On the Characterizations of Some Strongly Bounded Operators on C(K, X) Spaces

Abstract

Suppose X and Y are Banach spaces, K is a compact Hausdorff space, and $C(K, X)$ is the Banach space of all continuous X-valued functions (with the supremum norm). We will study some strongly bounded operators $T:C(K, X)\to Y$ with representing measures $m:\Sigma \to L(X,Y)$, where $L(X,Y)$ is the Banach space of all operators $T:X\to Y$ and $\Sigma$ is the $\sigma$-algebra of Borel subsets of K. The classes of operators that we will discuss are the Grothendieck, p-limited, p-compact, limited, operators with completely continuous, unconditionally converging, and p-converging adjoints, compact, and absolutely summing. We give a characterization of the limited operators (resp. operators with completely continuous, unconditionally converging, p-convergent adjoints) in terms of their representing measures.

1. Introduction

If K is a compact Hausdorff space and X is a Banach space, then $C(K,X)$ is the Banach space of all continuous X-valued functions defined on K (endowed with the supremum norm), and $\Sigma$ is the $\sigma$-algebra of Borel subsets of K. The topological dual of $C(K,X)$ can be identified with the space $rcabv(\Sigma ,X^{\ast })$ of all $X^{\ast }$-valued, regular countably additive Borel measures on K of bounded variation, endowed with the variation norm.

For any Banach space Y and any continuous linear function $T:C(K,X)\to Y$, there is a vector measure $m:\Sigma \to L(X,Y^{\ast \ast })$ of finite semi-variation ([1], [2] (p. 182)) such that

$$T\left(f\right)={\int }_{K}fdm,f\in C(K,X),\parallel T\parallel =\tilde{m}\left(K\right),$$

where $\tilde{m}$ denotes the semi-variation of m. This vector measure m is called the representing measure of T. We denote the correspondence $m\leftrightarrow T$. For $E\in \Sigma$, the semi-variation $\tilde{m}\left(E\right)$ is given by

$$\begin{array}{cc}\hfill \tilde{m}\left(E\right)=& sup\{\parallel \sum _{i=1}^{n}m\left(E_i\right)\left(x_i\right)\parallel :E_i\in \Sigma ,E_i\subset E,{\left\{E_i\right\}}_{i=1}^n\mathrm{pairwise}\mathrm{disjoint},\hfill \\ \hfill & x_i\in B_X,i=1,\dots n,n\in \mathbb{N}\}.\hfill \end{array}$$

A representing measure m is called strongly bounded (or $\tilde{m}$ is continuous at ∅) if $\left(\tilde{m}\left(A_i\right)\right)\to 0$ for every decreasing sequence $\left(A_i\right)\to \varnothing$ in $\Sigma$, or equivalently, if there exists a control measure for $\tilde{m}$, that is, a positive countably additive regular Borel measure $\lambda$ on K such that ${lim}_{\lambda \left(E\right)\to 0}\tilde{m}\left(E\right)=0$. An operator $m\leftrightarrow T:C(K,X)\to Y$ is called strongly bounded if m is strongly bounded [1].

The reader should note that if $m\leftrightarrow T$, then $m\left(A\right)x=T^{\ast \ast }\left({\chi }_{A}x\right)$, for each $A\in \Sigma$, $x\in X$, where ${\chi }_A$ denotes the characteristic function of a set A. Let $B(\Sigma ,X)$ denote the space of all bounded, $\Sigma$-measurable functions on K with separable range in X and the sup norm. If $f\in B(\Sigma ,X)$, then ${\int }_{K}fdm$ is well-defined and defines an extension $\hat{T}:B(\Sigma ,X)\to Y^{\ast \ast }$ of T given by

$$\hat{T}\left(f\right)={\int }_{K}fdm,f\in B(\Sigma ,X),$$

which is just the restriction to $B(\Sigma ,X)$ of the bitranspose $T^{\ast \ast }$ of T; e.g., see [3] (Theorem 3), [4] (p. 83). By [3] (Theorem 2), $\hat{T}$ maps $B(\Sigma ,X)$ into Y if and only if the representing measure m of T is $L(X,Y)$-valued. If $T:C(K,X)\to Y$ is strongly bounded, then m is $L(X,Y)$-valued [1] (Theorem 4.4), and thus $\hat{T}:B(\Sigma ,X)\to Y$. If T is unconditionally converging, then m is strongly bounded [5].

Several authors studied the properties of $\hat{T}$; e.g., [3,6,7,8,9,10,11,12]. In these papers it has been proved that $T:C(K,X)\to Y$ is weakly compact, compact, Dunford–Pettis, Dieudonné, unconditionally converging, strictly singular, strictly cosingular, weakly p-compact, limited, pseudo weakly compact, and Dunford–Pettis p-convergent if and only if its extension $\hat{T}:B(\Sigma ,X)\to Y$ has the same property.

We show that if $T:C(K,X)\to Y$ is a strongly bounded operator and $\hat{T}:B(\Sigma ,X)\to Y$ is its extension, then T is Grothendieck (resp. p-limited, has a p-limited adjoint) if and only if $\hat{T}$ is Grothendieck (resp. p-limited, has a p-limited adjoint). It is shown that if $m\leftrightarrow T:C(K,X)\to Y$ is a strongly bounded operator and T is Grothendieck, then $m\left(A\right):X\to Y$ is Grothendieck for each $A\in \Sigma$. If we additionally assume that K is a dispersed compact Hausdorff space, then a strongly bounded operator $m\leftrightarrow T:C(K,X)\to Y$ is Grothendieck whenever $m\left(A\right):X\to Y$ is Grothendieck, for each $A\in \Sigma$. We also show that if $m\leftrightarrow T:C(K,X)\to Y$ is p-limited (resp. p-compact, has a p-limited adjoint), then $m\left(A\right)$ is p-limited (resp. p-compact, has a p-limited adjoint) for each $A\in \Sigma$.

We give characterizations of strongly bounded limited operators (resp. operators with completely continuous, unconditionally converging, and p-convergent adjoints), compact and absolutely summing operators in terms of their representing measures. The almost Dunford–Pettis and almost DP p-convergent operators were introduced in [9,13]. We introduce and study the almost limited (resp. operators with almost completely continuous, unconditionally converging, p-convergent adjoints), almost compact, and almost absolutely summing operators. We use the same ideas and techniques as in [7,9,13].

2. Definitions and Notation

Throughout this paper, X and Y will denote Banach spaces. The unit ball of X will be denoted by $B_X$, and $X^{\ast }$ will denote the continuous linear dual of X. An operator $T:X\to Y$ will be a continuous and linear function.

A Banach space X has the Grothendieck property if every $w^{\ast }$-convergent sequence in $X^{\ast }$ is weakly convergent [2] (p. 179). An operator $T:X\to Y$ is called a Grothendieck operator if $T^{\ast }$ takes $w^{\ast }$-null sequences in $Y^{\ast }$ to weakly null sequences in $X^{\ast }$ [14].

A subset A of X is called a Grothendieck set if every operator $T:X\to c_0$ maps A onto a relatively weakly compact set [15].

Definition 1. 

A bounded subset A of a Banach space X is called a limited (resp. Dunford–Pettis ($DP$)) subset of X if every $w^{\ast }$-null (resp. weakly null) sequence $\left(x_n^{\ast }\right)$ in $X^{\ast }$ tends to 0 uniformly on A, i.e.,

$$\underset{x\in A}{sup}\left|x_n^{\ast }\left(x\right)\right|\to 0.$$

An operator $T:X\to Y$ is limited [16] if $T\left(B_X\right)$ is limited.

A series $\sum x_n$ of elements of X is weakly unconditionally convergent (wuc) if $\sum |x^{\ast }\left(x_n\right)|<\infty$, for each $x^{\ast }\in X^{\ast }$.

An operator $T:X\to Y$ is unconditionally converging if it maps weakly unconditionally convergent series to convergent ones.

An operator $T:X\to Y$ is completely continuous (Dunford–Pettis) if it maps weakly Cauchy sequences to norm convergent ones.

Definition 2. 

A bounded subset A of a Banach space X (resp. $X^{\ast }$) is called a $\left(V^{\ast }\right)$-subset of X (resp. a $\left(V\right)$-subset of $X^{\ast }$) provided that

$$\underset{x\in A}{sup}|x_n^{\ast }\left(x\right)|\to 0(\mathit{resp}.\underset{x^{\ast }\in A}{sup}|x^{\ast }\left(x_n\right)|\to 0),$$

for each wuc series $\sum x_n^{\ast }$ in $X^{\ast }$ (resp. wuc series $\sum x_n$ in X ).

Definition 3. 

A bounded subset A of $X^{\ast }$ is called an L-subset of $X^{\ast }$ if each weakly null sequence $\left(x_n\right)$ in X tends to 0 uniformly on A, i.e., ${sup}_{x^{\ast }\in A}\left|x^{\ast }\left(x_n\right)\right|\to 0$.

For $1\le p<\infty$, $p^{\prime }$ denotes the conjugate of p. If $p=1$, $c_0$ plays the role of ${\ell }_{p^{\prime }}$. The unit vector basis of ${\ell }_p$ will be denoted by $\left(e_n\right)$.

Let $1\le p<\infty$. We denote by ${\ell }_p\left(X\right)$ the Banach space of all p-summable sequences with the norm

$$\parallel \left(x_n\right){\parallel }_p=(\sum _{n=1}^{\infty }\parallel x_n{{\parallel }^p)}^{1/p}$$

and by $c_0\left(X\right)$ the space of all norm null sequences.

Let $1\le p<\infty$. A sequence $\left(x_n\right)$ in X is called weakly p-summable if $\left(x^{\ast }\left(x_n\right)\right)\in {\ell }_p$ for each $x^{\ast }\in X^{\ast }$ [17] (p. 32), [18] (p. 134). Let ${\ell }_p^w\left(X\right)$ denote the set of all weakly p-summable sequences in X. Let $c_0^w\left(X\right)$ be the space of weakly null sequences in X. If $p=\infty$, then we consider $c_0^w\left(X\right)$ instead of ${\ell }_{\infty }^w\left(X\right)$.

The space ${\ell }_p^w\left(X\right)$ is a Banach space with the norm

$$\parallel \left(x_n\right){\parallel }_p^w=sup\left\{\right(\sum _{n=1}^{\infty }|x^{\ast }\left(x_n\right){|}^p{)}^{1/p}:x^{\ast }\in B_{X^{\ast }}\}.$$

If $p<q$, then ${\ell }_p^w\left(X\right)\subseteq {\ell }_q^w\left(X\right)$. Furthermore, the unit vector basis of ${\ell }_{p^{\prime }}$ is weakly p-summable for all $1<p<\infty$. The weakly 1-summable sequences are precisely the weakly unconditionally convergent (wuc) series and the weakly ∞-summable sequences are precisely weakly null sequences.

We recall the following isometries: ${\ell }_p^w\left(X\right)\simeq L({\ell }_{p^{\prime }},X)$ for $1<p<\infty$; ${\ell }_p^w\left(X\right)\simeq L(c_0,X)$ if $p=1$ [17] (Proposition 2.2, p. 36). Let these isometries be denoted by E; $E:{\ell }_p^w\left(X\right)\to L({\ell }_{p^{\prime }},X)$ (resp. $E:{\ell }_1^w\left(X\right)\to L(c_0,X)$); $x=\left\{x_n\right\}\to E_x$, where $E_x\left({\alpha }_n\right)={\sum }_n{\alpha }_n{x}_n$, $\left({\alpha }_n\right)\in {\ell }_{p^{\prime }}$, if $1<p<\infty$ ($\left({\alpha }_n\right)\in c_0$, if $p=1$).

Let $1\le p<\infty$. A sequence $\left(x_n^{\ast }\right)$ in $X^{\ast }$ is called ${\mathit{weak}}^{\ast }$ p-summable if $\left(x_n^{\ast }\left(x\right)\right)\in {\ell }_p$ for each $x\in X$ [19]. Let ${\ell }_p^{w^{\ast }}\left(X^{\ast }\right)$ denote the set of all ${\mathrm{weak}}^{\ast }$ p-summable sequences in $X^{\ast }$. This is a Banach space with the norm

$$\parallel \left(x_n^{\ast }\right){\parallel }_p^{w^{\ast }}=sup\left\{\right(\sum _{n=1}^{\infty }|x_n^{\ast }\left(x\right){|}^p{)}^{1/p}:x\in B_X\},$$

The map $\left(x_i^{\ast }\right)\to L_{\left(x_i^{\ast }\right)}$, where $L_{\left(x_i^{\ast }\right)}\left(x\right)=\left(x_i^{\ast }\left(x\right)\right)$, identifies ${\ell }_p^{w^{\ast }}\left(X^{\ast }\right)$ and $L(X,{\ell }_p)$ isometrically for all $1<p<\infty$. The spaces ${\ell }_p^{w^{\ast }}\left(X^{\ast }\right)$ and ${\ell }_p^w\left(X^{\ast }\right)$ are the same for $1\le p<\infty$.

Let $1\le p\le \infty$. An operator $T:X\to Y$ is called p-convergent if T maps weakly p-summable sequences into norm null sequences [9]. The set of all p-convergent operators $T:X\to Y$ is denoted by $C_p(X,Y)$.

The 1-convergent operators are precisely the unconditionally converging operators and the ∞-convergent operators are precisely the completely continuous operators. If $p<q$, then $C_q(X,Y)\subseteq C_p(X,Y)$.

Definition 4. 

Let $1\le p\le \infty$. A bounded subset A of X is called a p-$\left(V^{\ast }\right)$ set [20] (or weakly-p-Dunford–Pettis set [21]) if ${sup}_{x\in A}\left|x_n^{\ast }\left(x\right)\right|\to 0$, for every weakly p-summable (weakly null for $p=\infty$) sequence $\left(x_n^{\ast }\right)$ in $X^{\ast }$.

A bounded subset A of $X^{\ast }$ is called a p-$\left(V\right)$ set [20] (or weakly-p-L-set [21]) if ${sup}_{x^{\ast }\in A}\left|x^{\ast }\left(x_n\right)\right|$$\to 0$, for every weakly p-summable (weakly null for $p=\infty$) sequence $\left(x_n\right)$ in X.

The 1-$\left(V^{\ast }\right)$ subsets of X are precisely the $\left(V^{\ast }\right)$-sets and the ∞-$\left(V^{\ast }\right)$ subsets of X are precisely the DP sets. If $p<q$, then a q-$\left(V^{\ast }\right)$ set is a p-$\left(V^{\ast }\right)$ set, since ${\ell }_p^w\left(X^{\ast }\right)\subseteq {\ell }_q^w\left(X^{\ast }\right)$. The 1-$\left(V\right)$ subsets of $X^{\ast }$ are precisely the $\left(V\right)$-sets and the ∞-$\left(V\right)$ subsets of $X^{\ast }$ are precisely the L-sets. If $p<q$, then a q-$\left(V\right)$ set is a p-$\left(V\right)$ set.

Definition 5. 

A subset A of X is p-limited ($1\le p<\infty$) [22] if for every ${\mathit{weak}}^{\ast }$ (weak) p-summable sequence $\left(x_n^{\ast }\right)$ in $X^{\ast }$, there exists $\left({\alpha }_n\right)\in {\ell }_p$ such that $|x_n^{\ast }\left(x\right)|\le {\alpha }_n$ for all $x\in A$ and $n\in \mathbb{N}$.

An operator $T:X\to Y$ is called p-limited if $T\left(B_X\right)$ is a p-limited set in Y [22].

Definition 6. 

Let $1\le p\le \infty$. A subset A of X is relatively p-compact [23] if there is a p-summable (resp. a norm null sequence, if $p=\infty$) sequence $\left(x_n\right)$ in X such that $K\subset E_x\left(B_{{\ell }_{p^{\prime }}}\right)$ (${\ell }_{p^{\prime }}={\ell }_1$, if $p=\infty$).

The ∞-compact sets are precisely the compact sets and p-compact sets are q-compact if $1\le p\le q\le \infty$.

An operator $T:X\to Y$ is called p-compact if $T\left(B_X\right)$ is a p-compact set in Y.

An operator $T:X\to Y$ is p-summing if $\left(T{x}_n\right)\in {\ell }_p\left(Y\right)$ whenever $\left(x_n\right)\in {\ell }_p^w\left(X\right)$ [17] (p. 34), [24] (p. 59). An operator $T:X\to Y$ is absolutely summing if T carries wuc series into absolutely convergent series. The absolutely summing operators coincide with the 1-summing operators.

A topological space S is called dispersed (or scattered) if every nonempty closed subset of S has an isolated point [25]. A compact Hausdorff space K is dispersed if and only if ${\ell }_1\cancel{\hookrightarrow }C\left(K\right)$ [26].

3. Main Results

The connection between an operator $T:C(K,X)\to Y$ and its representing measure has been intensely studied (e.g., [1,3,5,7,13,27,28]). In this section we study some strongly bounded operators, including Grothendieck operators, p-limited operators, p-compact operators, limited operators (resp. operators with completely continuous, unconditionally converging, p-convergent adjoints), and absolutely summing operators. We characterize limited operators (resp. operators with completely continuous, unconditionally converging, p-convergent adjoints) in terms of their representing measures, extending previous results from [11,12]. We also introduce some new classes of operators, such as the almost limited operators (resp. operators with almost completely continuous, almost unconditionally converging, almost p-convergent adjoints) and show that if K is dispersed, these operators coincide with the limited operators (resp. operators with completely continuous, unconditionally converging, p-convergent adjoints).

The following lemma provides a characterization of Grothendieck operators.

Lemma 1 

([14] (Lemma 1.3), [29] (Proposition 1)). Let X and Y be Banach spaces and $T:X\to Y$ be an operator. Then the following conditions are equivalent:

(i) 

$T:X\to Y$ is Grothendieck.

(ii) 

For any operator $S:Y\to c_0$, $ST$ is weakly compact.

(iii) 

For any bounded subset A of X, $T\left(A\right)$ is a Grothendieck set.

As noted in the introduction, if $m\leftrightarrow T:C(K,X)\to Y$ is strongly bounded, then m is $L(X,Y)$-valued and its extension $\hat{T}$ maps $B(\Sigma ,X)$ into Y.

Every p-summing operator $m\leftrightarrow T:C(K,X)\to Y$ is strongly bounded. Indeed, T is weakly compact and completely continuous [18] (Corollary 6.20, p. 148), [17] (Theorem 2.17, p. 50), and thus it is strongly bounded. Every p-limited operator $m\leftrightarrow T:C(K,X)\to Y$ is strongly bounded. Indeed, every p-limited operator is weakly compact by [30] (Proposition 2.1), and thus it is unconditionally converging and strongly bounded [5]. Similarly, every operator $m\leftrightarrow T:C(K,X)\to Y$ with p-limited adjoint is strongly bounded. Every p-compact operator is compact (since every relatively p-compact set is relatively compact); hence, every p-compact operator $m\leftrightarrow T:C(K,X)\to Y$ is strongly bounded.

Observation 1. 

(i) An operator $T:C(K,X)\to Y$ is p-summing if and only if its extension $\hat{T}:B(\Sigma ,X)\to Y$ is p-summing. Indeed, if $T:C(K,X)\to Y$ is p-summing, then $T^{\ast \ast }$ is p-summing [17] (Proposition 2.19, p. 50), and thus $\hat{T}$ is p-summing (since it is the restriction of a p-summing operator).

(ii) An operator $T:C(K,X)\to Y$ is p-compact if and only if its extension $\hat{T}:B(\Sigma ,X)\to Y$ is p-compact. Indeed, if $T:C(K,X)\to Y$ is p-compact, then $T^{\ast \ast }$ is p-compact [31] (Corollary 3.6), and thus $\hat{T}$ is p-compact.

An operator $T:X\to Y$ is p-summing if and only if $T^{\ast }$ is p-limited [32] (Theorem 2). An operator $T:X\to Y$ is p-limited if and only if $T^{\ast }$ is p-summing [30] (Theorem 3.1).

Theorem 1. 

Suppose that $T:C(K,X)\to Y$ is a strongly bounded operator and $\hat{T}:B(\Sigma ,X)\to Y$ is its extension. Then the following assertions hold:

(i) 

T is Grothendieck if and only if $\hat{T}$ is Grothendieck.

(ii) 

$T^{\ast }$ is p-limited if and only if ${\hat{T}}^{\ast }$ is p-limited.

(iii) 

T is p-limited if and only if $\hat{T}$ is p-limited.

Proof. 

(i) We show that if $T:C(K, X)\to Y$ is Grothendieck, then $\hat{T}$ is Grothendieck. Let $S:Y\to c_0$ be an operator. Note that $S\hat{T}:B(\Sigma ,X)\to c_0$. If $f\in B(\Sigma ,X)$, then

$${\left(ST\right)}^{\ast \ast }\left(f\right)=S^{\ast \ast }{T}^{\ast \ast }\left(f\right)=S^{\ast \ast }\hat{T}\left(f\right)=S\hat{T}\left(f\right),$$

and $S\hat{T}$ is the extension of $ST$ to $B(\Sigma ,X)$. Since T is Grothendieck, $ST$ is weakly compact (by Lemma 1). Therefore its extension $S\hat{T}$ is weakly compact (Introduction, p. 2; [3] (Theorem 6)), and thus $\hat{T}$ is Grothendieck (by Lemma 1).

(ii) It follows from Observation 1 and [32] (Theorem 2).

(iii) If $T:C(K, X)\to Y$ is p-limited, then $T^{\ast }$ is p-summing [30] (Theorem 3.1). Then $T^{\ast \ast \ast }$ is p-summing [17] (Proposition 2.19, p. 50), so $T^{\ast \ast }$ is p-limited [30] (Theorem 3.1). Thus $\hat{T}$ is p-limited. □

Corollary 1. 

Suppose that $m\leftrightarrow T:C(K,X)\to Y$ is an operator.

(i) 

If T is strongly bounded and Grothendieck, then $m\left(A\right):X\to Y$ is Grothendieck for each $A\in \Sigma$.

(ii) 

If $T^{\ast }$ is p-limited, then for each $A\in \Sigma$, $m{\left(A\right)}^{\ast }:Y^{\ast }\to X^{\ast }$ is p-limited.

(iii) 

If T is p-limited, then for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is p-limited.

(iv) 

If T is p-compact, then for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is p-compact.

Proof. 

We will only consider the case of Grothendieck operators. The other proofs are similar. If $A\in \Sigma$, $A\ne \varnothing$, define ${\theta }_A:X\to B(\Sigma ,X)$ by ${\theta }_A\left(x\right)={\chi }_{A}x$. Then ${\theta }_A$ is an isomorphic isometric embedding of X into $B(\Sigma ,X)$ and $\hat{T}{\theta }_A=m\left(A\right)$. By Theorem 1, $\hat{T}$ is Grothendieck, and thus $m\left(A\right)$ is Grothendieck. □

We note that by Corollary 1 (iii) and [30] (Theorem 3.1), if $m\leftrightarrow T:C(K,X)\to Y$ is an operator such that $T^{\ast }$ is p-summing, then for each $A\in \Sigma$, $m{\left(A\right)}^{\ast }:Y^{\ast }\to X^{\ast }$ is p-summing.

If $T:C(K,X)\to Y$ is an operator, $\overline{K}$ is a metrizable compact space, and $\pi :K\to \overline{K}$ a continuous map which is onto, we will call $\overline{K}$ a quotient of K. The map $\overline{\pi }:C\left(\overline{K}\right)\to C\left(K\right)$ given by $\overline{\pi }\overline{f}=\overline{f}\pi$ defines an isometric embedding of $C\left(\overline{K}\right)$ into $C\left(K\right)$. Let $\overline{T}:C(\overline{K},X)\to Y$ be the operator defined by $\overline{T}\left(\overline{f}\right)=T\left(\overline{f}\pi \right)$, where $\overline{f}\in C(\overline{K},X)$ and $\pi :K\to \overline{K}$ is the canonical mapping.

Lemma 2. 

An operator $T:C(K,X)\to Y$ is Grothendieck if and only if, for each metrizable quotient $\overline{K}$ of K, the operator $\overline{T}:C(\overline{K},X)\to Y$ defined as above is Grothendieck.

Proof. 

Suppose that $T:C(K, X)\to Y$ is Grothendieck and $\overline{K}$ is a metrizable quotient space of K. Then $\overline{T}$ is Grothendieck.

Conversely, let $T:C(K, X)\to Y$ be an operator and let $\left(f_n\right)$ be a sequence in the unit ball of $C(K, X)$. It is known (see [3]) that there exists a metrizable quotient space $\overline{K}$ of K and a sequence $\left(\overline{f_n}\right)$ in $C(\overline{K},X)$ defined by $\overline{f_n}\left(\pi \left(t\right)\right)=f_n\left(t\right)$ for all $t\in K$ and $n\in \mathbb{N}$. Define $\overline{T}:C(\overline{K},X)\to Y$ by $\overline{T}\left(\overline{f}\right)=T\left(\overline{f}\pi \right)$, where $\pi :K\to \overline{K}$ is the canonical mapping. By assumption, $\overline{T}$ is Grothendieck. Then $\left(\overline{T}\left(\overline{f_n}\right)\right)=\left(T\left(f_n\right)\right)$ is Grothendieck. □

Lemma 3. 

Let H be a bounded subset of X. If for each $ϵ>0$ there is a Grothendieck subset $H_{ϵ}$ of X so that $H\subseteq H_{ϵ}+ϵB_X$, then H is a Grothendieck set.

Proof. 

Let $S:X\to c_0$ be an operator. Without loss of generality assume $\parallel S\parallel \le 1$. Since $S\left(H\right)\subseteq S\left(H_{ϵ}\right)+ϵB_{c_0}$ and $S\left(H_{ϵ}\right)$ is weakly compact (by Lemma 1), $S\left(H\right)$ is weakly compact by a result of Grothendieck ([7] (Lemma 6), [24] (p. 227)). By Lemma 1, H is a Grothendieck set. □

Theorem 2. 

Suppose that K is a dispersed compact Hausdorff space and $m\leftrightarrow T:C(K,X)\to Y$ is a strongly bounded operator. Then T is Grothendieck if and only if $m\left(A\right):X\to Y$ is Grothendieck, for each $A\in \Sigma$.

Proof. 

Suppose $m\leftrightarrow T:C(K, X)\to Y$ is strongly bounded. If T is Grotendieck, then for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is Grothendieck by Corollary 1.

Conversely, suppose that $m\leftrightarrow T:C(K, X)\to Y$ is a strongly bounded operator and $m\left(A\right):X\to Y$ is Grothendieck for each $A\in \Sigma$. From Lemma 2,7] (Lemma 5), and the fact that a quotient space of a dispersed space is dispersed ([25] (8.5.3)), we can suppose without loss of generality that K is metrizable. Since K is dispersed and metrizable, K is countable ([25] (8.5.5)). Suppose that $K=\{t_i:i\in \mathbb{N}\}$. Let $\left(f_n\right)$ be a sequence in the unit ball of $C(K, X)$. For each $i\in \mathbb{N}$, the set $\{f_n\left(t_i\right):n\in \mathbb{N}\}$ is bounded in X and $m\left(\left\{t_i\right\}\right):X\to Y$ is Grothendieck. Then the set

$$H_i=\{m\left(\left\{t_i\right\}\right)\left(f_n\left(t_i\right)\right):n\in \mathbb{N}\}$$

is Grothendieck, for each $i\in \mathbb{N}$. Let $A_i=\{t_j:j>i\}$, $i\in \mathbb{N}$. Then $\left(A_i\right)$ is a decreasing sequence of sets. Let $ϵ>0$. Since m is strongly bounded, there is a $k\in \mathbb{N}$ such that $\tilde{m}\left(A_k\right)<ϵ$. For each $n\in \mathbb{N}$,

$$T\left(f_n\right)={\int }_K{f}_{n}dm=\sum _{i=1}^{k}m\left(\left\{t_i\right\}\right)\left(f_n\left(t_i\right)\right)+{\int }_{A_k}{f}_{n}dm.$$

Furthermore, $\parallel {\int }_{A_k}{f}_{n}dm\parallel \le \tilde{m}\left(A_k\right)<ϵ$. Therefore,

$$T\left(f_n\right)\in H_1+H_2+\dots +H_k+ϵB_Y.$$

Since $H_1+H_2+\dots +H_k$ is a Grothendieck set, the set $\{T\left(f_n\right):n\in \mathbb{N}\}$ is Grothendieck, by Lemma 3. Thus T is Grothendieck. □

We recall the following well-known result ([4] (Sec 13, Theorem 5)).

Theorem 3. 

Let λ be a positive Radon measure on K. If $m\in rcabv(\Sigma ,X^{\ast })$ is λ-continuous, then there exists a function $g:K\to X^{\ast }$ such that

(i) 

$\langle x,g\rangle$ is a λ-integrable function for every $x\in X$.

(ii) 

For every $x\in X$ and $A\in \Sigma$,

$$\langle x,m\left(A\right)\rangle ={\int }_A\langle x,g\rangle d\lambda$$

(iii) 

$\left|g\right|$ is λ-integrable and

$$\tilde{m}\left(A\right)={\int }_A\left|g\right|d\lambda ,$$

for every $A\in \Sigma$, where $\left|g\right|\left(t\right)=\parallel g\left(t\right)\parallel$, for $t\in K$.

By [13] (Lemma 1.3) the operator $T:C(K,X)\to Y$ with representing measure m is strongly bounded if and only if there exists a positive Radon measure $\lambda$ on K such that

(i) $T^{\ast }\left(y^{\ast }\right)$ is $\lambda$-continuous for every $y^{\ast }\in Y^{\ast }$, and

(ii) If $g_{y^{\ast }}$ is the element corresponding to $T^{\ast }\left(y^{\ast }\right)$ by Theorem 3, then the set $\left\{\right|g_{y^{\ast }}|:y^{\ast }\in B_{Y^{\ast }}\}$ is relatively weakly compact in $L^1\left(\lambda \right)$.

In this case, $\lambda$ is a control measure for m.

In our next result we will need the following lemma.

Lemma 4. 

(i) ([12] (Lemma 1)) If $T:X\to Y$ is an operator, then $T\left(B_X\right)$ is a $DP$ subset of Y if and only if $T^{\ast }:Y^{\ast }\to X^{\ast }$ is completely continuous.

(ii) ([12] (Lemma 1)) If $T:X\to Y$ is an operator, then $T\left(B_X\right)$ is a $\left(V^{\ast }\right)$-subset of Y if and only if $T^{\ast }:Y^{\ast }\to X^{\ast }$ is unconditionally converging.

(iii) ([21] (Theorem 14)) Let $1\le p<\infty$. If $T:X\to Y$ is an operator, then $T\left(B_X\right)$ is a p-$\left(V^{\ast }\right)$ set if and only if $T^{\ast }:Y^{\ast }\to X^{\ast }$ is p-convergent.

The following result is motivated by [13] (Theorem 1.4).

Theorem 4. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ and let $1<p<\infty$. The following are equivalent:

(a) 

T is limited (resp. $T^{\ast }$ is completely continuous, unconditionally converging, p-convergent).

(b) 

For every bounded sequence $\left(f_n\right)$ in $C(K,X)$ and every $w^{\ast }$-null (resp. weakly null, weakly 1-summable, weakly p-summable) sequence $\left(y_n^{\ast }\right)$ in $B_{Y^{\ast }}$, we have

$$\underset{n\to \infty }{lim}{\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda =0,$$

where $g_n$ is the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3.

Proof. 

We give the proof for limited operators; the other cases are similar when using Lemma 4.

$\left(a\right)\Rightarrow \left(b\right)$ Suppose T is limited. Let $\left(f_n\right)$ be a sequence in the unit ball of $C(K, X)$ and $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $B_{Y^{\ast }}$. For each n, let ${\varphi }_n$ be a scalar continuous function on K such that $\parallel {\varphi }_n\parallel \le 1$ and

$${\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda \le {\int }_K{\varphi }_n\langle f_n,g_n\rangle d\lambda +{\displaystyle \frac{1}{n}}.$$

Note that $\left({\varphi }_n{f}_n\right)$ is in the unit ball of $C(K, X)$. Recall that $T^{\ast }\left(y^{\ast }\right)=m_{y^{\ast }}\in rcabv(\Sigma ,X^{\ast })$, $y^{\ast }\in Y^{\ast }$. By Theorem 3, $\langle T\left(f\right),y_n^{\ast }\rangle ={\int }_K\langle g_n,f\rangle d\lambda$, for all $f\in C(K, X)$. Then

$$\begin{array}{cc}\hfill {\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda & \le {\int }_K\langle {\varphi }_n{f}_n,g_n\rangle d\lambda +{\displaystyle \frac{1}{n}}\hfill \\ \hfill & =\langle T\left({\varphi }_n{f}_n\right),y_n^{\ast }\rangle +{\displaystyle \frac{1}{n}}\to 0,\hfill \end{array}$$

since T is limited.

$\left(b\right)\Rightarrow \left(a\right)$ Let $\left(f_n\right)$ be a sequence in the unit ball of $C(K, X)$ and $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $Y^{\ast }$. Without loss of generality assume $\left(y_n^{\ast }\right)$ is in $B_{Y^{\ast }}$. Then

$$|\langle T\left(f_n\right),y_n^{\ast }\rangle |=|{\int }_K\langle f_n,g_n\rangle d\lambda |\le {\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda \to 0,$$

and thus T is limited. □

A positive Radon measure is discrete if every set of positive measure contains an atom ([33] (Ch 2, Sect. 8)). In this case, the measure is of the form ${\sum }_i{a}_i{\delta }_{t_i}$, with $\sum |a_i|<\infty$. In particular, it is concentrated on a countable set of its atoms.

Theorem 5. 

Let $1<p<\infty$ and let $m\leftrightarrow T:C(K,X)\to Y$ such that

(a) 

m is strongly bounded and admits a discrete control measure λ.

(b) 

For every $A\in \Sigma$, $m\left(A\right):X\to Y$ is limited (resp. $m{\left(A\right)}^{\ast }$ is completely continuous, unconditionally converging, p-convergent).

Then T is limited (resp. $T^{\ast }$ is completely continuous, unconditionally converging, p-convergent).

Proof. 

We will prove the result for limited operators. The other cases are similar. Let $\left(x_n\right)$ be a sequence in $B_X$ and $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $Y^{\ast }$. Without loss of generality assume $\left(y_n^{\ast }\right)$ is in $B_{Y^{\ast }}$. For each $n\in \mathbb{N}$, let $g_n$ be the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3. For every $A\in \Sigma$, $m\left(A\right):X\to Y$ is limited, and thus

$$\underset{n\to \infty }{lim}|{\int }_A\langle x_n,g_n\rangle d\lambda |=\underset{n\to \infty }{lim}\left|\langle m\left(A\right)\left(x_n\right),y_n^{\ast }\rangle \right|=0.$$

Then

$$\underset{n\to \infty }{lim}\langle x_n,g_n\left(t\right)\rangle =0,$$

for every $t\in K$ such that $\lambda \left(t\right)>0$.

Let $\left(f_n\right)$ be a sequence in the unit ball of $C(K, X)$. Then for every $t\in K$, $\left(f_n\left(t\right)\right)$ is in $B_X$, and

$$\underset{n\to \infty }{lim}\langle f_n\left(t\right),g_n\left(t\right)\rangle =0,$$

for every $t\in K$ such that $\lambda \left(t\right)>0$.

The sequence $\left(f_n\right)$ is bounded and the set $\left\{\right|g_n|:n\in \mathbb{N}\}$ is uniformly integrable (since it is relatively weakly compact in $L^1\left(\lambda \right)$ [2] (p. 76)). By Vitali’s Theorem we obtain

$$\underset{n\to \infty }{lim}{\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda =0.$$

Therefore, T is limited by Theorem 4. □

Remark 1. 

Every Radon measure on a compact dispersed space K is discrete [33] (Ch. 2, Sect. 8).

Corollary 2 

([12] (Theorem 11), [11] (Theorem 22)). Suppose that K is a dispersed compact Hausdorff space and $m\leftrightarrow T:C(K,X)\to Y$ is a strongly bounded operator. Let $1<p<\infty$. Then T is limited (resp. $T^{\ast }$ is completely continuous, unconditionally converging, p-convergent) if and only if for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is limited (resp. $m{\left(A\right)}^{\ast }$ is completely continuous, unconditionally converging, p-convergent).

Proof. 

Suppose that for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is limited (resp. $m{\left(A\right)}^{\ast }$ is completely continuous, unconditionally converging, p-convergent). Then T is limited (resp. $T^{\ast }$ is completely continuous, unconditionally converging, p-convergent) by Theorem 5 and Remark 1.

If T is limited (resp. $T^{\ast }$ is completely continuous, unconditionally converging, p-convergent), then for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is limited (resp. $m{\left(A\right)}^{\ast }$ is completely continuous, unconditionally converging, p-convergent) by [12] (Corollary 3), [11] (Corollary 15). □

Theorem 6. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. The following are equivalent:

(a) 

T is compact.

(b) 

For every sequence $\left(f_n\right)$ in the unit ball of $C(K,X)$ and every sequence $\left(y_n^{\ast }\right)$ in $B_{Y^{\ast }}$, we have

$$\underset{k\to \infty }{lim}{\int }_K\left|\langle f_{n_k},g_{n_k}-g_{m_k}\rangle \right|d\lambda =0,$$

for every increasing sequence $\left\{n_k\right\}$ and $\left\{m_k\right\}$ in $\mathbb{N}$, where $g_n$ is the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3.

Proof. 

$\left(a\right)\Rightarrow \left(b\right)$ Suppose T is compact. Let $\left(f_n\right)$ be a sequence in the unit ball of $C(K, X)$, $\left(y_n^{\ast }\right)$ be a sequence in $B_{Y^{\ast }}$, and let $\left\{n_k\right\}$ and $\left\{m_k\right\}$ be two increasing sequences in $\mathbb{N}$. For each k, let ${\varphi }_{n_k}$ be a scalar continuous function on K such that $\parallel {\varphi }_{n_k}\parallel \le 1$ and

$${\int }_K\left|\langle f_{n_k},g_{n_k}-g_{m_k}\rangle \right|d\lambda \le {\int }_K{\varphi }_{n_k}\langle f_{n_k},g_{n_k}-g_{m_k}\rangle d\lambda +{\displaystyle \frac{1}{n_k}}.$$

Note that $\left({\varphi }_{n_k}{f}_{n_k}\right)$ is in the unit ball of $C(K, X)$. Then

$$\begin{array}{cc}\hfill {\int }_K\left|\langle f_{n_k},g_{n_k}-g_{m_k}\rangle \right|d\lambda & \le {\int }_K\langle {\varphi }_{n_k}{f}_{n_k},g_{n_k}-g_{m_k}\rangle d\lambda +{\displaystyle \frac{1}{n_k}}\hfill \\ \hfill & =\langle T\left({\varphi }_{n_k}{f}_{n_k}\right),y_{n_k}^{\ast }-y_{m_k}^{\ast }\rangle +{\displaystyle \frac{1}{n_k}}\hfill \\ \hfill & \le \parallel T^{\ast }(y_{n_k}^{\ast }-y_{m_k}^{\ast })\parallel +{\displaystyle \frac{1}{n_k}}\to 0,\hfill \end{array}$$

since $T^{\ast }$ is compact.

$\left(b\right)\Rightarrow \left(a\right)$ Let $\left(y_n^{\ast }\right)$ be a sequence in $B_{Y^{\ast }}$, and let $\left\{n_k\right\}$ and $\left\{m_k\right\}$ be two increasing sequences in $\mathbb{N}$. Choose $\left(f_{n_k}\right)$ to be a sequence in the unit ball of $C(K, X)$ such that $\parallel T^{\ast }(y_{n_k}^{\ast }-y_{m_k}^{\ast })\parallel \le 2|\langle T^{\ast }(y_{n_k}^{\ast }-y_{m_k}^{\ast }),f_{n_k}\rangle |$ for each k. Then

$$\begin{array}{cc}\hfill \parallel T^{\ast }(y_{n_k}^{\ast }-y_{m_k}^{\ast })\parallel & \le 2|\langle T\left(f_{n_k}\right),y_{n_k}^{\ast }-y_{m_k}^{\ast }\rangle |=2|{\int }_K\langle f_{n_k},g_{n_k}-g_{m_k}\rangle d\lambda |\hfill \\ \hfill & \le 2{\int }_K\left|\langle f_{n_k},g_{n_k}-g_{m_k}\rangle \right|d\lambda \to 0.\hfill \end{array}$$

Therefore $T^{\ast }$, and thus T is compact. □

Theorem 7. 

Let $m\leftrightarrow T:C(K,X)\to Y$ such that

(a) 

m is strongly bounded and admits a discrete control measure λ.

(b) 

For every $A\in \Sigma$, $m\left(A\right):X\to Y$ is compact.

Then T is compact.

Proof. 

The proof is similar to that of Theorem 5. □

By Theorem 7 and Remark 1, we obtain the following result.

Corollary 3 

([12] (Remark 1)). Suppose that K is a dispersed compact Hausdorff space $m\leftrightarrow T:C(K,X)\to Y$ is a strongly bounded operator. Then T is compact if and only if for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is compact.

A Banach space X does not contain ${\ell }_1$ if and only if, for every Banach space Y, every completely continuous operator $T:X\to Y$ is compact [34] (p. 377), [35] (Theorem 1) if and only if every L-subset of $X^{\ast }$ is relatively compact if and only if every DP subset of $X^{\ast }$ is relatively compact [35].

Corollary 4. 

Suppose K is a dispersed compact Hausdorff space and $1\le p<\infty$. Then the following statements hold:

(i) 

([36] (Theorem 3.1.2)) X contains no copy of ${\ell }_1$ if and only if $C(K,X)$ contains no copy of ${\ell }_1$.

(ii) 

If every p-$\left(V\right)$ subset of $X^{\ast }$ is relatively compact, then every p-$\left(V\right)$ subset of $C{(K,X)}^{\ast }$ is relatively compact.

(iii) 

If every p-limited set of $X^{\ast }$ is relatively compact, then every p-limited subset of $C{(K,X)}^{\ast }$ is relatively compact.

Proof. 

(i) Suppose X contains no copy of ${\ell }_1$ and $m\leftrightarrow T:C(K, X)\to Y$ is completely continuous. Then for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is completely continuous [7]. Since ${\ell }_1\cancel{\hookrightarrow }X$, $m\left(A\right)$ is compact. Then T is compact by Corollary 3, and thus ${\ell }_1\cancel{\hookrightarrow }C(K,X)$. The converse is clear, since X is a subspace of $C(K, X)$.

(ii) Suppose every p-$\left(V\right)$ subset of $X^{\ast }$ that is relatively compact $m\leftrightarrow T:C(K, X)\to Y$ is p-convergent. Then T is strongly bounded and for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is p-convergent [9] (Proposition 2.1). Since every p-$\left(V\right)$ subset of $X^{\ast }$ is relatively compact, $m\left(A\right)$ is compact [21] (Theorem 21). Then T is compact, by Corollary 3. Thus every p-$\left(V\right)$ subset of $C{(K,X)}^{\ast }$ is relatively compact [21] (Theorem 21).

(iii) Suppose every p-limited subset of $X^{\ast }$ is relatively compact and $m\leftrightarrow T:C(K, X)\to Y$ is p-summing. Then for each $A\in \Sigma$, $m\left(A\right):X\to Y$ is p-summing [27] (Theorem 4.1 (iii)). By [32] (Theorem 4), $m\left(A\right)$ is compact. Then T is compact, and thus every p-limited subset of $C{(K,X)}^{\ast }$ is relatively compact [32] (Theorem 4). □

Theorem 8. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. The following are equivalent:

(a) 

T is absolutely summing.

(b) 

For every wuc series ${\sum }_n{f}_n$ in $C(K,X)$ and every sequence $\left(y_n^{\ast }\right)$ in $B_{Y^{\ast }}$, we have

$$\sum _n{\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda <\infty ,$$

where $g_n$ is the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3.

Proof. 

$\left(a\right)\Rightarrow \left(b\right)$ Suppose T is absolutely summing.

Let ${\sum }_n{f}_n$ be a wuc series in $C(K, X)$ and let $\left(y_n^{\ast }\right)$ be a sequence in $B_{Y^{\ast }}$. Let ${\varphi }_n$ be a scalar continuous function on K such that $\parallel {\varphi }_n\parallel \le 1$ and

$${\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda \le {\int }_K{\varphi }_n\langle f_n,g_n\rangle d\lambda +{\displaystyle \frac{1}{2^n}}.$$

Note that ${\sum }_n{\varphi }_n{f}_n$ is wuc in $C(K, X)$. Then

$$\begin{array}{cc}\hfill \sum _n{\int }_K\left|\langle f_n,g_n\rangle \right|d\lambda & \le \sum _n{\int }_K\langle {\varphi }_n{f}_n,g_n\rangle d\lambda +\sum _n{\displaystyle \frac{1}{2^n}}\hfill \\ \hfill & =\sum _n\langle T\left({\varphi }_n{f}_n\right),y_n^{\ast }\rangle +\sum _n{\displaystyle \frac{1}{2^n}}\hfill \\ \hfill & \le \sum _n\parallel T\left({\varphi }_n{f}_n\right)\parallel +\sum _n{\displaystyle \frac{1}{2^n}}<\infty ,\hfill \end{array}$$

since T is absolutely summing.

$\left(b\right)\Rightarrow \left(a\right)$ Let $\sum f_n$ be a wuc series in $C(K, X)$. Let $\left(y_n^{\ast }\right)$ be a sequence in $B_{Y^{\ast }}$ such that $\parallel T\left(f_n\right)\parallel =|\langle T\left(f_n\right),y_n^{\ast }\rangle |$ for each n. Then

$$\sum _n\parallel T\left(f_n\right)\parallel =\sum _n|\langle T\left(f_n\right),y_n^{\ast }\rangle |=\sum _n|{\int }_K\langle f_n,g_n\rangle d\lambda |\le \sum _n{\int }_K|\langle f_n,g_n\rangle |d\lambda <\infty ,$$

and thus T is absolutely summing. □

In the following, we introduce and study the almost limited (resp. absolutely summing, compact) operators, and operators with almost completely continuous (resp. unconditionally converging, p-convergent) adjoints, $1<p<\infty$.

Let $1<p<\infty$. Let $T:C(K,X)\to Y$ be a strongly bounded operator. We say that T is almost limited (resp. $T^{\ast }$ is almost completely continuous, almost unconditionally converging, almost p-convergent) if for every bounded sequence $\left(x_n\right)$ in X and every bounded sequence $\left({\varphi }_n\right)$ in $C\left(K\right)$, $\left(T\left({\varphi }_n{x}_n\right)\right)$ is a limited set (resp. DP set, $\left(V^{\ast }\right)$-set, p-$\left(V^{\ast }\right)$ set).

Let $T:C(K,X)\to Y$ be a strongly bounded operator. We say that T is almost absolutely summing if, for every wuc series $\sum x_n$ in X and every bounded sequence $\left({\varphi }_n\right)$ in $C\left(K\right)$, we have ${\sum }_n\parallel T\left({\varphi }_n{x}_n\right)\parallel <\infty$.

Let $T:C(K,X)\to Y$ be a strongly bounded operator. We say that T is almost compact if, for every bounded sequence $\left(x_n\right)$ in X and every bounded sequence $\left({\varphi }_n\right)$ in $C\left(K\right)$, we have $\left(T\left({\varphi }_n{x}_n\right)\right)$ is a relatively compact set.

Every limited (resp. absolutely summing, compact) operator $T:C(K,X)\to Y$ is almost limited (resp. almost absolutely summing, almost compact); every operator with completely continuous (resp. unconditionally converging, p-convergent) adjoint has an almost completely continuous (resp. unconditionally converging, p-convergent) adjoint.

The following result is motivated by [13] (Theorem 1.9).

Theorem 9. 

Let $1<p<\infty$ and let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. The following are equivalent:

(a) 

T is almost limited (resp. $T^{\ast }$ is almost completely continuous, almost unconditionally converging, almost p-convergent).

(b) 

For every bounded sequence $\left(x_n\right)$ in X and every $w^{\ast }$-null (resp. weakly null, weakly 1-summable, weakly p-summable) sequence $\left(y_n^{\ast }\right)$ in $B_{Y^{\ast }}$, we have

$$\underset{n\to \infty }{lim}{\int }_K\left|\langle x_n,g_n\rangle \right|d\lambda =0,$$

where $g_n$ is the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3.

Proof. 

We give the proof for limited operators; the other cases are similar.

$\left(a\right)\Rightarrow \left(b\right)$ Suppose T is almost limited. Let $\left(x_n\right)$ be a sequence in $B_X$ and $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $B_{Y^{\ast }}$. Let ${\varphi }_n$ be a scalar continuous function on K such that $\parallel {\varphi }_n\parallel \le 1$ and

$${\int }_K\left|\langle x_n,g_n\rangle \right|d\lambda \le {\int }_K{\varphi }_n\langle x_n,g_n\rangle d\lambda +{\displaystyle \frac{1}{n}}.$$

Then

$$\begin{array}{cc}\hfill {\int }_K\left|\langle x_n,g_n\rangle \right|d\lambda & \le {\int }_K\langle {\varphi }_n{x}_n,g_n\rangle d\lambda +{\displaystyle \frac{1}{n}}\hfill \\ \hfill & =\langle T\left({\varphi }_n{x}_n\right),y_n^{\ast }\rangle +{\displaystyle \frac{1}{n}}\to 0,\hfill \end{array}$$

since T is almost limited.

$\left(b\right)\Rightarrow \left(a\right)$ Let $\left(x_n\right)$ be a sequence in $B_X$ and $\left({\varphi }_n\right)$ be a bounded sequence in $C\left(K\right)$. Without loss of generality suppose $\parallel {\varphi }_n\parallel \le 1$. Let $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $B_{Y^{\ast }}$. Then

$$|\langle T\left({\varphi }_n{x}_n\right),y_n^{\ast }\rangle |=|{\int }_K\langle {\varphi }_n{x}_n,g_n\rangle \left|d\lambda \right|\le {\int }_K|\langle {\varphi }_n{x}_n,g_n\rangle |d\lambda \le {\int }_K\left|\langle x_n,g_n\rangle \right|d\lambda \to 0,$$

and thus T is almost limited. □

Corollary 5. 

Let $1<p<\infty$ and let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. If T is almost limited (resp. $T^{\ast }$ is almost completely continuous, almost unconditionally converging, almost p-convergent), then $m\left(A\right):X\to Y$ is limited (resp. $m{\left(A\right)}^{\ast }$ is completely continuous, unconditionally converging, p-convergent) for each $A\in \Sigma$.

Proof. 

We give the proof for almost limited operators; the other cases are similar. Suppose T is almost limited. Let $A\in \Sigma$, $\left(x_n\right)$ be a sequence in $B_X$, and let $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $B_{Y^{\ast }}$. By Theorem 9,

$$|\langle y_n^{\ast },m\left(A\right)x_n\rangle |=|{\int }_A\langle x_n,g_n\rangle d\lambda |\le {\int }_A\left|\langle x_n,g_n\rangle \right|d\lambda \to 0.$$

Thus, $\left(m\left(A\right)x_n\right)$ is limited. □

Corollary 6. 

Let $1<p<\infty$. Let K be a dispersed compact Hausdorff space and $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator. Then T is limited (resp. $T^{\ast }$ is completely continuous, unconditionally converging, p-convergent) if and only if T is almost limited (resp. $T^{\ast }$ is almost completely continuous, almost unconditionally converging, almost p-convergent).

Proof. 

We give the proof for limited operators; the other cases are similar. Every limited operator $T:C(K, X)\to Y$ is almost limited. Conversely, suppose T is almost limited. By Corollary 5, $m\left(A\right):X\to Y$ is limited, for each $A\in \Sigma$. Therefore, T is limited by Corollary 2. □

Theorem 10. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. The following are equivalent:

(a) 

T is almost absolutely summing.

(b) 

For every wuc series $\sum x_n$ in X and every sequence $\left(y_n^{\ast }\right)$ in $B_{Y^{\ast }}$, we have

$$\sum _n{\int }_K\left|\langle x_n,g_n\rangle \right|d\lambda <\infty ,$$

where $g_n$ is the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3.

Proof. 

The proof is similar to that of Theorem 9, using the fact that for every wuc series $\sum x_n$ in X and every bounded sequence $\left({\varphi }_n\right)$ in $C\left(K\right)$, the series $\sum {\varphi }_n{x}_n$ is wuc in $C(K, X)$. □

Corollary 7. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. If T is almost absolutely summing, then $m\left(A\right):X\to Y$ is absolutely summing for each $A\in \Sigma$.

Proof. 

The proof is similar to that of Corollary 5. □

Theorem 11. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. The following are equivalent:

(a) 

T is almost compact.

(b) 

For every bounded sequence $\left(x_n\right)$ in X and every sequence $\left(y_n^{\ast }\right)$ in $B_{Y^{\ast }}$, we have

$$\underset{k\to \infty }{lim}{\int }_K\left|\langle x_{n_k},g_{n_k}-g_{m_k}\rangle \right|d\lambda =0,$$

for every increasing sequences $\left\{n_k\right\}$ and $\left\{m_k\right\}$ in $\mathbb{N}$, where $g_n$ is the function corresponding to $T^{\ast }\left(y_n^{\ast }\right)$ by Theorem 3.

Proof. 

The proof is similar to that of Theorem 9. □

Corollary 8. 

Let $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator whose representing measure m has a control measure λ. If T is almost compact, then $m\left(A\right):X\to Y$ is compact for each $A\in \Sigma$.

Proof. 

The proof is similar to that of Corollary 5. □

Corollary 9. 

Let K be a dispersed compact Hausdorff space and $m\leftrightarrow T:C(K,X)\to Y$ be a strongly bounded operator. Then T is compact if and only if T is almost compact.

Proof. 

The proof is similar to that of Corollary 6. □

If $T:C(K,X)\to Y$ is linear and continuous, then there is a unique linear, conttinuous map $S:C\left(K\right)\to L(X,Y)$ such that $T\left(\varphi x\right)=S\varphi \left(x\right)$, for $\varphi \in C\left(K\right)$, $x\in X$ [4] (III. 19. 2). Let $Lim(X,Y)$ denote the set of all limited operators $T:X\to Y$, and let $U^d(X,Y)$ denote the set of all operators $T:X\to Y$ such that $T^{\ast }$ is unconditionally converging.

We note that if $T:C(K,X)\to Y$ is limited (resp. $T^{\ast }$ is unconditionally converging), then for each $\varphi \in C\left(K\right)$, the operator $S\varphi :X\to Y$ is limited (resp. ${\left(S\varphi \right)}^{\ast }$ is unconditionally converging).

The following result is motivated by [28] (Theorem 11 (ii)).

Corollary 10. 

Let K be a dispersed compact Hausdorff space and let $T:C(K,X)\to Y$ be a strongly bounded operator.

(i) 

If $S:C\left(K\right)\to Lim(X,Y)$ is limited, then T is limited.

(ii) 

If $S:C\left(K\right)\to U^d(X,Y)$ has an unconditionally converging adjoint, then T has an unconditionally converging adjoint.

Proof. 

(i) Let $\left(x_n\right)$ be a sequence in $B_X$, $\left({\varphi }_n\right)$ be in the unit ball of $C\left(K\right)$, and $\left(y_n^{\ast }\right)$ be a $w^{\ast }$-null sequence in $B_{Y^{\ast }}$. If $T:X\to Y$ is limited, then $T^{\ast }:Y^{\ast }\to X^{\ast }$ is $w^{\ast }$-norm sequentially continuous, and $|\langle x_n\otimes y_n^{\ast },T\rangle |\le \parallel T^{\ast }\left(y_n^{\ast }\right)\parallel \to 0$. Then $(x_n\otimes y_n^{\ast })$ is $w^{\ast }$-null in ${\left(Lim(X,Y)\right)}^{\ast }$,

$$|\langle S\left({\varphi }_n\right),x_n\otimes y_n^{\ast }\rangle |=|\langle T\left({\varphi }_n{x}_n\right),y_n^{\ast }\rangle |\to 0,$$

and thus T is almost limited. Therefore, T is limited by Corollary 6.

(ii) Let $\left(x_n\right)$ be a sequence in $B_X$, $\left({\varphi }_n\right)$ be in the unit ball of $C\left(K\right)$, and $\sum y_n^{\ast }$ be a wuc series in $Y^{\ast }$. Suppose $T\in U^d(X,Y)$ and $\sum |\langle x_n\otimes y_n^{\ast },T\rangle |$ is not convergent. Let $ϵ>0$. For each n, there are increasing sequences $\left(p_n\right)$ and $\left(q_n\right)$ in $\mathbb{N}$ with $p_1<q_1<p_2<q_2<\dots ,$ such that for each n,

$$\sum _{i=p_n}^{q_n}\left|\langle T,x_i\otimes y_i^{\ast }\rangle \right|>ϵ.$$

Then

$$\sum _{i=p_n}^{q_n}\parallel T^{\ast }\left(y_i^{\ast }\right)\parallel >\sum _{i=p_n}^{q_n}\left|\langle T^{\ast }\left(y_i^{\ast }\right),x_i\rangle \right|>ϵ,$$

which contradicts the fact that $T^{\ast }$ is unconditionally converging. Then $(x_n\otimes y_n^{\ast })$ is $w^{\ast }$-1-summable, and thus weakly 1-summable in ${\left(U^d(X,Y)\right)}^{\ast }$.

Since $S^{\ast }$ is unconditionally converging, $\left(S\left({\varphi }_n\right)\right)$ is a $\left(V^{\ast }\right)$-set in $U^d(X,Y)$, and thus,

$$|\langle S\left({\varphi }_n\right),x_n\otimes y_n^{\ast }\rangle |=|\langle T\left({\varphi }_n{x}_n\right),y_n^{\ast }\rangle |\to 0.$$

Therefore $\left(T\left({\varphi }_n{x}_n\right)\right)$ is a $\left(V^{\ast }\right)$-set in Y, and $T^{\ast }$ is almost unconditionally converging. Then $T^{\ast }$ is unconditionally converging by Corollary 6. □

4. Conclusions and Future Works

In Lemma 3 it was shown that if H is a bounded subset of X and for each $ϵ>0$ there is a Grothendieck subset $H_{ϵ}$ of X so that $H\subseteq H_{ϵ}+ϵB_X$, then H is a Grothendieck set. It would be interesting to know whether a similar result is true for p-limited (resp. p-compact) sets; that is, if H is a bounded subset of X and for each $ϵ>0$ there is a p-limited (resp. p-compact) subset $H_{ϵ}$ of X so that $H\subseteq H_{ϵ}+ϵB_X$, then H is a p-limited (resp. p-compact) set.

In Theorem 2 it was proved that if K is a dispersed compact Hausdorff space and $m\leftrightarrow T:C(K,X)\to Y$ is a strongly bounded operator, then T is Grothendieck whenever $m\left(A\right):X\to Y$ is Grothendieck, for each $A\in \Sigma$. It would be interesting to know whether T is p-limited (resp. p-compact) whenever $m\left(A\right):X\to Y$ is p-limited (resp. p-compact), for each $A\in \Sigma$.