : X −→ ` k m kid : X −→ ` k m n n n 1 n 2 and we can replace the log-term by 1, if X0 has ﬁnite cotype. Further

n n n X X 0 n kid : Xn −→ `1 k = sup |zi| = ei = λX× (n) = X× λ (n) kzkX ≤1 i=1 i=1 X

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

1/2 n n 1/2 n (1) kid : Xn −→ `2 k ≤ kid : Xn −→ `1 k = λX (n) n (2) kid : Xn −→ `2 k ≤ kX,→ `2k < ∞ (if X ⊂ `2)

1/2 n n (3) kid : Xn −→ `2 k ≤ c (if X is 2-convex) λX (n) It follows 1/r 1/s ∗ 1 1 zn ≤ K log n (2.27) λX (n) n with

1 −1 0 1 −1 (1) r = 2(1 + m ) , s = r = 2(1 − m ) 1 −1 (2) r = m , s = (1 − m ) 1 −1 (3) r = 1 , s = 2(1 − m ) and hence the claim by lemma 2.43 in the case where X is regular. If X0 has ﬁnite cotype, we can replace the log-term in (2.27) by 1 and also get what we want. For the remaining cases we deduce from (2.27) that

0 1 1/s z∗ ≤ K0 log n n n with

0 1 −1 (1) s = 2(1 − m ) 0 1 −1 (2) s = (1 − 2m ) A conjecture 93

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

For the ﬁrst part of this statement also note that

m dom P (`∞) ⊂ ` 1 −1 (2.28) 2(1− m ) +ε for all ε > 0 by lemma 2.50 and if X and Y are Banach sequence spaces with X ⊂ Y then dom Pm(X) ⊂ dom Pm(Y ) (see also remark 2.11). Note that if R is a bounded Reinhardt domain in a Banach sequence space X then every continuous m-homogeneous polynomial on X deﬁnes a bounded m holomorphic function on R. Hence dom H∞(R) ⊂ dom P (X) for all m and we can deduce the upper inclusions of theorem 2.44 from theorem 2.51. Together with theorem 2.49 we can determine dom Pm(X) almost precisely in the 2-convex case. Corollary 2.52. Let X be a 2-convex, regular and symmetric Banach sequence space. Then

m X · ` 1 −1 ⊂ dom P (X) ⊂ X · ` 1 −1 2(1− m ) 2(1− m ) +ε for all ε > 0.

A conjecture

From corollary 2.22 and theorem 2.51 we get

m 1 `1 ⊂ dom P (X) ⊂ X m · ` 1 −1 (1− m ) +ε for all ε > 0 if X is regular and contained in `2. We presume that in analogy m 1 to corollary 2.52 dom P (X) is nearly identical with X m · ` 1 −1 . (1− m ) Conjecture 2.53. Let X be a symmetric Banach sequence space contained in `2. Then 1 m X m · ` 1 −1 ⊂ dom P (X) (1− m )

For `p-spaces we know the following.

Example 2.54. We consider X = `p, 1 ≤ p ≤ ∞. Deﬁne qm by 1 = 1 + 1 (1 − 1 ). Then qm p 2 m (1) For p = ∞ :

m ` 1 −1 ⊂ dom P (`∞) ⊂ ` 1 −1 2(1− m ) 2(1− m ) +ε 94 2. Domains of convergence

(2) For 2 ≤ p < ∞ : m `qm ⊂ dom P (`p) ⊂ `qm,∞

(3) For 1 ≤ p ≤ 2 :

m 0 0 `max{1,qm} ⊂ dom P (`p) ⊂ `(mp ) ,∞

1 −1 Note that max{1, qm} = qm if and only if p ≥ 2(1 + m ) .

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

m `(mp0)0 ⊂ dom P (`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 theorem 2.49. P α m Let P = cαz ∈ P (`p). For 2 ≤ p ≤ ∞ our argument gives

X α X α sup |cαz | = sup sup |cα(xy) | z∈B x∈B y∈B `qm `p ` 1 |α|=m 2(1− m )−1 |α|=m

X α (2.30) ≤ cm sup sup cα(xz) x∈B z∈B `p c0 |α|=m

≤ cm kP k where the constant cm from the result of Bohnenblust and Hille only depends on m and not on the polynomial P (see also [48] for an estimate of the constant). For p = 1 we have α m X α X α m α sup |cαz | = sup |cα| m α |z | z∈B z∈B m α `1 |α|=m `1 |α|=m X m! ≤ sup kc zαk em |zα| α B`1 z∈B α! `1 |α|=m (2.31) X m! ≤ em kP k |zα| = em kP k kzkm α! `1 |α|=m ≤ em kP k .

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

(N) Theorem 2.55. Let cα ∈ C, α ∈ N0 . The following are equivalent: A conjecture 95

αα |α| (N) (1) (cα |α||α| r )α ∈ `∞(N0 ) for all r > 0. P α (2) (N) cαz is bounded on bounded subsets of `1. α∈N0 P α (3) (N) |cαz | is bounded on bounded subsets of `1. α∈N0 P α In this case f(z) := (N) cαz deﬁnes a holomorphic function of bounded α∈N0 type on `1 with X r + ε sup |c zα| ≤ kfk α e(r+ε)B`1 z∈rB` ε 1 (N) α∈N0 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 define m κm(Xn,Yn) to be the best constant c > 0 such that for all P ∈ P (Xn)

X α m sup |cα(P )z | ≤ c kP kP (Xn) z∈B Yn |α|=m

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 space X and f be a holomorphic mapping on R. Then

dom(f) −→ C Mf : P α z (N) |cα(f)|z α∈N0 is called the majorant function of f.

In a Banach sequence space X

X α X α sup |cα(P )z | = sup |cα(P )|z z∈B z∈B Xn |α|=m Xn |α|=m

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

m m kMP kP (Yn) ≤ c kP kP (Xn)

m for all P ∈ P (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 following statements are equivalent:

(1) Y ⊂ dom Pm(X) 96 2. Domains of convergence

(2) supn κm(Xn,Yn) < ∞

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

kMP kPm(Y ) ≤ c kP kPm(X) (2.32) holds for every P ∈ Pm(X).

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

m X α Lk,n := {P ∈ P (X) | sup |cα(P )z | ≤ n} 1 z∈ k BY |α|=m

m m is closed and we have P (X) = ∪k,nLk,n : For P ∈ P (X) and n ∈ N the set X α Hn = {z ∈ Y | |cα(P )z | ≤ n} |α|=m is closed and by the assumption we have Y = ∪nHn, hence by Baire’s theorem there must be an n such that Hn has non empty interior. As in the proof of 1 theorem 2.83 we conclude that k BY ⊂ Hn for some natural number k, i.e.

X α sup |cα(P )z | ≤ n 1 z∈ k BY |α|=m and thus P ∈ Lk,n.

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

X α 2n sup |cα(P )z | ≤ 1 s z∈ k BY |α|=m or 2nkm kMP k m ≤ P (Y ) s m for all P ∈ BPm(X), thus supν κm(Xν,Yν) ≤ 2nk /s.

Now let us assume that (2) is true and deﬁne c := supn κm(Xn,Yn). Let m n P ∈ P (X) and Pn := P |span{e1,...,en}. Then cα(P ) = cα(Pn) for α ∈ N0 and hence

X α m m m |cα(P )z | ≤ kMPnkP (Yn) ≤ κm(Xn,Yn) kPnkP (Xn) ≤ c kP kP (X) n α∈N0 |α|=m for all n and all z ∈ BY which implies

kMP kPm(Y ) ≤ c kP kPm(X) A conjecture 97

n since every ﬁnite subset A ⊂ [|α| = m] is contained in N0 for some n. In m particular, we have BY ⊂ dom P (X) and by m-homogeneity it follows (1).

0 0 Finally suppose c is a constant satisfying kMP kPm(Y ) ≤ c kP kPm(X) for all m P α m ˜ P ∈ P (X). Take P = n cαz ∈ P (Xn) and deﬁne P (z) := α∈N0 ,|α|=m P α n (N) c˜αz , z ∈ X wherec ˜α := cα if α ∈ N0 and zero otherwise. Then α∈N0

0 0 m ˜ m ˜ m m kMP kP (Yn) = kMP kP (Y ) ≤ c kP kP (X) = c kP kP (Xn) and hence c ≤ c0.

Thus our conjecture (2.29) for `p-spaces is equivalent to

n n sup κm(`p , `(mp0)0 ) < ∞ n

n n for 1 < p < 2. To decide this we investigate the asymptotic of κm(`p , `q ) for general p and q. As we can see from the calculations (2.30) and (2.31) we n n n n m have supn κm(`p , `qm ) ≤ cm and supn κm(`1 , `1 ) ≤ e .

We are going to identify κm(Xn,Yn) with another constant which is easier for 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 n X X εiµiT xi ≤ c µixi i=1 i=1 for all µ1, . . . , µn ∈ K and ε1, . . . , εn ∈ K with |εi| ≤ 1 and call it the unconditional constant of T with respect to (x1, . . . , xn).

If x1, . . . , xn are linearly independent, E = F and T = idE is the identity map then χ(idE, (x1, . . . , xn)) = χ((x1, . . . , xn)) is the unconditional basis constant 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 ) for a 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 the triangle inequality on L(E,F ). 98 2. Domains of convergence

Remark 2.60. Let E, F , G, H be Banach spaces, (xi)i∈I a ﬁnite family of elements xi ∈ E and T ∈ L(E,F ), S ∈ L(F,G), R ∈ L(G, H):

E −→T F −→S G −→R H

Let I0 := { i ∈ I | T xi 6= 0 }. Then

χ(RST, (xi)i∈I ) ≤ kRk χ(S, (T xi)i∈I0 ) kT k

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 that xi = 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 index set 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 ﬁnite families in E and R,S ∈ L(E,F ). Then

χ(idF , (Rxi)i∈I ∪ (Syj)j∈J ) ≤ (kRk + kSk) 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 E and I ∈ L(E,F ) be an isomorphism. Then

−1 −1 a kI k ≤ χ(I, (xs)s∈S) ≤ b kIk where

a = max{χ(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| = α m m }. Then the monomials (z )α∈J ∗(m,n) form a basis of P (Xn). For T ∈ m m L(P (Xn), P (Yn)) we deﬁne

α χM (T ) := χ(T, (z )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

m m κm(Xn,Yn) = χM (id : P (Xn) −→ P (Yn)) A conjecture 99

P α m Proof. Let P = |α|=m cαz ∈ P (Xn). Deﬁne εα := |cα|/cα if cα 6= 0 and 1 otherwise. Then

X α kMP kPm(Y ) = εαcαz m n P (Yn) |α|=m m m m ≤ χM (id : P (Xn) −→ P (Yn))kP kP (Xn) hence κm(Xn,Yn) ≤ χM (id). On the other hand, given scalars cα and εα with P α m |εα| ≤ 1 we have P := |α|=m cαz ∈ P (Xn) and

X α X α εαcαz m ≤ |cαz | m = kMP kPm(Y ) P (Yn) P (Yn) n |α|=m |α|=m

m ≤ κm(Xn,Yn) kP kP (Xn) which implies the other inequality.

Now our problem is to ﬁnd asymptotic estimates for

m n m n χM (id : P (`p ) −→ P (`q ))

m m,s 0 and we use lemma 1.1 and the identiﬁcation P (Xn) = ⊗εs (Xn) for Banach sequence spaces X to shift the problem to full tensor products.

m n Recall that (ei)i∈M(m,n) is a basis of ⊗ K and (σm(ej))j∈J (m,n) a basis of ⊗m,sKn (see the preliminaries, page 16). In particular, if X is a Banach 0 sequence space and ek denote the coeﬃcient functionals with respect to the 0 0 0 standard unit vectors ek (and ej = ej1 ⊗ · · · ⊗ ejm for j ∈ J (m, n)) then 0 m 0 m 0 (σm(ej))j∈J (m,n) is a basis of ⊗ (Xn) and the identiﬁcation ⊗εs (Xn) = m 0 α ∗ P (Xn) is given by σm(ej) = z where j ∈ J (m, n) and α ∈ J (m, n) such that 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 arbitrary symmetric and full m-fold tensor products of Banach sequence spaces: For m m m,s m,s norms α, β on ⊗ Xn and ⊗ Yn and αs, βs on ⊗ Xn and ⊗ Yn, respectively and operators T ∈ L(⊗mX , ⊗mY ) and S ∈ L(⊗m,sX , ⊗m,sY ) let α n β n αs n βs n

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 we m,s n m n complete the elements σm(ej) ∈ ⊗ K ⊂ ⊗ K , j ∈ J (m, n) with elements m ei, i ∈ M(m, n) to a basis of ⊗ K. 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. 100 2. Domains of convergence

Proof. Let X X λjσm(ej) + λjej = 0 j∈J (m,n) j∈Γ with λj ∈ K. Denote [j] the set of all permutations of j and |j| = |[j]|. Using [21, lemma 1] we get

X X X λj X X λj λ σ (e ) + λ e = e + + λ e j m j j j |j| j |j| i i j∈J (m,n) j∈Γ j∈J (m,n) j∈J (m,n) i∈[j]\{j} and this implies λj = 0 for all j ∈ J (m, n) and then also λi = −λj/|j| = 0 for 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

χ (id : ⊗m,sX −→ ⊗m,sY ) χ (id : ⊗mX −→ ⊗mY ) M αs n βs n M α n β n

Proof. Writing idn : Xn −→ Yn we have the commutative diagram

⊗m id ⊗m,sX n - ⊗m,sY αs n βs n 6 ιm σm ? ⊗m id m n - m ⊗α Xn ⊗β Yn and using remark 2.60,(2.33) and remark 2.61 together with proposition 2.64 we obtain

m,s m χM (⊗ idn) = χ(σm ◦ ⊗ idn ◦ιm, (σm(ej))j∈J (m,n)) m ≤ kσmk χ(⊗ idn, (ιm ◦ σm(ej))j∈J (m,n))kιmk m ≤ kσmkkιmk χ(⊗ idn, (ιm ◦ σm(ej))j∈J (m,n) ∪ (ej)j∈Γ) m ≤ kσmkkιmk (kσmkkιmk + 1) χ(⊗ idn, (ej)j∈M(m,n)) m ≤ cm χM (⊗ idn)

m 2 m 2 with cm ≤ (m /m!) ((m /m!) + 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

m m Pi m ⊗ idn m Ii m ⊗α XNn −→ ⊗α Xn −−−−→⊗β Yn −→ ⊗β 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

m m χM (Ii ◦ ⊗ idn ◦Pi) = χ(Ii ◦ ⊗ idn ◦Pi, (ej)j∈M(m,Nn)) m ≤ kIikkPik χ(⊗ idn, (Piej)j∈J0 ) m ≤ χ(⊗ idn, (ej)j∈M(m,n)) m = χM (⊗ idn) where kIik = 1 = kPik 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 natural inclusion I : YN1 −→ YN2 and projection P : XN2 −→ XN1 . Then

⊗m id m N1 - m ⊗α XN1 ⊗β YN1 6 ⊗mI ⊗mP ? ⊗m id m N2 - m ⊗α XN2 ⊗β YN2 is commutative and hence m m m m χM (⊗ idN1 ) = χ(⊗ P ◦ ⊗ idN2 ◦ ⊗ I, (ej)j∈M(m,N1)) m m ≤ (kP kkIk) χ(⊗ idN2 , (ej)j∈M(m,N2))

= χM (idN2 )

m by remark 2.60 and (2.33) since ⊗ I(ej) = ej. In particular, c2 = 1 is adequate. (iv) As in the proof of proposition 1.2 we use the factorization

⊗m id m n - m ⊗α Xn ⊗β Yn 6 I := σm◦I1⊗···⊗Im P := m! P1⊗···⊗Pm◦ ιm ? ⊗m,s id ⊗m,sX mn+k - ⊗m,sY αs mn+k βs mn+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 m m,s χM (⊗ idn) = χ(P ◦ ⊗ idmn+k ◦I, (ej)j∈M(m,n)) m,s ≤ kP kkIk χ(⊗ idmn+k, (σm(ej))j∈J (m,mn+k)) m,s = kP kkIk χM (⊗ idmn+k) 102 2. Domains of convergence

mm 2 with kP kkIk ≤ ( m! ) =: c3. m m,s Now lemma 1.1 implies χM (⊗ idn) ≤ c χM (⊗ idn) for all n ≥ m with mm 2 m c ≤ ( m! ) (m + 1) , and this ﬁnishes the proof. Corollary 2.66. Let X and Y be symmetric Banach sequence spaces. Then

m m m 0 m 0 χM (id : P (Xn) −→ P (Yn)) χM (id : ⊗ε (Xn) −→ ⊗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 theorem 2.49.

Lemma 2.67. Let 1 ≤ p ≤ q ≤ r ≤ ∞ and pr, qr ∈ [1, ∞] such that 1 = 1 − 1 and 1 = 1 − 1 . Then pr p r qr q r

m n m n m n m n χM (id : ⊗ε `pr −→ ⊗ε `qr ) ≤ χM (id : ⊗ε `p −→ ⊗ε `q )

n n n n n n Proof. We are simply using the fact that B` 0 = B`r B` 0 and B` 0 = B`r B` 0 pr p qr q 0 0 0 0 by H¨older’sinequality since 1/pr = 1/r + 1/p and 1/qr = 1/r + 1/q .

Let I := M(m, n) and µi, εi ∈ K with |εi| ≤ 1 for i ∈ I. Then X X εi µi ei m n = sup εi µi x1(i1) ··· xm(im) ⊗ε `qr xj ∈B n i∈I ` 0 i∈I qr X = sup sup εi µi y1(i1) ··· ym(im) z1(i1) ··· zm(im) yj ∈B`n zj ∈B`n r q0 i∈I X = sup εi µi y1(i1) ··· ym(im)ei ⊗m`n y ∈B n ε q j `r i∈I

m n m n X ≤ χM (id : ⊗ε `p −→ ⊗ε `q ) sup µi y1(i1) ··· ym(im)ei ⊗m`n y ∈B n ε p j `r i∈I

m n m n X = χM (id : ⊗ε `p −→ ⊗ε `q ) µi ei m n ⊗ε `pr i∈I

Theorem 2.68. Let 1 ≤ p ≤ ∞. Then

m n m n χM (id : ⊗ε `p −→ ⊗ε `∞) ≤ 1

1 −1 1 1 1 1 If 1 ≤ p ≤ 2(1 − m ) and q such that q = p − 2 (1 − m ) then √ m n m n m−1 χM (id : ⊗ε `p −→ ⊗ε `q ) ≤ ( 2) A conjecture 103

Proof. Factorization (see remark 2.60) gives

m n m n χM (id : ⊗ε `p −→ ⊗ε `∞) m n m n m n m n ≤ χM (id : ⊗ε `∞ −→ ⊗ε `∞) kid : ⊗ε `p −→ ⊗ε `∞k

nm nm nm n n m = χ(id : `∞ −→ `∞ , (ek)k=1) kid : `p −→ `∞k = 1

1 −1 in the general case and for p = 1, q = 2(1 + m ) we have

m n m n χM (id : ⊗ε `1 −→ ⊗ε `q )

nm m n nm nm nm m n m n ≤ kid : `q −→ ⊗ε `q k χ(id : `q −→ `q , (ek)k=1) kid : ⊗ε `1 −→ ⊗q `q k m n m n ≤ kid : ⊗ε `1 −→ ⊗ε `q k √ ≤ ( 2)m−1

m n m n The boundedness of kid : ⊗ε `1 −→ ⊗ε `q k in n is a restatement of the Bohnenblust and Hille result (theorem 2.47), the constant was proved by Kaijser [48] with the help of Khintchin’s inequality. The remaining cases 1 −1 1 1 1 1 1 < p < 2(1 − m ) , q = p − 2 (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 variables on a probability space (Ω, µ). Then  Z m(1/p−1/2)+1/2 X n p ≤ 2 ri(ω)ei m n dµ ≤ cm ⊗ε `p 1/p Ω i∈M(m,n) n p ≥ 2

This is proved in [21, p. 138]. Theorem 2.70. Let 1 ≤ p, q ≤ ∞. Then  n(m−1)/2 p ≤ q ≤ 2   n(m−1)/q 2 ≤ q m n m n  p χM (id : ⊗ε `p −→ ⊗ε `q ) ≺ m(1/q−1/p+1/2)−1/2 n q ≤ p ≤ 2   m/q−1/p 2 n q ≤ p and  nm(1/q−1/p−1/2)−1/2 1 ≤ p ≤ 2 m n m n  χM (id : ⊗ε `p −→ ⊗ε `q ) nm/q−1/p 2 ≤ p ≤ ∞ 104 2. Domains of convergence

In particular,  nm(1/q−1/p−1/2)−1/2 1 ≤ p ≤ 2 m n m n  χM (id : ⊗ε `p −→ ⊗ε `q ) nm/q−1/p 2 ≤ p ≤ ∞ if p ≥ q.

(See p. 26 for the notation ≺ and .)

Proof. In [21] it is proved that  n(m−1)/2 r ≤ 2 m n m n  χM (⊗ε `r ) χM (⊗ε `r ) n(m−1)/r r ≥ 2 hence m n m n (m−1)/ min{p,q,2} an := max{χM (⊗ε `p ), χM (⊗ε `q )} n

m n m n (m−1)/ max{p,q,2} bn := min{χM (⊗ε `p ), χM (⊗ε `q )} n Together with  1 p ≤ q m n m n n n m  kid : ⊗ε `p −→ ⊗ε `q k = kid : `p −→ `q k = nm(1/q−1/p) p ≥ q  nm(1/q−1/p) p ≤ q −1 m n m n −1 n n −m  kid : ⊗ε `p −→ ⊗ε `q k = kid : `q −→ `p k = 1 p ≥ q this implies the upper estimate and  nm(1/q−1/p+1/2)−1/2 p ≤ 2  q  nm/q−1/p 2 ≤ p ≤ q m n m n  χM (id : ⊗ε `p −→ ⊗ε `q ) n(m−1)/2 q ≤ 2  p  n(m−1)/q 2 ≤ q ≤ p since −1 −1 m n m n an kid k ≤ χM (id : ⊗ε `p −→ ⊗ε `q ) ≤ bn kidk by lemma 2.62. For q ≤ p this is not optimal. We will get a better estimate 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 n m/q X m X X n = ei n = ei m n = ri(ω) ri(ω) ei m n `q ⊗ε `q ⊗ε `q i=1 i∈M(m,n) i∈M(m,n)

m n m n X ≤ χM (id : ⊗ε `p −→ ⊗ε `q ) ri(ω) ei m n ⊗ε `p i∈M(m,n) The domain of convergence of the polynomials 105 for all ω ∈ Ω, hence Z m/q X n ≤ χM (id) ri(ω)ei m n dµ ⊗ε `p Ω i∈M(m,n) 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 correctness of our conjecture (2.29), it is enough to prove

m n m n sup χM (id : ⊗ε `p −→ ⊗ε `m p) < ∞ n for p > 2. Our theorem 2.70 only shows

m n m n (1−1/m)1/p 1 ≤ χM (id : ⊗ε `p −→ ⊗ε `m p) ≤ n

The domain of convergence of the polynomials

Let X be a Banach sequence space. Then [ P(X) := span Pm(X) m is the space of all continuous polynomials on X. Every non-zero polynomial P ∈ P(X) can be written as a ﬁnite sum

P = P0 + P1 + ... + Pn

m of uniquely determined homogeneous polynomials Pm ∈ P (X), m = 1, . . . , n with 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 dom P(X). Theorem 2.71. Let X be a Banach sequence space. Then

`1 ∪ X · `2 ⊂ dom P(X)

Proof. Since every continuous polynomial on X is holomorphic and bounded on every bounded subset of X we have

dom H∞(R) ⊂ dom P(X) for every bounded Reinhardt domain R in X and theorem 2.25 implies `1 ⊂ dom P(X). On the other hand by (2.12) [ \ dom P(X) = dom Pm(X) = dom Pm(X) (2.34) m m and theorem 2.49 implies the second part of the claim. 106 2. Domains of convergence

Using (2.34) we can deduce upper inclusions for dom P(X) from the results of section 5.

Theorem 2.72. Let X be a symmetric Banach sequence space and ε > 0. Then dom P(X) ⊂ `2+ε If X is regular, then 1 dom P(X) ⊂ X 2 · `2+ε

Proof. By theorem 2.51 (1)(a) and (2.34) we have

dom P(X) ⊂ ` 1 −1 2(1− m ) +ε for all m and this implies the ﬁrst statement. The second one follows in the 1 −1 same way from theorem 2.51 (1)(b): We choose m such that 2(1 − m ) < 2 + ε/2. Then

1 (1+ 1 ) 1 dom P(X) ⊂ X 2 m · ` 1 −1 ε ⊂ X 2 · `2+ε 2(1− m ) + 2 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 ⊂ dom P(X) ⊂ `1+ε

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

X · `2 ⊂ dom P(X) ⊂ X · `2+ε

Proof. We have already proved the lower inclusions (theorem 2.71). The upper inclusion of (1) follows from theorem 2.51 (2)(b) since the Banach 1 sequence space X m 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 two continuous polynomials on X are equal if they are equal on R and we can consider 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 functions 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 ⊂ dom F(R) ⊂ `1+ε ∩ R

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

(X · `2) ∩ R ⊂ dom F(R) ⊂ (X · `2+ε) ∩ R

Proof.

dom H∞(R) ⊂ dom F(R) ⊂ dom P(X)

If X is symmetric and R is bounded then under the assumptions of theorem 2.44 on X we see that dom H∞(R) and R ∩ dom P(X) have the same description. So these sets are identical “up to an ε”.

Problem 2.75. ? dom H∞(R) = R ∩ dom P(X)

The exceptional role of `1

In [53] Lempert treated the problem of solving the ∂-equation in an open subset 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 consider monomial expansions of holomorphic functions as were investigated by Ryan in [70], also for specifically the space `1. This space seems to play a special role in infinite dimensional holomorphy. Lempert gave part of an answer to his question by explaining how his approach depends on the structure of `1. He pointed out that, as in finite dimensions the geometric series is an important tool, he based his arguments on the convergence of an appropriate series in infinitely many variables, basically

X |α||α| |zα| (2.35) αα (N) α∈N0 This series converges if and only if the series

X |α|! |zα| (2.36) α! (N) α∈N0 converges. To understand this, note that we have |α|!/α! ≤ |α||α|/αα and from the proof of lemma 2.14 (which uses the arguments of Lempert) we can 108 2. Domains of convergence

see that for every z ∈ B`1 we can ﬁnd a w ∈ B`1 and a constant c > 0 such that |α||α| |α|! |zα| ≤ c |wα| αα α!

(N) for every α ∈ N0 . Thus (2.35) converges if and only if z ∈ B`1 since this is true for (2.36). In this case we have

∞ ∞ X |α|! α X X n! α X n 1 |z | = |z | = kzk`1 = α! α! 1 − kzk`1 (N) n=1 |α|=n n=1 α∈N0

Lempert also remarked that the series (2.35) dominates P|zα| which con- N Q verges if and only if z ∈ `1 ∩ D with limit n 1/(1 − |zn|).

P |α||α| α Note that the series (N) α z defines a holomorphic function on B`1 α∈N0 α since the convergence of (2.35) is, as Lempert proved, even uniform on compact subsets of B`1 . More generally Lempert gave a necessary and sufficient P α condition on the coefficients cα such that f(z) = cαz defines a holomorphic function on rB`1 , extending the work of Ryan. Our formulation is a slight extension of that result.

(N) Theorem 2.76. Let cα ∈ C, α ∈ N0 and r > 0. The following conditions are equivalent:

(1) (c αα ξα) ∈ ` ( (N)) for all ξ ∈ rB+ . α |α||α| ∞ N0 c0 (2) (c αα ξα) ∈ c ( (N)) for all ξ ∈ rB+ . α |α||α| 0 N0 c0 P α (3) cαz is bounded on compact subsets of rB`1 . P α (4) |cαz | is bounded on compact subsets of rB`1 . P α (5) cαz is compactly convergent on rB`1 . P α (6) |cαz | is compactly convergent on rB`1 .

P α In this case f(z) := α∈ (N) cαz , z ∈ rB`1 deﬁnes a holomorphic function N0 P with normally convergent Taylor series expansion m Pmf(0).

(N) By c0(N0 ) we mean the space of all families (λα) (N) which can be ar- α∈N0 (N) bitrarily small (in absolute value) outside finite subsets of N0 , i.e. for all (N) ε > 0 we can find a finite subset A ⊂ N0 such that

sup |λα| < ε α/∈A The exceptional role of `1 109

P α Proof. Obviously (6) y (4) y (1) and (6) y (5). If cαz converges uni- P α formly on compact subsets of rB`1 then f(z) := cαz deﬁnes a continuous mapping on rB`1 , thus (5) y (3). The mapping f is even holomorphic: Let x ∈ rB`1 , y ∈ `1 and choose ε > 0 such that K := x + εDy ⊂ rB`1 . Then K P α is a compact subset of rB`1 and this implies that cα(x + λy) converges uniformly in λ on εD. Hence λ f(x + λy) is holomorphic on εD. Thus f is G-holomorphic.

+ (N) n We now prove (3) y (1). Let ξ ∈ rBc0 and α ∈ N0 , say α ∈ N . Denote β (N) cβ,ξ := cβξ for β ∈ N0 . Then

X β gn(z) := cβ,ξz n β∈N0 deﬁnes a bounded holomorphic function on B n and using the Cauchy in- `1 equalities (2.2) we get

αα X |c | ξα = kc zαk ≤ kg k = c zβ α |α| α,ξ B`n n B`n β ξB n |α| 1 1 `1 (N) β∈N0 X β ≤ cβz ξB`1 (N) β∈N0

+ + + α + (1) y (2) follows from Bc0 = Bc0 · Bc0 and the fact that ξ −→ 0 for ξ ∈ Bc0 : Let ε > 0. Choose n0 such that ξn < ε for n > n0. Since ξi < 1 we can ﬁnd k α k0 with ri < ε for k > k0 and i = 1, . . . , n0. Then ξ < ε outside the ﬁnite ( ) n0 N N set A = [0, k0] × {0} ⊂ N0 . Finally (2) y (6) is a consequence of proposition 2.15 which also yields that

X X α |cαz | K < ∞ m |α|=m

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 arbitrary Reinhardt domains R ⊂ X by corollary 2.20. In the proof of theorem 2.76 we have deduced the normal convergence of the Taylor series from proposition 2.15. But this is always true for holomorphic functions 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 a Banach space E and

∞ X f(ξ + x) = Pmf(ξ)(x) m=0 110 2. Domains of convergence the Taylor expansion of f in ξ ∈ U convergent for all x in an open, balanced subset B ⊂ E with ξ + B ⊂ U. Then the Taylor series converges normally in B, i.e. we have ∞ X kPmf(ξ)kK < ∞ m=0 for all compact subsets K ⊂ B.

Proof. Let K ⊂ B be compact. We can assume that K is balanced (see remark 2.17). Since B is balanced and open we have

[ 1 K ⊂ B = B t t>1 and by compactness of K there is a t > 1 such that tK ⊂ B. By the Cauchy inequalities (0.13) and m-homogeneity we have for all m ∈ N

−m −m kPmf(ξ)kK = t kPmf(ξ)ktK ≤ t kfkξ+tK and we conclude

∞ X kfkξ+tK kP f(ξ)k ≤ |f(ξ)| + m K t − 1 m=0

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): For + ξ ∈ rBc0 the function fξ defined by fξ(z) := f(ξz), z ∈ B`1 is holomorphic α on B`1 with monomial coefficients cα(fξ) = cα(f)ξ and since

α α α kz kB = (2.37) `1 |α||α| it follows α α α α |cα(f)| ξ = kcα(fξ)z kB ≤ kfξkB = kfkξB < ∞ |α||α| `1 `1 `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 monomial P α expansion series but the convergence is in general not normal, i.e. kcαz kK does in general not converge for every compact subset K ⊂ rB`1 : Deﬁne cen := 1 and cα := 0 for α∈ / {en | n ∈ N}. Then α α α sup |cα| |α| ξ = sup ξn < ∞ α |α| The exceptional role of `1 111

+ for all ξ ∈ Bc0 , but X αα X |c | ξα = ξ = ∞ α |α||α| n (N) n α∈N0 for ξn = 1/(2n).

+ The following characterization is immediate from the fact that (ξB`1 ) ξ∈rBc0 is a fundamental system of compact subsets of rB`1 .

(N) Remark 2.78. Let cα ∈ C, α ∈ N0 and r > 0. The following are equivalent:

P α (1) cαz converges normally on rB`1 . (2) (c αα ξα) ∈ ` ( (N)) for all ξ ∈ rB+ . α |α||α| 1 N0 c0

So the situation for holomorphic functions on `1 is nice, but not as nice as in ﬁnite dimensions.

Now we give an answer to Lempert’s question by proving that `1 is nearly the only Banach sequence space X in which the situation dom H∞(R) = R for some bounded Reinhardt domain R ⊂ X is possible.

Theorem 2.79. Let X be a symmetric Banach sequence space. If X contains a bounded Reinhardt domain R such that dom H∞(R) is absorbant, in particular, if dom H∞(R) = R, then

`1 ⊂ X ⊂ `1+ε for all ε > 0.

Proof. By corollary 2.41 we have dom H∞(R) ⊂ `2+ε and thus X ⊂ `2+ε for all ε > 0 since by the assumption for all z ∈ X there is a λ > 0 such that λz ∈ dom H∞(R). But now, since the inclusion X ⊂ `2+ε is automatically n continuous, factorization through `2+ε gives

n 1/2−1/(2+ε) kid : Xn −→ `2 k ≤ n for all ε > 0. Thus the estimate in proposition 2.40 implies that dom H∞(R) and hence X is contained in `1+ε for all ε > 0.

The assumption of absorbance in the preceding theorem is not so much weaker than dom H∞(R) = R. In fact, it is equivalent to say that dom H∞(R) contains a zero neighbourhood (see theorem 2.83). This raises the following question. 112 2. Domains of convergence

Problem 2.80. Can we take ε = 0 in theorem 2.79 ?

In the previous section we asked if dom H∞(R) = R ∩ dom P(X) since they have the same description under our assumptions on the underlying Banach sequence space X. What if we assume dom P(X) to be absorbant? We ﬁrst consider homogeneous polynomials.

Proposition 2.81. Let X be a symmetric Banach sequence space such that dom Pm(X) is absorbant. Then

`1 ⊂ X ⊂ ` 1 −1 (1− m ) +ε for all ε > 0.

Proof. Theorem 2.51 (1)(a) and the absorbance of dom Pm(X) imply

X ⊂ ` 1 −1 for all ε > 0. Let s > 2 and z ∈ X. Without loss of 2(1− m ) +ε generality we can assume that z ∈ dom Pm(X). Then we use the estimate n n in lemma 2.50 and factorize both id : Xn −→ `1 and id : Xn −→ `2 through n ` 1 −1 to get s(1− m ) √ 2m log n 1 1 1 1 1 1 1 ∗ (1 − s (1 − m )) m ( 2 − s (1 − m ))(1 − m ) zn ≺ n n n √ 2m log n = ( 1 + 1 )(1 − 1 ) n 2 s m

1 1 −1 This implies z ∈ ` 1 −1 for all ε > 0, where ts = ( + ) ↓ 1 for s ↓ 2 (1− m ) ts+ε 2 s and hence the result.

This immediately implies the following result.

Theorem 2.82. Let R be a Reinhardt domain in a symmetric Banach sequence space X and P(X) ⊂ F(R) ⊂ H(R). If dom F(R) is absorbant then

`1 ⊂ X ⊂ `1+ε for all ε > 0.

Proof. We have

dom F(R) ⊂ dom P(X) ⊂ dom Pm(X) implying that dom Pm(X) is absorbant and hence by proposition 2.81

`1 ⊂ X ⊂ ` 1 −1 (1− m ) +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 this case, we know that we cannot take ε = 0. In [28] an example of a Lorentz sequence space X = d(w, 1) diﬀerent from `1 is given such that for every m the monomials with respect to the standard basis vectors en form an unconditional basic sequence of Pm(X). In other words:

m m sup χM (P (Xn)) = sup χM (idP (Xn)) < ∞ n n

This implies X ⊂ dom Pm(X) by theorem 2.58 and remark 2.63.

Domains of convergence, Bohr radii and unconditionality

Our main result in this section gives a good insight into the connection between domains of convergence, Bohr radii and unconditionality. Additionally we will have a short look at the border case X = `∞ .

Theorem 2.83. Let X and Y be Banach sequence spaces, R ⊂ X be a bounded Reinhardt domain and F(R) a closed subspace of H∞(R) which contains the space P(X) of all polynomials on X. The following are equivalent:

(1) tBY ⊂ dom F(R) for some t > 0. (2) dom F(R) absorbs Y , i.e. for every z ∈ Y there is a t > 0 such that tz ∈ dom F(R). (3) There is an 0 < r < 1 and a constant λ > 0 such that

X α sup |cα(f)z | ≤ λkfkR (2.38) z∈rBY (N) α∈N0

for all f ∈ F(R). (4) There is a constant c > 0 such that for all m

m m m sup χM (id : P (Xn) −→ P (Yn)) ≤ c n

Before we start with the proof note that, for example, condition (2) implies that Y ⊂ X. Condition (3) says that there exists some sort of infinite dimensional Bohr radius (for λ = 1, X = Y , R = BX and F(R) = H∞(R) this would be exactly the existence of the Bohr radius in the infinite dimensional Reinhardt domain BX ). Condition (4) is a stronger version of condition (2) given in theorem 2.58 for fixed m (see also remark 2.63). 114 2. Domains of convergence

Proof. Of course (1) implies (2). We now prove that (2) implies (3). We will use a Baire argument. For natural numbers m and n the sets

X α Fm,n = {f ∈ F | sup |cα(f)z | ≤ n} z∈ 1 B m Y (N) α∈N0 are closed in F := F(R) since the cα are continuous linear functionals on H∞(R) by the Cauchy integration formulas (2.1) and we have F = ∪m,nFm,n: Let f ∈ F . Since dom F absorbs Y by the assumption, we have Y = ∪m,nHm,n with X z α H = {z ∈ Y | |c (f) | ≤ n} m,n α m (N) α∈N0 and these sets are closed in Y : If (zν) is a sequence in Hm,n converging (N) to z ∈ Y then it converges coordinatewise. A ﬁnite subset A ⊂ N0 is k α α1 αk contained in N0 for some k and hence zν = zν(1) ··· zν(k) converges to α P z α z for every α ∈ A which implies that the ﬁnite sum α∈A|cα m | as the α P zν limit of α∈A|cα m | 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. 1 we can find z ∈ Hm,n and k ∈ N such that z + k BY ⊂ Hm,n and this even 1 1 implies k B ⊂ Hm,n: Let u ∈ k BY . Define θ = (θn) by θn := zn/|zn| if zn 6= 0 1 and zero otherwise. Then θ|u| ∈ k BY since BY is a Reinhardt domain and for 1 w := θ(|z| + |u|) = z + θ|u| ∈ z + B k Y 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 that int(Fm,n) 6= ∅, which means f0 + sBF ⊂ Fm,n for some f0 ∈ Fm,n and s > 0. So if f ∈ BF then 1 c (f) = (c (f + sf) − c (f )) α s α 0 α 0

(N) for α ∈ N0 implies

X α 2n sup |cα(f)z | ≤ =: λ z∈ 1 B s m Y (N) α∈N0 which is (3) with r = 1/m. If (3) is true then, in particular, for all m and all P ∈ Pm(X)

X α −1 m −1 m sup |cα(P )z | ≤ (r λ) kP kR ≤ (r λs) kP k z∈B Y |α|=m Domains of convergence, Bohr radii and unconditionality 115

if R ⊂ sBX (without loss of generality λ ≥ 1). By theorem 2.58 and remark 2.63 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

X α m m sup |cα(P )z | ≤ c kP k ≤ (ct) kP kR z∈B Y |α|=m 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

X α m X −1 α m |cα(f)z | = r |cα(Pmf(0))(r z) | ≤ (rct) kPmf(0)kR |α|=m |α|=m

m ≤ (rct) kfkR by (2.6) and the Cauchy inequalities (0.13) and hence

∞ X α X X α |cα(f)z | = |c0(f)| + |cα(f)z |

(N) m=1 |α|=m α∈N0 ∞ X m ≤ |c0(f)| + kfkR (rct) < ∞ m=1

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 satisfying one of the conditions (1)–(4). Then there is a constant c > 0 such that for all m and n

n n X 0 1/2 1/(2m) X 0 (m+1)/(2m) ei Y 0 ≤ c m (log n) ei X0 (2.39) i=1 i=1 In particular, if X = Y then for all ε > 0 there is a constant c > 0 such that for all n n X 0 1+ε ei X0 ≤ c (log log n) (2.40) i=1

Proof. We use (2.21),

n n n X X 0 kid : Xn −→ `1 k = sup |zi| = ei X0 z∈B Xn i=1 i=1 116 2. Domains of convergence and (2.24) to estimate

n n m X 0 m X 0 ei Y 0 = ei Y 0 i=1 i=1

X 0 X m! α = |j|e m = z m j P (Yn) P (Yn) n α! j∈J (m,n) α∈N0 |α|=m √ r m m X m! α ≤ m! χM (id : P (Xn) −→ P (Yn)) ± z m P (Xn) n α! α∈N0 |α|=m p 3 3m−1 n n m−1 ≤ m! m 2 log n κm(Xn,Yn) kid : Xn −→ `1 kkid : Xn −→ `2 k n p 3 3m−1 X 0 (m+1)/2 ≤ c m! m 2 log n κm(Xn,Yn) ei X0 i=1

m m m This gives the ﬁrst claim since m! ≤ m and χM (id : P (Xn) −→ P (Yn)) ≤ cm by the assumption with a constant c independent of m and n. If X = Y then it follows n X m 1 0 m−1 m−1 ei X0 ≤ c m (log n) i=1 m So if ε > 0 we choose m = [log log n] such that m−1 < 1 + ε. Then

n 1 log log n X 0 1+ε log log n−2 1+ε log log n−2 ei X0 ≤ c (log log n) (log n) = c (log log n) e i=1 ≤ c0 (log log n)1+ε

Lemma 2.85. Let X be a symmetric Banach sequence space satisfying that for all ε > 0 there is a constant c > 0 such that for all n

1−ε λX (n) ≥ c n

Then X ⊂ `p for all p > 1.

Proof. Let p > 1 and choose ε > 0 with p(1 − ε) > 1. Then for x ∈ X we have n n 1−ε ∗ ∗ X ∗ X ∗ ∗ n xn ≤ cε λX (n) xn = cε xnei ≤ cε xi ei ≤ cε kx k = cε kxk i=1 i=1

∗ 1−ε ∗ Hence xn ≤ cε kxk 1/n , which implies that x and by symmetry also x is an element of `p.

We can now improve theorem 2.79. Domains of convergence, Bohr radii and unconditionality 117

Theorem 2.86. Let X be a symmetric Banach sequence space. If there is a bounded Reinhardt domain R ⊂ X and a closed subspace F(R) ⊂ H∞(R) containing P(X) such that dom F(R) is absorbant, then n λ (n) X (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) n 0 × P and λX (n) = i=1 ei X0 holds. For the last part apply lemma 2.85.

Now let R be a bounded Reinhardt domain in a symmetric Banach sequence space X. From the preceding section we know that if dom H∞(R) is absorbant then X ⊂ `1+ε for all ε > 0. In view of theorem 2.44 and example 2.46 (a) a natural question is the following: If dom H∞(R) absorbs `2−ε for all ε > 0, how close is X to `∞ ? Proposition 2.87. Let R be a bounded Reinhardt domain in a symmetric Banach sequence space X. Then the following are equivalent:

(1) dom H∞(R) absorbs `2−ε for all ε > 0.

(2) `p ⊂ X for all 1 ≤ p < ∞.

(3) `2−ε ⊂ X and `2−ε ∩ R ⊂ dom H∞(R) for all ε > 0.

(4) For every ε > 0 there is a λ > 0 such that λB`2−ε ⊂ dom H∞(R).

Proof. We start with (1) implies (2): Suppose (1) is true. Let ε > 0 and 1 1 p > 2 such that 2−ε + p = 1. Then by proposition 2.84

n n n 1 (m+1)/(2m) p X X 0 1/2 1/(2m) X 0 n = ei = e 0 ≤ c m (log n) e 0 `p i (`2−ε) i X i=1 i=1 i=1

and thus m m+1 m+1 m 2 1 1 1 m+1 p λ × (n) ≥ n X c log n m m 2 ε Now choose m = [log n] and n large enough such that m+1 p ≥ 1 − 2 and ε n 2 / log n ≥ 1. Then

1 1 1− ε e 1−ε λ × (n) ≥ e n 2 ≥ n X c log n c

× 0 Hence, given p < ∞, by lemma 2.85 it follows X ⊂ `p0 with p > 1 the dual 1 1 0 × exponent such that p + p0 = 1. Then by duality (recall (Xn) = (X )n from chapter 1)

n × n × kid : `p −→ Xnk = kid : (X )n −→ `p0 k ≤ kX ,→ `p0 k 118 2. Domains of convergence

for all n, and thus `p ⊂ X. Now suppose that (2) holds and we are going to prove (3). Let 0 < ε < 1 and p > 0 such that 1 = 1 + 1 . Using the fact that set inclusion of Banach ε 2−ε 2 pε sequence spaces implies continuity of the inclusion mapping we get from the assumption 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) ⊂ dom H∞(`pε ∩ R) ⊂ dom H∞(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 continuous. We choose t > 0 such that tBX ⊂ R and then λ := t/k`2−ε ,→ Xk satisﬁes λB`2−ε ⊂ `2−ε ∩ R ⊂ dom H∞(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 dom H∞(BX ) is absorbant, then for every ε > 0 there is a λ > 0 such that

× λ B`2−ε ⊂ dom H∞(BX )

× Proof. By theorem 2.79 we have X ⊂ `1+ε for all ε > 0 which implies `p ⊂ X × for 1 ≤ p < ∞, since (`q) = `q0 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

1 ≤ i ≤ m ei ⊗ fj , 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) basis 2 for, say, the projective tensor product E⊗˜ πF . In other words: N has to be ordered 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, 1) ← (3, 2) ← (3, 3) ··· ......

Let us write i1 ··· in for (i1, . . . , in). Then the order is

11 12 22 21 13 23 33 32 31 (3.2) 14 24 34 44 43 42 41 . .

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 a ˜ m simple way to π-tensor products ⊗π,i=1Ei of m Banach spaces for arbitrary m in [69]. 120 3. Tensor and s-tensor orders

One may also ask the same question for symmetric tensor products: Given E a Banach space with basis (en) and an s-tensor norm α, do the symmetrized tensors 1 X σ (e ) = e ⊗ · · · ⊗ e , j ∈ J (m) (3.3) m j m! jσ(1) jσ(m) σ∈Sm

˜ m,s m in some order form a basis for ⊗α E ? (Recall J (m) = {j ∈ N | j1 ≤ ... ≤ jm}.) This is of particular interest since certain symmetric tensor products can be identified with spaces of homogeneous polynomials. For ˜ m,s 0 m example ⊗εs E = Pappr(E) is the space of approximable polynomials in the 0 0 0 way that the functional σm(ej) = ej1 ⊗ · · · ⊗ ejm , j ∈ J (m) is identified with α Q∞ 0 αi the monomial z = i=1(ei) , αi = |{k | jk = i}|. 0 Thus if the coefficient functionals (en) with respect to the basis (en) of E form a basis of the dual E0, then the problem to find an order which makes 0 (N) (σm(ej))j∈J (m) a basis is the same as to order the multiindices α ∈ N0 such α m that the monomials z form a basis for Pappr(E). The latter was solved in [34].

In this chapter we investigate in a systematic way orders ν on Nm such that, given any tensor norm α and Banach spaces Ei with basis (eij)j∈N, the elements m m ⊗i=1eiji , j ∈ N ˜ m ordered by ν form a basis for ⊗α,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 a ˜ m,s Banach space E form a basis for the symmetric tensor product ⊗α E, α any s-tensor norm. The key is decompositions in intervals. We first define decomposition numbers for orders on Nm and J (m) and prove some important properties related to them. Using these numbers we then give strong necessary conditions of purely combinatorial nature which permit to check whether a given order is a tensor order (s-tensor order) and to generate examples in an easy way. Then we give three examples. The third one solves a problem raised in [69] (which was again posed in [44]). The next step is to relate the results to spaces of homogeneous polynomials and finally we prove that the monomial expansion series of any polynomial on a Banach space with basis always converges conditionally pointwise on the whole space. Very recently a preprint [44] was brought to our attention that deals with the same kind of problem we consider here, though not in a systematic way as we do it. The main aim of the paper [44] is to construct one example of what we call an s-tensor order here. They managed to do this going through full tensor products. The order they got is the same as was used in [34] for the special 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 the proof reduces to a few lines with better constants than in [44]. This chapter was worked out without knowledge of [44], but we have added some remarks concerning that paper in the right places.

Orders on Nm

During this chapter k, l, m, n are always natural numbers. For vectors of natural numbers we use the letters i and j.

m In this chapter we write Am for N or J (m). By an order on Am we mean a bijective mapping ν : N −→ Am. Important will be minimal decompositions of 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.

m I = [a, b] = {i ∈ N | a ≤ i ≤ b} = [a1, b1] × ... × [am, bm] where a, b ∈ Nm, a ≤ b or I = ∅. If not stated otherwise, “≤” stands for the natural partial order.

Deﬁnition 3.1. For A ⊂ Nm and r ⊂ Nm × Nm a partial order we deﬁne

n [ ρr(A) := min{n | A = [ai, bi]r} i=1 and n [ ρr,d(A) := min{n | A = [ai, bi]r disjoint union} i=1

We only use this deﬁnition with r being the natural partial order or r being a 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 disjoint intervals whose number can be controlled.

m n−1 Remark 3.2. Let I1,...,In be intervals in N . Then there is k ≤ (3m) m and there are disjoint intervals J1,...,Jk ⊂ N such that

n k [ [ Ii = Ji i=1 i=1

Proof. This is proved by induction on n. The case n = 1 is obvious. 122 3. Tensor and s-tensor orders

m For the induction step ﬁrst note that if I and J are intervals in N then J ∩{I can be written as a union of at most 2m disjoint intervals. This follows easily with induction since

m−1 {I = {I1 × N ∪ 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 k0 ≤ (3m)n−1 and disjoint m n k0 intervals J1,...,Jk0 ⊂ N such that ∪i=1Ii = ∪i=1Ji. It follows that

n+1 k0 [ [ Ii = In+1 ∪ (Ji ∩ {In+1) i=1 i=1 is a union of k ≤ 2m k0 + 1 ≤ (3m)n disjoint intervals.

Corollary 3.3. For all A ⊂ Nm

ρ(A)−1 ρ(A) ≤ ρd(A) ≤ (3m)

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 m+n [ A × B := A ×i B i∈J (n,m+1)

The following is immediate from the deﬁnitions.

Remark 3.5.

(1) For A ⊂ Nm, B ⊂ Nn and C ⊂ Nk we have

m+n m+n+k m+n+k n+k (A × B) × C = A × (B × C)

(2) If A ⊂ Nm and B ⊂ N then

m+n n n Sm+n(A × B ) = (SmA) × B 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 × · · · × Im is an interval in Nm+n. In particular if k ≤ l then l [ m+1 m+1 I × {r} = I × [k, l] (3.4) r=k can be written as the union of m + 1 intervals in Nm+1. m n Remark 3.6. Let I ⊂ N and J ⊂ N be intervals such that Ii ∩ Jj = ∅ for all i, j. Then m+n m + n ρ (I × J) ≤ d n

Proof. The assumption Ik ∩Jl = ∅ for all k, l assures that the intervals I ×i J, i ∈ J (n, m + 1) are pairwise disjoint. Thus

m+n m + n ρ (I × J) ≤ |J (n, m + 1)| = d n

m Notation 3.7. Let ν be an order on Am and r be a partial order on N . Then we deﬁne

 m supn ρr(ν[1, n]) if Am = N |ν|r := supn ρr(Smν[1, n]) if Am = J (m) and  m supn ρr,d(ν[1, n]) if Am = N |ν|r,d := 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 r m being the natural partial order on N . 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|−1 |ν| ≤ |ν|d ≤ (3m) (3.5)

This deﬁnition is motivated by the characterization of sum-preserving permutations π : N −→ N in the case m = 1 (then |π| = |π|d). It seems that the following result for X = K ﬁrst appeared in [1]. The charactarization is very natural and was rediscovered in [65, lemma 1]. For completeness we give a proof combining arguments from [65] and a nice idea of V. Ogranovich from 1985, which can be found in the book [47, p. 141]. 124 3. Tensor and s-tensor orders

Remark 3.8. Let X be a Banach space and π : N −→ N a permutation. Then the following are equivalent: P P (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

N P E = {x ∈ X | xn convergent in X} with norm n X kxk = sup xi X n i=1 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 kTπk ≥ |π|. For arbitrary n we have b [ π[1, n] = [ni, mi] (3.6) i=1 with natural numbers n1 ≤ m1 < n2 ≤ m2 < . . . < nb ≤ mb and b =

ρ(π[1, n]). Choose e ∈ X with kek = 1 and deﬁne x ∈ E by xmi := e, xmi+1 := −e for i = 1, . . . , b and xk := 0 otherwise. Then kxk = 1 and hence

n b mi X X X X kTπk ≥ kTπxk ≥ xπ(i) X = xi X = xk X i=1 i∈π[1,n] i=1 k=ni

= kb ekX = b

This implies kTπk ≥ |π| and hence (2). P 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 X xi − x < ε i=1

−1 Then we have for all n ≥ max π [1, n0] that [1, n0] ⊂ π[1, n] and hence we can write π[1, n] as a union of intervals as in (3.6) with n1 = 1 and m1 ≥ n0. It follows

n m1 b mi X X X X xπ(i) − x ≤ xπ(i) − x + xk ≤ (2|π| − 1)ε

i=1 i=1 i=2 k=ni Orders on Nm 125

The ﬁrst characterization of convergence-preserving permutations was given by Kronrod [50, 51] and at the same time by Levi [55]. They both used the same condition, which can be easily seen to be equivalent to the criterion given above (see for example [71]).

m Remark 3.9. Let ν be an order on N . Then |ν|d ≥ m.

m Proof. Let Mm := [1m, 2m] = {i ∈ N | kik∞ ≤ 2} (see (3.12) for the notation). Then we have

m |Mm ∩ I| ∈ {0, 1, 2, 4,..., 2 } (3.7) for every interval I in Nm. This is obvious for m = 1 and if it is true for some m deﬁne Mm+1,k := {i ∈ Mm+1 | im+1 = k} for k = 1, 2. Then, given an interval I ⊂ Nm+1,

m |Mm+1 ∩ I| = |Mm+1,k ∩ I| = |Mm ∩ I| ∈ {0, 1, 2, 4,..., 2 } for some k ∈ {1, 2} or

m+1 |Mm+1 ∩ I| = 2|Mm+1,k ∩ I| = 2|Mm ∩ I| ∈ {0, 2, 4,..., 2 }

0 where k ∈ {1, 2} = {k, k } is chosen such that |Mm+1,k ∩ I| ≥ |Mm+1,k0 ∩ I|, since i, j ∈ I with im+1 = 1, jm+1 = 2 implies (i, 2), (j, 1) ∈ I. Here we have used notation (3.11), see page 134.

¿From (3.7) it follows that if I1,...,In are pairwise disjoint intervals with

n [ m |Mm ∩ Ii| = 2 − 1 i=1 then it must be n ≥ m. (Check by induction that if m−1 Pn m a1, . . . , an ∈ {1, 2, 4,..., 2 } with i=1 ai = 2 − 1 then there are natural numbers 1 = n1 < n2 < . . . < nm = n + 1 such that

nk+1−1 X k−1 ai = 2

i=nk for k = 1, . . . , m.) We choose n with m |ν[1, n] ∩ Mm| = 2 − 1 to ﬁnish the proof.

−1 If ν1 and ν2 are orders on Am, i.e. bijective mappings N −→ Am, then ν2 ν1 is a permutation on N. 126 3. Tensor and s-tensor orders

Notation 3.10. Let ν1 and ν2 be orders on Am. We deﬁne

−1 ν1 ≺ ν2 :⇐⇒ |ν2 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 if ν1 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 can always construct an order ν2 with

ν2 ≺ ν1 6≺ ν2

This is due to the fact that the sum-preserving permutations on N do form a semigroup, but not a group (see [55, p. 582] or [65, theorem 1]). Thus we can take π : N −→ N bijective with |π| < ∞ and |π−1| = ∞ and deﬁne ν2 := 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 increasing 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 − sup{i ∈ N | π(ni) ≤ nk} for all k ≥ k0.

Proof. First observe that

ak := sup{i ∈ N | π(ni) − π(n1) ≤ nk} ≤ sup{i ∈ 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 to Ss prove. Otherwise we choose intervals Ii ⊂ N with π[1, nk] = i=1 Ii and max Ii < min Ii+1 for i = 1, . . . , s − 1. Using that (π(nr))r is increasing we can 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

s X nk = |π[1, nk]| = |Ii| ≥ π(nrs−(s−1)) − π(n1) i=1

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 ⊂ N and ϕ : N −→ N bijective with

sup n − ϕ(n) = ∞ n∈I

−1 −1 as well as a sequence (in) in Am such that ν1 (in) ≤ ν2 (iϕ(n)) for inﬁnitely many n ∈ I. Then

ν1 6≺ ν2

−1 −1 Proof. Deﬁne nk := ν1 (ik) and π := ν2 ν1. Then by assumption we have π(nϕ(k)) ≥ nk and hence ak := sup{i ∈ N | π(ni) ≤ nk} ≤ ϕ(k) or k − ak ≥ k − ϕ(k) for inﬁnitely many k ∈ I. It follows from the assumption on ϕ that the sequence (k − ak)k must be unbounded and we can apply lemma 3.12 to get |π| = ∞.

If ν is an order on Nm then there is a unique injective mapping ι : N ,→ N with 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 the elements 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 will often write [i, j]s and i ≤s j whenever it is clear from the context where the order is coming from. 128 3. Tensor and s-tensor orders

Theorem 3.14. Let ν be an order on J (m) with kνk < ∞, where k · k = | · | or | · |d. Then there is an order ω on Nm with (1) kωk ≤ kνk + 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

k k [ [ Smν[1, k] = Smν(r) = ω[nr−1, nr − 1] = ω[1, nk − 1] r=1 r=1 for all k gives |ν|ω = 1.

Further given n ∈ N there is a unique k with nk−1 ≤ n < nk and thus

ρˆ(ω[1, n]) ≤ ρˆ(ω[1, nk−1 − 1]) + nk − nk−1 =ρ ˆ(Smν[1, k − 1]) + |Smν(k)| ≤ |ν| + m! withρ ˆ = ρ or ρd, implying kωk ≤ kνk + m!.

m For restrictions of orders ν on N 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 be expressed using the following notation (see also [47, p. 141] for the case m = 1).

m Notation 3.15. Let r be a partial order on N andρ ˆ = ρr or ρr,d. For an order ν on Am we deﬁne

 m supk≤l ρˆ(ν[k, l]) if Am = N ρˆ(ν) := supk≤l ρˆ(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 a total 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 −1 |ν1| ≤ ρ(ν2)|ν2 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 ρˆ(ν) < ∞, where ρˆ = ρ or ρd. Then there is an order ω on Nm with (1) ρˆ(ω) ≤ ρˆ(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 and π := ν−1ω. Then

|ν|s|ν = sup ρ(π[1, nk]) , ρν(ν|s) = sup ρ(π[nl, nk]) k l≤k and ρ(π[1, n]) ≤ ρ(π[1, nk]) + m! if nk ≤ n < nk+1 (here (nk) is the sequence of natural numbers connected with 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!) 130 3. Tensor and s-tensor orders Orders for tensor products

Deﬁnition 3.20.

(1) An order ν on Nm is called tensor order if whenever we take Banach spaces Ei with basis (eij)j∈N, i = 1, . . . , m and a tensor norm α of order ˜ m m m then (ˆeν(n))n∈N is a basis for ⊗α,i=1Ei, where we denotee ˆj := ⊗i=1eiji for j ∈ Nm. (2) An order ν on J (m) is called s-tensor order if for every Banach space E with basis (en) and every s-tensor norm α (of order m) the sequence ˜ m,s (σm(eν(n)))n∈N is a basis of ⊗α E.

If ν1 and ν2 are orders on Am with ν1 ≺ ν2 and ν2 is a tensor order (s-tensor order) then by deﬁnition of the relation ≺ and remark 3.8 it is immediate that also ν1 is a tensor order (s-tensor order). In particular, if ν1 and ν2 are equivalent, 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, . . . , m and α is a tensor norm of order m then we have

m b(ˆeν(n)) ≤ |ν|d 2 b(e1j) ··· b(emj) for the basis constants.

0 Proof. Let eij be the coeﬃcient functionals and

n X 0 Pin = eij ⊗ eij j=1 the n-th projection of the basis (e ) in E . Denotee ˆ0 := ⊗m e0 ∈ ij j∈N i j i=1 iji ˜ m 0 m (⊗α,i=1Ei) for j ∈ N . Then

0 eˆi(ˆej) = δij

m 0 ˜ m for i, j ∈ N , i.e. ((ˆeν(n)), (ˆeν(n)))n∈N is a biorthogonal system in ⊗α,i=1Ei. m m Hence it is enough to show that span{ eˆj | j ∈ N } is dense in ⊗α,i=1Ei and that the projections

n X 0 Pn = eˆν(k) ⊗ eˆν(k) , n ∈ N k=1 Orders for tensor products 131 are uniformly bounded (see e.g. [56]). m To see the ﬁrst, take z ∈ ⊗i=1Ei. By linearity we can assume z = x1⊗· · ·⊗xm. Then

zn := P1n x1 ⊗ · · · ⊗ Pmn xm n X 0 0 m = hx1, e1j1 i · · · hxm, emjm ie1j1 ⊗ · · · ⊗ em jm ∈ span{ eˆj | j ∈ N } j1,...,jm=1 converges to z because the tensor product mapping is continuous relative to any tensor norm. For the second claim the main observation is that by the m-linearity of the 0 tensor product mapping an elemente ˆj ⊗ eˆj can be written as the tensor product 0 0 (e1j1 ⊗ e1j1 ) ⊗ ... ⊗ (emjm ⊗ emjm ) of one dimensional projections (check with elementary tensors). r Let n ∈ N. By the assumption we have a decomposition ν[1, n] = ∪i=1[ai, bi] in r ≤ |ν|d disjoint intervals, hence n r X 0 X X 0 Pn = eˆν(k) ⊗ eˆν(k) = eˆj ⊗ eˆj

k=1 i=1 j∈[ai,bi]

r bi(1) bi(m) X X X 0 0 = ··· (e1j1 ⊗ e1j1 ) ⊗ ... ⊗ (emjm ⊗ emjm ) i=1 j1=ai(1) jm=ai(m)

r bi(1) bi(m) X X 0 X 0 = e1j ⊗ e1j ⊗ · · · ⊗ emj ⊗ emj

i=1 j=ai(1) j=ai(m)

r X = (P1bi(1) − P1ai(1)−1) ⊗ · · · ⊗ (Pmbi(m) − Pmai(m)−1) i=1 By the metric mapping property of α this implies

m kPnk ≤ |ν|d 2 b(e1j) ··· b(emj) with b(eij) = supnkPink 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 of order m then we have mm b(σ (e )) ≤ |ν| 2 (2 b(e ))m m ν(n) d m! n for the basis constants. 132 3. Tensor and s-tensor orders

m m Proof. We have that span{ei | i ∈ N } is dense in ⊗α E and so is m span{σm(ej) | j ∈ J (m)} = σm(span{ei | i ∈ N }) m,s m ˜ m,s in ⊗αs E = σm(⊗α E) (and hence in ⊗αs E).

For j ∈ J (m) let µm(j) be the number of all permutations σ ∈ Sm with σj = j and σ (e0 ) := 1 P e0 ⊗ · · · ⊗ e0 . Then we have for m j m! σ∈Sm jσ(1) jσ(m) i, j ∈ J (m) 1 X hσ (e ), σ (e0 )i = he , e0 i · · · he , e0 i m i m j m! iσ(1) j1 iσ(m) jm σ∈Sm  µm(i)/m! i = j = 0 otherwise

j∈Smν|s[1,n] j∈Smν|s[1,n] and arguing in the same way as in the proof of theorem 3.21 we obtain

r X 0 X X 0 ej ⊗ ej = ej ⊗ ej

j∈Smν|s[1,n] i=1 j∈[ai,bi]

r X = (Pbi(1) − Pai(1)−1) ⊗ · · · ⊗ (Pbi(m) − Pai(m)−1) i=1 with r ≤ |ν|d, where n X 0 Pn = ej ⊗ ej j=1 denotes the n-th projection of the basis (ej)j∈N in E. Hence m kPs,nk ≤ |ν|d kσmk kιmk (2 supkPnk) n

Recall that a basis in a Banach space is called shrinking if the coordinate functionals 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 an m order on N . Let αs be an s-tensor norm of order m and α an equivalent ˜ m tensor norm such that (eν(n))n∈N is a basis for ⊗α E. If |ν| | < ∞ then (σ (e )) is a basis for ⊗˜ m,sE, which is shrinking s ν m ν|s(n) n∈N αs whenever (eν(n))n∈N is. Moreover

b(σm(eν|s(n))) ≤ |ν|s|ν 2b(eν(n)) kσmkkιmk ˜ m,s ˜ m ˜ m where ιm : ⊗αs E −→ ⊗α E is the natural injection and σm : ⊗α E −→ ˜ m,s ⊗αs E the natural projection.

m For an order ν on N satisfying |ν|s|ν < ∞ we will say that it has the restriction property.

˜ m,s Proof. We have seen in the proof of theorem 3.22 that span{σm(ej)} = ⊗αs E and we only have to show the ﬁniteness of the basis constant. For the n-th projection we have

j∈Smν|s[1,n] i=1 j∈ν[ni,mi] with n1 ≤ m1 < n2 ≤ m2 < . . . < nt ≤ mt and t ≤ |ν|s|, and it follows

t t X X 0 X Ps,n = σm ◦ ej ⊗ ej ◦ ιm = σm ◦ Pmi − Pni−1 ◦ ιm

i=1 j∈ν[ni,mi] i=1

Pk 0 with Pk = l=1 eν(l) ⊗ eν(l) the k-th basis projection of the basis (eν(l))l∈N in ˜ m ⊗α E, P0 := 0. We conclude

kPs,nk ≤ kσmk|ν|s|ν 2b(eν(n)) kιmk

Finally, if the basis (e ) is even shrinking, then, given x0 ∈ (⊗˜ m,sE)0 ν(n) n∈N αs 0 0 ˜ m 0 and ε > 0, we have y := x ◦ σm ∈ (⊗α E) and hence we can ﬁnd n0 such that 0 ky ◦ (id −Pn)k < ε for all n ≥ n0. Choose k0 such that

Smν|s[1, k0] ⊃ ν[1, n0]

Then for all n ≥ k0 an optimal decomposition

t [ Smν|s[1, n] = ν[ni, mi] i=1 134 3. Tensor and s-tensor orders

with 1 = n1 ≤ m1 < n2 ≤ m2 < . . . < nt ≤ mt and t = ρν(Smν|s[1, n]) satisﬁes m1 ≥ n0. Hence from the representation

t X Pn,s = σm ◦ Pm1 + Pmi − Pni−1 ◦ ιm i=2 it follows

t 0 0 X x ◦ (ids −Ps,n) = x ◦ σm ◦ id −Pm1 − ( Pmi − Pni−1) ◦ ιm i=2

t 0 X 0 = y ◦ (id −Pm1 ) − y ◦ (Pmi − Pni−1) ◦ ιm i=2 and thus 0 kx ◦ (ids −Ps,n)k < kιmk (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-tensor norm αs of order m with equivalent tensor norm α, then

b(σm(eν|s(n))) ≤ |ν|s|ν 2b(eν(n)) kσmkkιmk ˜ m and if the basis (eν(n))n∈N of ⊗α E is shrinking, then also (σm(eν|s(n)))n∈N is ˜ m,s a shrinking basis of ⊗αs E.

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}. m Our examples will be orders νm on Am = N or J (m) deﬁned for every m. ν1 will always be the natural order on N. To compare elements with respect to these orders we write i ≤νm j or i ≤ν j or simply i ≤ j if the necessary informations are clear from the context. In this case we don’t m −1 make a diﬀerence between an element j ∈ N and its position νm (j), which allows to compare vectors of arbitrary length. For example j ≤ j means −1 −1 −1 νm−1(j) ≤ νm (j) or we can write jm ≤ j, i.e. jm ≤ νm (j) or νm(jm) ≤ j if you want. This will be very convenient in example 3.

m Writing vectors i = (i1, . . . , im) ∈ N we leave brackets and commas away as often as possible as is done in (3.2), and we use the notation km for the vector of length m with all entries equal to the natural number k. For example

m 1nkm−n = (1,..., 1, k, . . . , k) ∈ N (3.12) | {z } | {z } n m−n

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 order on 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 Lamadrid order (3.2) restricted to J (2).

Remark 3.26. The order νm on J (m) deﬁned in 3.25 is an s-tensor order with m |νm|d ≤ 2

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)

m+1 Sm+1[1m+1, j]s = [1m+1, (jm+1 − 1)m+1] ∪ (Sm[1m, j]s) × {jm+1} (3.13)

(This already implies that νm is an s-tensor order with |νm| ≤ (m + 1)! .) 136 3. Tensor and s-tensor orders

Now ﬁx m. Given j ∈ J (m), j 6= 1m we deﬁne

In := [1m−n, (jm−n − 1)m−n] for n = 0, . . . , m0 := max{r ∈ N0 | jm−r > 1} and ( [1m−m0−1, 1m−m0−1] if m0 < m − 1 Im0+1 := ∅ if m0 = m − 1

Then by (3.13) and remark 3.5 (1) we obtain

m +1 [0 m Sm[1m, j]s = I0 ∪ In × {(jm−n+1, . . . , jm)} (3.14) n=1 The sets on the right side are pairwise disjoint. This is due to the fact that m a vector from In × {(jm−n+1, . . . , jm)} has no component equal to jm−n.

Using remark 3.6 it follows (ρd(∅) := 0)

m +1 m X0 m X m ρ (S [1 , j] ) ≤ 1 + ρ (I × {(j , . . . , j )}) ≤ = 2m d m m s d n m−n+1 m n n=1 n=0

Thus if E is a Banach space with normalized basis (en) and α an s-tensor ˜ m,s norm then the basis constant Cm of the basis (σm(eν(n))) in ⊗α E can be estimated by mm 4m 2 m! (theorem 3.22). This is better than [44] where they obtain an estimate worse than m−1 X m C ≤ C m n n n=0 √ m But then supm Cm need not be ﬁnite. Using the equality

m+1 Sm+1[i, j]s = (Sm[i, (im+1)m]s) × {im+1}

m+1 ∪ [1m, (jm+1 − 1)m] × [im+1 + 1, jm+1 − 1]

m+1 ∪ (Sm[1m, j]s) × {jm+1}

m we can get ρ(νm) ≤ m!3 as a rough estimate, but this can be considerably improved. Examples 137

Remark 3.27. The order νm on J (m) deﬁned in 3.25 satisﬁes

m ρd(νm) ≤ (4 + m!) 2m (3m − 1)

Proof. Let ωm be the order 3.28 deﬁned in example 2. Then ωm|s = νm. We use remarks 3.16, 3.19, corollary 3.30 and theorem 3.31 to estimate m ρ (ν ) ≤ ρ (ν ) ρ (ω ) ≤ 4(|ω | | + m!) (3m − 1) d m ωm m d m m s ωm 2 ≤ (4m + m!) 2m (3m − 1)

By theorems 3.14 and 3.18 we can extend the order 3.25 to a tensor orderν ˜m m m with |ν˜m|d ≤ 2 + m! and ρd(˜νm) ≤ (4 + m!) 2m (3m − 1) + 2m!. The order ν˜m could be constructed in the way that every element of J (m) is followed by all its permutations in some order (see the proof of theorem 3.14). In the case 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 (3.15) 233 323 332 333 114 141 411 124 214 142 241 412 421 224 242 422 134 314 143 341 413 431 . .

This shows that the extension theorems 3.14 and 3.18 actually are very strong results (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 are symmetric. Example 2

For this example we use the following notation: For i ∈ Nm we deﬁne 138 3. Tensor and s-tensor orders

µ(i) := max{k | ik = kik∞ } ◦ 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) kik∞ < kjk∞

or (2) kik∞ = kjk∞ and (a) µ(i) < µ(j) or (b) µ(i) = µ(j) and i◦ ≤ j◦

In the case m = 2 we get

11 21 12 22 31 32 13 23 33 (3.16) 41 42 43 14 24 34 44 . . which is not far away from the order of Gelbaum and Gil de Lamadrid. It uses the same principle of successively “walking down the inner sides of the upper left hand blocks”. Let’s have a short look at the situation m = 3. Examples 139

111

211

121 221

112 212 122 222

311 321 312 322

131 231 132 232 331 332

113 213 123 223 (3.17) 313 323 133 233 333

411 421 412 422 431 432 413 423 433

141 241 142 242 341 342 143 243 343 441 442 443

114 214 124 224 314 324 134 234 334 414 424 434 144 244 344 444

. .

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 example 1, obtained in such a way that every element j ∈ J (m) is followed by all its permutations (as for example in (3.15)) is not equivalent to the order νm 140 3. Tensor and s-tensor orders 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.

m Theorem 3.29. The order νm on N 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

m m I (k) := {j ∈ N | kjk∞ ≤ k} and m m Ir (k) := {j ∈ N | kjk∞ = k , µ(j) = r} m m m for r = 1, . . . , m. By the deﬁnition of the order I (k), I1 (k+1),...,Im (k+1) are consecutive νm-intervals, and we have

m I (k) = [1m, km] = [1m, km]νm and m Ir (k + 1) = [(1r−1, k + 1, 1m−r), ((k + 1)r, km−r)]νm as well as

r m [ m I (k) ∪ Ii (k + 1) = [1m, ((k + 1)r, km−r)] = [1m, ((k + 1)r, km−r)]νm i=1 for all k, m ∈ N and r = 0, . . . , m.

Further deﬁne for a νm-interval I = νm[k, n]

∗ ρ (I) := sup ρd(νm[k, l]) k≤l≤n

Then ∗ m |νm|d = sup ρ (I (k)) k and it will be enough to show that

∗ m ρ (I (k)) ≤ m Am(k) holds for all natural numbers m and k. We prove by induction that

∀ m : Am(k) A(k) Examples 141

m is true for all k. We start with I (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 to prove 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

m+1 m+1 m+1 [ m+1 I (k + 1) = I (k) ∪ Ir (k + 1) r=1 with ρ∗(Im+1(k)) ≤ m + 1 by the induction hypothesis A(k). Further

◦ m+1 m m I1 (k + 1) = {j ∈ N | kjk∞ ≤ k} = I (k) ◦ ◦ (here A for a subset A ⊂ Nm+1 of course means the set {j | j ∈ A}) and ◦ m+1 m m Ir (k + 1) = I (k) ∪ {j ∈ N | kjk∞ = k + 1 , µ(j) ≤ r − 1}

r−1 m [ m = I (k) ∪ Ii (k + 1) i=1

= [1m, ((k + 1)r−1, km−r+1)]νm

m for r = 2, . . . , m + 1 are initial νm-intervals contained in I (k + 1). m+1 Now observe that for a subset A ⊂ N with |µ(A)| = 1 and kjk∞ = l for some l ∈ N and all j ∈ A the equality ◦ ρd(A) = ρd(A ) holds. This implies

◦ ∗ m+1 ∗ m+1 ∗ m ρ (Ir (k + 1)) = ρ (Ir (k + 1) ) ≤ ρ (I (k + 1)) ≤ m for r = 1, . . . , m + 1 by the induction hypothesis Am(k + 1). Thus for all η = 0, . . . , m + 1 we have

η m+1 m+1 [ m+1 I (k + 1) = I (k) ∪ Ir (k + 1) = [1m+1, ((k + 1)η, km+1−η)] r=1 with ρ∗(Im+1(k)) ≤ m + 1 142 3. Tensor and s-tensor orders and ∗ m+1 ρ (Ir (k + 1)) ≤ m for r = 1, . . . , m + 1. But then it must be

ρ∗(Im+1(k + 1)) ≤ m + 1

m Corollary 3.30. For the order νm on N deﬁned in 3.28 we have m ρ (ν ) ≤ (3m − 1) d m 2

m 0 Proof. Let i, j ∈ N with i ≤νm j and m0 = µ(i), m0 = µ(j), k = kik∞, 0 k = kjk∞. Then using the notation of the preceding proof we have

[i, j]νm = [i, (km0 , (k − 1)m−m0 )]νm

0 0 m k −1 m m0−1 [ m [ [ m [ m 0 ∪ Ir (k) ∪ Ir (l) ∪ Ir (k ) (3.18) r=m0+1 l=k+1 r=1 r=1

0 ∪ [(1 0 , k , 1 0 ), j] m0−1 m−m0 νm For any natural numbers 1 < u < u0 and 1 ≤ r ≤ m we have

m 0 0 r−1 0 0 m−r Ir (u ) = [1, u ] × {u } × [1, u − 1] = [1, u0]r−1 × {u0} × [1, u − 1]m−r

m [ ∪ [1, u0]r−1 × {u0} × [1, u0 − 1]n−r−1 × [u, u0 − 1] × [1, u − 1]m−n n=r+1

¿From this it can be deduced that

0 0 k −1 m m0−1 [ [ m [ m 0 Ir (l) ∪ Ir (k ) l=k+1 r=1 r=1

m0 −1 [0 = [1, k0]n−1 × [k + 1, k0] × [1, k]m−n n=1 m [ 0 m0 −1 0 n−m0 0 m−n ∪ [1, k ] 0 × [1, k − 1] 0 × [k + 1, k − 1] × [1, k] 0 n=m0

Further we have

◦ [i, (km0 , (k − 1)m−m0 )]νm = [i , (km0−1, (k − 1)m−m0 )]νm−1 ×m0 {k} Examples 143 and 0 ◦ 0 [(1 0 , k , 1 0 ), j] = [1 , j ] × 0 {k } m0−1 m−m0 νm m−1 νm−1 m0

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 m X m ρ (ν ) ≤ (3n − 2) = (3m − 1) d m 2 n=1

We already know from example 1 that the restriction νm|s of the order deﬁned in 3.28 is an s-tensor order. We will now prove that it even has the restriction property.

m Theorem 3.31. The order νm on N deﬁned in 3.28 satisﬁes

m |νm|s|νm ≤ 4

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 as there) we have

m +1 [0 m Sm[1m, j]s = I0 ∪ In × {(jm−n+1, . . . , jm)} n=1

Since I0 = [1m, (jm − 1)m] = [1m, (jm − 1)m]νm and m [ In × {(jm−n+1, . . . , jm)} = [1m−n ×i bn, (jm−n − 1)m−n ×i bn]νm i∈J (n,m) with bn = (jm−n+1, . . . , jm) for n = 1, . . . , m0 (and m [ Im0+1 × {(jm−m0 , . . . , jm)} = {1m−m0−1 ×i (jm−m0 , . . . , jm)}

i∈J (m0+1,m) if m0 < m − 1) it follows

m +1 m +1 X0 X0 m + n − 1 ρ (S [1 , j] ) ≤ 1 + |J (n, m)| = 1 + νm m m s n n=1 n=1 m X 2m ≤ = 4m n n=0 144 3. Tensor and s-tensor orders

Example 3 We have seen that the orders of example 1 and 2 are related by ≺. We will 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 the restriction property.

In [69] an extension of the order (3.2), say ν2, of Gelbaum and Gil de Lamadrid 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. 111 112 122 121 113 123 223 222 221 114 124 224 214 213 212 211 (3.20) 115 125 225 215 135 134 133 132 131 . .

m Thus if (jk) are the elements of N ordered by νm and we order (jk, n)k,n∈N by m+1 ν2 then we obtain the same sequence as if we order N by νm+1. It is then ˜ m ˜ ˜ m+1 immediate from the associativity (⊗π,i=1Ei)⊗πEm+1 = ⊗π,i=1Ei of the π- tensor product that this order works for π-tensor products of arbitrary length and moreover for every m-fold α-tensor product, whenever α is associative. However there are tensor norms which are not associative. Actually most of the 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

2 For the order ν2 on N we have |ν2| = |ν2|d = 2 (see (3.2)). Examples 145

m Theorem 3.33. The order νm on N deﬁned in 3.32 is a tensor order with

m−2 |νm| ≤ 2 · 6 and m−1 ρ(νm) ≤ 6 for m ≥ 2.

Proof. For m ≥ 2 we can write

t t t [ [ [i νm+1[1, n] = νm[ni, mi] × [ki, li] = [aij, bij] × [ki, li]

i=1 i=1 ri=1 with t ≤ |ν2| = 2 and ti ≤ ρ(νm), where ni ≤ mi, ki ≤ li are natural numbers m and aij ≤ bij are elements of N (use (3.19)). This implies |νm+1| ≤ 2 ρ(νm). The same argument gives

ρ(νm+1) ≤ ρ(ν2)ρ(νm)

m−1 m−2 for m ≥ 2, hence by induction ρ(νm) ≤ ρ(ν2) and |νm| ≤ 2 ρ(ν2) for all 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 ∈ 2 N with i ≤ν2 j then there are unique natural numbers k ≤ n such that i ∈ 1, k and j ∈ 1, n . k 1 ν2 n 1 ν2 If k = n then  i1 j1 1 n  , if ≤2 i, j ≤2  n n n n  n n n n [i, j]ν2 = j , i if n ≤2 i, j ≤2 1  2 1 i1 n n n n  , ∪ , if i ≤2 ≤2 j n n j2 n n If k < n then

k Sn−1 1 r 1 [i, j]ν = i, ∪ , ∪ , j 2 1 ν2 r=k+1 r 1 ν2 n ν2

By the case that we have just considered, the ﬁrst and the last ν2-interval in this decomposition can be represented by a union of maximal two intervals in N2 (with the natural partial order). So it will be enough to prove that for all natural numbers k ≤ n

Sn 1, r = 1, n ∪ k, n (3.21) r=k r 1 ν2 k n 1 k−1 We ﬁx k and proceed by induction on n. For any natural number n we have

1, n = 1, n ∪ n, n n 1 ν2 n n 1 n−1 146 3. Tensor and s-tensor orders

(see (3.2)). Suppose now that (3.21) is true for some n ≥ k. We have

1 n+1 1 n n+1 n+1 1 n+1 k , n+1 = k , n ∪ k , n ∪ n+1 , n+1 and k n+1 k n n+1 n+1 1 , k−1 = 1 , k−1 ∪ 1 , k−1 Hence 1 n+1 k n+1 k , n+1 ∪ 1 , k−1 1 n k n 1 n+1 n+1 n+1 = k , n ∪ 1 , k−1 ∪ n+1 , n+1 ∪ 1 , n = Sn 1, r ∪ 1 , n+1 r=k r 1 ν2 n+1 1 ν2 = Sn+1 1, r r=k r 1 ν2

m Remark 3.34. The order νm on N deﬁned in 3.32 does not have the restriction property for m ≥ 3.

−1 2 Proof. Let p := (νm|s) . Then p(1m−1, k) = 1 + (k − 1) and p(k, 1m−1) = k2(m−1). It follows

2m−3 3 p(n + 1, 1m−1) − p(n, 1m−1) ≥ n ≥ n ≥ m! p(1m−1, k)

> |Sm[1m, (1m−1, k)]νm|s |

2 1/3 for all n ≥ m!(1 + (k − 1) ) =: ck. Thus if k − n ≥ ck and Sr m Sm[1m, (1m−1, k)]νm|s = i=1[ai, bi]νm with ai, bi ∈ N , then the elements

(k − l + 1, 1m−1) ∈ Sm[1m, (1m−1, k)]νm|s must be in diﬀerent intervals for l = 1, . . . , n + 1 since the distance of two diﬀerent elements’ position exceeds the number of elements of the set Sm[1m, (1m−1, k)]νm|s . This implies

2 k→∞ ρν(Smνm|s[1, 1 + (k − 1) ]) ≥ [k − ck] + 1 ≥ k − ck −−−→∞ where [k − ck] means the biggest integer not exceeding k − ck.

m Remark 3.35. The order νm on N deﬁned in 3.32 and its restriction νm|s on J (m) are essentially diﬀerent to the orders given in example 1 and 2 if m ≥ 3.

m Proof. Let ω1,m be any extension to N of the order 3.25 deﬁned in example 1 such that each element j ∈ J (m) is followed by all its permutations (see for example (3.15) in the case m = 3). Further let ω2,m be the order 3.28 on Nm given in example 2.

Then ω1|s = ω2|s is the order deﬁned in 3.25 of example 1 and it is not hard to check that Examples 147

−1 m ω1 (1m−1k) = (k − 1) + 1 −1 m ω1 (km) = k −1 m m−1 ω2 (1m−1k) = k − k + 1 −1 m ω2 (km) = k −1 Pk −1 (ω2|s) (km) = i=1(ω2|s) (im−1) −1 −1 (ω2|s) (1m−1k) = (ω2|s) ((k − 1)m) + 1

Thus for ω = ωi, i = 1, 2 we have

−1 −1 m ω (1m−1k) ∼ ω (km) ∼ k −1 −1 m (ω|s) (1m−1k) ∼ (ω|s) (km) ∼ k asymptotically in k. On the other hand

−1 2 ν (1m−1k) = (k − 1) + 1 −1 −1 Pν (km−1) −1 2 2m−1 ν (km) = i=1 (2i−1)−(k−1) = (ν (km−1)) −(k−1) ∼ k and for the restriction deﬁne

M(km) := [1m, km]ν|s

M(km) := [1m, km]ν \ [1m, km]ν|s

−1 and A(km) := |M(km)|. Then (ν|s) (km) can be written as

−1 −1 (ν|s) (km) = ν (km) − A(km) and since

−1 X A(km) = A(km−1) ν (km−1) + (kjk∞ − 1)

j∈M(km−1)\{km−1}

−1 −1 −1 ≤ A(km−1) ν (km−1) + (ν|s) (km−1) · ν (km−2) it follows

−1 −1 −1 −1 (ν|s) (km) ≥ (ν (km−1) − ν (km−2)) (ν|s) (km−1) − (k − 1)

2m−2 −1 2m−1 k (ν|s) (km−1) k

−1 −1 We have (ν|s) (km) ≤ ν (km) and hence

−1 2m−1 (ν|s) (km) ∼ k

−1 2 In a similar way we obtain (ν|s) (1m−1k) ∼ k .

Apply corollary 3.13 to see that ωi,m 6≺ νm 6≺ ωi,m and ωi,m|s 6≺ νm|s 6≺ ωi,m|s for i = 1, 2 and m ≥ 3. 148 3. Tensor and s-tensor orders

We will now prove that the restriction of νm is an s-tensor order (theorem 3.56). This needs some preparations. The following results should bring some light into the structure of the order, as far as we need it for the proof of 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 respect to 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 and if 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 0 0 0 0 0 0 0 0 (i) if i , j ∈ J (m) with i < j then (i , im) < (j , jm)

Proof. (a) If j ≤ jm+1 then i ≤ j implies

im+1 < jm+1 and i < jm+1 (3.22) or im+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) or jm+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 deﬁnition 3.32 (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 by deﬁnition 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 = j since 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 which implies (1m, r) ≤ i by deﬁnition 3.32 (1)(b). If i > im+1 then im+1 ≤ r = i and 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 obtain r = 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. 2 (d) From (c) we have j ≥ (1m, m(j)) = (m(j) − 1) + 1 ≥ m(j), where the last 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 implies j = 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 have m(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 we must be in the second case and hence im = jm ≤ jm+1 ≤ im+1 = im gives im+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 or i = 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 = j and i ≤ m(i) ≤ m(j) = j which implies i ≤ j by deﬁnition 3.32 (2)(b). 150 3. Tensor and s-tensor orders

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. 0 0 0 0 0 0 (i) By (e) we have m(i , im) = i < j = m(j , jm) which by (b) implies 0 0 0 0 (i , im) < (j , jm). Remark 3.38. Let i ∈ Nm and j = min{i < j0 | j0 ∈ J (m)}. Then im ≤ jm + 1 and jm + 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 definition 0 m 0 0 3.32. Thus if we define j ∈ N by jm := jm + 1 and jn := jn for n < m 0 0 0 0 then jm < im ≤ j = j and i ≤ j = j , implying i < j (definition 3.32 (2)(b); 0 0 0 0 i = j cannot be since jm < im). Further we have jm < jm < j and j = j which implies j0 < j, and hence i < j0 < j with j0 ∈ J (m), since j ∈ J (m), 0 0 jn = jn for n < m and jm ≤ jm. 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 then using lemma 3.37(a) it follows

2 jm ≤ m(j) ≤ m(1m−1, m(i) + 1) = m(i) + 1 < (m(i) − 1) + 1

= (1m−1, m(i)) ≤ i

If i = 1m−221 then j = 1m3 and jm + 1 = 4 = i. (0) Notation 3.39. For j ∈ J (m) we deﬁne j := 1m and

(n) j := max{i ≤s j | in = ... = im} Remark 3.40. Let j ∈ J (m + 1) with j 6= j(m). Then

j(m) = max{i ∈ 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 hence 0 m(i, im) < m(j) which implies i := (i, im) < j by lemma 3.37(b) and thus 0 (m) (m) 0 0 i ≤ j by the deﬁnition of j , since im = im+1. Then applying lemma 3.37(a) and (e) we get i = m(i0) ≤ m(j(m)) = j(m). Examples 151

Lemma 3.41.

(n) (n) (a) If j ∈ J (m + 1) with jm = jm+1 then j = (j) for all n ≤ m. (b) If j ∈ J (m + 1) then j(n) = (j(k))(n) for n ≤ k ≤ m. (n) (c) If j ∈ J (m) and n ≤ m then jm ≤ jm.

(n) (n) Proof. (a) If j = 1m+1 then j = 1m = (j) .

(n) (n) (n) If j = 1m−122 then j = j and hence j = j = (j) for n = m and (n) (n) (n) j = 1m+1, j = 1m−12, thus j = 1m = (j) for n < m. (n) If j 6= 1m+1, 1m−122 then j > jm+1 by lemma 3.37(e). Thus j ≤ j implies (n) (n) (n) (n) (n) j ≤ j and it follows j ≤ (j) =: i. Since jm = jm+1 and jm = jm+1 (n) (n) we obtain j ≤ (i, im) ≤ j by lemma 3.37(i) from which we conclude j = (n) (n) (n) (i, im) (by the deﬁnition of j ) and hence j = i = (j) . (n) (k) 0 (n) 0 (n) 0 0 0 (b) j ≤ j =: j ≤ j implies j = (j ) ≤ j with jm = jm+1 (since k ≤ m) and hence j(n) = (j0)(n) = (j(k))(n) by (a). (0) (c) This is proved by induction. For m = 1 we have j1 = 1 ≤ j = j1. Now suppose (c) is true for an m ∈ N. Let j ∈ J (m + 1) and n ≤ m + 1 and let us assume that j(n) < j. Then it must be n ≤ m.

(m) We ﬁrst consider the case n = m. We have j < j, in particular jm < jm+1.

(m) (m) If j ≤ jm+1 then jm+1 ≤ m(j ) ≤ m(j) = jm+1 (lemma 3.37(a),(e)).

If j > jm+1 then we verify j = min{j(m) < j0 | j0 ∈ J (m)} =: j∗

(m) (m) (m) (m) to apply remark 3.38: jm+1 = jm ≤ jm + 1 ≤ jm+1. ¿From jm = jm+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 jm ≤ j < j (lemma ∗ ∗ ∗ ∗ 3.37(d)), hence m(j , jm) < j = m(j) from which it follows (j , jm) ≤ j by ∗ ∗ (m) (m) lemma 3.37(b) and thus (j , jm) ≤ j by the definition of j . But lemma (m) ∗ ∗ (m) ∗ ∗ 3.37(i) gives j < (j , jm) since j < j (by the definition of j ). This is a contradiction. Hence it must be j = j∗, and this proves the case n = m. Now let n < m. Define j0 := j(m). Then using (b) (with k = m) we get (n) 0 (n) 0 (n) (n) j = (j ) ≤ j ∈ J (m), hence by induction hypothesis jm+1 = jm = 0 (n) 0 (m) (n) (m) (m) (j )m ≤ jm = jm . It follows jm+1 ≤ jm = jm+1 ≤ jm+1 by what we’ve just proved. Notation 3.42. For j ∈ J (m) and n = 0, . . . , m + 1 we define

( (n) jm if n ≤ m tn(j) := m(j) + 1 if n = m + 1 152 3. Tensor and s-tensor orders and for n = 1, . . . , m + 1

( (n−1) (n) (n−1) (n) min{im | j

~~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 ≤ m with tn(j) ≤ r < tn+1(j).~~

Proof. First we prove tn(j) ≤ tn+1(j). We have tm(j) = jm ≤ m(j) + 1 = (n) (n+1) (n) tm+1(j) and for n < m from j ≤ j (by deﬁnition) it follows j = (n+1) (n) (n+1) max{i ≤s j | in = ... = im}, hence tn(j) = jm ≤ jm = tn+1(j) by lemma 3.41 (c). Further we have

~~( (n) (n+1) (n) (n+1) min{im | j~~

~~= tn+1(j)~~

(n) (n+1) Finally for tn(j) ≤ tn+1,1(j) we only need to check the case where j < j and n + 1 < m since otherwise we have tn+1(j) = tn+1,1(j) by the deﬁnition of tn+1,1(j) (and we have already seen that tn(j) ≤ tn+1(j)). In this case (n) (n+1) 0 (n) 0 (n+1) let j

~~(n) (n+1) tn+1,1(j) = min{im+1 | j~~

~~Proof. We use induction. If m = 1 then it must be n = 0 and we have j(0) = 11 and j(1) = kk for some k ≥ 2. It follows~~

~~t1,1(j) = min{i2 | 11~~

~~Let n + 1 ≤ m and j ∈ J (m + 1) with j(n) < j(n+1). We have to prove that~~

~~(n) (n+1) tn+1,1(j) = min{im+1 | j~~

~~(n) holds. Obviously “ ≤ ” is true. For the remaining inequality let j~~

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 generality (m) that jm = jm+1 (otherwise take j instead of j). Using lemma 3.41(a) and (n) (n+1) lemma 3.37(h) we get (j)

~~If i < im+1 then we deﬁne r := m(i) = im+1.~~

~~(n) ∗ (n+1) ∗ ∗ If νm(r) ∈ J (m) then j < i := νm(r) ≤ j with im ≤ i = νm(r) = im+1 (lemma 3.37(d)).~~

(n) (n+1) If νm(r) ∈/ J (m) then j < νm(r) < j implies that ∗ 0 0 (n) ∗ (n+1) ∗ i := min{νm(r) < j | j ∈ J (m)} satisﬁes j < i ≤ j and im < ∗ νm(r) = im+1 by remark 3.38 (since νm(r) < i , νm(r) ∈/ J (m) it must be ∗ i 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

~~(n) (n+1) tn+1,1(j) = min{im | j~~

~~(n) (n+1) (n) 0 Proof. To see “≤” take j~~

~~0 (n) 0 (n+1) im+1 = im ≥ min{im | j~~

~~Lemma 3.46. Let i, j ∈ J (m + 1) with im+1 < k ≤ jm+1 and i < j such 0 0 0 that im < im+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 it must be m(i) < m(j), since otherwise j > jm+1 implies i = j and hence jm+1 ≤ im+1 by deﬁnition 3.32(2)(b), and j ≤ jm+1 implies i < jm+1 by deﬁnition 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), 0 0 0 implying i ≤ (i, im) =: i < j by lemma 3.37(g) and (b), with im = im = im+1, which contradicts the assumptions.~~

~~Thus it must be i ≤ im+1. Then i < (1m, m(i) + 1) ≤ (1m, k) by lemma 3.37(c) and since m(i) + 1 = im+1 + 1 ≤ k.~~

~~(n) Lemma 3.47. Let n + 1 ≤ m, j ∈ J (m + 1), tn+1,1(j) < tn+1(j) and j~~

~~Proof. We prove by induction. For m = 1 it must be n = 0. Then t1,1(j) = (1) (1) (1) min{ i2 | 11~~

~~(n) (n+1) Let n + 1 ≤ m, j ∈ J (m + 1) with tn+1,1(j) < tn+1(j) and j~~

~~0 (m) If n + 1 = m then we deﬁne i := (1m−1, k, k). With k ≤ tm(j) = jm+1 = (m) (m) jm ≤ m(j ) and lemma 3.37(c) we have~~

~~(m) (m) (1m−1, k) ≤ (1m−1, m(j )) ≤ j~~

~~0 (m) and hence i = (1m−1, k, k) ≤ j by lemma 3.37(i).~~

~~(m) (m) Further i ≤ j implies i ≤ j by lemma 3.37(h) and we have im = im+1 = (m) (m) (m) (m) l < tm(j) = jm+1 = jm , hence i < j with im = l < k ≤ jm . It cannot Examples 155~~

~~00 (m) 00 00 ∗ 00 00 (m) be i ≤s i~~

~~Thus we can apply lemma 3.46 and obtain i < (1m−1, k) and hence i < 0 (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). (n) (n+1) Then for all tn+1,1(j) ≤ k ≤ tn+1(j) there is an element j~~

~~(n+1) (n+1) max{ i ≤s j | im+1 = k } = max{ i ≤s j | 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) = (1) 11 < j = rr (hence r ≥ 2) and t1,1(j) = min{i2 | 11

~~Let m ≥ 2 and assume that the statement (with k = tn+1,1(j)) is true for m − 1.~~

~~(n) Let n + 1 ≤ m and j ∈ J (m + 1) with tn+1,1(j) < tn+1(j). Let j~~

~~0 (n) (n+1) Suppose k = m. Then im < im+1. We have j < m(i) ≤ j by lemma 3.37(f), (a) and (e).~~

~~(n) (n+1) (n) 0 (n+1) If i ≥ im+1 then j < i ≤ j which implies j~~

~~tn+1,1(j) = tn+1,1(j) = k < tn+1(j) = tn+1(j) 156 3. Tensor and s-tensor orders~~

~~(n) (n+1) and (j)~~

~~(n+1) (n+1) max{i ≤s j | im+1 = k} ≤ max{i ≤s j | in+1 = ... = im+1 = k} and the remaining inequality is obvious.~~

~~(n) In particular, if n < m and j ∈ J (m) with tn,1(j) < tn(j) = jm then there (n) (n) is an i ≤s j with in = ... = im = jm − 1.~~

~~Notation 3.49. For j ∈ J (m) and n = 1, . . . , m + 1 we denote~~

~~ (n) max{i ≤ j(n) | i = ... = i = j − 1} if n < m, t (j) < t (j)  s n m m n,1 n j(n,2) := 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 3.48 or remark 3.44 that~~

~~(n) (n+1) (n+1) (n+1,2) j < max{i ≤s j | in+1 = ... = im = jm − 1} = j 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~~

~~(n+1,2) (n+1) (n+1) (a) j = max{i ≤s j | im+1 = jm+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~~

(k) (k) tn+1,1(j ) = tn+1,1(j) < tn+1(j) = tn+1(j) = tn+1(j ) and (k)(n+1,2) (n+1) (n+1) j = max{i ≤s j | im = jm − 1} We have j(n+1,2) ≤ j(n+1) (remark 3.50) which implies j(n+1,2) ≤ j(k) by (n+1,2) (n+1) (n+1,2) (k)(n+1,2) lemma 3.37(h). Since jm = jm − 1 it follows j ≤ j .

(n+1) (n+1) 0 For the other inequality take i ≤s j with im = jm − 1. Then i := (n+1) 0 (n+1) (n+1) (i, im) ≤ j by lemma 3.37(i) with im+1 = jm − 1 = jm+1 − 1. With (a) it follows i0 ≤ j(n+1,2) and then i = i0 ≤ 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 the statement be true for some m ∈ N and take i ∈ J (m + 1) and k ∈ N with km+1 < i. By lemma 3.37(f) we have m(km+1) < m(i). Thus if i ≤ im+1 then k ≤ m(km+1) < m(i) = im+1 and if i > im+1 then km = m(km+1) < m(i) = i which 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 also im ≤ im+1 ≤ m(i) ≤ m(km+1). Thus m(ϕr(i), im) ≤ m(km+1), and again applying lemma 3.37(g) we conclude ϕ(i) = (ϕr(i), im) ≤ km+1.

~~0 Proposition 3.55. Let i, j ∈ J (m), r < im and n, n ≤ m such that tn(j) ≤ r < tn+1(j) and tn0 (j) ≤ im < tn0+1(j).~~

~~(n) If (1) i ≤ j and tn(j) ≤ r < tn+1,1(j) 158 3. Tensor and s-tensor orders~~

~~(n+1,2) or (2) i ≤ j and tn+1,1(j) ≤ r < tn+1(j) then~~

~~(n0) (a) tn0 (j) ≤ im < tn0+1,1(j) implies ϕr(i) ≤ j (n0+1,2) (b) tn0+1,1(j) ≤ im < tn0+1(j) implies ϕr(i) ≤ j~~

~~Proof. First note that by remark 3.50 the numbers n and n0 exist since in any case we have i ≤ j and hence r < im ≤ m(i) ≤ m(j), and it must be 0 n ≤ n since r ≤ im.~~

We proceed by induction. For m = 1 we have t0(j) = 1, t1(j) = t1,1(j) = (0) (1) (1,2) (2,2) t2,1(j) = j, t2(j) = j + 1 and j = 1, j = j = j = j. We show that neither (1) nor (2) is satisﬁed, i.e. there is nothing to prove and the (0) statement is true. If n = 0 it would be (1) 1 = t0(j) ≤ r < i ≤ j = 1 (1,2) or (2) j = t1,1(j) ≤ r < i ≤ j = 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 hence j = 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), 0 r < im+1 and n, n ≤ m+1 such that tn(j) ≤ r < tn+1(j) and tn0 (j) ≤ im+1 < tn0+1(j).~~

~~(n) (1) i ≤ j and tn(j) ≤ r < tn+1,1(j):~~

~~(n0) (a) If tn0 (j) ≤ im < tn0+1,1(j), then we have to prove ϕr(i) ≤ j .~~

~~First suppose r ≥ im. Then ϕr(i) = (i, r).~~

0 If n = m + 1 then also n = m + 1 and hence tm+1(j) ≤ r < im+1 < tm+2(j) (n0) and j = 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 have jm+1 ≤ r < im+1 ≤ j and i ≤ j which shows that (i, r) ≤ j (deﬁnition 3.32(2)(b)).

~~(n) (n) If n ≤ m then r < im+1 and i ≤ j imply m(i, r) ≤ m(i) ≤ m(j ) by lemma 3.37(a) and applying lemma 3.37(g) and remark 3.50 (n ≤ n0) we get (i, r) ≤ j(n) ≤ j(n0).~~

~~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 in particular 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, since jm+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) =: j0 and it follows~~

~~0 0 (m) (m) tm(j ) = jm = jm = jm+1 = tm(j) ≤ r < im 0 0 0 ≤ m(i) ≤ m(j ) < tm+1(j ) = tm+1,1(j )~~

0 so that from the induction hypothesis we get ϕr(i) ≤ j . We also have im ≤ 0 0 0 (m) (m) m(j ) ≤ j and hence m(ϕr(i), im) ≤ j = j = m(j ). An application of 0 (m) (n0) lemma 3.37(g) and remark 3.50 (m = n ≤ n ) gives (ϕr(i), im) ≤ j ≤ j . If n < m deﬁne j0 := j(m). Then from lemma 3.41(b) and corollary 3.45 we 0 0 0 0 get 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) = (j0)(n) (again lemma 3.41(b)). By remark 00 0 0 3.43 there is a unique n ≤ m with tn00 (j ) ≤ im < tn00+1(j ), and it must be 00 0 0 n ≤ n ≤ n since r < im and tn00 (j) = tn00 (j ) ≤ im ≤ im+1.

0 0 0 (n00) (n00) (n00) Now if tn00 (j ) ≤ im < tn00+1,1(j ) then ϕr(i) ≤ (j ) = j = m(j ) by induction hypothesis, lemma 3.41(b) and lemma 3.37(e). Also im ≤ m(i) ≤ m(j(n)) ≤ m(j(n00)) by lemma 3.37(a) and remark 3.50, hence using lemma (n00) (n0) 3.37(g) and remark 3.50 we conclude (ϕr(i), im) ≤ j ≤ j .

0 0 00 If tn00+1,1(j ) ≤ im < tn00+1(j ) then it must be n < m (since by deﬁnition 0 0 00 0 tm+1,1(j ) = tm+1(j )) and n < n (using lemma 3.41(b) and corollary 3.45 0 we have tn00+1,1(j) = tn00+1,1(j ) ≤ im). By induction hypothesis and remark 0 (n00+1,2) 0 (n00+1) (n00+1) (n00+1) 3.50 ϕr(i) ≤ (j ) ≤ (j ) = j = m(j ) together with (n) (n00+1) (n00+1) (n0) im ≤ m(i) ≤ m(j ) ≤ m(j ) gives ϕr(i) = (ϕr(i), im) ≤ j ≤ j (again lemma 3.37(g) and remark 3.50).

~~0 (b) If tn0+1,1(j) ≤ im+1 < tn0+1(j) then it must be n < m, since by deﬁnition 0 tn0+1,1(j) = tn0+1(j) for n ≥ m.~~

~~(n0+1,2) We have to prove that ϕr(i) ≤ j .~~

~~(n) (n) If r ≥ im then as in case (a) m(i, r) ≤ m(i) ≤ m(j ) implies (i, r) ≤ j ≤ j(n0) ≤ j(n0+1,2).~~

~~0 (m) 0 If r < im then again as in case (a) for j := j we have tn(j ) = tn(j) ≤ 0 (n) 0 (n) 0 0 r < tn+1,1(j) = tn+1,1(j ), i ≤ j = (j ) and tn00 (j ) ≤ im < tn00+1(j ) with n ≤ n00 ≤ n0.~~

~~0 0 If tn00 (j ) ≤ im < tn00+1,1(j ) then exactly as in case (a) it follows ϕr(i) = (n00) (n0) (n0+1,2) (ϕr(i), im) ≤ j ≤ j ≤ j .~~

0 0 0 (n00+1,2) (n00+1,2) (n00+1,2) If tn00+1,1(j ) ≤ im < tn00+1(j ) then ϕr(i) ≤ (j ) = j = m(j ) (n) (n00+1,2) by induction hypothesis, im ≤ m(i) ≤ m(j ) ≤ m(j ), hence ϕr(i) = (n00+1,2) (n0+1,2) (ϕr(i), im) ≤ j ≤ j by lemma 3.37(g).

~~(n+1,2) (2) i ≤ j 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. 160 3. Tensor and s-tensor orders~~

0 (a) If tn0 (j) ≤ im+1 < tn0+1,1(j) then n < n since otherwise tn0+1,1(j) = tn+1,1(j) ≤ r < im+1 < tn0+1,1(j). (n0) We have to prove that ϕr(i) ≤ j . (n+1,2) (n+1,2) If r ≥ im then i ≤ j and r < im+1 imply m(i, r) ≤ m(i) ≤ m(j ) 0 (n+1,2) (n0) and since n + 1 ≤ m and n + 1 ≤ n it follows ϕr(i) = (i, r) ≤ j ≤ j by lemma 3.37(g) and remark 3.50.

0 (m) 0 If r < im then we deﬁne j := j . Since n < m we have tn+1,1(j ) = tn+1,1(j) 0 0 0 (n+1,2) and tn+1(j ) = tn+1(j), thus tn+1,1(j ) ≤ r < tn+1(j ). Further i ≤ j implies i ≤ j(n+1,2) = (j0)(n+1,2) and there is a unique n ≤ n00 ≤ n0 with 0 0 tn00 (j ) ≤ im < tn00+1(j ) (see case (1)(a)). 0 0 0 If tn00 (j ) ≤ im < tn00+1,1(j ) and n ≤ m then by induction hypothesis ϕr(i) ≤ 0 (n00) 0 (n0) (n0) (n0) (n+1,2) (n0) (j ) ≤ (j ) = j = m(j ) and im ≤ m(i) ≤ m(j ) ≤ m(j ) 0 (n0) (since n < n ) and hence ϕr(i) = (ϕr(i), im) ≤ j by lemma 3.37(g). 0 0 (n00) 0 If n = m + 1 then the induction hypothesis gives ϕr(i) ≤ (j ) ≤ j = (m) (m) (n+1,2) (m) j = m(j ) and we have im ≤ m(i) ≤ m(j ) ≤ m(j ) (since n < m). Again an application of lemma 3.37(g) gives ϕr(i) = (ϕr(i), im) ≤ j(m) ≤ j = j(n0).

~~0 0 00 00 0 If tn00+1,1(j ) ≤ im < tn00+1(j ) then it must be n < m and n < n .~~

0 (n00+1,2) (n00+1,2) (n00+1,2) Then ϕr(i) ≤ (j ) = j = m(j ) by induction hypothesis (n+1,2) (n00+1,2) and also im ≤ m(i) ≤ m(j ) ≤ m(j ). From lemma 3.37(g) we (n00+1,2) (n0) 00 0 obtain ϕr(i) = (ϕr(i), im) ≤ j ≤ j since n < n . 0 (b) If tn0+1,1(j) ≤ im+1 < tn0+1(j) then n < m. (n0+1,2) We have to prove that ϕr(i) ≤ j is true. (n+1,2) (n0+1,2) If r ≥ im then m(i, r) ≤ m(i) ≤ m(j ) ≤ m(j ) implies ϕr(i) = (i, r) ≤ j(n0+1,2) since n0 < m.

0 (m) If r < im then as in the corresponding case (1)(b) for j := j we have 0 0 (n+1,2) 0 (n+1,2) tn+1,1(j ) = tn+1,1(j) ≤ r < tn+1(j) = tn+1(j ), i ≤ j = (j ) and 0 0 00 0 tn00 (j ) ≤ im < tn00+1(j ) for some n ≤ n ≤ n . 0 (n00+1,2) 0 (n0+1,2) In any case the induction hypothesis implies ϕr(i) ≤ (j ) ≤ (j ) = (n0+1,2) (n0+1,2) 0 (n+1,2) j = m(j ) (since n < m) and also im ≤ m(i) ≤ m(j ) ≤ (n0+1,2) (n0+1,2) m(j ), hence ϕr(i) = (ϕr(i), im) ≤ j by lemma 3.37(g).

~~m Theorem 3.56. The restriction νm|s on J (m) of the order νm on N deﬁned in 3.32 is an s-tensor order with~~

~~m 2 |νm|s| ≤ 2 (m!)~~

~~Proof. We use induction. For m = 1 we have the natural order on N, thus 1 |ν |s| = 1. Examples 161~~

~~m m 2 Now suppose |ν |s| ≤ 2 (m!) for some m ∈ N and take j ∈ J (m + 1). It is enough to prove that~~

~~Sm+1[1m+1, j]s~~

~~m+1 tn+1,1−1 tn+1−1 m+1 m+1 [ [ (n) [ (n+1,2) = (Sm[1m, j ]s) × {r} ∪ (Sm[1m, j ]s) × {r}~~

~~n=1 r=tn r=tn+1,1 with natural numbers tn ≤ tn+1,1 ≤ tn+1, n = 1, . . . m + 1. Then we have~~

~~ϕ1(n) ϕ2(n) (n) [ (n) (n+1,2) [ (n) Sm[1m, j ]s = Ii ,Sm[1m, j ]s = Ji i=1 i=1~~

~~(n) (n) m m with (natural order) intervals Ii and Ji in N and ϕi ≤ |ν |s| for i = 1, 2 (n) (n) m+1 and hence by (3.4) we can ﬁnd intervals Iij and Jij ⊂ N such that~~

~~Sm+1[1m+1, j]s~~

~~m+1 ϕ1(n) ϕ2(n) [ [ (n) m+1 [ (n) m+1 = Ii × [tn, tn+1,1 − 1] ∪ Ji × [tn+1,1, tn+1 − 1] n=1 i=1 i=1~~

~~m+1 ϕ1(n) m+1 ϕ2(n) m+1 [ [ [ (n) [ [ (n) = Iij ∪ Jij n=1 i=1 j=1 i=1 j=1~~

~~2 2 m This is a union of (m+1) (ϕ1(n)+ϕ2(n)) ≤ 2 (m+1) |ν |s| intervals, hence~~

~~2 m m+1 2 ρ(Sm+1[1m+1, j]s) ≤ 2 (m + 1) |ν |s| ≤ 2 ((m + 1)!) by induction hypothesis.~~

~~To prove the formula for Sm+1[1m+1, j]s we use several steps.~~

~~First deﬁne  {(i, r) ∈ J (m + 1) | i < m(j)} r < jm+1 Ar := {(i, r) ∈ J (m + 1) | i < j} jm+1 ≤ r ≤ m(j)~~

~~Then~~

~~m(j) [ (1) [1m+1, j]s = Ar r=1 162 3. Tensor and s-tensor orders~~

~~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 [ = {(i, r) ∈ J (m + 1) | i < jm+1 = m(j)} r=1~~

~~m(j) [ ∪ {(i, r) ∈ J (m + 1) | i ≤ j}~~

~~r=jm+1 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 [ = {(i, r) ∈ J (m + 1) | i < j = m(j)} r=1~~

~~m(j) [ ∪ {(i, r) ∈ J (m + 1) | i ≤ j}~~

~~r=jm+1~~

~~Next we deﬁne~~

~~tn+1,1(j)−1 [ (n) Bn := {(i, r) ∈ J (m + 1) | i ≤ j }~~

~~r=tn(j)~~

~~tn+1(j)−1 [ ∪ {(i, r) ∈ J (m + 1) | i ≤ j(n+1,2)}~~

~~r=tn+1,1(j) for n = 0, . . . , m + 1, and we claim~~

~~m(j) m+1 [ [ (2) Ar = Bn r=1 n=0 We have~~

~~m(j) m(j) [ [ Bm+1 = {(i, r) ∈ J (m + 1) | i ≤ j} = Ar~~

~~r=tm+1(j) r=jm+1~~

~~Stm+1(j)−1 Sm and it remains to prove that r=1 Ar = n=0 Bn.~~

“⊂”: 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

(n) If tn(j) ≤ r < tn+1,1(j) then we have to prove i ≤ j . 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) =: j0 ∈ J (m) and by remark 3.50 (and lemma 3.41(b)) there is a unique 0 ≤ k ≤ m − 1 with (k) 0 (k) 0 (k+1) (k+1) j = (j ) ~~j(n). Then it must be k ≥ n and hence~~

~~0 (k) 0 (k+1) im ≥ min{im | j~~

~~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 (n+1) (n+1) (n+1) (n+1) (n+1) i ≤ j . Further we have jm+1 − 1 < jm+1 ≤ m(j ) = j , hence (n+1) (n+1) 0 (n+1) (n+1) m(i, jm+1 − 1) ≤ j , implying i := (i, jm+1 − 1) ≤ j by lemma 0 (n+1) 3.37(g) and i ∈ J (m + 1) since im ≤ r < tn+1(j) = jm+1 . It follows

~~0 00 (n+1) 00 (n+1) (n+1,2) i ≤ max{i ≤s j | im+1 = j − 1} = j~~

~~(lemma 3.51(a)) and thus i = i0 ≤ m(i0) ≤ 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).~~

(n) If i ≤ j 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 then (n) (n+1) it must be n < m since tn(j) < tn+1(j), hence j < j which implies i ≤ j(n) = m(j(n)) < m(j(n+1)) ≤ m(j) (lemma 3.37(f)).

(n+1,2) If i ≤ j and tn+1,1(j) ≤ r < tn+1(j) then n < m, in particular r < (n+1,2) tm(j) ≤ tm+1(j) = jm+1. Further tn+1.1(j) < tn+1(j) implies j < 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

~~tn+1,1(j)−1 tn+1(j)−1 [ (n) [ (n+1,2) Cn := [1m, j ]s × {r} ∪ [1m, j ]s × {r}~~

~~r=tn(j) r=tn+1,1(j) for n = 0, . . . , m + 1 we now claim~~

~~m+1 m+1 [ [ (3) Sm+1 Bn = Sm+1 Cn n=0 n=0 164 3. Tensor and s-tensor orders~~

Obviously we have Bn ⊂ Cn and thus “⊂” holds. To prove “⊃” it will be enough to show that for every n ≤ m + 1 and 0 (i, r) ∈ Cn with r < im we have (ϕr(i), im) ∈ Bn0 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 j0 = j(n) or j(n+1,2) we have j0 ≤ j and hence 0 0 j ≤ m(j ) ≤ m(j), implying im ≤ i ≤ m(j) (lemma 3.37(a), (d)). Thus by 0 remark 3.43 there is n ≤ m + 1 such that tn0 (j) ≤ im < tn0+1(j). We claim that (ϕr(i), im) ∈ Bn0 , i.e. we have to prove

~~(n0) (a) tn0 (j) ≤ im < tn0+1,1(j) implies ϕr(i) ≤ j~~

~~(n0+1,2) (b) tn0+1,1(j) ≤ im < tn0+1(j) implies ϕr(i) ≤ j If n0 < m then with j0 := j(m) we have~~

~~0 0 0 tk(j ) = tk(j) for k = n, n + 1, n , n + 1 (3.26) as well as~~

~~0 tk+1,1(j ) = tk+1,1(j) for k = n, n0 (3.27) (j0)(k) = j(k), (j0)(k+1,2) = j(k+1,2)~~

~~0 0 by lemma 3.41(b) and corollary 3.45. It follows tn(j ) ≤ r < tn+1(j ) and 0 tn0 (j ) ≤ im < tn0+1(j), and since (i, r) ∈ Cn, r < im we have~~

(n) 0 (n) 0 0 i ≤ j = (j ) and tn(j ) ≤ r < tn+1,1(j ) (3.28) (n+1,2) 0 (n+1,2) 0 0 or i ≤ j = (j ) and tn+1,1(j ) ≤ r < tn+1(j ) An application of proposition 3.55 gives  0 (n0) (n0) 0 0 (j ) = j if tn0 (j ) ≤ im < tn0+1,1(j ) ϕr(i) ≤ 0 (n0+1,2) (n0+1,2) 0 0 (j ) = j if tn0+1,1(j ) ≤ im < tn0+1(j ) which by (3.26) and (3.27) is what we want.

~~0 If n = m then tn0+1,1(j) = tn0+1(j) and there is only the one case (a) tm(j) ≤ (m) im < tm+1(j), i.e. we have to prove that ϕr(i) ≤ j .~~

0 (m) 0 0 If n < m then again with j := j we have tn(j ) ≤ r < tn+1(j ) and (3.28). 0 00 In particular im ≤ m(i) ≤ m(j ) and thus there is a unique n ≤ n ≤ m with 0 0 00 tn00 (j ) ≤ im < tn00+1(j ) by remark 3.43. If n < m then by proposition 3.55 0 (n+1,2) 0 00 0 we have ϕr(i) ≤ (j ) ≤ j and if n = m = n then proposition 3.55 0 0 (m) gives ϕr(i) ≤ j . Thus in any case ϕr(i) ≤ j = j .

~~(m) 0 If n = m then we have tm(j) ≤ r < im < tm+1(j). Since i ≤ j =: j 0 0 0 it follows im ≤ m(i) ≤ m(j ) < tm+1(j ) and hence tm(j ) = tm(j) ≤ r < Examples 165~~

~~0 im < tm+1(j ) such that again we can apply proposition 3.55 and obtain 0 (m) ϕr(i) ≤ j = j .~~

~~0 If n = m + 1 then again tn0+1,1(j) = tn0+1(j) and there is only case (a) tm+1(j) ≤ im < tm+2(j). We have to prove ϕr(i) ≤ j.~~

0 (m) If n ≤ m then in the same way as for the case n = m we see that ϕr(i) ≤ j . (m) (m) Then ϕr(i) ≤ j = m(j ) ≤ 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 that jm+1 = tm+1(j) ≤ im ≤ i < m(j). In the second case, note that it must be im ≤ m(j), since im < tm+2(j) = m(j) + 1 by assumption. Thus we have jm+1 = tm+1(j) ≤ im < m(j) or m(j) = im ≤ i ≤ j.

~~0 If n = m + 1 = n then i ≤ j and we have to prove that ϕr(i) ≤ j. This follows from~~

~~tm(j) = jm ≤ jm+1 = tm+1(j) ≤ r < im ≤ m(i) ≤ m(j) < tm+1(j) by proposition 3.55.~~

~~Sm+1 Now from (1), (2) and (3) we obtain Sm+1[1m+1, j]s = Sm+1 n=0 Cn and using remark 3.5 (2) it follows~~

~~(4) Sm+1[1m+1, j]s =~~

~~m+1 tn+1,1(j)−1 tn+1(j)−1 m+1 m+1 [ [ (n) [ (n+1,2) (Sm[1m, j ]s) × {r} ∪ (Sm[1m, j ]s) × {r}~~

~~n=0 r=tn(j) r=tn+1,1(j) This formula can be improved. We claim that~~

~~(5) Sm+1[1m+1, j]s =~~

~~m+1 tn+1,1−1 tn+1−1 m+1 m+1 [ [ (n) [ (n+1,2) (Sm[1m, j ]s) × {r} ∪ (Sm[1m, j ]s) × {r}~~

~~n=1 r=tn r=tn+1,1 holds with t1 := 1, tn := tn(j) and tn,1 := tn,1(j) for n = 2, . . . , m + 2. To understand this ﬁrst note that~~

~~t1(j)−1 0 [ (1) B0 = B0 := {(i, r) ∈ J (m + 1) | i ≤ j } r=1~~

(1) (1) (1,2) (0) If j < 2m+1 then j = 1m+1 = j = j , t1,1(j) = 1 = t1(j) and thus 0 B0 = ∅ = B0. (1) (0) (1,2) (1) If j ≥ 2m+1 then t1,1(j) = 2 ≤ t1(j), and j ≤ j ≤ j implying (0) (1,2) (1) 0 j ≤ j ≤ j gives B0 ⊂ B0. 166 3. Tensor and s-tensor orders

(1) For the other inclusion take (i, r) ∈ J (m+1) with i ≤ j and 1 ≤ r < t1(j). If r = 1 then (i, r) = 1m+1 ∈ B0. (1,2) (1) If r > 1 then 2 ≤ r < t1(j) and it is enough to prove i ≤ j . If j = 2m+1 (1,2) (1) (1) (1) then j = j and we are done. If j > 2m+1 then j = km+1 for some (1,2) k ≥ 3 and j = (k − 1)m+1, and 2 = t1,1(j) ≤ r ≤ t1(j) − 1 = k − 1. (1,2) Suppose i > j = (k − 1)m. Then im > k − 1 by lemma 3.52 and thus (1,2) r ≥ im > k − 1, a contradiction. Hence it must be i ≤ j . 0 Sm+1 Then using (2) it follows Sm+1[1m+1, j]s = Sm+1(B0 ∪ n=1 Bn) and we claim that m+1 m+1 0 [ 0 [ Sm+1 B0 ∪ Bn = Sm+1 C0 ∪ Cn n=1 n=1 0 St1(j)−1 (1) where C0 := r=1 [1m, j ]s × {r}. By (3) it is enough to prove m+1 0 0 [ C0 ⊂ Sm+1 B0 ∪ Bn n=1 To understand this observe that (1) (1) im ≤ i ≤ j = m(j ) ≤ m(j) < tm+2(j) (1) implies M := max{im | i ≤ j } < tm+2(j) and thus M m+1 [ (1) 0 [ {(i, r) ∈ J (m + 1) | i ≤s j } ⊂ B0 ∪ Bn r=1 n=1 (1) Hence it is enough to prove that if i ≤ j and 1 ≤ r < t1(j) with r < im SM (1) then (ϕr(i), im) ∈ Sm+1( r=1{(i, r) ∈ J (m + 1) | i ≤s j }), which is the (1) case if and only if ϕr(i) ≤ j . This is true, since we can apply remark 3.54 (1) with k := t1(j) = jm+1. We have 0 C0 ∪ C1

~~t1(j)−1 t2,1(j)−1 t2(j)−1 [ (1) [ (1) [ (2,2) = [1m, j ]s × {r} ∪ [1m, j ]s × {r} ∪ [1m, j ]s × {r}~~

~~r=1 r=t1(j) r=t2,1(j)~~

~~t2,1−1 t2−1 [ (1) [ (2,2) = [1m, j ]s × {r} ∪ [1m, j ]s × {r}~~

~~r=1 r=t2,1 and then m+1 0 [ Sm+1[1m+1, j]s = Sm+1 C0 ∪ Cn n=1 t −1 t −1 m+1 n+1,1 n+1 [ [ (n) [ (n+1,2) = Sm+1 [1m, j ]s × {r} ∪ [1m, j ]s × {r}~~

~~n=1 r=tn r=tn+1,1 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 with basis can be identiﬁed with the dual space of the symmetric projective tensor ˜ m,s 0 product (⊗πs E) in such a way that the monomial

~~∞ ∞ α Y 0 αi (N) X z = (ei) , α ∈ N0 , |α| = αi = m i=1 i=1~~

0 with respect to the basis (en) on E (with en the coeﬃcient functionals) cor- 0 0 0 responds to the functional σm(ej) = ej1 ⊗ · · · ⊗ ejm with j ∈ J (m) and αi = |{k | jk = i}|. Thus ordering J (m) means ordering the monomials of order m. For example the s-tensor order given in example 1 can be expressed as follows:

~~(N) For α ∈ N0 deﬁne n(α) := max{k | αk 6= 0} and α by  αn − 1 n = n(α) αn := αn otherwise~~

~~Then for multiindices α, β of order m we have~~

~~α ≤ β ⇐⇒ (1) n(α) < n(β) (3.29) or (2) n(α) = n(β) and α ≤ β~~

Theorem 3.57. Let E be a Banach space with shrinking basis (en). Then the α associated monomials (z )|α|=m ordered by an s-tensor order form a basis for m m 0 Pappr(E) and Pnuc(E). In particular: If we additionally assume that E has m the Radon-Nikod´ymproperty, then the monomials form a basis for Pint(E).

~~For the approximable polynomials with the order (3.29) this is due to Dimant and Dineen [34, proposition 10]. See also [36], proposition 4.4 (and 2.8).~~

˜ m,s 0 m ˜ m,s 0 m Proof. We have ⊗εs E = Pappr(E) (see [36], p. 104) and ⊗πs E = Pnuc(E) isometrically since E0 has the approximation property. The nuclear polynomials on E can be identiﬁed with the integral ones if E0 has the Radon- Nikod´ymproperty (see [39], 4.3 and 4.5 for the last two statements). 168 3. Tensor and s-tensor orders

Next we prove that if the Banach space E has not only a shrinking basis but also the Dunford-Pettis property, the monomials with respect to the basis even form a basis for the whole space Pm(E) of all m-homogeneous polynomials (with norm kpk = sup |p(z)|), as Ryan claimed in his thesis [69]. z∈BE This is based on his ideas. Nevertheless we cannot work with an inductive argument as he suggested, and some multilinear generalizations have to be investigated. Recall that a Banach space E is said to have the Dunford-Pettis property if every weakly compact operator T from E into some Banach space (the image of the unit ball is relatively weakly compact) is fully complete, i.e. it transforms weakly convergent into norm convergent sequences. A reﬂexive inﬁnite 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. 0 (2) If (xn) is a weak Cauchy sequence in E and (xn) is a weak zero sequence 0 0 in E then (hxn, xni) is a zero sequence. 0 (3) If (xn) is a weak zero sequence in E and (xn) is a weak Cauchy sequence 0 0 in E then (hxn, xni) is a zero sequence. 0 (4) If (xn) is a weak Cauchy sequence in E and (xn) a weak Cauchy se- 0 0 quence in E then (hxn, xni) is a Cauchy sequence.

Lemma 3.59. Let E1,...,Em be Banach spaces with Dunford-Pettis prop- i erty, F be a Banach space and (xk)k∈N weak Cauchy sequences in Ei for m 1 m i = 1, . . . , m. If φ ∈ L (E1,...,Em; F ) then φ(xk, . . . , xk ) is a weak k∈N Cauchy sequence in F .

Proof. First we take F = K and prove by induction. The case m = 1 is clear, 0 since φ ∈ E1 is uniformly weakly continuous. Now let m ∈ N and assume that every continuous m-linear functional on Qm 0 i=1 Ei maps σ(Ei,Ei)-Cauchy sequences to a Cauchy sequence. m+1 For φ ∈ L (E1,...,Em+1) we then deﬁne the m-linear and continuous mapping Qm E −→ E0 ψ : i=1 i m+1 x φ(x, · ) 00 m 00 00 Then we have x ◦ ψ ∈ L (E1,...,Em) for every x ∈ Em+1, hence by the 00 1 m 00 00 induction hypothesis (x ◦ ψ)(xk, . . . , xk ) k is Cauchy for every x ∈ Em+1, 1 m 0 m+1 i.e. ψ(xk, . . . , xk ) k is weak Cauchy in Em+1. Since (xk )k is weak Cauchy 1 m+1 in Em+1 the sequence φ(xk, . . . , xk ) 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 y0 ◦ φ for every functional y0 on 1 m F , so φ(xk, . . . , xk ) k is weak Cauchy in F .~~

Corollary 3.60. Let E1,...,Em be Banach spaces with the Dunford-Pettis m i property, ϕ ∈ L (E1,...,Em) and (xk)k weak Cauchy sequences, i = 1, . . . , m, 1 m one of which converges weakly to zero. Then ϕ(xk, . . . , xk ) k is a zero sequence.

~~m Proof. The case m = 1 is obvious, so let m ≥ 2. We can assume that (xk )k is a weak zero sequence. The mapping~~

~~Qm−1 E −→ E0 φ : i=1 i m x ϕ(x, · )~~

~~1 m−1 is (m−1)-linear and continuous, so lemma 3.59 assures that φ(xk, . . . , xk ) k 0 is weak Cauchy in Em. Hence the Dunford-Pettis property of Em gives the claim.~~

~~With corollary 3.60 at hand the following multilinear extension of [69, proposition 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 ˜ m m m (ˆeν(n))n∈N is a basis for ⊗π,i=1Ei and ν[1, k ] = [1, k] holds for every k ∈ N. Then (ˆeν(n))n∈N is shrinking.

~~See deﬁnition 3.20 for the notatione ˆν(n).~~

˜ m Proof. We deﬁne X := ⊗π,i=1Ei, I := idX and let Pn be the n-th coordinate projection of the basis (ˆeν(k))k∈N. Suppose the basis is not shrinking. Taking into account the fact that (I − 0 m Pn)X ⊂ (I − Pl)X for n ≥ l, we can choose a ϕ ∈ X = L (E1,...,Em) and a δ > 0 such that kϕ|(I−Pn)X k ≥ 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~~

m m m X I − Pkm = ⊗i=1Ii − ⊗i=1Pi,k = Ri,k i=1 Pm and hence δ ≤ i=1kϕ|Ri,kX k for every k. So there must be an index i0 ∈ {1, . . . , m} such that δ kϕ|R X k ≥ i0,k m 170 3. Tensor and s-tensor orders

~~˜ m for inﬁnitely many k. Note that Ri0,kX = ⊗π,i=1Fi with  P E i = 1, . . . , i − 1  i,k i 0 Fi = (Ii − Pi,k)Ei i = i0  Ei i = i0 + 1, . . . , m~~

~~i Then there are sequences (xk)k in Fi ∩ BEi such that δ |ϕ(x1, . . . , xm)| ≥ k k 2m~~

0 0 for all k. Since Ei is separable, (BEi , σ(Ei,Ei)) is a precompact metric space, i 0 and we can assume that (xk)k is σ(Ei,Ei)-Cauchy for i = 1, . . . , m. We have i0 xk ∈ (Ii0 − Pi0,k)Ei0 and hence

~~i0 0 i0 0 0 |hxk , x i| = |hxk , x ◦ (Ii0 − Pi0,k)i| ≤ kx ◦ (Ii0 − Pi0,k)k −→ 0~~

0 0 for every functional x ∈ Ei0 , since the basis (ei0j)j of Ei0 is shrinking. i0 But then (xk )k is a weak zero sequence in Ei0 and corollary 3.60 gives 1 m ϕ(xk, . . . , xk ) −→ 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 is the restriction ν|s of the order ν given in example 2. By proposition 3.61 it ˜ m follows that (eν(n))n∈N is a shrinking basis of ⊗π E and hence theorem 3.31 together with corollary 3.24 imply that (σm(eν|s(n)))n∈N is a shrinking basis ˜ m,s of ⊗πs E.

~~Pointwise convergence~~

Theorem 3.63. Let E be a Banach space with basis (en) and ν be an order on J (m) such that (σ (e )) is a basis for ⊗˜ m,sE. Then the monomials m ν(n) n∈N πs m with respect to the basis (en) in order coming from ν form a basis of P (E) in the topology of pointwise convergence.

~~0 Proof. The coeﬃcient functionals xn in a Banach space X with basis (xn) build a weak*-basis of the dual space X0, since for ϕ ∈ X0 and P∞ 0 x = n=1 xn(x)xn ∈ X we have~~

~~∞ X 0 ϕ(x) = ϕ(xn)xn(x) n=1 Pointwise convergence 171~~

˜ m,s By assumption the family (σm(ej))j∈J (m) ordered by ν gives a basis for ⊗πs E 0 with coeﬃcient functionals |j|σm(ej), where |j| is the number of elements i ∈ Nm with σi = j for some permutation σ (see the proof of theorem 3.22). ˜ m,s So if ϕ is the linear functional on ⊗πs E corresponding to a polynomial P ∈ Pm(E) and z ∈ E we have

~~m X 0 m P (z) = ϕ(σm(⊗ z)) = ϕ(σm(ej)) |j| σm(ej)(σm(⊗ z)) (3.30) j∈J (m)~~

~~(summation in order ν) and we prove that the series on the right side is the monomial expansion of P in z.~~

m The elements ei = ei1 ⊗ · · · ⊗ eim , i ∈ N (ordered by some tensor order) ˜ m 0 0 0 form a basis for ⊗π E with coeﬃcient functionals ei = ei1 ⊗ · · · ⊗ eim . We have 0 m 0 0 ei(⊗ z) = ei1 (z) ··· eim (z) = zi1 ··· zim m ˜ m so the element ⊗ z in ⊗π E has the basis representation

~~m X ⊗ z = zi1 ··· zim ei m i∈N where summation is carried out in some tensor order . Then~~

~~m X σm(⊗ z) = zi1 ··· zim σm(ei) m i∈N 0 and applying the coeﬃcients functionals |j|σm(ej) of the basis (σm(ej))j∈J (m) ˜ m,s of ⊗πs E we get~~

~~0 m |j|σm(ej)(σm(⊗ z)) = |j| zj1 ··· zjm (3.31) since ( 0 1 ∃ permutation σ : σi = j h|j|σm(e ), σm(ei)i = j 0 otherwise The equation~~

~~X α m X cα(P )u = P (u) = ϕ(σm(⊗ u)) = |j|uj1 ··· ujm ϕ(σm(ej)) |α|=m j∈J (m) for u ∈ span{en} implies~~

~~cα(P ) = |j|ϕ(σm(ej)) if αi = |{k | jk = i}|, and this together with (3.31) and (3.30) gives~~

~~X X α P (z) = |j|zj1 ··· zjm ϕ(σm(ej)) = cα(P )z j∈J (m) |α|=m where both series are summed up in the order given by ν. 172 3. Tensor and s-tensor orders~~

Now we extend the result to the space PN (E) of all polynomials of degree less or equal than N. Let us deﬁne J (0) = {()}, where the “empty sequence” () is some element not contained in N(N). For i ∈ N we deﬁne i := (). By an order on J (0) we mean the mapping ν0 : {1} −→ J (0).

~~N Given orders νm on J (m) for m = 0,...,N, by ⊕m=0νm we ﬁx some order SN SN ν on m=0 J (m) (i.e. ν : N −→ m=0 J (m) bijective) whose restriction to J (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) Im ϕi ∩ Im ϕj = ∅ 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 deﬁning the “empty sequence” () to be the smallest element: () ≤ j for all j ∈ N(N). SN Deﬁnition 3.64. For i, j ∈ 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 . . .. .~~

~~3 13 113 ··· 1N−13 33 133 ··· 1N−233 3N~~

~~23 123 ··· 1N−223 233 ··· 1N−3233 .. . 223 ··· 1N−3223 . . ··· .. . . . 2N−233~~

~~2N−13~~

~~. .~~

~~SN In this way we get an order on m=0 J (m) whose restriction to J (m) is an s-tensor order for m = 1,...,N.~~

SN (N) The set m=0 J (m) corresponds to the set {α ∈ N0 | |α| ≤ N} in the way described at the beginning of the last section, where the empty sequence is mapped on the zero sequence. Deﬁning the zero sequence to be the smallest (N) element, the order on {α ∈ N0 | |α| ≤ N} can be described with (3.29). Let us identify the sets J (m) and [|α| = m] in the usual way. Then the next result is an immediate consequence of theorem 3.63.

Corollary 3.65. Let N ∈ N and νm be an s-tensor order on J (m) for m = 1,...,N. Then for every Banach space E with basis (en) the monomials N with respect to the basis ordered by ⊕m=0νm form a basis of PN (E) in the topology of pointwise convergence.

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~~Name Christopher Prengel Born 28.10.1975 in Hanover, Germany Nationality german~~

Education 1995 School-leaving 1995–96 Civilian service 1996 Preparatory course for music studies at the conservatory in Hanover 1997–2002 Student at the University of Oldenburg 1997 Studies of music and mathematics to become a teacher (Gymnasiallehramt) 1998–2002 Studies of mathematics 2002 Diploma in mathematics 2002–05 Candidate for a doctor of science at the University of 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 mathematics at the technical college (Fachhochschule) Oldenburg/Ostfriesland/Wilhelmshaven in Oldenburg and Bremen 2003–05 GradF¨oGStipendium 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 expansions of holomorphic functions (preprint), with A. Defant and M. Maestre