The g_E-Approximation Property Determined by the Banach Space E = ℓ_q(ℓ_p)

Abstract

We study the $g_E$-approximation property for the Banach space $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$, which is an extension of Saphar’s p-approximation property. We establish some characterizations of the $g_E$-approximation property using the space of E-summing operators, which is an extension of the space of p-summing operators.

1. Introduction

One of the most important properties in Banach space theory is the approximation property (AP) which was systematically investigated by Grothendieck [1]. It not only deserves to be studied individually, but also provides descriptions of various properties in the Banach space theory. The main notion of this paper originates from the AP, the injective tensor norm $\epsilon$, and the projective tensor norm π. We say that a Banach space X has the AP if for every compact subset K of X and every $ϵ>0$, there exists a finite rank operator $S:X\to X$ such that ${sup}_{x\in K}\parallel Sx-x\parallel \le ϵ$. Throughout this paper, Banach spaces will be denoted by X and Y over $\mathbb{R}$ or $\mathbb{C}$, with dual spaces $X^{*}$ and $Y^{*}$ and the closed unit ball of X will be denoted by $B_X$. Let $X\otimes Y$ be the algebraic tensor product of X and Y. For basic backgrounds of tensor products, we refer to ([2], Chapter 1). For $u\in X\otimes Y$,

$$\epsilon (u;X,Y):=sup\left\{|\sum _{n=1}^m{x}^{*}(x_n)y^{*}(y_n)|:x^{*}\in B_{X^{*}},y^{*}\in B_{Y^{*}}\right\},$$

where ${\sum }_{n=1}^m{x}_n\otimes y_n$ is any representation of u, and

$$\pi (u;X,Y):=inf\left\{\sum _{n=1}^m\parallel x_n\parallel \parallel y_n\parallel :u=\sum _{n=1}^m{x}_n\otimes y_n,m\in \mathbb{N}\right\}.$$

The normed space $X\otimes Y$ equipped with a norm $\alpha$ will be denoted by $X{\otimes }_{\alpha }Y$ and its completion is $X{\widehat{\otimes }}_{\alpha }Y$. Grothendieck [1] proved that X has the AP if and only if for every Banach space Y, the canonical inclusion map

$$J_{\pi }:Y{\widehat{\otimes }}_{\pi }X\hookrightarrow Y{\widehat{\otimes }}_{\epsilon }X$$

is injective (cf. ([3], Theorem 5.6)).

Let $\alpha$ be a general tensor norm. For general backgrounds of the theory of tensor norms, we refer to [2,3]. In view of the criterion of the AP, naturally, one may say that X has the $\alpha$-approximation property ($\alpha$-AP) if for every Banach space Y, the inclusion map

$$J_{\alpha }:Y{\widehat{\otimes }}_{\alpha }X\hookrightarrow Y{\widehat{\otimes }}_{\epsilon }X$$

is injective (cf. ([3], Section 21.7)). Note that the Banach space Y can be replaced by dual spaces (see ([3], Proposition 21.7(4))). For every tensor norm $\alpha$, it is well known that X has the $\alpha$-AP if X has the AP (cf. ([3], Proposition 21.7(1))).

For $1\le p<\infty$, let

$${\mathcal{\ell }}_p(X):=\left\{{(x_n)}_n \mathrm{in} X:\sum _{n=1}^{\infty }{\parallel x_n\parallel }^p<\infty \right\}$$

with the norm ${\parallel {(x_n)}_n\parallel }_{{\mathcal{\ell }}_p(X)}:={({\sum }_{n=1}^{\infty }{\parallel x_n\parallel }^p)}^{1/p}$, let

$${\mathcal{\ell }}_p^w(X):=\left\{{(x_n)}_n \mathrm{in} X:\sum _{n=1}^{\infty }{|x^{*}(x_n)|}^p<\infty \mathrm{for}\mathrm{each} x^{*}\in X^{*}\right\}$$

with the norm ${\parallel {(x_n)}_n\parallel }_{{\mathcal{\ell }}_p^w(X)}:={sup}_{x^{*}\in B_{X^{*}}}({\sum }_{n=1}^{\infty }|x^{*}(x_n){{|}^p)}^{1/p}$ and let

$$c_0(X):=\{{(x_n)}_n \mathrm{in} X:\underset{n\to \infty }{lim}\parallel x_n\parallel =0\}$$

with the norm ${\parallel {(x_n)}_n\parallel }_{c_0(X)}:={sup}_{n\in \mathbb{N}}\parallel x_n\parallel$. The main subject of this paper comes from the tensor norm $g_p$ which is due to Chevet and Saphar [4,5]. For $u\in X\otimes Y$ and $1/p+1/p^{*}=1$,

$$g_p(u;X,Y):=inf\left\{{\parallel {(x_n)}_{n=1}^m\parallel }_{{\mathcal{\ell }}_p(X)}{\parallel {(y_n)}_{n=1}^m\parallel }_{{\mathcal{\ell }}_{p^{*}}^w(Y)}:u=\sum _{n=1}^m{x}_n\otimes y_n,m\in \mathbb{N}\right\}.$$

Since $g_1=\pi$ (cf. ([2], Proposition 6.6)), a Banach space X has the $g_1$-AP if and only if X has the AP. The $g_p$-AP was systematically investigated by Saphar [6]. For $1\le p<\infty$, an operator $T:X\to Y$ is (absolutely) p-summing if there exists a $C>0$ such that for every finite sequence ${(x_n)}_{n=1}^m$ in X,

$${\parallel {(T{x}_n)}_{n=1}^m\parallel }_{{\mathcal{\ell }}_p(Y)}\le C{\parallel {(x_n)}_{n=1}^m\parallel }_{{\mathcal{\ell }}_p^w(X)}.$$

For p-summing operators, we refer to [3,7,8]. We denote the space of p-summing operators from X to Y by ${Ƥ}_p(X,Y)$. For $T\in {Ƥ}_p(X,Y)$, we see that ${(T{x}_n)}_n\in {\mathcal{\ell }}_p(Y)$ for every ${(x_n)}_n\in {\mathcal{\ell }}_p^w(X)$ (cf. ([7], Proposition 2.1)). Then, we can define the locally convex topology, which will be denoted by ${\tau }_{wp,p}$, on ${Ƥ}_p(X,Y)$ generated by the seminorms

$$\parallel {(T{x}_n)}_n{\parallel }_{{\mathcal{\ell }}_p(Y)}$$

for every ${(x_n)}_n\in {\mathcal{\ell }}_p^w(X)$. The following theorem was proved in ([9], Theorem 2.9), where $Ƒ(X,Y)$ is the space of finite rank operators from X to Y.

Theorem 1. 

Let $1<p<\infty$. Then, the following statements are equivalent.

(a) 

X has the $g_p$-AP.

(b) 

For every Banach space Y, ${Ƥ}_{p^{*}}(X,Y)={\overline{Ƒ(X,Y)}}^{{\tau }_{w{p}^{*},p^{*}}}$.

(c) 

For every separable reflexive Banach space Z, ${Ƥ}_{p^{*}}(X,Z)={\overline{Ƒ(X,Z)}}^{{\tau }_{w{p}^{*},p^{*}}}$.

The main goal of this paper is to extend Theorem 1 to a more general setting. The above concepts are basically defined from the classical sequence spaces ${\mathcal{\ell }}_p$ and $c_0$, where the sequences of standard unit vectors are 1-unconditional Schauder bases. We consider those concepts defined from any Banach space having an 1-unconditional Schauder basis. Throughout this paper, E is a Banach space having an 1-unconditional Schauder basis ${(e_n)}_n$, ${(e_n^{*})}_n$ is the sequence of biorthogonal functionals for ${(e_n)}_n$ and $E_{*}:=\overline{\mathrm{span}}{\left\{e_n^{*}\right\}}_{n=1}^{\infty }$. If ${(e_n)}_n$ is shrinking, then $E_{*}=E^{*}$. For basic backgrounds of Schauder bases, we refer to ([10], Chapter 4). Let

$$E(X):=\left\{{(x_n)}_n \mathrm{in} X:\sum _{n=1}^{\infty }\parallel x_n\parallel e_n \mathrm{converges}\mathrm{in} E\right\}$$

with the norm ${\parallel {(x_n)}_n\parallel }_{E(X)}:=\parallel {\sum }_{n=1}^{\infty }\parallel x_n\parallel e_n{\parallel }_E$ and let

$$E^w(X):=\left\{{(x_n)}_n \mathrm{in} X:\sum _{n=1}^{\infty }{x}^{*}(x_n)e_n \mathrm{converges}\mathrm{in} E \mathrm{for}\mathrm{each} x^{*}\in X^{*}\right\}$$

with the norm ${\parallel {(x_n)}_n\parallel }_{E^w(X)}:={sup}_{x^{*}\in B_{X^{*}}}\parallel {\sum }_{n=1}^{\infty }{x}^{*}(x_n)e_n{\parallel }_E$. In [11], the generalization of $g_p$ was introduced as follows: For $u\in X\otimes Y$, let

$$g_E(u;X,Y):=inf\left\{\parallel {(x_n)}_{n\in F}{\parallel }_{E(X)}{\parallel {(y_n)}_{n\in F}\parallel }_{E_{*}^w(Y)}:u=\sum _{n\in F}{x}_n\otimes y_n, \mathrm{finite} F\subset \mathbb{N}\right\}.$$

We see that $g_{{\mathcal{\ell }}_p}=g_p$$(1\le p<\infty )$ and $g_{c_0}=g_{\infty }$. It was shown in ([11], Theorem 1) that $g_E$ is a tensor norm for the case $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$$(1\le p,q\le \infty )$, which is the Banach space of ${\mathcal{\ell }}_q$ direct sum of ${\mathcal{\ell }}_p$’s. When p or $q=\infty$, we will consider $c_0$ instead of ${\mathcal{\ell }}_{\infty }$. Note that the sequence of standard unit vectors in ${\mathcal{\ell }}_q({\mathcal{\ell }}_p)$ is an 1-unconditional Schauder basis and ${\mathcal{\ell }}_p({\mathcal{\ell }}_p)={\mathcal{\ell }}_p$. According to the definition in [12], an operator $T:X\to Y$ is (absolutely) E-summing if there exists a $C>0$ such that for every finite subset F of $\mathbb{N}$ and every sequence ${(x_n)}_{n\in F}$ in X,

$$\parallel {(T{x}_n)}_{n\in F}{\parallel }_{E(Y)}\le C{\parallel {(x_n)}_{n\in F}\parallel }_{E^w(X)}.$$

In view of the definition of p-summing operator, we see that the p-summing operator is exactly the ${\mathcal{\ell }}_p$-summing operator. We will denote by ${Ƥ}_E(X,Y)$ the collection of all E-summing operators from X to Y. If ${(e_n)}_n$ is boundedly complete, then for $T\in {Ƥ}_E(X,Y)$, ${(T{x}_n)}_n\in E(Y)$ for every ${(x_n)}_n\in E^w(X)$ (see Lemma 1(b)). Then, we can define the locally convex topology, which will be denoted by ${\tau }_{E^w,E}$, on ${Ƥ}_E(X,Y)$ generated by the seminorms

$$\parallel {(T{x}_n)}_n{\parallel }_{E(Y)}$$

for every ${(x_n)}_n\in E^w(X)$. In the present paper, we have

Theorem 2.

Let $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$$(1<p,q<\infty )$. Then, the following statements are equivalent.

(a) 

X has the $g_E$-AP.

(b) 

For every Banach space Y,

$${Ƥ}_{E^{*}}(X,Y)={\overline{Ƒ(X,Y)}}^{{\tau }_{{(E^{*})}^w,E^{*}}}.$$

(c) 

For every separable reflexive Banach space Z,

$${Ƥ}_{E^{*}}(X,Z)={\overline{Ƒ(X,Z)}}^{{\tau }_{{(E^{*})}^w,E^{*}}}.$$

We use the argument in [9] to prove Theorem 2. Predominantly, we apply the Hahn–Banach separation theorem for locally convex topologies. In order to do this, in Section 2, we represent some dual spaces of ${Ƥ}_E(X,Y)$ to prove Theorem 2 in Section 3, additionally, for the other cases $E=c_0({\mathcal{\ell }}_p)$$(1\le p\le \infty )$, $E={\mathcal{\ell }}_p(c_0)$$(1\le p\le \infty )$, $E={\mathcal{\ell }}_1({\mathcal{\ell }}_p)$$(1\le p\le \infty )$, $E={\mathcal{\ell }}_p({\mathcal{\ell }}_1)$$(1\le p\le \infty )$, we obtain some similar results.

2. Some Dual Spaces of the Space of E-Summing Operators

In this section, we represent some dual spaces of ${Ƥ}_E(X,Y)$ equipped with our topologies. In order to do this, we need a subspace of $E^w(X)$ with the same norm. Let

$$E^u(X):=\left\{{(x_n)}_n \mathrm{in} X:\underset{l\to \infty }{lim}\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{n\ge l}{x}^{*}(x_n)e_n{\parallel }_E=0\right\}.$$

Lemma 1. 

Let $T\in {Ƥ}_E(X,Y)$. Then, the following statements hold.

(a) 

If ${(x_n)}_n\in E^u(X)$, then ${(T{x}_n)}_n\in E(Y)$.

(b) 

If ${(e_n)}_n$ is boundedly complete and ${(x_n)}_n\in E^w(X)$, then ${(T{x}_n)}_n\in E(Y)$.

Proof. 

Let $C>0$ be the E-summing constant of T.

(a): If ${(x_n)}_n\in E^u(X)$, then

$$\begin{array}{cc}\hfill \underset{n,m\to \infty }{lim}\parallel \sum _{l=n}^m{\parallel T{x}_l\parallel }_Y{e}_l{\parallel }_E& \le \underset{n,m\to \infty }{lim}C\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{l=n}^m{x}^{*}(x_l)e_l{\parallel }_E\hfill \\ \hfill & \le C\underset{n\to \infty }{lim}\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{l\ge n}{x}^{*}(x_l)e_l{\parallel }_E=0.\hfill \end{array}$$

Hence, ${(T{x}_n)}_n\in E(Y)$.

(b): If ${(x_n)}_n\in E^w(X)$, then

$$\begin{array}{cc}\hfill \underset{m\in \mathbb{N}}{sup}\parallel \sum _{n=1}^m{\parallel T{x}_n\parallel }_Y{e}_n{\parallel }_E& \le \underset{m\in \mathbb{N}}{sup}C\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{n=1}^m{x}^{*}(x_n)e_n{\parallel }_E\hfill \\ \hfill & \le C\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{n=1}^{\infty }{x}^{*}(x_n)e_n{\parallel }_E<\infty .\hfill \end{array}$$

Since ${(e_n)}_n$ is boundedly complete, ${(T{x}_n)}_n\in E(Y)$. □

Proposition 1.

Suppose that ${(e_n)}_n$ is boundedly complete and shrinking. Then, we have

$${({Ƥ}_E(X,Y),{\tau }_{E^w,E})}^{*}=\left\{f(T)=\sum _{n=1}^{\infty }{y}_n^{*}(T{x}_n):{(x_n)}_n\in E^w(X),{(y_n^{*})}_n\in E^{*}(Y^{*})\right\}.$$

Proof. 

If ${(e_n)}_n$ is shrinking, then ${(e_n^{*})}_n$ is also an 1-unconditional Schauder basis for $E^{*}$. Suppose that for every $T\in {Ƥ}_E(X,Y)$,

$$f(T)=\sum _{n=1}^{\infty }{y}_n^{*}(T{x}_n),$$

where ${(x_n)}_n\in E^w(X)$ and ${(y_n^{*})}_n\in E^{*}(Y^{*})$. Then, for every $T\in {Ƥ}_E(X,Y)$, we have

$$\begin{array}{cc}\hfill |f(T)|& \le \sum _{n=1}^{\infty }\parallel y_n^{*}\parallel \parallel T{x}_n\parallel \hfill \\ \hfill & =\left(\sum _{n=1}^{\infty }\parallel y_n^{*}\parallel e_n^{*}\right)\left(\sum _{n=1}^{\infty }\parallel T{x}_n\parallel e_n\right)\hfill \\ \hfill & \le \parallel {(y_n^{*})}_n{\parallel }_{E^{*}(Y^{*})}{\parallel {(T{x}_n)}_n\parallel }_{E(Y)}.\hfill \end{array}$$

Hence, $f\in {({Ƥ}_E(X,Y),{\tau }_{E^w,E})}^{*}$.

Conversely, suppose that $f\in {({Ƥ}_E(X,Y),{\tau }_{E^w,E})}^{*}$. Then, there exists ${(x_n)}_n\in E^w(X)$ such that

$$|f(T)|\le \parallel {(T{x}_n)}_n{\parallel }_{E(Y)}$$

for every $T\in {Ƥ}_E(X,Y)$. Let us consider the linear subspace

$$\{{(T{x}_n)}_n:T\in {Ƥ}_E(X,Y)\}$$

of $E(Y)$ and the linear functional $\phi$ on $\{{(T{x}_n)}_n:T\in {Ƥ}_E(X,Y)\}$ given by

$$\phi ({(T{x}_n)}_n)=f(T).$$

We see that $\phi$ is well defined and linear, and $\parallel \phi \parallel \le 1$. Let $\widehat{\phi }\in E{(Y)}^{*}=E^{*}(Y^{*})$ be a Hahn–Banach extension of $\phi$. Let

$$\widehat{\phi }:={(y_n^{*})}_n.$$

Then, for every $T\in {Ƥ}_E(X,Y)$, we have

$$f(T)=\widehat{\phi }({(T{x}_n)}_n)=\sum _{n=1}^{\infty }{y}_n^{*}(T{x}_n).$$

□

We need another topology to obtain similar results for the other cases in the introduction. By Lemma 1(a), we can define the locally convex topology, which will be denoted by ${\tau }_{E^u,E}$, on ${Ƥ}_E(X,Y)$ generated by the seminorms

$$\parallel {(T{x}_n)}_n{\parallel }_{E(Y)}$$

for every ${(x_n)}_n\in E^u(X)$. As in the proof of Proposition 1, we have

Proposition 2.

Suppose that ${(e_n)}_n$ is shrinking. Then, we have

$${({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}=\left\{f(T)=\sum _{n=1}^{\infty }{y}_n^{*}(T{x}_n):{(x_n)}_n\in E^u(X),{(y_n^{*})}_n\in E^{*}(Y^{*})\right\}.$$

Lemma 2.

If ${(x_n)}_n\in E^u(X)$, then

$$\underset{l\to \infty }{lim}sup\left\{\sum _{n\ge l}|x^{*}(x_n)e^{*}(e_n)|:x^{*}\in B_{X^{*}},e^{*}\in B_{E^{*}}\right\}=0.$$

Proof. 

Let $ϵ>0$ be given. Then, there exists an $l\in \mathbb{N}$ such that

$$\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{n\ge l}{x}^{*}(x_n)e_n{\parallel }_E \le ϵ.$$

Let $x^{*}\in B_{X^{*}}$ and let $e^{*}\in B_{E^{*}}$. Then

$$\begin{array}{cc}\hfill \sum _{n\ge l}|x^{*}(x_n)e^{*}(e_n)|& =\sum _{n\ge l}{\gamma }_n{x}^{*}(x_n)e^{*}(e_n) (|{\gamma }_n|=1)\hfill \\ \hfill & \le \parallel \sum _{n\ge l}{\gamma }_n{x}^{*}(x_n)e_n{\parallel }_E\hfill \\ \hfill & = \parallel \sum _{n\ge l}{x}^{*}(x_n)e_n{\parallel }_E\le ϵ.\hfill \end{array}$$

Consequently,

$$sup\left\{\sum _{n\ge l}|x^{*}(x_n)e^{*}(e_n)|:x^{*}\in B_{X^{*}},e^{*}\in B_{E^{*}}\right\}\le ϵ.$$

This completes the proof.

□

Corollary 1.

If ${(x_n)}_n\in E^u(X)$, then for every permutation σ of $\mathbb{N}$,

$$\underset{l\to \infty }{lim}\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{n\ge l}{x}^{*}(x_{\sigma (n)})e_{\sigma (n)}{\parallel }_E=0.$$

Proof. 

Let $\sigma$ be a permutation of $\mathbb{N}$. Let $ϵ>0$ be given. Then, by Lemma 2, there exists an $m\in \mathbb{N}$ such that

$$sup\left\{\sum _{n\ge m}|x^{*}(x_n)e^{*}(e_n)|:x^{*}\in B_{X^{*}},e^{*}\in B_{E^{*}}\right\}\le ϵ.$$

Choose an $l\in \mathbb{N}$ so that $n\ge l$ implies $\sigma (n)\in \{m,m+1,\dots \}$. Then,

$$\underset{x^{*}\in B_{X^{*}}}{sup}\parallel \sum _{n\ge l}{x}^{*}(x_{\sigma (n)})e_{\sigma (n)}{\parallel }_E\le sup\left\{\sum _{n\ge m}|x^{*}(x_n)e^{*}(e_n)|:x^{*}\in B_{X^{*}},e^{*}\in B_{E^{*}}\right\}\le ϵ.$$

□

The following lemma is well known. Since its proof is standard, we omit the proof.

Lemma 3.

Let K be a collection of sequences of positive numbers. If ${lim}_{l\to \infty }{sup}_{{(k_j)}_j\in K}{\sum }_{j\ge l}{k}_j=0$, then there exists a sequence ${(b_j)}_j$ of real numbers with $b_j\nearrow \infty$ and $b_j>1$ for all j such that

$$\underset{l\to \infty }{lim}\underset{{(k_j)}_j\in K}{sup}\sum _{j\ge l}{k}_j{b}_j=0.$$

Now, we consider the cases that ${(e_n)}_n$ is not shrinking.

Proposition 3.

Let $E={\mathcal{\ell }}_p({\mathcal{\ell }}_1)$$(1\le p\le \infty )$. Then, we have

$${({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}$$

$$=\left\{f(T)=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }{y}_{nk}^{*}(T{x}_{nk}):{(x_{nk})}_{n,k}\in E^u(X),{(y_{nk}^{*})}_{n,k}\in {\mathcal{\ell }}_{p^{*}}(c_0)(Y^{*})\right\}.$$

Proof. 

Suppose that for every $T\in {Ƥ}_E(X,Y)$,

$$f(T)=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }{y}_{nk}^{*}(T{x}_{nk}),$$

where ${(x_{nk})}_{n,k}\in E^u(X)$ and ${(y_{nk}^{*})}_{n,k}\in {\mathcal{\ell }}_{p^{*}}(c_0)(Y^{*})$. Then, for every $T\in {Ƥ}_E(X,Y)$, we have

$$\begin{array}{cc}\hfill |f(T)|& \le \sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\parallel y_{nk}^{*}\parallel \parallel T{x}_{nk}\parallel \hfill \\ \hfill & \le \sum _{n=1}^{\infty }\underset{k}{sup}\parallel y_{nk}^{*}\parallel \sum _{k=1}^{\infty }\parallel T{x}_{nk}\parallel \hfill \\ \hfill & \le {\left(\sum _{n=1}^{\infty }(\underset{k}{sup}\parallel y_{nk}^{*}{\parallel )}^{p^{*}}\right)}^{1/p^{*}}{\left(\sum _{n=1}^{\infty }{\left(\sum _{k=1}^{\infty }\parallel T{x}_{nk}\parallel \right)}^p\right)}^{1/p}\hfill \\ \hfill & ={\left(\sum _{n=1}^{\infty }(\underset{k}{sup}\parallel y_{nk}^{*}{\parallel )}^{p^{*}}\right)}^{1/p^{*}}{\parallel {(T{x}_{nk})}_{n,k}\parallel }_{E(Y)}.\hfill \end{array}$$

Hence, $f\in {({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}$.

Conversely, suppose that $f\in {({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}$. Then, there exists ${(x_{nk})}_{n,k}\in E^u(X)$ such that

$$|f(T)|\le \parallel {(T{x}_{nk})}_{n,k}{\parallel }_{E(Y)}$$

for every $T\in {Ƥ}_E(X,Y)$. We consider the following sequence of rectangular array:

$$\begin{array}{cccccccc}{x}_{11}& \to & x_{12}& & x_{13}& ···& x_{1n}& ···\\ & & \downarrow & & \downarrow & & \downarrow \\ x_{21}& \leftarrow & x_{22}& & x_{23}& ···& x_{2n}& ···\\ & & & & \downarrow & & \downarrow \\ x_{31}& \leftarrow & x_{32}& \leftarrow & x_{33}& ···& x_{3n}& ···\\ & & & & & & \downarrow \\ & ⋮& & ⋮& & ···& ⋮& ···\\ & & & & & & \downarrow \\ x_{n1}& \leftarrow & ···& \leftarrow & x_{n(n-1)}& \leftarrow & x_{nn}& ···\\ & ⋮& & ⋮& & ⋮& & ⋮\end{array}$$

Let ${(u_m)}_m$ be the above sequence. By Corollary 1, ${(u_m)}_m\in E^u(X)$. By Lemmas 2 and 3, there exists a sequence ${({\beta }_m)}_m$ of positive numbers with ${\beta }_m>1$ and ${lim}_{m\to \infty }{\beta }_m=\infty$ such that ${({\beta }_m{u}_m)}_m\in E^u(X)$. Let ${({\alpha }_{nk})}_{n,k}$ be the rearranging sequence of ${({\beta }_m)}_m$. Then, we see that for every $n\in \mathbb{N}$,

$$\underset{k\to \infty }{lim}1/{\alpha }_{nk}=0.$$

Now, let us consider the linear subspace

$$\{{(T{x}_{nk})}_{n,k}:T\in {Ƥ}_E(X,Y)\}$$

of $E(Y)$ and the linear functional $\phi$ on $\{{(T{x}_{nk})}_{n,k}:T\in {Ƥ}_E(X,Y)\}$ given by

$$\phi ({(T{x}_{nk})}_{n,k})=f(T).$$

We see that $\phi$ is well defined and linear, and $\parallel \phi \parallel \le 1$. Let $\widehat{\phi }\in E{(Y)}^{*}={\mathcal{\ell }}_{p^{*}}({\mathcal{\ell }}_{\infty })(Y^{*})$ be a Hahn–Banach extension of $\phi$. Let

$$\widehat{\phi }:={(y_{nk}^{*})}_{n,k}.$$

Then, for every $T\in {Ƥ}_E(X,Y)$, we have

$$f(T)=\widehat{\phi }({(T{x}_{nk})}_{n,k})=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }{y}_{nk}^{*}(T{x}_{nk})=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }(1/{\alpha }_{nk})y_{nk}^{*}(T{\alpha }_{nk}{x}_{nk}).$$

Since ${({\beta }_m{u}_m)}_m\in E^u(X)$, by Corollary 1, ${({\alpha }_{nk}{x}_{nk})}_{n,k}\in E^u(X)$. Since for every $n\in \mathbb{N}$,

$$\underset{k\to \infty }{lim}\parallel (1/{\alpha }_{nk})y_{nk}^{*}\parallel =0$$

and

$$\sum _{n=1}^{\infty }(\underset{k}{sup}\parallel (1/{\alpha }_{nk})y_{nk}^{*}{\parallel )}^{p^{*}}\le \sum _{n=1}^{\infty }(\underset{k}{sup}\parallel y_{nk}^{*}{\parallel )}^{p^{*}}<\infty ,$$

${((1/{\alpha }_{nk})y_{nk}^{*})}_{n,k}\in {\mathcal{\ell }}_{p^{*}}(c_0)(Y^{*})$. This completes the proof. □

Proposition 4.

Let $E={\mathcal{\ell }}_1({\mathcal{\ell }}_p)$$(1\le p\le \infty )$. Then, we have

$${({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}$$

$$=\left\{f(T)=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }{y}_{nk}^{*}(T{x}_{nk}):{(x_{nk})}_{n,k}\in E^u(X),{(y_{nk}^{*})}_{n,k}\in c_0({\mathcal{\ell }}_{p^{*}})(Y^{*})\right\}.$$

Proof. 

Suppose that for every $T\in {Ƥ}_E(X,Y)$,

$$f(T)=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }{y}_{nk}^{*}(T{x}_{nk}),$$

where ${(x_{nk})}_{n,k}\in E^u(X)$ and ${(y_{nk}^{*})}_{n,k}\in c_0({\mathcal{\ell }}_{p^{*}})(Y^{*})$. Then, for every $T\in {Ƥ}_E(X,Y)$, we have

$$\begin{array}{cc}\hfill |f(T)|& \le \sum _{n=1}^{\infty }\sum _{k=1}^{\infty }\parallel y_{nk}^{*}\parallel \parallel T{x}_{nk}\parallel \hfill \\ \hfill & \le \sum _{n=1}^{\infty }{\left(\sum _{k=1}^{\infty }{\parallel y_{nk}^{*}\parallel }^{p^{*}}\right)}^{1/p^{*}}{\left(\sum _{k=1}^{\infty }{\parallel T{x}_{nk}\parallel }^p\right)}^{1/p}\hfill \\ \hfill & \le \underset{n}{sup}{\left(\sum _{k=1}^{\infty }{\parallel y_{nk}^{*}\parallel }^{p^{*}}\right)}^{1/p^{*}}{\parallel {(T{x}_{nk})}_{n,k}\parallel }_{E(Y)}.\hfill \end{array}$$

Hence, $f\in {({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}$.

Conversely, suppose that $f\in {({Ƥ}_E(X,Y),{\tau }_{E^u,E})}^{*}$. Then, there exists ${(x_{nk})}_{n,k}\in E^u(X)$ such that

$$|f(T)|\le \parallel {(T{x}_{nk})}_{n,k}{\parallel }_{E(Y)}$$

for every $T\in {Ƥ}_E(X,Y)$. As in the proof of Propositon 2, let ${(u_m)}_m$ be the sequence of rectangular array of ${(x_{nk})}_{n,k}$. By Corollary 1, ${(u_m)}_m\in E^u(X)$. By Lemmas 2 and 3, there exists a sequence ${({\beta }_m)}_m$ of positive numbers such that ${\beta }_m>1$, ${lim}_{m\to \infty }{\beta }_m=\infty$ and ${({\beta }_m{u}_m)}_m\in E^u(X)$. Let ${({\alpha }_{nk})}_{n,k}$ be the rearranging sequence of ${({\beta }_m)}_m$. Then, we see that

$$\underset{n\to \infty }{lim}\underset{k}{sup}1/{\alpha }_{nk}=0.$$

Consider the linear subspace

$$\{{(T{x}_{nk})}_{n,k}:T\in {Ƥ}_E(X,Y)\}$$

of $E(Y)$ and the linear functional $\phi$ on $\{{(T{x}_{nk})}_{n,k}:T\in {Ƥ}_E(X,Y)\}$ given by

$$\phi ({(T{x}_{nk})}_{n,k})=f(T).$$

Then, $\phi$ is well defined and linear, and $\parallel \phi \parallel \le 1$. Let $\widehat{\phi }\in E{(Y)}^{*}={\mathcal{\ell }}_{\infty }({\mathcal{\ell }}_{p^{*}})(Y^{*})$ be a Hahn–Banach extension of $\phi$. Let

$$\widehat{\phi }:={(y_{nk}^{*})}_{n,k}.$$

Then, for every $T\in {Ƥ}_E(X,Y)$, we have

$$f(T)=\widehat{\phi }({(T{x}_{nk})}_{n,k})=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }{y}_{nk}^{*}(T{x}_{nk})=\sum _{n=1}^{\infty }\sum _{k=1}^{\infty }(1/{\alpha }_{nk})y_{nk}^{*}(T{\alpha }_{nk}{x}_{nk}).$$

Since ${({\beta }_m{u}_m)}_m\in E^u(X)$, by Corollary 1, ${({\alpha }_{nk}{x}_{nk})}_{n,k}\in E^u(X)$. Since

$$\underset{n\to \infty }{lim}{\left(\sum _{k=1}^{\infty }{\parallel (1/{\alpha }_{nk})y_{nk}^{*}\parallel }^{p^{*}}\right)}^{1/p^{*}}\le \underset{n}{sup}{\left(\sum _{k=1}^{\infty }{\parallel y_{nk}^{*}\parallel }^{p^{*}}\right)}^{1/p^{*}}\underset{n\to \infty }{lim}\underset{k}{sup}1/{\alpha }_{nk}=0,$$

${((1/{\alpha }_{nk})y_{nk}^{*})}_{n,k}\in c_0({\mathcal{\ell }}_{p^{*}})(Y^{*})$. This completes the proof. □

3. Proofs of Main Results

We begin with the following lemma to prove our main results.

Lemma 4.

If ${(x_n)}_n\in E(X)$ and ${(y_n)}_n\in E_{*}^w(Y)$, then the series ${\sum }_{n=1}^{\infty }{x}_n\otimes y_n$ converges in $X{\widehat{\otimes }}_{g_E}Y$.

Proof. 

Since

$$\underset{l,m\to \infty }{lim}{g}_E\left(\sum _{n=l}^m{x}_n\otimes y_n;X,Y\right)\le \underset{l,m\to \infty }{lim}\parallel \sum _{n=l}^m\parallel x_n\parallel e_n{\parallel }_E\underset{y^{*}\in B_{Y^{*}}}{sup}\parallel \sum _{n=1}^{\infty }{y}^{*}(y_n)e_n^{*}{\parallel }_{E_{*}}=0,$$

the assertion follows. □

We also need some representation of the $g_E$-tensor element.

Lemma 5

([11], Proposition 5). Let $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$$(1\le p,q\le \infty )$. If $u\in X{\widehat{\otimes }}_{g_E}Y$, then there exist ${(x_n)}_n\in E(X)$ and ${(y_n)}_n\in E_{*}^u(Y)$ such that

$$\sum _{n=1}^{\infty }{x}_n\otimes y_n=u$$

unconditionally converges in $X{\widehat{\otimes }}_{g_E}Y$.

The following lemma is essentially due to ([3], Lemma 21.9) which considers the case $E={\mathcal{\ell }}_p$.

Lemma 6.

Let $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$$(1\le p,q\le \infty )$. If $u\in Y{\widehat{\otimes }}_{g_E}X$, then there exist a separable reflexive Banach space Z, which is a linear subspace of Y, and a $v\in Z{\widehat{\otimes }}_{g_E}X$ such that

$$j\otimes i{d}_X(v)=u,$$

where $j:Z\to Y$ is the inclusion and $i{d}_X:X\to X$ is the identity map.

Proof. 

By Lemma 5, there exist ${(y_n)}_n\in E(Y)$ and ${(x_n)}_n\in E_{*}^u(X)$ such that ${\sum }_{n=1}^{\infty }{y}_n\otimes x_n=u$ converges in $Y{\widehat{\otimes }}_{g_E}X$. We can find a sequence ${({\beta }_n)}_n$ with ${\beta }_n>1$ such that ${lim}_{n\to \infty }{\beta }_n=\infty$ and ${({\beta }_n{y}_n)}_n\in E(Y)$. Then, the balanced closed convex hull $\overline{bco}{\{y_n/({\beta }_n\parallel y_n\parallel )\}}_n$ is a compact subset of $B_Y$.

Now, it is well known that there exists a separable reflexive Banach space Z, which is a linear subspace of Y, and $\overline{bco}\{y_n/({\beta }_n\parallel y_n{\parallel )\}}_n\subset B_Z$ (cf. [13]). Since

$$\underset{l\to \infty }{lim}\parallel \sum _{n\ge l}\parallel y_n{\parallel }_Z{e}_n{\parallel }_E\le \underset{l\to \infty }{lim}\parallel \sum _{n\ge l}{\parallel {\beta }_n{y}_n\parallel }_Y{e}_n{\parallel }_E=0,$$

${(y_n)}_n\in E(Z)$. Hence, by Lemma 4, $v:={\sum }_{n=1}^{\infty }{y}_n\otimes x_n$ converges in $Z{\widehat{\otimes }}_{g_E}X$ and $j\otimes i{d}_X(v)=u$. □

Now, we can extend ([3], p. 282, Proposition 1) which considers the case $E={\mathcal{\ell }}_p$.

Proposition 5.

Let $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$$(1\le p,q\le \infty )$. Then, X has the $g_E$-AP if (and only if) for every separable reflexive Banach space Z, the inclusion map

$$J_{g_E}:Z{\widehat{\otimes }}_{g_E}X\hookrightarrow Z{\widehat{\otimes }}_{\epsilon }X$$

is injective.

Proof. 

To show that X has the $g_E$-AP from our assumption, let Y be a Banach space. Assume that $u=0$ in $Y{\widehat{\otimes }}_{\epsilon }X$ for $u\in Y{\widehat{\otimes }}_{g_E}X$. We should show that $u=0$ in $Y{\widehat{\otimes }}_{g_E}X$. By Lemma 6, there exist a separable reflexive Banach space Z, which is a linear subspace of Y, and a $v\in Z{\widehat{\otimes }}_{g_E}X$ such that

$$j\otimes i{d}_X(v)=u.$$

Let us consider the maps

$$J_{g_E}:Z{\widehat{\otimes }}_{g_E}X\hookrightarrow Z{\widehat{\otimes }}_{\epsilon }X \mathrm{and} j\otimes i{d}_X:Z{\widehat{\otimes }}_{\epsilon }X\to Y{\widehat{\otimes }}_{\epsilon }X.$$

Since $j\otimes i{d}_X$ is injective, we see that $v=0$ in $Z{\widehat{\otimes }}_{\epsilon }X$. By our assumption, $v=0$ in $Z{\widehat{\otimes }}_{g_E}X$. Hence, $u=0$ in $Y{\widehat{\otimes }}_{g_E}X$. □

We need a duality relationship between $g_E$ and ${Ƥ}_E$ to prove our main results. We will assume that $g_E$ and $g_{E_{*}}$ are finitely generated tensor norms (cf. ([11], Proposition 1)). A finitely generated tensor norm is uniquely associated with a maximal Banach operator ideal (cf. ([3], Section 17.3)). Let ${\alpha }^{\prime }$ be the dual tensor norm of a tensor norm $\alpha$ (cf. ([3], Section 15)) and let ${\alpha }^t$ be the transposed tensor norm of $\alpha$. Then, the adjoint tensor norm is defined by

$${\alpha }^{*}:={({\alpha }^{\prime })}^t={({\alpha }^t)}^{\prime }.$$

If $\alpha$ is finitely generated, then ${\alpha }^{\prime }$, ${\alpha }^t$ and ${\alpha }^{*}$ are all finitely generated and ${({\alpha }^{\prime })}^{\prime }=\alpha$. Let $[{\mathcal{A}}^{adj},\parallel ·{\parallel }_{{\mathcal{A}}^{adj}}]$ be the adjoint ideal of a Banach operator ideal $[\mathcal{A},\parallel ·{\parallel }_{\mathcal{A}}]$. If a tensor norm $\alpha$ is associated with a Banach operator ideal $[\mathcal{A},\parallel ·{\parallel }_{\mathcal{A}}]$, then $[{\mathcal{A}}^{adj},\parallel ·{\parallel }_{{\mathcal{A}}^{adj}}]$ is a maximal Banach operator ideal associated with ${\alpha }^{*}$ (cf. ([3], Section 17.9)). For $T\in {Ƥ}_E$, let ${\parallel T\parallel }_{{Ƥ}_E}:=infC$ which is the infimum taken over all such inequalities in the introduction. Then, it was shown in ([12], Theorem 3.2) that $[{Ƥ}_E,\parallel ·{\parallel }_{{Ƥ}_E}]$ is a maximal Banach operator ideal.

Proposition 6.

$g_{E_{*}}^{*}$ is associated with $[{Ƥ}_E,\parallel ·{\parallel }_{{Ƥ}_E}]$.

Proof. 

It was shown in ([11], Corollary 1) that $g_E$ is associated with the ideal $[{\mathcal{N}}_E,\parallel ·{\parallel }_{{\mathcal{N}}_E}]$ of E-nuclear operators and that $[{\mathcal{N}}_{E_{*}}^{adj}{,\parallel ·\parallel }_{{\mathcal{N}}_{E_{*}}^{adj}}]=[{Ƥ}_E{,\parallel ·\parallel }_{{Ƥ}_E}]$ in ([13], Theorem 3.6 i)). Hence, the assertion follows. □

Lemma 7

([3], Theorem 17.5). Let $[\mathcal{A},\parallel ·{\parallel }_{\mathcal{A}}]$ be the maximal Banach operator ideal associated with a finitely generated tensor norm α. Then, for all Banach spaces X and Y,

$${(X{\otimes }_{{\alpha }^{\prime }}Y)}^{*}=\mathcal{A}(X,Y^{*})$$

holds isometrically with the dual action $[T,{\sum }_{n=1}^m{x}_n\otimes y_n]={\sum }_{n=1}^m(T{x}_n)(y_n)$.

Corollary 2.

For all Banach spaces X and Y,

$${(Y{\otimes }_{g_{E_{*}}}X)}^{*}={Ƥ}_E(X,Y^{*})$$

holds isometrically.

Proof. 

Since ${Ƥ}_E$ is associated with $g_{E_{*}}^{*}$, by Lemma 7,

$${(Y{\otimes }_{g_{E_{*}}}X)}^{*}={(X{\otimes }_{g_{E_{*}}^t}Y)}^{*}={(X{\otimes }_{{(g_{E_{*}}^{*})}^{\prime }}Y)}^{*}={Ƥ}_E(X,Y^{*})$$

hold isometrically. □

We are now ready to prove Theorem 2.

Proof 

(Proof of Theorem 2). (b)⇒(c) is trivial.

(a)⇒(b): Let Y be a Banach space and let $T\in {Ƥ}_{E^{*}}(X,Y)$. In order to use the Hahn–Banach separation theorem, let $f\in {({Ƥ}_{E^{*}}(X,Y),{\tau }_{{(E^{*})}^w,E^{*}})}^{*}$ be such that $f(S)=0$ for every $S\in Ƒ(X,Y)$. Then, it will show that $f(T)=0$. By Proposition 1, there exist ${(x_n)}_n\in {(E^{*})}^w(X)$ and ${(y_n^{*})}_n\in E(Y^{*})$ such that

$$f(R)=\sum _{n=1}^{\infty }{y}_n^{*}(R{x}_n)$$

for every $R\in {Ƥ}_{E^{*}}(X,Y)$. Then, we see that ${\sum }_{n=1}^{\infty }{y}_n^{*}(y)x_n$ for every $y\in Y$ since $f(S)=0$ for every $S\in Ƒ(X,Y)$. By Lemma 4, ${\sum }_{n=1}^{\infty }{y}_n^{*}\otimes x_n\in Y^{*}{\widehat{\otimes }}_{g_E}X$. Since X has the $g_E$-AP,

$$\sum _{n=1}^{\infty }{y}_n^{*}\otimes x_n=0 \mathrm{in} Y^{*}{\widehat{\otimes }}_{g_E}X.$$

By Corollary 2,

$${Ƥ}_{E^{*}}(X,Y^{**})={(Y^{*}{\widehat{\otimes }}_{g_E}X)}^{*}$$

holds isometrically. Since $i_{Y}T\in {Ƥ}_{E^{*}}(X,Y^{**})$, where $i_Y:Y\to Y^{**}$ is the canonical isometry,

$$0=\sum _{n=1}^{\infty }(i_{Y}T{x}_n)(y_n^{*})=\sum _{n=1}^{\infty }{y}_n^{*}(T{x}_n)=f(T).$$

Consequently, $T\in {\overline{Ƒ(X,Y)}}^{{\tau }_{{(E^{*})}^w,E^{*}}}$.

(c)⇒(a): We use Proposition 5. Let Z be a separable reflexive Banach space. Then, by (c),

$${Ƥ}_{E^{*}}(X,Z^{**})={\overline{Ƒ(X,Z^{**})}}^{{\tau }_{{(E^{*})}^w,E^{*}}}.$$

Let $u\in Z^{*}{\widehat{\otimes }}_{g_E}X$ with $u=0$ in $Z^{*}{\widehat{\otimes }}_{\epsilon }X$. By Lemma 5, there exist ${(z_n^{*})}_n\in E(Z^{*})$ and ${(x_n)}_n\in {(E^{*})}^u(X)$ such that

$$u=\sum _{n=1}^{\infty }{z}_n^{*}\otimes x_n.$$

To show that $u=0$ in $Z^{*}{\widehat{\otimes }}_{g_E}X$, let $T\in {Ƥ}_{E^{*}}(X,Z^{**})={(Z^{*}{\widehat{\otimes }}_{g_E}X)}^{*}$. Then, we will show that the dual action $[T,u]=0$. Now, let us consider $f:={\sum }_{n=1}^{\infty }{i}_{Z^{*}}(z_n^{*})(· x_n)$. Then, by Proposition 1,

$$f\in {({Ƥ}_{E^{*}}(X,Z^{**}),{\tau }_{{(E^{*})}^w,E^{*}})}^{*}.$$

Since for every $S:={\sum }_{k=1}^m{x}_k^{*}\otimes z_k^{**}\in Ƒ(X,Z^{**})$,

$$f(S)=\sum _{n=1}^{\infty }{i}_{Z^{*}}(z_n^{*})(S{x}_n)=\sum _{n=1}^{\infty }\sum _{k=1}^m{x}_k^{*}(x_n)z_k^{**}(z_n^{*})=\sum _{k=1}^m\sum _{n=1}^{\infty }{x}_k^{*}(x_n)z_k^{**}(z_n^{*})=0$$

and $T\in {\overline{Ƒ(X,Z^{**})}}^{{\tau }_{{(E^{*})}^w,E^{*}}}$,

$$0=f(T)=\sum _{n=1}^{\infty }{i}_{Z^{*}}(z_n^{*})(T{x}_n)=\sum _{n=1}^{\infty }(T{x}_n)(z_n^{*}).$$

Hence, $[T,u]=0$. □

Now, let us consider the other cases. The following theorem was proved in ([9], Theorem 2.6).

Theorem 3. 

The following statements are equivalent.

(a) 

X has the $g_{\infty }$-AP.

(b) 

For every Banach space Y, ${Ƥ}_1(X,Y)={\overline{Ƒ(X,Y)}}^{{\tau }_{{\mathcal{\ell }}_1^u,{\mathcal{\ell }}_1}}$.

(c) 

For every separable reflexive Banach space Z, ${Ƥ}_1(X,Z)={\overline{Ƒ(X,Z)}}^{{\tau }_{{\mathcal{\ell }}_1^u,{\mathcal{\ell }}_1}}$.

We use Propositions 2–4 to prove the other cases which extend Theorem 3.

Theorem 4. 

Let $E=c_0({\mathcal{\ell }}_p)$$(1\le p\le \infty )$, $E={\mathcal{\ell }}_p(c_0)$$(1\le p\le \infty )$, $E={\mathcal{\ell }}_1({\mathcal{\ell }}_p)$$(1\le p<\infty )$ or $E={\mathcal{\ell }}_p({\mathcal{\ell }}_1)$$(1\le p<\infty )$. Then, the following statements are equivalent.

(a) 

X has the $g_E$-AP.

(b) 

For every Banach space Y,

$${Ƥ}_{E_{*}}(X,Y)={\overline{Ƒ(X,Y)}}^{{\tau }_{{(E_{*})}^u,E_{*}}}.$$

(c) 

For every separable reflexive Banach space Z,

$${Ƥ}_{E_{*}}(X,Z)={\overline{Ƒ(X,Z)}}^{{\tau }_{{(E_{*})}^u,E_{*}}}.$$

Proof. 

Proposition 4 and Proposition 3, respectively, are used to prove the cases $E=c_0({\mathcal{\ell }}_p)$$(1\le p\le \infty )$ and $E={\mathcal{\ell }}_p(c_0)$$(1\le p\le \infty )$, respectively. Proposition 2 is used to prove the cases $E={\mathcal{\ell }}_1({\mathcal{\ell }}_p)$$(1\le p<\infty )$ and $E={\mathcal{\ell }}_p({\mathcal{\ell }}_1)$ $(1\le p<\infty )$. We use the proof of Theorem 2 and only prove the case $E={\mathcal{\ell }}_p(c_0)$.

(b)⇒(c) is trivial.

(a)⇒(b): Let Y be a Banach space and let $T\in {Ƥ}_{E_{*}}(X,Y)$. In order to use the Hahn–Banach separation theorem, let $f\in {({Ƥ}_{E_{*}}(X,Y),{\tau }_{{(E_{*})}^u,E_{*}})}^{*}$ be such that $f(S)=0$ for every $S\in Ƒ(X,Y)$. Since $E_{*}={\mathcal{\ell }}_{p^{*}}({\mathcal{\ell }}_1)$, by Proposition 3, there exist ${(x_n)}_n\in {(E_{*})}^u(X)$ and ${(y_n^{*})}_n\in E(Y^{*})$ such that

$$f(R)=\sum _{n=1}^{\infty }{y}_n^{*}(R{x}_n)$$

for every $R\in {Ƥ}_{E_{*}}(X,Y)$. Then, we see that ${\sum }_{n=1}^{\infty }{y}_n^{*}(y)x_n$ for every $y\in Y$. By Lemma 4, ${\sum }_{n=1}^{\infty }{y}_n^{*}\otimes x_n\in Y^{*}{\widehat{\otimes }}_{g_E}X$. Since X has the $g_E$-AP,

$$\sum _{n=1}^{\infty }{y}_n^{*}\otimes x_n=0 \mathrm{in} Y^{*}{\widehat{\otimes }}_{g_E}X.$$

By Corollary 2,

$${Ƥ}_{E_{*}}(X,Y^{**})={(Y^{*}{\widehat{\otimes }}_{g_E}X)}^{*}$$

holds isometrically. Since $i_{Y}T\in {Ƥ}_{E_{*}}(X,Y^{**})$,

$$0=\sum _{n=1}^{\infty }(i_{Y}T{x}_n)(y_n^{*})=\sum _{n=1}^{\infty }{y}_n^{*}(T{x}_n)=f(T).$$

Consequently, $T\in {\overline{Ƒ(X,Y)}}^{{\tau }_{{(E_{*})}^u,E_{*}}}$.

(c)⇒(a): We use Proposition 5. Let Z be a separable reflexive Banach space. Then, by (c),

$${Ƥ}_{E_{*}}(X,Z^{**})={\overline{Ƒ(X,Z^{**})}}^{{\tau }_{{(E_{*})}^u,E_{*}}}.$$

Let $u\in Z^{*}{\widehat{\otimes }}_{g_E}X$ with $u=0$ in $Z^{*}{\widehat{\otimes }}_{\epsilon }X$. By Lemma 5, there exist ${(z_n^{*})}_n\in E(Z^{*})$ and ${(x_n)}_n\in {(E_{*})}^u(X)$ such that

$$u=\sum _{n=1}^{\infty }{z}_n^{*}\otimes x_n.$$

To show that $u=0$ in $Z^{*}{\widehat{\otimes }}_{g_E}X$, let $T\in {Ƥ}_{E_{*}}(X,Z^{**})={(Z^{*}{\widehat{\otimes }}_{g_E}X)}^{*}$.

Now, consider $f:={\sum }_{n=1}^{\infty }{i}_{Z^{*}}(z_n^{*})(· x_n)$. Then, by Proposition 3,

$$f\in {({Ƥ}_{E_{*}}(X,Z^{**}),{\tau }_{{(E_{*})}^u,E_{*}})}^{*}.$$

As in the proof of Theorem 2, for every $S\in Ƒ(X,Z^{**})$, $f(S)=0$. Hence,

$$0=f(T)=\sum _{n=1}^{\infty }{i}_{Z^{*}}(z_n^{*})(T{x}_n)=\sum _{n=1}^{\infty }(T{x}_n)(z_n^{*}).$$

This completes the proof. □

4. Discussion

This work is general and natural extensions of some results about the $g_p$-AP. There have been much more investigations about the $g_p$-AP and the ideal of p-summing operators. We expect that those results can be extended to the $g_E$-AP and the ideal of E-summing operators. Moreover, we can consider the $g_E$-AP and the ideal of E-summing operators for the case $E={\mathcal{\ell }}_q({\mathcal{\ell }}_p)$ as the following subjects.

1. Some relationships between the $g_E$-AP and the $g_p$-AP, and between the deal of E-summing operators and the deal of p-summing operators.

2. A investigation of the $w_E$-AP and the ideal of E-dominated operators which were introduced in [11,12].