Cohen strongly p-summing holomorphic mappings on Banach spaces

Abstract

Let E and F be complex Banach spaces, U be an open subset of E and $1\le p\le \infty .$ We introduce and study the notion of a Cohen strongly p-summing holomorphic mapping from U to F,  a holomorphic version of a strongly p-summing linear operator. For such mappings, we establish both Pietsch Domination/Factorization Theorems and analyse their linearizations from (the canonical predual of ) and their transpositions on Concerning the space formed by such mappings and endowed with a natural norm we show that it is a regular Banach ideal of bounded holomorphic mappings generated by composition with the ideal of strongly p-summing linear operators. Moreover, we identify the space with the dual of the completion of tensor product space endowed with the Chevet–Saphar norm $g_p.$

1 Introduction

The linear theory of absolutely summing operators between Banach spaces was initiated by Grothendieck [11] in 1950 with the introduction of the concept of 1-summing operator. In 1967, Pietsch [22] defined the class of absolutely p-summing operators for any $p>0$ and established many of their fundamental properties.

The nonlinear theory for such operators started with Pietsch [23] in 1983. Since then, the idea of extending the theory of absolutely p-summing operators to other settings has been developed by various authors, namely, the polynomial, multilinear, Lipschitz and holomorphic settings (see, for example, [1, 2, 7, 8, 19, 27, 28]).

Summability for holomorphic mappings was first considered by Matos in a series of papers (see e.g. [13, 14]). Our approach in this paper is different from that of Matos. Moreover, strong p-summability in the sense of Dimant [7] was also addressed for subspaces of holomorphic mappings as polynomials and multilinear mappings under the name of factorable strongly p-summing (see [20, 24, 25]). In these papers, it was proved that the ideal of factorable strongly p-summing polynomials (multilinear mappings) coincides with the ideal formed by composition with p-summing linear operators. Ideals of polynomial mappings were also studied by Floret and García [9, 10].

In 1973, Cohen [5] introduced the concept of a strongly p-summing linear operator to characterize those operators whose adjoints are absolutely $p^*$-summing operators, where $p^*$ denotes the conjugate index of $p\in (1,\infty ].$ Influenced by this class of operators, we introduce and study a new concept of summability in the category of bounded holomorphic mappings, which yields the called Cohen strongly p-summing holomorphic mappings.

We now describe the contents of the paper. Let E and F be complex Banach spaces, U be an open subset of E and $1\le p\le \infty .$ We denote by the Banach space of all bounded holomorphic mappings from U to F,  equipped with the supremum norm. In particular, stands for the space It is known that is a dual Banach space whose canonical predual, denoted is the norm-closed linear subspace of generated by the evaluation functionals at the points of U.

In Sect. 1, we fix the notation and recall some results on the space essentially, a remarkable linearization theorem due to Mujica [16] which is a key tool to establish our results.

In Sect. 2, we show that the space of all Cohen strongly p-summing holomorphic mappings denoted and equipped with a natural norm is a regular Banach ideal of bounded holomorphic mappings. Furthermore, with

The elements of the tensor product of two linear spaces can be viewed as linear mappings or bilinear forms (see [26, Section 1.3]). Following this idea, in Sect. 3 we introduce the tensor product $\Delta (U)\otimes F$ as a space of linear functionals on the space and equip this space with the known Chevet–Saphar norms $g_p$ and $d_p.$

Section 4 addresses the duality theory: the space is canonically isometrically isomorphic to the dual of the completion of the tensor product space In particular, we deduce that is a dual space.

Pietsch [22] established a Domination/Factorization Theorem for p-summing linear operators between Banach spaces. Characterizing previously the elements of the dual space of $\Delta (U)\otimes _{g_p} F,$ we present for Cohen strongly p-summing holomorphic mappings both versions of Pietsch Domination Theorem and Pietsch Factorization Theorem in Sects. 5 and 6, respectively.

Moreover, in Sect. 5, we prove that a mapping $f:U\rightarrow F$ is Cohen strongly p-summing holomorphic if and only if Mujica’s linearization is a strongly p-summing operator. Several interesting applications of this fact are obtained.

In addition, we show that the ideal is generated by composition with the ideal of strongly p-summing linear operators, that is, every mapping admits a factorization in the form $f=T\circ g,$ for some complex Banach space G,  and Moreover, coincides with $\inf \{d_p(T)\left\| g\right\| _{\infty }\},$ where the infimum is extended over all such factorizations of f,  and, curiously, this infimum is attained at Mujica’s factorization of f. We also show that every factors through a Hilbert space whenever F is reflexive, and establish some inclusion and coincidence properties of spaces

These results represent advances in the research program initiated by Aron et al. [4] on the factorization of bounded holomorphic mappings in terms of an element of an operator ideal and a bounded holomorphic mapping.

Finally, we analyse holomorphic transposition of their elements and prove that every member of has relatively weakly compact range that becomes relatively compact whenever F is reflexive. We thus contribute to the study of holomorphic mappings with relatively (weakly) compact range, begun by Mujica [16] and continued in [12].

2 Notation and preliminaries

Throughout this paper, unless otherwise stated, E and F will denote complex Banach spaces and U an open subset of E.

We first introduce some notation. As usual, $B_E$ denotes the closed unit ball of E. For two vector spaces E and F,  L(E, F) stands for the vector space of all linear operators from E into F. In the case that E and F are normed spaces, represents the normed space of all bounded linear operators from E to F endowed with the canonical norm of operators. In particular, the algebraic dual $L(E,{\mathbb {K}})$ and the topological dual are denoted by $E^{\prime }$ and $E^*,$ respectively. For each $e\in E$ and $e^*\in E^{\prime },$ we frequently will write $\langle e^*,e\rangle$ instead of $e^*(e).$ We denote by $\kappa _E$ the canonical isometric embedding of E into $E^{**}$ defined by $\left\langle \kappa _E(e),e^*\right\rangle =\left\langle e^*,e\right\rangle$ for $e\in E$ and $e^*\in E^*.$ For a set $A\subseteq E,$ ${\textrm{co}}(A)$ denotes the convex hull of A.

We now recall some concepts and results of the theory of holomorphic mappings on Banach spaces.

Theorem 1.1

(See [18, 7 Theorem] and [15, Theorem 8.7]) Let E and F be complex Banach spaces and let U be an open set in E. For a mapping $f:U\rightarrow F,$ the following conditions are equivalent : 

• (i) For each $a\in U,$ there is an operator such that

  $$\begin{aligned} \lim _{x\rightarrow a}\frac{f(x)-f(a)-T(x-a)}{\left\| x-a\right\| }=0. \end{aligned}$$

• (ii) For each $a\in U,$ there exist an open ball $B(a,r)\subseteq U$ and a sequence of continuous m-homogeneous polynomials $(P_{m,a})_{m\in {\mathbb {N}}_0}$ from E into F such that

  $$\begin{aligned} f(x)=\sum _{m=0}^\infty P_{m,a}(x-a), \end{aligned}$$

  where the series converges uniformly for $x\in B(a,r).$

• (iii) f is G-holomorphic (that is,  for all $a\in U$ and $b\in E,$ the map $\lambda \mapsto f(a+\lambda b)$ is holomorphic on the open set $\{\lambda \in {\mathbb {C}}:a+\lambda b\in U\})$ and continuous. $\square$

A mapping $f:U\rightarrow F$ is said to be holomorphic if it verifies the equivalent conditions in Theorem 1.1. The mapping T in condition (i) is uniquely determined by f and a,  and is called the differential of f at a and denoted by Df(a).

A mapping $f:U\rightarrow F$ is locally bounded if f is bounded on a suitable neighborhood of each point of U. Given a Banach space E,  a subset $N\subseteq B_{E^*}$ is said to be norming for E if the function

$$\begin{aligned} N(x)=\sup \left\{ \left| x^*(x)\right| :x^*\in N\right\} \quad (x\in E) \end{aligned}$$

defines the norm on E.

If $U\subseteq E$ and $V\subseteq F$ are open sets, will represent the set of all holomorphic mappings from U to V. We will denote by the linear space of all holomorphic mappings from U into F and by the subspace of all such that f(U) is bounded in F. When $F={\mathbb {C}},$ then we will write

It is easy to prove that the linear space equipped with the supremum norm:

is a Banach space. Let denote the norm-closed linear hull in of the set $\left\{ \delta (x):x\in U\right\}$ of evaluation functionals defined by

In [16, 17], Mujica established the following properties of

Theorem 1.2

[16, Theorem 2.1] Let E be a complex Banach space and let U be an open set in E.

• (i) is isometrically isomorphic to via the evaluation mapping given by

  

• (ii) The mapping defined by $g_U(x)=\delta (x)$ is holomorphic with $\left\| g_U(x)\right\| =1$ for all $x\in U.$

• (iii) For each complex Banach space F and each mapping there exists a unique operator such that $T_f\circ g_U=f.$ Furthermore,  $\left\| T_f\right\| =\left\| f\right\| _{\infty }.$

• (iv) The mapping $f\mapsto T_f$ is an isometric isomorphism from onto

• (v) [16, Corollary 4.12] (see also [17, Theorem 5.1]). consists of all functionals of the form $\gamma =\sum _{i=1}^{\infty }\lambda _i\delta (x_i)$ with $(\lambda _i)_{i\ge 1}\in \ell _1$ and $(x_i)_{i\ge 1}\in U^\mathbb {N}.$ Moreover,  $\left\| \gamma \right\| =\inf \left\{ \sum _{i=1}^{\infty }\left| \lambda _i\right| \right\}$ where the infimum is taken over all such representations of $\gamma .$ $\square$

3 Cohen strongly p-summing holomorphic mappings

Let E and F be Banach spaces and $1\le p\le \infty .$ Let us recall [6] that an operator is p-summing if there exists a constant $C\ge 0$ such that, regardless of the natural number n and regardless of the choice of vectors $x_1,\ldots ,x_n$ in E,  we have the inequalities:

$$\begin{aligned} \left( \sum _{i=1}^n\left\| T(x_i)\right\| ^p\right) ^{\frac{1}{p}}&\le C \sup _{x^*\in B_{E^*}}\left( \sum _{i=1}^n\left| x^*(x_i)\right| ^p\right) ^{\frac{1}{p}}\quad&\text {if}\ {}&1\le p<\infty , \\ \max _{1\le i\le n}\left\| T(x_i)\right\|&\le C \sup _{x^*\in B_{E^*}}\left( \max _{1\le i\le n}\left| x^*(x_i)\right| \right) \quad&\text {if}\ {}&p=\infty . \end{aligned}$$

The infimum of such constants C is denoted by $\pi _p(T)$ and the linear space of all p-summing operators from E into F by $\Pi _p(E,F).$

The analogous notion for holomorphic mappings could be introduced as follows.

Definition 2.1

Let E and F be complex Banach spaces, let U be an open subset of E,  and let $1\le p\le \infty .$ A holomorphic mapping $f:U\rightarrow F$ is said to be p-summing if there exists a constant $C\ge 0$ such that for all $n\in {\mathbb {N}}$ and $x_1,\ldots ,x_n\in U,$ we have

We denote by the infimum of such constants C,  and by the set of all p-summing holomorphic mappings from U into F.

p-Summing holomorphic mappings are of little interest to us as with for all and furthermore the subclass of p-summing holomorphic mappings that we will study in this paper includes this case.

Let $1\le p\le \infty$ and let $p^*$ denote the conjugate index of p given by

$$\begin{aligned} p^*=\left\{ \begin{array}{ll} \frac{p}{p-1} &{}\quad \text {if}\ 1<p<\infty , \\ \infty &{} \quad \text {if}\ p=1, \\ 1 &{} \quad \text {if}\ p=\infty . \end{array} \right. \end{aligned}$$

In [5], Cohen introduced the following subclass of p-summing operators between Banach spaces: an operator is strongly p-summing if there exists a constant $C\ge 0$ such that for all $n\in {\mathbb {N}},$ $x_1,\ldots ,x_n\in E$ and $y^*_1,\ldots ,y^*_n\in F^*,$ we have

$$\begin{aligned} \sum _{i=1}^n\left| \left\langle y^*_i,T(x_i)\right\rangle \right|&\le C\left( \sum _{i=1}^n\left\| x_i\right\| \right) \sup _{y^{**}\in B_{F^{**}}}\left( \max _{1\le i\le n}\left| y^{**}(y^*_i)\right| \right) \quad&\text {if}\quad&p=1, \\ \sum _{i=1}^n\left| \left\langle y^*_i,T(x_i)\right\rangle \right|&\le C\left( \sum _{i=1}^n\left\| x_i\right\| ^p\right) ^{\frac{1}{p}} \sup _{y^{**}\in B_{F^{**}}}\left( \sum _{i=1}^n\left| y^{**}(y^*_i)\right| ^{p^*}\right) ^{\frac{1}{p^*}}\quad&\text {if}\quad&1<p<\infty , \\ \sum _{i=1}^n\left| \left\langle y^*_i,T(x_i)\right\rangle \right|&\le C\left( \max _{1\le i\le n}\left\| x_i\right\| \right) \sup _{y^{**}\in B_{F^{**}}}\left( \sum _{i=1}^n\left| y^{**}(y^*_i)\right| \right) \quad&\text {if}\quad&p=\infty . \end{aligned}$$

The infimum of such constants C is denoted by $d_p(T),$ and the space of all strongly p-summing operators from E into F by If $p=1,$ we have

We now introduce a version of this concept in the setting of holomorphic mappings.

Definition 2.2

Let E and F be complex Banach spaces, let U be an open subset of E,  and let $1\le p\le \infty .$ A holomorphic mapping $f:U\rightarrow F$ is said to be Cohen strongly p-summing if there exists a constant $C\ge 0$ such that for all $n\in {\mathbb {N}},$ $\lambda _1,\ldots ,\lambda _n\in {\mathbb {C}},$ $x_1,\ldots ,x_n\in U$ and $y^*_1,\ldots ,y^*_n\in F^*,$ we have

$$\begin{aligned} \sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle y^*_i,f(x_i)\right\rangle \right|&\le C\left( \sum _{i=1}^n\left| \lambda _i\right| \right) \sup _{y^{**}\in B_{F^{**}}}\left( \max _{1\le i\le n}\left| y^{**}(y^*_i)\right| \right) \quad&\text {if}\quad&p=1, \\ \sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle y^*_i,f(x_i)\right\rangle \right|&\le C\left( \sum _{i=1}^n\left| \lambda _i\right| ^p\right) ^{\frac{1}{p}} \sup _{y^{**}\in B_{F^{**}}}\left( \sum _{i=1}^n\left| y^{**}(y^*_i)\right| ^{p^*}\right) ^{\frac{1}{p^*}}\quad&\text {if}\quad&1<p<\infty , \\ \sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle y^*_i,f(x_i)\right\rangle \right|&\le C\left( \max _{1\le i\le n}\left| \lambda _i\right| \right) \sup _{y^{**}\in B_{F^{**}}}\left( \sum _{i=1}^n\left| y^{**}(y^*_i)\right| \right) \quad&\text {if}\quad&p=\infty . \end{aligned}$$

We denote by the infimum of such constants C,  and by the set of all Cohen strongly p-summing holomorphic mappings from U into F.

The introduction of the scalars $\lambda _i$ in the previous definition is justified by the assertion (v) of Theorem 1.2. Proposition 2.5 shows that

The concept of an ideal of bounded holomorphic mappings is inspired by the analogous one for bounded linear operators between Banach spaces [26, Section 8.2].

Definition 2.3

An ideal of bounded holomorphic mappings (or simply, a bounded-holomorphic ideal) is a subclass of the class of all bounded holomorphic mappings such that for each complex Banach space E,  each open subset U of E and each complex Banach space F,  the components

satisfy the following properties:

• (I1) is a linear subspace of

• (I2) For any and $y\in F,$ the mapping $g\cdot y:x\mapsto g(x)y$ from U to F is in

• (I3) The ideal property: If H, G are complex Banach spaces, V is an open subset of H,  and then $S\circ f\circ h$ is in

A bounded-holomorphic ideal is said to be normed (Banach) if there exists a function such that for every complex Banach space E,  every open subset U of E and every complex Banach space F,  the following conditions are satisfied:

• (N1) is a normed (Banach) space with for all

• (N2) for every and $y\in F,$

• (N3) If H, G are complex Banach spaces, V is an open subset of H,  and then

A normed bounded-holomorphic ideal is said to be regular if for any we have that with whenever

The following class of bounded holomorphic mappings appears involved in Definition 2.3.

Lemma 2.4

Let and $y\in F.$ The mapping $g\cdot y:U\rightarrow F,$ given by $(g\cdot y)(x)=g(x)y,$ belongs to with $\left\| g\cdot y\right\| _{\infty }=\left\| g\right\| _{\infty }\left\| y\right\| .$ $\square$

We are now ready to establish the main result of this section.

Proposition 2.5

is a regular Banach ideal of bounded holomorphic mappings. Furthermore,  with

Proof

We will only prove the case $1<p<\infty .$ The cases $p=1$ and $p=\infty$ follow similarly.

(N1) We first show that with for all Indeed, given we have

for all $x\in U$ and $y^*\in F^*.$ By Hahn–Banach Theorem, we obtain that for all $x\in U.$ Hence with

Let Given $n\in {\mathbb {N}},$ $\lambda _1,\ldots ,\lambda _n\in {\mathbb {C}},$ $x_1,\ldots ,x_n\in U$ and $y^*_1,\ldots ,y^*_n\in F^*,$ we have

Using the two inequalities above, we obtain

This tells us that with

Let $\lambda \in {\mathbb {C}}$ and Given $n\in {\mathbb {N}},$ $\lambda _i\in {\mathbb {C}},$ $x_i\in U$ and $y^*_i\in F^*$ for $i=1,\ldots ,n,$ we have

and thus with This implies that if $\lambda =0.$ For $\lambda \ne 0,$ we have hence and so

Moreover, if and then $\left\| f\right\| _{\infty }=0$ by (N1), and so $f=0.$ Thus, is a normed space.

To prove that is complete, it suffices to prove that every absolutely convergent series is convergent. So let $(f_n)_{n\in {\mathbb {N}}}$ be a sequence in such that is convergent. Since for all $n\in {\mathbb {N}}$ and is a Banach space, then $\sum _{n\in {\mathbb {N}}}f_n$ converges in to a function Given $m\in {\mathbb {N}},$ $x_1,\ldots ,x_m\in U,$ $y^*_1,\ldots ,y^*_m\in F^*$ and $\lambda _1,\ldots ,\lambda _m\in {\mathbb {C}},$ we have

for all $n\in {\mathbb {N}},$ and by taking limits with $n\rightarrow \infty$ yields

Hence with Moreover, we have

for all $n\in {\mathbb {N}},$ and thus f is the -limit of the series $\sum _{n\in {\mathbb {N}}}f_n.$

(N2) Let and $y\in F.$ If $y=0,$ there is nothing to prove. Assume $y\ne 0.$ By Lemma 2.4, Given $n\in {\mathbb {N}},$ $x_1,\ldots ,x_n\in U,$ $y^*_1,\ldots ,y^*_n\in F^*$ and $\lambda _1,\ldots ,\lambda _n\in {\mathbb {C}},$ we have

$$\begin{aligned}&\sum _{i=1}^n \left| \lambda _i\right| \left| \left\langle y^*_i,(g\cdot y)(x_i)\right\rangle \right| \\&\quad =\sum _{i=1}^n \left| \lambda _i\right| \left| g(x_i)\right| \left| \left\langle y^*_i,y\right\rangle \right| \\&\quad \le \left\| g\right\| _{\infty }\left\| y\right\| \sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle y^*_i,\frac{y}{\left\| y\right\| }\right\rangle \right| \\&\quad \le \left\| g\right\| _{\infty }\left\| y\right\| \left( \sum _{i=1}^n\left| \lambda _i\right| ^p\right) ^{\frac{1}{p}}\left( \sum _{i=1}^n\left| \left\langle y^*_i,\frac{y}{\left\| y\right\| }\right\rangle \right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&\quad =\left\| g\right\| _{\infty }\left\| y\right\| \left( \sum _{i=1}^n\left| \lambda _i\right| ^p\right) ^{\frac{1}{p}}\left( \sum _{i=1}^n\left| \left\langle \kappa _F\left( \frac{y}{\left\| y\right\| }\right) ,y^*_i,\right\rangle \right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&\quad \le \left\| g\right\| _{\infty }\left\| y\right\| \left( \sum _{i=1}^n\left| \lambda _i\right| ^p\right) ^{\frac{1}{p}}\sup _{y^{**}\in B_{F^{**}}}\left( \sum _{i=1}^n\left| y^{**}(y^*_i)\right| ^{p^*}\right) ^{\frac{1}{p^*}} \end{aligned}$$

by applying the Hölder inequality, and therefore with Conversely, by applying what was proved in (N1), we have

(N3) Let H, G be complex Banach spaces, V be an open subset of H,  and We can suppose $S\ne 0.$ Given $n\in {\mathbb {N}},$ $x_1,\ldots ,x_n\in U,$ $y^*_1,\ldots ,y^*_n\in G^*$ and $\lambda _1,\ldots ,\lambda _n\in {\mathbb {C}},$ we have

and therefore with

We now prove that the ideal is regular. Let and assume that Given $n\in {\mathbb {N}},$ $x_1,\ldots ,x_n\in U,$ $y^*_1,\ldots ,y^*_n\in F^*$ and $\lambda _1,\ldots ,\lambda _n\in {\mathbb {C}},$ we have

and thus with The reverse inequality follows from (N3).

Finally, we have seen in (N1) that with for all For the converse, let If $f=0,$ there is nothing to prove. Assume $f\ne 0.$ Given $n\in {\mathbb {N}},$ $x_1,\ldots ,x_n\in U,$ $y^*_1,\ldots ,y^*_n\in F^*$ and $\lambda _1,\ldots ,\lambda _n\in {\mathbb {C}},$ we have

$$\begin{aligned} \sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle y^*_i,f(x_i)\right\rangle \right|&=\left\| f\right\| _{\infty }\sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle \kappa _F\left( \frac{f(x_i)}{\left\| f\right\| _{\infty }}\right) ,y^*_i\right\rangle \right| \\&\le \left\| f\right\| _{\infty }\left( \sum _{i=1}^n\left| \lambda _i\right| \right) \max _{1\le i\le n}\left( \sup _{y^{**}\in B_{F^{**}}}\left| y^{**}(y^*_i)\right| \right) \\&=\left\| f\right\| _{\infty }\left( \sum _{i=1}^n\left| \lambda _i\right| \right) \sup _{y^{**}\in B_{F^{**}}}\left( \max _{1\le i\le n}\left| y^{**}(y^*_i)\right| \right) , \end{aligned}$$

and therefore with $\square$

4 The tensor product $\Delta (U)\otimes F$

We introduce $\Delta (U)\otimes F$ as a space of linear functionals on

Definition 3.1

Let E and F be complex Banach spaces and let U be an open subset of E. For each $x\in U,$ let be the linear functional defined by

Let $\Delta (U)$ be the linear subspace of spanned by the set $\left\{ \delta (x):x\in U\right\} .$

For any $x\in U$ and $y\in F,$ let be the linear functional given by

We define the tensor product $\Delta (U)\otimes F$ as the linear subspace of spanned by the set

$$\begin{aligned} \left\{ \delta (x)\otimes y:x\in U,\, y\in F\right\} . \end{aligned}$$

We say that $\delta (x)\otimes y$ is an elementary tensor of $\Delta (U)\otimes F.$ Note that each element u in $\Delta (U)\otimes F$ is of the form $u=\sum _{i=1}^n\lambda _i(\delta (x_i)\otimes y_i),$ where $n\in {\mathbb {N}},$ $\lambda _i\in {\mathbb {C}},$ $x_i\in U$ and $y_i\in F$ for $i=1,\ldots ,n.$ This representation of u is not unique. It is worth noting that each element u of $\Delta (U)\otimes F$ can be represented as $u=\sum _{i=1}^n \delta (x_i)\otimes y_i$ since $\lambda (\delta (x)\otimes y)=\delta (x)\otimes (\lambda y).$

As a straightforward consequence from Definition 3.1, we describe the action of a tensor u in $\Delta (U)\otimes F$ on a function f in :

Lemma 3.2

Let $u=\sum _{i=1}^n \lambda _i\delta (x_i)\otimes y_i\in \Delta (U)\otimes F$ and Then

$$\begin{aligned} u(f)=\sum _{i=1}^n\lambda _i\left\langle f(x_i),y_i\right\rangle . \end{aligned}$$

$\Box$

The following characterization of the zero tensor of $\Delta (U)\otimes F$ follows immediately from [26, Proposition 1.2].

Proposition 3.3

If $u=\sum _{i=1}^n\delta (x_i)\otimes y_i\in \Delta (U)\otimes F,$ the following are equivalent : 

• (i) $u=0.$

• (ii) $\sum _{i=1}^ng(x_i)\phi (y_i)=0$ for every and $\phi \in B_{F^*}.$ $\Box$

By Definition 3.1, $\Delta (U)\otimes F$ is a linear subspace of In fact, we have:

Proposition 3.4

forms a dual pair,  where the bilinear form $\left\langle \cdot ,\cdot \right\rangle$ associated to the dual pair is given by

$$\begin{aligned} \left\langle u,f\right\rangle =\sum _{i=1}^n \lambda _i\left\langle f(x_i),y_i\right\rangle \end{aligned}$$

for $u=\sum _{i=1}^n\lambda _i \delta (x_i)\otimes y_i\in \Delta (U)\otimes F$ and

Proof

Since $\langle u,f\rangle =u(f)$ by Lemma 3.2, it is immediate that $\langle \cdot ,\cdot \rangle$ is a well-defined bilinear map from to ${\mathbb {C}.}$ On the one hand, if $u\in \Delta (U)\otimes F$ and $\langle u,f\rangle =0$ for all then $u=0$ follows easily from Proposition 3.3, and thus separates points of $\Delta (U)\otimes F.$ On the other hand, if and $\langle u,f\rangle =0$ for all $u\in \Delta (U)\otimes F,$ then $\left\langle f(x),y\right\rangle =\left\langle \delta (x)\otimes y,f\right\rangle =0$ for all $x\in U$ and $y\in F,$ this means that $f=0$ and thus $\Delta (U)\otimes F$ separates points of $\square$

Since is a dual pair, we can identify with a linear subspace of $(\Delta (U)\otimes F)^{\prime }$ as follows.

Corollary 3.5

For each the functional $\Lambda _0(f):\Delta (U)\otimes F\rightarrow {\mathbb {C}},$ given by

$$\begin{aligned} \Lambda _0(f)(u)=\sum _{i=1}^n\lambda _i\left\langle f(x_i),y_i\right\rangle \end{aligned}$$

for $u=\sum _{i=1}^n \lambda _i\delta (x_i)\otimes y_i\in \Delta (U)\otimes F,$ is linear. We will say that $\Lambda _0(f)$ is the linear functional on $\Delta (U)\otimes F$ associated to f. Furthermore,  the map $f\mapsto \Lambda _0(f)$ is a linear monomorphism from into $(\Delta (U)\otimes F)^{\prime }.$

Proof

Let Note that $\Lambda _0(f)(u)=\left\langle u,f\right\rangle$ for all $u\in \Delta (U)\otimes F.$ It is immediate that $\Lambda _0(f)$ is a well-defined linear functional on $\Delta (U)\otimes F$ and that $f\mapsto \Lambda _0(f)$ from into $(\Delta (U)\otimes F)^{\prime }$ is a well-defined linear map. Finally, let and assume that $\Lambda _0(f)=0.$ Then $\left\langle u,f\right\rangle =0$ for all $u\in \Delta (U)\otimes F.$ Since $\Delta (U)\otimes F$ separates points of by Proposition 3.4, it follows that $f=0$ and this proves that $\Lambda _0$ is one-to-one. $\square$

Given two linear spaces E and F,  the tensor product space $E\otimes F$ equipped with a norm $\alpha$ will be denoted by $E\otimes _{\alpha } F,$ and the completion of $E\otimes _{\alpha } F$ by $E{\widehat{\otimes }}_{\alpha } F.$ If E and F are normed spaces, a cross-norm on $E\otimes F$ is a norm $\alpha$ such that $\alpha (x\otimes y)=\left\| x\right\| \left\| y\right\|$ for all $x\in E$ and $y\in F.$

Given two normed spaces E and F,  the projective norm $\pi$ on $E\otimes F$ (see [26, Chapter 2]) takes the following form on $\Delta (U)\otimes F$:

$$\begin{aligned} \pi (u)=\inf \left\{ \sum _{i=1}^n\left| \lambda _i\right| \left\| y_i\right\| :u=\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\right\} \quad (u\in \Delta (U)\otimes F), \end{aligned}$$

where the infimum is taken over all the representations of u as above.

We next see that, on the space $\Delta (U)\otimes F,$ the projective norm and the norm induced by the dual norm of the supremum norm of coincide.

Theorem 3.6

The linear space $\Delta (U)\otimes F$ is contained in Moreover,  $\pi (u)=H(u)$ for all $u\in \Delta (U)\otimes F,$ where H is the norm on $\Delta (U)\otimes F$ defined by

Proof

Let $\lambda \in {\mathbb {C}},$ $x\in U$ and $y\in F.$ Since $\lambda \delta (x)\otimes y$ is a linear map on and

$$\begin{aligned} \left| (\lambda \delta (x)\otimes y)(f)\right| =\left| \lambda \left\langle f(x),y\right\rangle \right| \le \left| \lambda \right| \left\| f(x)\right\| \left\| y\right\| \le \left| \lambda \right| \left\| f\right\| _{\infty }\left\| y\right\| \end{aligned}$$

for all then with $\left\| \lambda \delta (x)\otimes y\right\| \le \left| \lambda \right| \left\| y\right\| ,$ and thus

Let $u\in \Delta (U)\otimes F$ and let $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i$ be a representation of u. Since u is linear and

$$\begin{aligned} \left| u(f)\right| =\left| \sum _{i=1}^n\lambda _i\left\langle f(x_i),y_i\right\rangle \right| \le \sum _{i=1}^n\left| \lambda _i\right| \left\| f(x_i)\right\| \left\| y_i\right\| \le \left\| f\right\| _{\infty }\sum _{i=1}^n\left| \lambda _i\right| \left\| y_i\right\| \end{aligned}$$

for all we deduce that $H(u)\le \sum _{i=1}^n\left| \lambda _i\right| \left\| y_i\right\| .$ Since this holds for each representation of u,  it follows that $H(u)\le \pi (u).$ Hence, $H\le \pi .$ To prove that the reverse inequality, suppose by contradiction that $H(u_0)<1<\pi (u_0)$ for some $u_0\in \Delta (U)\otimes F.$ Denote $B=\{u\in \Delta (U)\otimes F:\pi (u)\le 1\}.$ Clearly, B is a closed and convex set in $\Delta (U)\otimes _\pi F.$ Applying the Hahn–Banach Separation Theorem to B and $\{u_0\},$ we obtain a functional $\eta \in (\Delta (U)\otimes _\pi F)^*$ such that

$$\begin{aligned} 1=\Vert \eta \Vert =\sup \{{\textrm{Re}}\,\eta (u):u\in B \} < {\textrm{Re}}\,\eta (u_0). \end{aligned}$$

Define $f:U\rightarrow F^*$ by $\langle f(x),y\rangle =\eta \left( \delta (x)\otimes y\right)$ for all $y\in F$ and $x\in U.$ It is easy to prove that f is well defined and with $\left\| f\right\| _{\infty }\le 1.$ Moreover, $u(f)=\eta (u)$ for all $u\in \Delta (U)\otimes F.$ Therefore $H(u_0)\ge |u_0(f)|\ge {\textrm{Re}}\,u_0(f)={\textrm{Re}}\,\eta (u_0),$ so $H(u_0)>1$ and this is a contradiction. $\square$

We now will define the Chevet–Saphar norms on the tensor product $E\otimes F.$ Let E and F be normed spaces and let $1\le p\le \infty .$ Given $u=\sum _{i=1}^n x_i\otimes y_i\in E\otimes F,$ denote

$$\begin{aligned} \left\| (x_1,\ldots ,x_n)\right\| _{\ell ^n_p(E)} =\left\{ \begin{array}{ll} {\left( \sum _{i=1}^n\left\| x_i\right\| ^p\right) ^{\frac{1}{p}}} &{} \quad \text {if } 1\le p<\infty , \\ \max _{1\le i\le n}\left\| x_i\right\| &{} \quad \text {if }p=\infty , \end{array}\right. \end{aligned}$$

and

$$\begin{aligned} \left\| (y_1,\ldots , y_n)\right\| _{\ell ^{n,w}_{p}(F)} =\left\{ \begin{array}{ll} \sup _{y^*\in B_{F^*}}\left( \sum _{i=1}^n\left| y^*(y_i)\right| ^p\right) ^{\frac{1}{p}} &{}\quad \text {if}\ 1\le p<\infty ,\\ \sup _{y^*\in B_{F^*}}\left( \max _{1\le i\le n}\left| y^*(y_i)\right| \right) &{}\quad \text {if } p=\infty . \end{array}\right. \end{aligned}$$

If $E=F={\mathbb {C}},$ we write $\ell ^n_p(E)=\ell ^n_p$ and $\ell ^{n,w}_{p^*}(F)=\ell ^{n,w}_{p^*}.$ According to [26, Section 6.2], the Chevet–Saphar norms are defined on $E\otimes F$ by

$$\begin{aligned} d_p(u)&=\inf \left\{ \left\| (x_1,\ldots ,x_n)\right\| _{\ell ^{n,w}_{p^*}(E)}\left\| (y_1,\ldots , y_n)\right\| _{\ell ^n_p(F)}\right\} , \\ g_p(u)&=\inf \left\{ \left\| (x_1,\ldots ,x_n)\right\| _{\ell ^n_p(E)}\left\| (y_1,\ldots , y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}\right\} , \end{aligned}$$

the infimum being taken over all representations of u as $u=\sum _{i=1}^nx_i\otimes y_i\in E\otimes F.$

Since $\left\| \delta (x)\right\| =1$ for all $x\in U,$ the norm $g_p$ on $\Delta (U)\otimes F$ takes the form:

$$\begin{aligned} g_p(u)=\inf \left\{ \left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left\| (y_1,\ldots , y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}:u=\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\right\} . \end{aligned}$$

Notice that $g_p$ is a cross-norm on $\Delta (U)\otimes F.$

We next show that $g_1$ on $\Delta (U)\otimes F$ is just the projective tensor norm $\pi .$

Proposition 3.7

$g_1(u)=\pi (u)$ for all $u\in \Delta (U)\otimes F.$

Proof

Let $u\in \Delta (U)\otimes F$ and let $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i$ be a representation of u. We have

$$\begin{aligned} \pi (u)&\le \sum _{i=1}^n |\lambda _i|\left\| y_i\right\| =\sum _{i=1}^n|\lambda _i|\left( \sup _{y^*\in B_{F^*}}\left| y^*(y_i)\right| \right) \\&\le \sum _{i=1}^n|\lambda _i|\max _{1\le i\le n}\left( \sup _{y^*\in B_{F^*}}\left| y^*(y_i)\right| \right) =\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_1}\left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{\infty }(F)}, \end{aligned}$$

and taking the infimum over all representations of u gives $\pi (u)\le g_1(u).$ For the reverse inequality, notice that $g_1(\lambda \delta (x)\otimes y)\le |\lambda |\left\| y\right\|$ for all $\lambda \in {\mathbb {C}},$ $x\in U$ and $y\in F.$ Since $g_1$ is a norm on $\Delta (U)\otimes F,$ it follows that

$$\begin{aligned} g_1(u) =g_1\left( \sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\right) \le \sum _{i=1}^ng_1\left( \lambda _i\delta (x_i)\otimes y_i\right) \le \sum _{i=1}^n\left| \lambda _i\right| \left\| y_i\right\| \end{aligned}$$

and taking the infimum over all representations of u yields $g_1(u)\le \pi (u).$ $\square$

5 Duality for Cohen strongly p-summing holomorphic mappings

We now show that the duals of the tensor product can be canonically identified as spaces of Cohen strongly p-summing holomorphic mappings.

Theorem 4.1

Let $1\le p\le \infty .$ Then is isometrically isomorphic to via the mapping defined by

$$\begin{aligned} \Lambda (f)(u)=\sum _{i=1}^n\lambda _i\left\langle f(x_i),y_i\right\rangle \end{aligned}$$

for and $u=\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\in \Delta (U)\otimes F.$ Furthermore,  its inverse is given by

$$\begin{aligned} \left\langle \Lambda ^{-1}(\varphi )(x),y\right\rangle =\left\langle \varphi ,\delta (x)\otimes y\right\rangle \end{aligned}$$

for $x\in U$ and $y\in F.$

Proof

We prove it for $1<p\le \infty .$ The case $p=1$ is similarly proved.

Let and let $\Lambda _0(f):\Delta (U)\otimes F\rightarrow {\mathbb {C}}$ be its associate linear functional. We claim that $\Lambda _0(f)\in (\Delta (U)\otimes _{g_p} F)^*$ with Indeed, given $u=\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\in \Delta (U)\otimes F,$ we have

and taking infimum over all the representations of u,  we deduce that Since u was arbitrary, then $\Lambda _0(f)$ is continuous on $\Delta (U) \otimes _{g_p} F$ with as claimed.

Since $\Delta (U)$ is a norm-dense linear subspace of and $g_p$ is a cross-norm on then $\Delta (U)\otimes F$ is a dense linear subspace of and therefore also of its completion Hence there is a unique continuous mapping $\Lambda (f)$ from to ${\mathbb {C}}$ that extends $\Lambda _0(f).$ Further, $\Lambda (f)$ is linear and $\left\| \Lambda (f)\right\| =\left\| \Lambda _0(f)\right\| .$

Let be the mapping so defined. Since the mapping is a linear monomorphism by Corollary 3.5, it follows easily that $\Lambda$ is so. To prove that $\Lambda$ is a surjective isometry, let and define $f_{\varphi }:U\rightarrow F^*$ by

$$\begin{aligned} \left\langle f_{\varphi }(x),y\right\rangle =\varphi (\delta (x)\otimes y)\quad \left( x\in U,\; y\in F\right) . \end{aligned}$$

Given $x\in U,$ the linearity of both $\varphi$ and the product tensor in the second variable yields that the functional $f_{\varphi }(x):F\rightarrow {\mathbb {C}}$ is linear, and since

$$\begin{aligned} \left| \left\langle f_{\varphi }(x),y\right\rangle \right| =\left| \varphi (\delta (x)\otimes y)\right| \le \left\| \varphi \right\| g_p(\delta (x)\otimes y) \le \left\| \varphi \right\| \left\| y\right\| \end{aligned}$$

for all $y\in F,$ we deduce that $f_{\varphi }(x)\in F^*$ with $\Vert f_{\varphi }(x)\Vert \le \left\| \varphi \right\| .$ Since x was arbitrary, we have that $f_{\varphi }$ is bounded with $\left\| f_{\varphi }\right\| _{\infty }\le \left\| \varphi \right\| .$

We now prove that $f_{\varphi }:U\rightarrow F^*$ is holomorphic. To this end, we first claim that, for every $y\in F,$ the function $f_y:U\rightarrow {\mathbb {C}}$ defined by

$$\begin{aligned} f_y(x)=\varphi (\delta (x)\otimes y)\quad (x\in U) \end{aligned}$$

is holomorphic. Let $a\in U.$ Since is holomorphic by Theorem 1.2, there exists such that

$$\begin{aligned} \lim _{x\rightarrow a}\frac{\delta (x)-\delta (a)-Dg_U(a)(x-a)}{\left\| x-a\right\| }=0. \end{aligned}$$

Consider the function $T(a):E\rightarrow {\mathbb {C}}$ given by

$$\begin{aligned} T(a)(x)=\varphi (Dg_U(a)(x)\otimes y)\quad \left( x\in E\right) . \end{aligned}$$

Clearly, $T(a)\in E^*$ and since

$$\begin{aligned}&f_y(x)-f_y(a)-T(a)(x-a)\\&\quad =\varphi (\delta (x)\otimes y)-\varphi (\delta (a)\otimes y)-\varphi (Dg_U(a)(x-a)\otimes y) \\&\quad =\varphi \left( (\delta (x)-\delta (a)-Dg_U(a)(x-a))\otimes y\right) , \end{aligned}$$

it follows that

$$\begin{aligned} \lim _{x\rightarrow a}\frac{f_y(x)-f_y(a)-T(a)(x-a)}{\left\| x-a\right\| }&=\lim _{x\rightarrow a}\frac{\varphi \left( (\delta (x)-\delta (a)-Dg_U(a)(x-a))\otimes y\right) }{\left\| x-a\right\| } \\&=\lim _{x\rightarrow a}\varphi \left( \frac{\delta (x)-\delta (a)-Dg_U(a)(x-a)}{\left\| x-a\right\| }\otimes y\right) \\&=\varphi (0\otimes y)=\varphi (0)=0. \end{aligned}$$

Hence, $f_y$ is holomorphic at a with $Df_y(a)=T(a),$ and this proves our claim. Now, notice that the set $\left\{ \kappa _F(y):y\in B_F\right\} \subseteq B_{F^{**}}$ is norming for $F^*$ since

$$\begin{aligned} \left\| y^*\right\| =\sup \left\{ \left| y^*(y)\right| :y\in B_F\right\} =\sup \left\{ \left| \kappa _F(y)(y^*)\right| :y\in B_F\right\} \end{aligned}$$

for every $y^*\in F^*,$ and that $\kappa _F(y)\circ f_{\varphi }=f_y$ for every $y\in F$ since

$$\begin{aligned} (\kappa _F(y)\circ f_{\varphi })(x)=\kappa _F(y)(f_{\varphi }(x))=\left\langle f_{\varphi }(x),y\right\rangle =\varphi (\delta (x)\otimes y)=f_y(x) \end{aligned}$$

for all $x\in U.$

We are now ready to show that $f_{\varphi }:U\rightarrow F^*$ is holomorphic. Indeed, let $a\in U$ and $b\in E.$ Denote $V=\left\{ \lambda \in {\mathbb {C}} :a+\lambda b\in U\right\} .$ Clearly, the mapping $h:V\rightarrow U$ given by $h(\lambda )=a+\lambda b$ is holomorphic. Since $f_{\varphi }\circ h$ is locally bounded and $\kappa _F(y)\circ (f_{\varphi }\circ h)=f_y\circ h$ is holomorphic on the open set $V\subseteq {\mathbb {C}}$ for all $y\in F,$ Proposition A.3 in [3] assures that $f_{\varphi }\circ h$ is holomorphic. This means that $f_{\varphi }$ is G-holomorphic but since it is also locally bounded, we deduce that $f_{\varphi }$ is continuous by [15, Proposition 8.6]. Now, we conclude that $f_{\varphi }$ is holomorphic by Theorem 1.1.

We now prove that To see this, take $n\in {\mathbb {N}},$ $\lambda _i\in {\mathbb {C}},$ $x_i\in U$ and $y^{**}_i\in F^{**}$ for $i=1,\ldots ,n.$ Let $\varepsilon >0$ and consider the finite-dimensional subspaces $V={\textrm{lin}}\{y^{**}_1,\ldots ,y^{**}_n\} \subseteq F^{**}$ and $W={\textrm{lin}}\{f_{\varphi }(x_1),\ldots ,f_{\varphi }(x_n) \}\subseteq F^{*}.$ The principle of local reflexivity [6, Theorem 8.16] gives us a bounded linear operator $T_{(\varepsilon , V,W)}:V\rightarrow F$ such that

• (i) $T_{(\varepsilon , V,W)}(y^{**})=y^{**}$ for every $y^{**}\in V\cap \kappa _F(F),$

• (ii) $(1-\varepsilon )\left\| y^{**}\right\| \le \left\| T_{(\varepsilon ,V,W)}(y^{**})\right\| \le (1+\varepsilon )\left\| y^{**}\right\|$ for every $y^{**}\in V,$

• (iii) $\left\langle y^*,T_{(\varepsilon , V,W)}(y^{**})\right\rangle =\left\langle y^{**},y^* \right\rangle$ for every $y^{**}\in V$ and $y^*\in W.$

Using (iii) and taking $y_i=T_{(\varepsilon , V,W)}(y_i^{**}),$ we first have

$$\begin{aligned} \left| \sum _{i=1}^n\lambda _i\left\langle y_i^{**},f_{\varphi }(x_i)\right\rangle \right|&=\left| \sum _{i=1}^n\lambda _i\left\langle f_{\varphi }(x_i),T_{(\varepsilon ,V,W)}(y_i^{**})\right\rangle \right| \\&=\left| \sum _{i=1}^n\lambda _i\left\langle f_{\varphi }(x_i),y_i\right\rangle \right| \\&=\left| \varphi \left( \sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\right) \right| \\&\le \left\| \varphi \right\| g_p\left( \sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\right) \\&\le \left\| \varphi \right\| \left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}. \end{aligned}$$

Since

$$\begin{aligned} \left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}&=\sup _{y^{*}\in B_{F^{*}}}\left( \sum _{i=1}^n\left| y^{*}(y_i)\right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&=\sup _{y^{*}\in B_{F^{*}}}\left( \sum _{i=1}^n\left| \left\langle y^{*},T_{(\varepsilon ,V,W)}(y^{**}_i)\right\rangle \right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&=\sup _{y^{*}\in B_{F^{*}}}\left( \sum _{i=1}^n\left| \left\langle \kappa _F(T_{(\varepsilon ,V,W)}(y^{**}_i)),y^{*}\right\rangle \right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&\le \left\| \kappa _F\circ T_{(\varepsilon , V,W)}\right\| \sup _{y^{*}\in B_{F^{*}}}\left( \sum _{i=1}^n\left| \left\langle y^{**}_i,y^{*}\right\rangle \right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&=\left\| T_{(\varepsilon , V,W)}\right\| \sup _{y^{*}\in B_{F^{*}}}\left( \sum _{i=1}^n\left| \left\langle \kappa _{F^*}(y^*),y^{**}_i\right\rangle \right| ^{p^*}\right) ^{\frac{1}{p^*}}\\&\le (1+\varepsilon )\sup _{y^{***}\in B_{F^{***}}}\left( \sum _{i=1}^n\left| y^{***}(y^{**}_i)\right| ^{p^*}\right) ^{\frac{1}{p^*}}\\&=(1+\varepsilon )\left\| (y^{**}_1,\ldots ,y^{**}_n)\right\| _{\ell ^{n,w}_{p^*}(F^{**})}, \end{aligned}$$

it follows that

$$\begin{aligned} \left| \sum _{i=1}^n\lambda _i\left\langle y_i^{**},f_{\varphi }(x_i)\right\rangle \right| \le \left\| \varphi \right\| \left\| (\lambda _1,\ldots ,\lambda _n) \right\| _{\ell ^n_p}(1+\varepsilon )\left\| (y^{**}_1,\ldots ,y^{**}_n) \right\| _{\ell ^{n,w}_{p^*}(F^{**})}. \end{aligned}$$

By the arbitrariness of $\varepsilon ,$ we deduce that

$$\begin{aligned} \left| \sum _{i=1}^n\lambda _i\left\langle y_i^{**},f_{\varphi }(x_i)\right\rangle \right| \le \left\| \varphi \right\| \left\| (\lambda _1,\ldots ,\lambda _n) \right\| _{\ell ^n_p}\left\| (y^{**}_1,\ldots ,y^{**}_n)\right\| _{\ell ^{n,w}_{p^*}(F^{**})}, \end{aligned}$$

and this implies that with

For any $u=\sum _{i=1}^n \lambda _i\delta (x_i)\otimes y_i\in \Delta (U)\otimes F,$ we get

$$\begin{aligned} \Lambda (f_{\varphi })(u) =\sum _{i=1}^n\lambda _i\left\langle f_{\varphi }(x_i),y_i\right\rangle =\sum _{i=1}^n\lambda _i\varphi (\delta (x_i)\otimes y_i) =\varphi \left( \sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\right) =\varphi (u). \end{aligned}$$

Hence $\Lambda (f_{\varphi })=\varphi$ on a dense subspace of and, consequently, $\Lambda (f_{\varphi })=\varphi ,$ which shows the last statement of the theorem. Moreover, and the theorem holds. $\square$

In particular, in view of Theorem 4.1 and taking into account Propositions 2.5, 3.6 and 3.7, we can identify the space with the dual space of

Corollary 4.2

is isometrically isomorphic to via the mapping given by

$$\begin{aligned} \Lambda (f)(u)=\sum _{i=1}^n\lambda _i\left\langle f(x_i),y_i\right\rangle \end{aligned}$$

for and $u=\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i\in \Delta (U)\otimes F.$ Furthermore,  its inverse is given by

$$\begin{aligned} \left\langle \Lambda ^{-1}(\varphi )(x),y\right\rangle =\left\langle \varphi ,\delta (x)\otimes y\right\rangle \end{aligned}$$

for $x\in U$ and $y\in F.$ $\square$

Remark 4.3

It is known (see [26, p. 24]) that if E and F are Banach spaces, then is isometrically isomorphic to $(E{\widehat{\otimes }}_\pi F)^*,$ via given by

$$\begin{aligned} \left\langle \Phi (T),\sum _{i=1}^n x_i\otimes y_i\right\rangle =\sum _{i=1}^n\left\langle T(x_i),y_i\right\rangle \end{aligned}$$

for and $\sum _{i=1}^n x_i\otimes y_i\in E\otimes F.$ Notice that the identification $\Lambda$ in Corollary 4.2 is just $\Phi \circ \Phi _0,$ where $\Phi _0:f\mapsto T_f$ is the isometric isomorphism from onto given in Theorem 1.2.

6 Pietsch domination for Cohen strongly p-summing holomorphic mappings

In [22], Pietsch established a domination theorem for p-summing linear operators between Banach spaces. To present a version of this theorem for Cohen strongly p-summing holomorphic mappings on Banach spaces, we first characterize the elements of the dual space of $\Delta (U)\otimes _{g_p} F.$

Theorem 5.1

Let $\varphi \in (\Delta (U)\otimes F)^{\prime },$ $C>0$ and $1<p\le \infty .$ The following conditions are equivalent : 

• (i) $\left| \varphi (u)\right| \le Cg_p(u)$ for all $u\in \Delta (U)\otimes F.$

• (ii) For any representation $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i$ of $u\in \Delta (U)\otimes F,$ we have

  $$\begin{aligned} \sum _{i=1}^n\left| \varphi (\lambda _i\delta (x_i)\otimes y_i)\right| \le Cg_p(u). \end{aligned}$$

• (iii) There exists a Borel regular probability measure $\mu$ on $B_{F^{*}}$ such that

  $$\begin{aligned} \left| \varphi (\lambda \delta (x)\otimes y)\right| \le C\left| \lambda \right| \left\| y\right\| _{L_{p^*}(\mu )} \end{aligned}$$

  for all $\lambda \in {\mathbb {C}},$ $x\in U$ and $y\in F,$ where

  $$\begin{aligned} \left\| y\right\| _{L_{p^*}(\mu )}=\left( \int _{B_{F^{*}}}\left| y^*(y)\right| ^{p^*}\ {\textrm{d}}\mu (y^*)\right) ^{\frac{1}{p^*}}. \end{aligned}$$

Proof

$(\text {i})\ \Rightarrow \ (\text {ii})$: Let $u\in \Delta (U)\otimes F$ and let $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i$ be a representation of u. It is elementary that the function $T:{\mathbb {C}}^n\rightarrow {\mathbb {C}}$ defined by

$$\begin{aligned} T(t_1,\ldots ,t_n)=\sum _{i=1}^n t_i\varphi (\lambda _i\delta (x_i)\otimes y_i),\quad \forall (t_1,\ldots ,t_n)\in {\mathbb {C}}^n \end{aligned}$$

is linear and continuous on $({\mathbb {C}}^n,\left\| \cdot \right\| _{\ell _{\infty }^n})$ with $\left\| T\right\| =\sum _{i=1}^n\left| \varphi (\lambda _i\delta (x_i)\otimes y_i)\right| .$

For any $(t_1,\ldots ,t_n)\in {\mathbb {C}}^n$ with $\left\| (t_1,\ldots ,t_n)\right\| _{\ell _{\infty }^n}\le 1,$ by (i) we have

$$\begin{aligned} \left| T(t_1,\ldots ,t_n)\right|&=\left| \sum _{i=1}^nt_i\varphi (\lambda _i\delta (x_i)\otimes y_i)\right| =\left| \varphi \left( \sum _{i=1}^nt_i\lambda _i\delta (x_i)\otimes y_i\right) \right| \\&\le C g_p\left( \sum _{i=1}^nt_i\lambda _i\delta (x_i)\otimes y_i\right) \\&\le C\left\| (t_1\lambda _1,\ldots ,t_n\lambda _n)\right\| _{\ell ^n_p}\left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{p^*}(F)} \\&\le C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}, \end{aligned}$$

and, therefore,

$$\begin{aligned} \sum _{i=1}^n\left| \varphi (\lambda _i\delta (x_i)\otimes y_i)\right| \le C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}. \end{aligned}$$

Taking infimum over all the representations of u,  we deduce that

$$\begin{aligned} \sum _{i=1}^n\left| \varphi (\lambda _i\delta (x_i)\otimes y_i)\right| \le C g_p(u). \end{aligned}$$

$(\text {ii})\ \Rightarrow \ (\text {iii})$: Let be the set of all Borel regular probability measures $\mu$ on $B_{F^*}.$ Clearly, it is a convex compact subset of $(C(B_{F^*})^*,w^*).$ Assume first $1<p<\infty .$ Let M be set of all functions from to ${\mathbb {R}}$ of the form

$$\begin{aligned}{} & {} f_{((\lambda _{i})_{i=1}^{n},(x_{i})_{i=1}^{n},(y_{i})_{i=1}^{n})}(\mu ) =\sum _{i=1}^{n}\left| \varphi (\lambda _{i}\delta _{U}(x_{i})\otimes y_{i})\right| \\{} & {} \quad -\left( \frac{C}{p}\left\| (\lambda _{i})_{i=1}^{n}\right\| _{\ell _{p}^{n}}^{p} +\frac{C}{p^*}\sum _{i=1}^{n}\left\| y_{i}\right\| _{L_{p^{*}}(\mu )}^{p^{*}}\right) , \end{aligned}$$

where $n\in {\mathbb {N}},$ $\lambda _{i}\in {\mathbb {C}},$ $x_{i}\in U$ and $y_{i}\in F$ for $i=1,\ldots ,n.$

It is easy check that M satisfies the three conditions of Ky Fan’s Lemma (see [6, 9.10]):

• (a) Each $f_{((\lambda _{i})_{i=1}^{n},(x_{i})_{i=1}^{n},(y_{i})_{i=1}^{n})}\in M$ is convex and lower semicontinuous.

• (b) If $g\in {\textrm{co}}(M),$ there is $f_{((\lambda _{i})_{i=1}^{n},(x_{i})_{i=1}^{n},(y_{i})_{i=1}^{n})}{\in }M$ with $g(\mu ){\le } f_{((\lambda _{i})_{i=1}^{n},(x_{i})_{i=1}^{n},(y_{i})_{i=1}^{n})} (\mu )$ for all

• (c) Each $f_{((\lambda _{i})_{i=1}^{n},(x_{i})_{i=1}^{n},(y_{i})_{i=1}^{n})}\in M$ has a value less or equal than 0.

By Ky Fan’s Lemma, there is a such that $f(\mu )\le 0$ for all $f\in M.$ In particular, we have

$$\begin{aligned} f_{(t\lambda ,x,t^{-1}y)}(\mu )=\left| \varphi (t\lambda \delta _{U}(x)\otimes t^{-1}y)\right| -\frac{C}{p}t^{p}\left| \lambda \right| ^{p}-\frac{C}{p^*}t^{-p^*}\left\| y\right\| _{L_{p^*}(\mu )}^{p^*}\le 0 \end{aligned}$$

for all $t\in {\mathbb {R}}^{+},$ $\lambda \in {\mathbb {C}},$ $x\in U$ and $y\in F.$ It follows that

$$\begin{aligned} \left| \varphi (\lambda \delta _{U}(x)\otimes y)\right| \le C\left( \frac{t^{p}\left| \lambda \right| ^{p}}{p}+\frac{t^{-p^{*}}\left\| y\right\| _{L_{p^*}(\mu )}^{p^*}}{p^*}\right) , \end{aligned}$$

and, applying again the aforementioned identity, we conclude that

$$\begin{aligned} \left| \varphi (\lambda \delta _{U}(x)\otimes y)\right| \le C\left| \lambda \right| \left\| y\right\| _{L_{p^*}(\mu )}. \end{aligned}$$

The case $p=\infty$ is similarly proved but without applying the cited identity and taking $C/p=0$ and $p^*=1.$

$(\text {iii})\ \Rightarrow \ (\text {i})$: Let $u\in \Delta (U)\otimes F$ and let $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y_i$ be a representation of u. Using (iii) and the Hölder inequality, we obtain

$$\begin{aligned} \left| \varphi (u)\right|&\le \sum _{i=1}^n\left| \varphi \left( \lambda _i\delta (x_i)\otimes y_i\right) \right| \le C\sum _{i=1}^n\left| \lambda _i\right| \left\| y_i\right\| _{L_{p^*}(\mu )}\\&\le C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left( \sum _{i=1}^n\left\| y_i\right\| ^{p^*}_{L_{p^*}(\mu )}\right) ^{\frac{1}{p^*}} \\&=C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left( \int _{B_{F^{*}}}\sum _{i=1}^n\left| y^*(y_i)\right| ^{p^*}\ d\mu (y^*)\right) ^{\frac{1}{p^*}} \\&\le C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left( \sup _{y^*\in B_{F^*}}\sum _{i=1}^n\left| y^*(y_i)\right| ^{p^*}\right) ^{\frac{1}{p^*}} \\&=C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left\| (y_1,\ldots ,y_n)\right\| _{\ell ^{n,w}_{p^*}(F)}, \end{aligned}$$

and taking infimum over all the representations of u,  we conclude that $\left| \varphi (u)\right| \le Cg_p(u).$ $\square$

We are now ready to present the announced result. Compare to [5, Theorem 2.3.1].

Theorem 5.2

(Pietsch Domination) Let $1<p\le \infty$ and The following conditions are equivalent : 

• (i) f is Cohen strongly p-summing holomorphic.

• (ii) For any $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y^*_i\in \Delta (U)\otimes F^*,$ we have

  

• (iii) There is a constant $C>0$ and a Borel regular probability measure $\mu$ on $B_{F^{**}}$ such that

  $$\begin{aligned} \left| \left\langle y^*,f(x)\right\rangle \right| \le C\left\| y^*\right\| _{L_{p^*}(\mu )} \end{aligned}$$

  for all $x\in U$ and $y^*\in F^*,$ where

  $$\begin{aligned} \left\| y^*\right\| _{L_{p^*}(\mu )}=\left( \int _{B_{F^{**}}}\left| y^{**}(y^*)\right| ^{p^*}d\mu (y^{**})\right) ^{\frac{1}{p^*}}. \end{aligned}$$

In this case,  is the minimum of all constants $C>0$ satisfying the preceding inequality.

Proof

$(\text {i})\ \Rightarrow \ (\text {ii})$ is immediate from Definition 2.2.

$(\text {ii})\ \Rightarrow \ (\text {iii})$: Clearly, Appealing to Corollary 3.5, consider its associate linear functional $\Lambda _0(\kappa _F\circ f):\Delta (U)\otimes F^*\rightarrow {\mathbb {C}}.$ Given $u=\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y^*_i\in \Delta (U)\otimes F^*,$ we have

by (ii). Since it holds for each representation of u,  we deduce that

By Theorem 5.1, there exists a Borel regular probability measure $\mu$ on $B_{F^{**}}$ such that

for all $x\in U$ and $y^*\in F^*.$ Moreover, belongs to the set of all constants $C>0$ satisfying the inequality in (iii).

$(\text {iii})\ \Rightarrow \ (\text {i})$: Given $x\in U$ and $y^*\in F^*,$ we have

$$\begin{aligned} \left| \Lambda _0(\kappa _F\circ f)(\delta (x)\otimes y^*)\right| =\left| \left\langle y^*,f(x)\right\rangle \right| \le \left\| y^*\right\| _{L_{p^*}(\mu )} \end{aligned}$$

by applying (iii). Now, Theorem 5.1 tells us that for any representation $\sum _{i=1}^n\lambda _i\delta (x_i)\otimes y^*_i$ of $u\in \Delta (U)\otimes F^*,$ we have

$$\begin{aligned} \sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle y^*_i,f(x_i)\right\rangle \right|&=\sum _{i=1}^n\left| \lambda _i\right| \left| \left\langle (\kappa _F\circ f)(x_i),y^*_i\right\rangle \right| =\sum _{i=1}^n\left| \Lambda _0(\kappa _F\circ f)(\lambda _i\delta (x_i)\otimes y^*_i)\right| \\&\le Cg_p( u) \le C\left\| (\lambda _1,\ldots ,\lambda _n)\right\| _{\ell ^n_p}\left\| (y^*_1,\ldots ,y^*_n)\right\| _{\ell ^{n,w}_{p^*}(F^*)}. \end{aligned}$$

Hence with This also shows the last assertion of the statement. $\square$

Remark 5.3

Theorem 5.2 is mainly a particular case of Theorem 4.6 in [21] since a Cohen strongly p-summing holomorphic mapping $(1<p<\infty )$ is an $R_1,R_2-S$-abstract $(p,p^*)$-summing mapping for $R_1:[0,1]\times U\times {\mathbb {C}}\rightarrow [0,\infty )$ defined by

$$\begin{aligned} R_1(t,x,\lambda )=|\lambda |, \end{aligned}$$

$R_2:B_{F^{**}}\times U\times F^*\rightarrow [0,\infty )$ given by

$$\begin{aligned} R_2(y^{**},x,y^*)=|y^{**}(y^*)|, \end{aligned}$$

and defined by

$$\begin{aligned} S(f,x,\lambda ,y^*)=|\lambda ||\langle y^*,f(x)\rangle |. \end{aligned}$$

This unified abstract version of Pietsch Domination Theorem has been used by several authors whenever trying to get a domination result in a very short way. Our proof is also short and appeals directly to Ky Fan’s Lemma as it was made to establish such an abstract version.

We now study the relationship between a Cohen strongly p-summing holomorphic mapping from U to F and its associate linearization from to F.

Theorem 5.4

Let $1<p\le \infty$ and The following conditions are equivalent : 

• (i) $f:U\rightarrow F$ is Cohen strongly p-summing holomorphic.

• (ii) is strongly p-summing.

In this case,  Furthermore,  the mapping $f\mapsto T_f$ is an isometric isomorphism from onto

Proof

$(\text {i})\ \Rightarrow \ (\text {ii})$: Assume that By Theorem 5.2, there is a constant $C>0$ and a Borel regular probability measure $\mu$ on $B_{F^{**}}$ such that $\left| \left\langle y^*,f(x)\right\rangle \right| \le C\left\| y^*\right\| _{L_{p^*}(\mu )}$ for all $x\in U$ and $y^*\in F^*.$

Let $y^*\in F^*$ and By Theorem 1.2, given $\varepsilon >0,$ we can take a representation $\sum _{i=1}^{\infty }\lambda _{i}\delta (x_i)$ of $\gamma$ such that $\sum _{i=1}^{\infty }\left| \lambda _i\right| \le \left\| \gamma \right\| +\varepsilon .$ We have

$$\begin{aligned} \left| \left\langle y^*,T_{f}(\gamma )\right\rangle \right|&=\left| \left\langle y^*,\sum _{i=1}^{\infty }\lambda _{i}T_{f}(\delta _{U}(x_{i}))\right\rangle \right| =\left| \left\langle y^*,\sum _{i=1}^{\infty }\lambda _{i}f(x_{i})\right\rangle \right| \\&\le \sum _{i=1}^{\infty }\left| \lambda _{i}\right| \left| \left\langle y^*,f(x_{i})\right\rangle \right| \\&\le C\left\| y^*\right\| _{L_{p^*}(\mu )}\sum _{i=1}^{\infty }\left| \lambda _{i}\right| \le C\left\| y^*\right\| _{L_{p^*}(\mu )}\left( \left\| \gamma \right\| +\varepsilon \right) . \end{aligned}$$

As $\varepsilon$ was arbitrary, it follows that

$$\begin{aligned} \left| \left\langle y^*,T_{f}(\gamma )\right\rangle \right| \le C\left\| y^*\right\| _{L_{p^*}(\mu )}\left\| \gamma \right\| . \end{aligned}$$

Taking infimum over all such constants C,  we have

by Theorem 5.2. It follows that

for all Therefore with by Pietsch Domination Theorem for strongly p-summing operators [5, Theorem 2.3.1].

$(\text {ii})\ \Rightarrow \ (\text {i})$: Assume that Given $x\in U$ and $y^*\in F^*,$ we have

$$\begin{aligned} \left| \left\langle y^*,f(x)\right\rangle \right| =\left| \left\langle y^*,T_{f}(\delta _{U}(x)\right\rangle \right| \le d_{p}(T_{f})\left\| y^*\right\| _{L_{p^*(\mu )}}\left\| \delta _{U}(x)\right\| =d_{p}(T_{f})\left\| y^*\right\| _{L_{p^*(\mu )}} \end{aligned}$$

by [5, Theorem 2.3.1] for some Borel regular probability measure $\mu$ on $B_{F^{**}}.$ It follows that with by Theorem 5.2.

Since for all to prove the last assertion of the statement, it suffices to show that the mapping $f\mapsto T_{f}$ from to is surjective. Indeed, take and then $T=T_{f}$ for some by Theorem 1.2. Hence and thus by the above proof. $\square$

The equivalence $(\text {i})\ \Leftrightarrow \ (\text {iii})$ of Theorem 5.2 admits the following reformulation.

Corollary 5.5

Let $1<p\le \infty$ and The following conditions are equivalent : 

• (i) $f:U\rightarrow F$ is Cohen strongly p-summing holomorphic.

• (ii) There exists a complex Banach space G and an operator such that

  $$\begin{aligned} \left| \left\langle y^*,f(x)\right\rangle \right| \le \left\| S^*(y^*)\right\| \quad (x\in U,\; y^*\in F^*). \end{aligned}$$

In this case,  is the infimum of all $d_p(S)$ with S satisfying (ii), and this infimum is attained at $T_f$ (Mujica’s linearization of f).

Proof

$(\text {i})\ \Rightarrow \ (\text {ii})$: If then with by Theorem 5.4. From Theorem 1.2, we infer that

$$\begin{aligned} \left| \left\langle y^*,f(x)\right\rangle \right| =\left| \left\langle y^*,T_{f}(\delta _{U}(x)\right\rangle \right| =\left| \left\langle (T_{f})^*(y^*),\delta _{U}(x)\right\rangle \right| \le \left\| (T_{f})^*(y^*)\right\| \end{aligned}$$

for all $x\in U$ and $y^*\in F^*.$

$(\text {ii})\ \Rightarrow \ (\text {i})$: Assume that (ii) holds. Then $S^*\in \Pi _{p^*}(F^*,G^*)$ with $\pi _{p^*}(S^*)=d_p(S)$ by [5, Theorem 2.2.2]. By Pietsch Domination Theorem for p-summing linear operators (see [6, Theorem 2.12]), there is a Borel regular probability measure $\mu$ on $B_{F^{**}}$ such that

$$\begin{aligned} \left\| S^*(y^*)\right\| \le \pi _{p^*}(S^*)\left\| y^*\right\| _{L_{p^*}(\mu )} \end{aligned}$$

for all $y^*\in F^*.$ For any $x\in U$ and $y^*\in F^*,$ it follows that

$$\begin{aligned} \left| \left\langle y^*,f(x)\right\rangle \right| \le \left\| S^*(y^*)\right\| \le \pi _{p^*}(S^*)\left\| y^*\right\| _{L_{p^*}(\mu )}. \end{aligned}$$

Hence, with by Theorem 5.2. $\square$

As a consequence of Theorem 5.4, an application of [4, Theorem 3.2] shows that the Banach ideal is generated by composition with the Banach operator ideal but we prefer to give here a proof to complete the information.

Corollary 5.6

Let $1<p\le \infty$ and The following conditions are equivalent : 

• (i) $f:U\rightarrow F$ is Cohen strongly p-summing holomorphic.

• (ii) There is a complex Banach space G,  and so that $f=T\circ g.$

In this case,  where the infimum is taken over all factorizations of f as in (ii), and this infimum is attained at $T_f\circ g_U$ (Mujica’s factorization of f).

Proof

$(\text {i})\ \Rightarrow \ (\text {ii})$: If we have $f=T_f\circ g_U,$ where is a complex Banach space, and by Theorems 1.2 and 5.4. Moreover,

$(\text {ii})\ \Rightarrow \ (\text {i})$: Assume $f=T\circ g$ with G,  g and T being as in (ii). Since $g=T_g\circ g_U$ by Theorem 1.2, it follows that $f=T\circ T_g\circ g_U$ which implies that $T_f=T\circ T_g,$ and thus by the ideal property of By Theorem 5.4, we obtain that with

and so by taking the infimum over all factorizations of f. $\square$

When F is reflexive, every factors through a Hilbert space as we see below.

Corollary 5.7

Let F be a reflexive complex Banach space. If then there exist a Hilbert space H,  an operator and a mapping such that $f=T\circ g.$

Proof

Assume that By Theorem 5.4, Hence by [5, Theorem 2.2.2]. By [6, Corollary 2.16 and Examples 2.9 (b)], there exist a Hilbert space H and operators $T_1\in \Pi _2(F^*,H)$ and such that $(T_f)^*=T_2\circ T_1.$

On the one hand, we have $(T_f)^{**}=(T_1)^*\circ (T_2)^*,$ where by [5, Theorem 2.2.2]. On the other hand, we have with $\kappa _F$ being bijective (since F is reflexive). Consequently, we obtain $f=T\circ g,$ where and $\square$

Applying Theorem 5.4 and [5, Theorem 2.4.1], we get useful inclusion relations.

Corollary 5.8

Let $1<p_1\le p_2\le \infty .$ If then and $\square$

These inclusion relations can become coincidence relations when $F^*$ has cotype 2 (see [6, pp. 217–221] for definitions and results on this class of spaces). Compare to [6, Corollary 11.16].

Corollary 5.9

Let $2<p\le \infty .$ If $F^*$ has cotype 2,  then and for all

Proof

By Corollary 5.8, we have with for all

For the converse, let Then with by Theorem 5.4. Hence with $\pi _2((T_{f})^*)=d_2(T_f)$ by [5, Theorem 2.2.2]. Then, by [6, Corollary 11.16], with $\pi _1((T_{f})^*)=\pi _2((T_f)^*).$ Hence, with $\pi _{p^*}((T_{f})^*)\le \pi _1((T_f)^*)$ by [6, Theorem 2.8]. Then, by [5, Theorem 2.2.2], with $d_p(T_{f})=\pi _{p^*}((T_f)^*).$

Finally, with by Theorem 5.4, and therefore $\square$

Given the transpose of f is the mapping defined by

$$\begin{aligned} f^t(y^*)=y^*\circ f\quad (y^*\in F^*). \end{aligned}$$

It is known (see [12, Proposition 1.6]) that with $\Vert f^t\Vert =\left\| f\right\| _{\infty }.$ Furthermore, $f^t=J^{-1}_U\circ (T_f)^*$ with being the identification established in Theorem 1.2.

The next result establishes the relation of a Cohen strongly p-summing holomorphic mapping $f:U\rightarrow F$ and its transpose Compare to [5, Theorem 2.2.2].

Theorem 5.10

Let $1<p\le \infty$ and Then if and only if In this case, 

Proof

Applying Theorem 5.4, [5, Theorem 2.2.2] and [6, 2.4 and 2.5], respectively, we have

In this case, $\square$

The study of holomorphic mappings with relatively (weakly) compact range was initiated by Mujica [16] and followed in [12].

Corollary 5.11

Let $1<p\le \infty .$

• (i) Every Cohen strongly p-summing holomorphic mapping $f:U\rightarrow F$ has relatively weakly compact range.

• (ii) If F is reflexive,  then every Cohen strongly p-summing holomorphic mapping $f:U\rightarrow F$ has relatively compact range.

Proof

If then by Theorem 5.10. Hence the linear operator $f^t$ is weakly compact and completely continuous by [6, 2.17]. Since $f^t$ is weakly compact, this means that f has relatively weakly compact range by [12, Theorem 2.7]. Since $f^t$ is completely continuous and $F^*$ is reflexive, it is known that $f^t$ is compact and, equivalently, f has relatively compact range by [12, Theorem 2.2]. $\square$

7 Pietsch factorization for Cohen strongly p-summing holomorphic mappings

We devote this section to the analogue of Pietsch Factorization Theorem for p-summing linear operators [6, Theorem 2.13] for the class of Cohen strongly p-summing holomorphic mappings. Recall that, for every Banach space F,  we have the canonical isometric injections $\kappa _F:F\rightarrow F^{**}$ and $\iota _F:F\rightarrow C\left( B_{F^*}\right)$ defined, respectively, by

$$\begin{aligned}&\left\langle \kappa _F(y),y^*\right\rangle =y^*(y)\quad \left( y\in F,\; y^*\in F^*\right) ,\\&\left\langle \iota _F(y),y^*\right\rangle =y^*(y)\quad \left( y\in F,\; y^*\in B_{F^*}\right) . \end{aligned}$$

Moreover, if $\mu$ is a regular Borel measure on $(B_{F^{**}},w^*),$ $j_{p}$ denotes the canonical map from $C\left( B_{F^*}\right)$ to $L_{p}\left( \mu \right) .$

Theorem 6.1

(Pietsch Factorization) Let $1<p\le \infty$ and The following conditions are equivalent : 

• (i) $f:U\rightarrow F$ is Cohen strongly p-summing holomorphic.

• (ii) There exist a regular Borel probability measure $\mu$ on $(B_{F^{**}},w^*),$ a closed subspace $S_{p^*}$ of $L_{p^*}(\mu )$ and a bounded holomorphic mapping $g:U\rightarrow (S_{p^*})^*$ such that the following diagram commutes : 

  

In this case, 

Proof

$(\text {i})\ \Rightarrow \ (\text {ii})$: Let Then by Theorem 5.10. By [6, Theorem 2.13], there exist a regular Borel probability measure $\mu$ on $(B_{F^{**}},w^*),$ a subspace $S_{p^*}:=\overline{j_{p^*}\left( i_{F^*}\left( F^*\right) \right) }$ of $L_{p^*}(\mu ),$ and an operator with $\left\| T\right\| =\Vert f^t\Vert$ such that the following diagram commutes:

Dualizing, we obtain

Let $g:=T^*\circ g_U.$ Clearly, with $\left\| g\right\| _{\infty }\le \left\| T\right\| ,$ and thus

Moreover, since $f^t=T\circ j_{p^*}\circ \iota _{F^*},$ we have

$$\begin{aligned} \kappa _F\circ f=(f^t)^*\circ g_U=(\iota _{F^*})^*\circ (j_{p^*})^*\circ T^*\circ g_U=(\iota _{F^*})^*\circ (j_{p^*})^*\circ g. \end{aligned}$$

$(\text {ii})\ \Rightarrow \ (\text {i})$: Since $\kappa _{F}\circ f=(\iota _{F^*})^*\circ (j_{p^*})^*\circ g,$ it follows that $f^{t}\circ (\kappa _F)^*=((\iota _{F^*})^*\circ (j_{p^*})^*\circ g)^t.$ As $(\kappa _{F})^*\circ \kappa _{F^*}=\textrm{id}_{F^*},$ we obtain that

$$\begin{aligned} f^t=((\iota _{F^*})^*\circ (j_{p^*})^*\circ g)^t\circ \kappa _{F^*}. \end{aligned}$$

Since $j_{p^*}\in \Pi _{p^*}(\iota _{F^*}(F^*),S_{p^*})$ (see [6, Examples 2.9]), then

by [5, Theorem 2.2.2]. Hence with

by the ideal property of Corollary 5.6 and [5, Theorem 2.2.2]. Applying Theorem 5.10 and the ideal property of $\Pi _p,$ we deduce that Again, Theorem 5.10 gives that with Moreover,

$\square$