Characterizations of continuous operators on $C_b(X)$ with the strict topology

Abstract

Let X be a completely regular Hausdorff space and $C_b(X)$ be the space of all bounded continuous functions on X, equipped with the strict topology $\beta$. We study some important classes of $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operators from $C_b(X)$ to a Banach space $(E,\Vert \cdot \Vert _E)$: $\beta$-absolutely summing operators, compact operators and $\beta$-nuclear operators. We characterize compact operators and $\beta$-nuclear operators in terms of their representing measures. It is shown that dominated operators and $\beta$-absolutely summing operators $T:C_b(X)\rightarrow E$ coincide and if, in particular, E has the Radon–Nikodym property, then $\beta$-absolutely summing operators and $\beta$-nuclear operators coincide. We generalize the classical theorems of Pietsch, Tong and Uhl concerning the relationships between absolutely summing, dominated, nuclear and compact operators on the Banach space C(X), where X is a compact Hausdorff space.

1 Introduction and preliminaries

The Riesz representation theorem plays a crucial role in the study of operators on the Banach space C(X) of continuous functions on a compact Hausdorff space X. Due to this theorem, different classes of operators on C(X) have been characterized in terms of their representing Radon vector measures.

Absolutely summing operators between Banach spaces have been the object of several studies (see [1, pp. 209–233] and [5, 8, 11, 27, 28, 31, 34]). It originates in the fundamental paper of Grothendieck [17] from 1953. Grothendieck’s inequality has equivalent formulation using the theory of absolutely summing operators (see [1, Theorem 8.3.1] and [4, 22]). In the multilinear case, it is also connected with the Bohnenblust–Hille and the Hardy–Littlewood inequalities (see [2]). There is a vast literature on absolutely summing operators from the Banach space C(X) to a Banach space E (see [1], [9, Chap. VI], [11, 34, 43]).

The concept of nuclearity in Banach spaces is due to Grothendieck [17, 18] and Ruston [33] and has the origin in Schwartz’s kernel theorem [18]. Many authors have studied nuclear operators between locally convex spaces (see [21, §17.3], [37, Chap. 3, §7], [46, p. 289]) and Banach spaces (see [9, Chap. VI], [11, 16] [46, p. 279]). If F is a Banach space, nuclear operators from the Banach space C(X, F) of F-valued continuous functions on a compact Hausdorff space X to E have been studied intensively by Popa [29], Saab [35], Saab and Smith [36]. In particular, a characterization of nuclear operators from C(X) to E in terms of their representing measures can be found in [9, Theorem 4, pp. 173–174], [34, Proposition 5.30], [43, Proposition 1.2].

The interplay between absolutely summing operators, dominated operators of Dinculeanu (see [12, §19], [13, §1]) and nuclear operators $T:C(X)\rightarrow E$ has been an interesting issue in operator theory. Pietsch [27, 2.3.4, Proposition, p. 41] proved that dominated operators and absolutely summing operators on the Banach space C(X) coincide. It is known that if in particular, E has the Radon–Nikodym property, then absolutely summing and nuclear operators $T:C(X)\rightarrow E$ coincide (see [9, Corollary 5, p. 174]). Moreover, Uhl [44, Theorem 1] showed that if, E has the Radon–Nikodym property, then every dominated operator $T:C(X)\rightarrow E$ is compact.

The aim of this paper is to extend these classical results to the setting, where X is a completely regular Hausdorff k-space.

Throughout the paper, we assume that $(X,{{\mathcal {T}}})$ is a completely regular Hausdorff space. By ${{\mathcal {K}}}$ we denote the family of all compact sets in X. Let ${{\mathcal {B}}}o$ denote the $\sigma$-algebra of Borel sets in X.

Let $C_b(X)$ (resp. $B({{\mathcal {B}}}o))$ denote the Banach space of all bounded continuous (resp. bounded ${{\mathcal {B}}}o$-measurable) scalar functions on X, equipped with the topology $\tau _u$ of the uniform norm $\Vert \cdot \Vert _\infty$. By ${{\mathcal {S}}}({{\mathcal {B}}}o)$ we denote the space of all ${{\mathcal {B}}}o$-simple scalar functions on X. Let $C_b(X)'$ stand for the Banach dual of $C_b(X)$.

Following [15, 37] and [45, Definition 10.4, p. 137] the strict topology $\beta$ on $C_b(X)$ is the locally convex topology determined by the seminorms

$$\begin{aligned} p_w(u):=\sup _{t\in X} w(t)|u(t)| \ \ \text{ for }\ \ u\in C_b(X), \end{aligned}$$

where w runs over the family ${{\mathcal {W}}}$ of all bounded functions $w:X\rightarrow [0,\infty )$ which vanish at infinity, that is, for every $\varepsilon \!>\!0$ there exists $K\!\in \!{{\mathcal {K}}}$ such that $\sup _{t\in X{\setminus }K} w(t)\le \varepsilon$. Let ${{\mathcal {W}}}_1:=\{w\in {{\mathcal {W}}}:0\le w\le \mathbb {1}_X\}$. For $w\in {{\mathcal {W}}}_1$ and $\eta >0$ let

$$\begin{aligned} U_w(\eta ):=\{u\in C_b(X):p_w(u)\le \eta \}. \end{aligned}$$

Note that the family $\{U_w(\eta ):w\in {{\mathcal {W}}}_1,\eta >0\}$ is a local base at 0 for $\beta$.

The strict topology $\beta$ on $C_b(X)$ has been studied intensively (see [15, 20, 38, 41, 45]). Note that $\beta$ can be characterized as the finest locally convex Hausdorff topology on $C_b(X)$ that coincides with the compact-open topology $\tau _c$ on $\tau _u$-bounded sets (see [41, Theorem 2.4]). The topologies $\beta$ and $\tau _u$ have the same bounded sets. This means that $(C_b(X),\beta )$ is a generalized DF-space (see [38, Corollary]), and it follows that $(C_b(X),\beta )$ is quasinormable (see [32, p. 422]). If, in particular, X is locally compact (resp. compact), then $\beta$ coincides with the original strict topology of Buck [6] (resp. $\beta =\tau _u$).

Recall that a countably additive scalar measure $\mu$ on ${{\mathcal {B}}}o$ is said to be a Radon measure if its variation $|\mu |$ is regular, that is, for every $A\in {{\mathcal {B}}}o$ and $\varepsilon >0$ there exist $K\in {{\mathcal {K}}}$ and $O\in {{\mathcal {T}}}$ with $K\subset A\subset O$ such that $|\mu |(O{\setminus }K)\le \varepsilon$. Let M(X) denote the Banach space of all scalar Radon measures, equipped with the total variation norm $\Vert \mu \Vert :=|\mu |(X)$.

The following characterization of the topological dual of $(C_b(X),\beta )$ will be of importance (see [15, Lemma 4.5]), [20, Theorem 2].

Theorem 1.1

For a linear functional $\Phi$ on $C_b(X)$ the following statements are equivalent:

• (i) $\Phi$ is $\beta$-continuous.

• (ii) There exists a unique $\mu \in M(X)$ such that

  $$\begin{aligned} \Phi (u)=\Phi _\mu (u)=\int _X u\,d\mu \ \text{ for } \ u\in C_b(X) \end{aligned}$$

and $\Vert \Phi _\mu \Vert '=|\mu |(X)$ for $\mu \in M(X)$ (here $\Vert \cdot \Vert '$ denotes the conjugate norm in $C_b(X)')$.

The following result will be useful (see [41, Theorem 5.1]).

Theorem 1.2

For a subset ${{\mathcal {M}}}$ of M(X) the following statements are equivalent:

• (i) $\sup _{\mu \in {{\mathcal {M}}}}|\mu |(X)<\infty$ and ${{\mathcal {M}}}$ is uniformly tight, that is, for each $\varepsilon >0$ there exists $K\in {{\mathcal {K}}}$ such that $\sup _{\mu \in {{\mathcal {M}}}}|\mu |(X\setminus K)\le \varepsilon$.

• (ii) The family $\{\Phi _\mu :\mu \in {{\mathcal {M}}}\}$ is $\beta$-equicontinuous.

Recall that a completely regular Hausdorff space $(X,{{\mathcal {T}}})$ is a k-space if any subset A of X is closed whenever $A\cap K$ is compact for all compact sets K in X. In particular, every locally compact Hausdorff space, every metrizable space and every space satisfying the first countability axiom is a k-space (see [14, Chap. 3, § 3]).

From now on, we will assume that $(X,{{\mathcal {T}}})$ is a k-space. Then, the space $(C_b(X),\beta )$ is complete (see [15, Theorem 2.4]).

We assume that $(E,\Vert \cdot \Vert _E)$ is a Banach space. Let $B_{E'}$ stand for the closed unit ball in the Banach dual $E'$ of E.

Recall that a bounded linear operator $T:C_b(X)\rightarrow E$ is said to be absolutely summing if there exists a constant $c>0$ such that for any finite set $\{u_1,\ldots ,u_n\}$ in $C_b(X)$,

$$\begin{aligned} \sum _{i=1}^n\Vert T(u_i)\Vert _E\le c\sup \left\{ \sum _{i=1}^n|\Phi (u_i)|:\Phi \in B_{C_b(X)'}\right\} . \end{aligned}$$

(1.1)

The infimum of number of $c>0$ satisfying (1.1) denoted by $\Vert T\Vert _{\text {as}}$ is called an absolutely summing norm of T.

It is known that a bounded linear operator $T:C_b(X)\rightarrow E$ is absolutely summing if and only if T maps unconditionally convergent series in $C_b(X)$ into absolutely convergent series in E (see [9, Definition 1, p. 161 and Proposition 2, p. 162]).

For $t\in X$, let $\delta _t$ stand for the point mass measure, that is, $\delta _t(A):=\mathbb {1}_A(t)$ for $A\in {{\mathcal {B}}}o$. Then $\delta _t\in M^+(X)$ and $\int _X u\,{\text {d}}\delta _t=u(t)$ for $u\in C_b(X)$. Clearly, $\Vert \delta _t\Vert =\delta _t(X)=1$.

Lemma 1.3

For a bounded linear operator $T:C_b(X)\rightarrow E$, the following statements are equivalent:

• (i) T is absolutely summing.

• (ii) There exists $c>0$ such that for any set $\{u_1,\ldots ,u_n\}$ in $C_b(X)$,

  $$\begin{aligned} \sum _{i=1}^n\Vert T(u_i)\Vert _E\le c\sup \left\{ \sum _{i=1}^n\bigg |\int _Xu_i\,{\text {d}}\mu \bigg |:\mu \in M(X),|\mu |(X)\le 1\right\} . \end{aligned}$$

Proof

(i)$\Rightarrow$(ii) There exists $c>0$ such that for any set $\{u_1,\ldots ,u_n\}$ in $C_b(X)$,

$$\begin{aligned} \sum _{i=1}^n\Vert T(u_i)\Vert _E\le c\sup \left\{ \sum _{i=1}^n|\Phi (u_i)|:\Phi \in B_{C_b(X)'}\right\} . \end{aligned}$$

Note that we have (see [1, p. 205]),

$$\begin{aligned} \sup \left\{ \sum _{i=1}^n|\Phi (u_i)|:\Phi \in B_{C_b(X)'}\right\} =\sup \left\{ \bigg \Vert \sum _{i=1}^n\varepsilon _iu_i\bigg \Vert _\infty :(\varepsilon _i)\in \{-1,1\}^n\right\} . \end{aligned}$$

Hence, we get,

$$\begin{aligned} \begin{aligned} \sum _{i=1}^n\Vert T(u_i)\Vert _E&\le c\sup \left\{ \bigg \Vert \sum _{i=1}^n\varepsilon _iu_i\bigg \Vert _\infty :(\varepsilon _i)\in \{-1,1\}^n\right\} \\&=c\sup \left\{ \bigg |\sum _{i=1}^n\varepsilon _iu_i(t)\bigg |:(\varepsilon _i)\in \{-1,1\}^n,t\in X\right\} \\&\le c\sup \left\{ \sum _{i=1}^n|u_i(t)|:t\in X\right\} =c\sup \left\{ \sum _{i=1}^n\bigg |\int _Xu_i\,{\text {d}}\delta _t\bigg |:t\in X\right\} \\&\le c\sup \left\{ \sum _{i=1}^n\bigg |\int _Xu_i\,{\text {d}}\mu \bigg |:\mu \in M(X),|\mu |(X)\le 1\right\} . \end{aligned} \end{aligned}$$

(ii)$\Rightarrow$(i) This is obvious. $\square$

The general theory of absolutely summing operators between locally convex spaces was developed by Pietsch [27].

Following [27, 1.2, pp. 23–24], we say that a sequence $(u_n)$ in $C_b(X)$ is $\beta$ -weakly summable if $\sum ^\infty _{n=1}|\int _X u_n\,{\text {d}}\mu | <\infty$ for every $\mu \in M(X)$. By $\ell ^1_w(C_b(X),\beta )$, we denote the linear space of all $\beta$-weakly summable sequences in $C_b(X)$.

Let $(u_n)\in \ell ^1_w(C_b(X),\beta )$. Then, in view of [27, 1.2.3, pp. 23–24] for each $w\in {{\mathcal {W}}}_1$ and $\eta >0$ there exists $\varrho _{w,\eta }>0$ such that

$$\begin{aligned} {{\mathcal {E}}}_{w,\eta }((u_n)):=\sup \left\{ \sum ^\infty _{n=1}\bigg |\int _X u_n\,{\text {d}}\mu \bigg |: \mu \in U_w(\eta )^0\right\} \le \varrho _{w,\eta },\, \end{aligned}$$

where $U_w(\eta )^0$ stands for the polar of $U_w(\eta )$ with respect to the pairing $\langle C_b(X),M(X)\rangle$. Then, ${{\mathcal {E}}}_{w,\eta }$ is a seminorm on $\ell ^1_w(C_b(X),\beta )$ and the family $\{{{\mathcal {E}}}_{w,\eta }:w\in {{\mathcal {W}}}_1,\,\eta >0\}$ generates the so-called ${{\mathcal {E}}}$ -topology on $\ell ^1_w(C_b(X),\beta )$ (see [27, 1.2.3]).

Let ${{\mathcal {F}}}({\mathbb {N}})$ denote the family of all finite sets in ${\mathbb {N}}$, the set of all natural numbers. By $\ell ^1_s(C_b(X),\beta )$ we denote the ${{\mathcal {E}}}$-closed subspace of $\ell ^1_w(C_b(X),\beta )$ consisting of all $\beta$ -summable sequences in $C_b(X)$ (see [27, 1.3]). In view of [27, Theorem 1.3.6] a sequence $(u_n)\in \ell ^1_s(C_b(X),\beta )$ if and only if the net $(s_M)_{M\in {{\mathcal {F}}}({\mathbb {N}})}$ of partial sums $s_M:=\sum _{i\in M} u_i$ forms a $\beta$-Cauchy sequence in $C_b(X)$, where ${{\mathcal {F}}}({\mathbb {N}})$ is directed by inclusion.

Let $\ell ^1(E)$ stand for the linear space of all absolutely summable sequences in E, i.e., $(e_n)\in \ell ^1(E)$ if $\sum ^\infty _{n=1}\Vert e_n\Vert _E<\infty$. Then, $\ell ^1(E)$ can be equipped with the norm $\pi _E((e_n)):=\sum ^\infty _{n=1} \Vert e_n\Vert _E$ (see [27, 1.4]).

According to [27, 2.1], we have

Definition 1.4

A $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator $T:C_b(X)\rightarrow E$ is said to be $\beta$ -absolutely summing if $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E<\infty$ whenever $(u_n)\in \ell ^1_s(C_b(X),\beta )$.

Recall that a linear operator $T:C_b(X)\rightarrow E$ is said to be $\beta$ -compact (resp. $\beta$ -weakly compact) if there exists a $\beta$-neighborhood V of 0 such that T(V) is a relatively norm compact (resp. relatively weakly compact) subset of E.

We will say that an operator $T:C_b(X)\rightarrow E$ is compact (resp. weakly compact) if T is $\tau _u$-compact (resp. $\tau _u$-weakly compact).

Proposition 1.5

Let $T:C_b(X)\rightarrow E$ be a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator. Then, the following statements are equivalent:

• (i) T is weakly compact (resp. compact).

• (ii) T is $\beta$-weakly compact (resp. $\beta$-compact).

Proof

(i)$\Rightarrow$(ii) Assume that (i) holds. Topologies $\beta$ and $\tau _u$ have the same bounded sets in $C_b(X)$, so T maps $\beta$-bounded sets onto relatively weakly compact (resp. norm compact) sets in E. Since the space $(C_b(X),\beta )$ is quasinormable, by the Grothendieck classical result (see [32, p. 429]), we obtain that T is $\beta$-weakly compact (resp. $\beta$-compact).

(ii)$\Rightarrow$(i) This is obvious because $\beta \subset \tau _u$. $\square$

Following [12, § 19, Section 3], [13, § 1, Section H] one can distinguish an important class of linear operators on $C_b(X)$.

Definition 1.6

A linear operator $T:C_b(X)\rightarrow E$ is said to be dominated if there exists $\mu \in M^+(X)$ such that

$$\begin{aligned} \Vert T(u)\Vert _E\le \int _X |u|\,{\text {d}}\mu \ \ \text{ for } \ \ u\in C_b(X). \end{aligned}$$

Then, we say that T is dominated by $\mu$.

According to [25, Proposition 3.1] we have.

Proposition 1.7

Every dominated operator $T:C_b(X)\rightarrow E$ is $(\beta ,\Vert \cdot \Vert _E)$-continuous and weakly compact.

Following [37, Chap. 3, §7] (see also [21, §17.3, p. 376]) and using Theorem 1.2 we have the following definition.

Definition 1.8

A linear operator $T:C_b(X)\rightarrow E$ is said to be $\beta$ -nuclear, if there exist a uniformly bounded and uniformly tight sequence $(\mu _n)$ in M(X), a bounded sequence $(e_n)$ in E and a sequence $(\lambda _n)\in \ell ^1$ such that

$$\begin{aligned} T(u)=\sum _{n=1}^\infty \lambda _n\Big (\int _X u\,{\text {d}}\mu _n\Big )e_n\ \ \text{ for }\ \ u\in C_b(X). \end{aligned}$$

(1.2)

If $T:C_b(X)\rightarrow E$ is $\beta$-nuclear operator, let us put

$$\begin{aligned} \Vert T\Vert _{\beta \text{- }{\text {nuc}}}:=\inf \Big \{\sum _{n=1}^\infty |\lambda _n||\mu _n|(X)\Vert e_n\Vert _E\Big \}, \end{aligned}$$

where the infimum is taken over all sequences $(\mu _n)$ in M(X), $(e_n)$ in E and $(\lambda _n)\in \ell ^1$ such that T admits a representation (1.2).

Every $\beta$-nuclear operator $T:C_b(X)\rightarrow E$ is $(\beta ,\Vert \cdot \Vert _E)$-continuous and $\beta$-compact (see [37, Chap. 3, §7, Corollary 1]).

In [24], the theory of integral representation of continuous operators on $C_b(X)$, equipped with the strict topology $\beta$ has been developed. Making use of the results of [24], we study $\beta$-absolutely summing operators, compact operators and $\beta$-nuclear operators $T:C_b(X)\rightarrow E$. We characterize compact operators and $\beta$-nuclear operators $T:C_b(X)\rightarrow E$ in terms of their representing measures (see Theorems 4.1 and 5.1 below). It is shown that dominated operators and $\beta$-absolutely summing operators $T:C_b(X)\rightarrow E$ coincide (see Corollary 3.4) and if, in particular, E has the Radon–Nikodym property, then $\beta$-absolutely summing and $\beta$-nuclear operators $T:C_b(X)\rightarrow E$ coincide (see Corollary 5.2). We prove that a natural kernel operator $T:C_b(X)\rightarrow C(K)$ is $\beta$-nuclear (see Theorem 6.3).

2 Integral representation

In this section, we collect basic concepts and facts concerning integral representation of operators on $C_b(X)$ that will be useful (see [24] for notation and more details).

Let $m:{{\mathcal {B}}}o\rightarrow E$ be a finitely additive measure. By |m|(A) (resp. $\Vert m\Vert (A))$, we denote the variation (resp. the semivariation) of m on $A\in {{\mathcal {B}}}o$ (see [9, Definition 4, p. 2]). Then, $\Vert m\Vert (A)\le |m|(A)$ for $A\in {{\mathcal {B}}}o$.

For $e'\in E'$, let

$$\begin{aligned} m_{e'}(A):=e'(m(A)) \ \text{ for } \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Then,

$$\begin{aligned} \Vert m\Vert (A)=\sup _{e'\in B_{E'}}|m_{e'}|(A), \end{aligned}$$

where $|m_{e'}|(A)$ stands for the variation of $m_{e'}$ on $A\in {{\mathcal {B}}}o$.

Recall that a countably additive measure $m:{{\mathcal {B}}}o\rightarrow E$ is called a Radon measure if its semivariation $\Vert m\Vert$ is regular, i.e., for each $A\in {{\mathcal {B}}}o$ and $\varepsilon >0$ there exist $K\in {{\mathcal {K}}}$ and $O\in {{\mathcal {T}}}$ with $K\subset A\subset O$ such that $\Vert m\Vert (O{\setminus }K)\le \varepsilon$ (see [24, Definition 3.3]).

We will need the following result (see [12, §15.6, Proposition 19]).

Lemma 2.1

Assume that $m:{{\mathcal {B}}}o\rightarrow E$ is a Radon measure and $|m|(X)<\infty$. Then, $|m|\in M^+(X)$.

Assume that $m:{{\mathcal {B}}}o\rightarrow E$ is a finitely additive measure with $\Vert m\Vert (X)<\infty$. Then, for every $v\in B({{\mathcal {B}}}o)$, one can define the so-called immediate integral $\int _X v\,{\text {d}}m\in E$ by

$$\begin{aligned} \int _X v\,dm:=\lim \int _X s_n\,{\text {d}}m, \end{aligned}$$

(2.1)

where $(s_n)$ is a sequence in ${{\mathcal {S}}}({{\mathcal {B}}}o)$ such that $\Vert s_n-v\Vert _\infty \rightarrow 0$ (see [9, p. 5], [13, § 1, Section G]). Then, for $v\in B({{\mathcal {B}}}o)$,

$$\begin{aligned} \bigg \Vert \int _X v\,{\text {d}}m\bigg \Vert _E\le \Vert v\Vert _\infty \,\Vert m\Vert (X). \end{aligned}$$

For $e'\in E'$, we have

$$\begin{aligned} e'\bigg (\int _X v\,{\text {d}}m\bigg )=\int _X v\,{\text {d}}m_{e'} \ \ \text{ for } \ \ v\in B({{\mathcal {B}}}o). \end{aligned}$$

(2.2)

Let $ca({{\mathcal {B}}}o)$ denote the Banach space of all countably additive scalar measures on ${{\mathcal {B}}}o$, equipped with the total variation norm $\Vert \mu \Vert :=|\mu |(X)$. For $\mu \in ca({{\mathcal {B}}}o)^+$, let ${{\mathcal {L}}}^1(\mu )$ denote the space of all $\mu$-integrable scalar functions on X, equipped with the seminorm $\Vert v\Vert _1:=\int _X|v|\,{\text {d}}\mu$ for $v\in {{\mathcal {L}}}^1(\mu )$. Then

$$\begin{aligned} C_b(X)\subset B({{\mathcal {B}}}o)\subset {{\mathcal {L}}}^1(\mu ). \end{aligned}$$

Assume that $m:{{\mathcal {B}}}o\rightarrow E$ is a countably additive measure of finite variation |m|, i.e., $|m|(X)<\infty$. Then $|m|\in ca({{\mathcal {B}}}o)^+$ (see [9, Proposition 9, p. 3]). Since ${{\mathcal {S}}}({{\mathcal {B}}}o)$ is $\Vert \cdot \Vert _1$-dense in ${{\mathcal {L}}}^1(|m|)$, for every

$$\begin{aligned} \int _X v\,{\text {d}}m:=\lim \int _X s_n\,{\text {d}}m, \end{aligned}$$

(2.3)

where $(s_n)$ is a sequence in ${{\mathcal {S}}}({{\mathcal {B}}}o)$ such that $\Vert s_n-v\Vert _1\rightarrow 0$ (see [13, § 2, Sect. D]).

Note that for $v\in B({{\mathcal {B}}}o)\subset {{\mathcal {L}}}^1(|m|)$, the integral $\int _X v\,dm$ defined in (2.3) coincides with the immediate integral defined in (2.1). We have

$$\begin{aligned} \bigg \Vert \int _X v\,{\text {d}}m\bigg \Vert _E\le \int _X |v|\,{\text {d}}|m| \ \ \text{ for }\ \ v\in {{\mathcal {L}}}^1(|m|). \end{aligned}$$

(2.4)

Hence, the corresponding integration operator $T_m:{{\mathcal {L}}}^1(|m|)$ $\rightarrow E$ given by

$$\begin{aligned} T_m(v):=\int _X v\,{\text {d}}m \ \ \text{ for } \ \ v\in {{\mathcal {L}}}^1(|m|) \end{aligned}$$

is $(\Vert \cdot \Vert _1,\Vert \cdot \Vert _E)$-continuous.

Let $C_b(X)'_\beta$ and $C_b(X)''_\beta$ denote the dual and the bidual of $(C_b(X),\beta )$. Since $\beta$-bounded subsets of $C_b(X)$ are $\tau _u$-bounded, the strong topology $\beta (C_b(X)'_\beta ,C_b(X))$ in $C_b(X)'_\beta$ coincides with the $\Vert \cdot \Vert '$-norm topology in $C_b(X)'$ restricted to $C_b(X)'_\beta$. Hence, we have $C_b(X)''_\beta =(C_b(X)'_\beta ,\Vert \cdot \Vert ')'$ and we get $\Psi \in C_b(X)''_\beta$$\Vert \Psi \Vert ''=\sup \{|\Psi (\Phi )|:\Phi \in C_b(X)'_\beta ,\Vert \Phi \Vert '\le 1\}$. Then, one can embed isometrically $B({{\mathcal {B}}}o)$ into $C_b(X)''_\beta$ by the mapping $\pi :B({{\mathcal {B}}}o)\rightarrow C_b(X)''_\beta$, where for $v\in B({{\mathcal {B}}}o)$,

$$\begin{aligned} \pi (v)(\Phi _\mu ):=\int _X v\,{\text {d}}\mu \ \ \text{ for } \ \ \mu \in M(X). \end{aligned}$$

Note that $C_b(X)'_\beta$ is a closed subspace of $(C_b(X)',\Vert \cdot \Vert ')$ (see [24, p. 847]).

Let $i_E:E\rightarrow E''$ stand for the canonical injection, that is, $i_E(e)(e'):=e'(e)$ for $e\in E$, $e'\in E'$. Let $j_E:i_E(E)\rightarrow E$ denote the left inverse of $i_E$, i.e., $j_E(i_E(e)):=e$ for $e\in E$.

Assume that $T:C_b(X)\rightarrow E$ is a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator. Then we can define the biconjugate mapping

$$\begin{aligned} T'':C_b(X)''_\beta \rightarrow E'' \end{aligned}$$

by putting $T''(\Psi )(e'):=\Psi (e'\circ T)$ for $\Psi \in C_b(X)''_\beta$ and $e'\in E'$. Then $T''$ is $(\Vert \cdot \Vert '',\Vert \cdot \Vert _{E''})$-continuous. Let

$$\begin{aligned} {\hat{T}}:=T''\circ \pi :B({{\mathcal {B}}}o)\rightarrow E''. \end{aligned}$$

Then, ${\hat{T}}$ is $(\Vert \cdot \Vert _\infty , \Vert \cdot \Vert _{E''})$-continuous.

For $A\in {{\mathcal {B}}}o$, let

$$\begin{aligned} {\hat{m}}(A):={\hat{T}}(\mathbb {1}_A). \end{aligned}$$

Hence, ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ is a finitely additive bounded measure (i.e., $\Vert {\hat{m}}\Vert (X)<\infty )$ and is called a representing measure of T. For every $e'\in E'$, let

$$\begin{aligned} {\hat{m}}_{e'}(A):={\hat{m}}(A)(e') \ \text{ for } \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Then for every $v\in B({{\mathcal {B}}}o)$, we have (see [24, Theorem 3.1])

$$\begin{aligned} {\hat{T}}(v)=\int _X v\,{\text {d}}{\hat{m}} \ \text{ and } \ {\hat{T}}(v)(e')= \int _X v\,{\text {d}}{\hat{m}}_{e'} \ \text{ for } \text{ every } \ e'\in E', \end{aligned}$$

where ${\hat{m}}_{e'}\in M(X)$ for every $e'\in E'$. From the general properties of the operator $T''$ it follows that ${\hat{T}}(C_b(X))\subset i_E(E)$ and

$$\begin{aligned} T(u)=j_E\big ({\hat{T}}(u)\big )=j_E\bigg (\int _X u\,{\text {d}}{\hat{m}}\bigg ) \ \text{ for } \ u\in C_b(X). \end{aligned}$$

(2.5)

According to [24, Theorem 4.2], we have the following characterization of $(\beta ,$ $\Vert \cdot \Vert _E)$-continuous weakly compact operators $T:C_b(X)\rightarrow E$.

Theorem 2.2

Let $T:C_b(X)\rightarrow E$ be a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator and ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ be its representing measure. Then the following statements are equivalent:

• (i) T is weakly compact.

• (ii) ${\hat{m}}(A)\in i_E(E)$ for every $A\in {{\mathcal {B}}}o$.

• (iii) ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ is a Radon measure.

• (iv) ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ is countably additive.

• (v) $T(u_n)\rightarrow 0$ whenever $(u_n)$ is a uniformly bounded sequence in $C_b(X)$ such that $u_n(t)\rightarrow 0$ for every $t\in X$.

• (vi) $T(u_n)\rightarrow 0$ whenever $(u_n)$ is a uniformly bounded sequence in $C_b(X)$ such that $\mathrm{supp}\,u_k\cap \mathrm{supp}\,u_n=\emptyset$ for $n\ne k$.

The following result will be useful.

Theorem 2.3

Let $T:C_b(X)\rightarrow E$ be a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator and ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ be its representing measure. Then the following statements hold:

• (i) If T is weakly compact, then $m:=j_E\circ {\hat{m}}:{{\mathcal {B}}}o\rightarrow E$ is a Radon measure and

  $$\begin{aligned} T(u)=\int _X u\,dm \ \ for \ \ u\in C_b(X). \end{aligned}$$

• (ii) If $|{{\hat{m}}}|(X)<\infty$, then T is weakly compact and ${\hat{m}}$ is a Radon measure with $|{{\hat{m}}}|\in M^+(X)$.

Proof

(i) See [24, Theorem 3.5] and Theorem 2.2.

(ii) Assume that $|{{\hat{m}}}|(X)<\infty$. Then ${{\hat{m}}}$ is strongly additive (see [9, Proposition 15, p. 7]) and hence the operator ${{\hat{T}}}: B({{\mathcal {B}}}o)\rightarrow E''$ is weakly compact (see [9, Theorem 1, p. 148]). Therefore, in view of (2.5), the operator $T:C_b(X)\rightarrow E$ is weakly compact and by Theorem 2.2, ${{\hat{m}}}$ is a Radon measure. Using Lemma 2.1, we get $|{{\hat{m}}}|\in M^+(X)$. $\square$

3 Absolutely summing operators

In this section, we characterize $\beta$-absolutely summing operators $T:C_b(X)\rightarrow E$ and show that $\beta$-absolutely summing operators and dominated operators on $C_b(X)$ coincide.

We will need the following lemma.

Lemma 3.1

For a sequence $(u_n)$ in $C_b(X)$, the following statements are equivalent:

• (i) $\sup \big \{\Vert \sum _{i\in M}\varepsilon _i \;u_i\Vert _\infty :\varepsilon _i=\pm 1, M\in {{\mathcal {F}}}({\mathbb {N}})\big \}<\infty$.

• (ii) $\sum ^\infty _{n=1}|\Phi (u_n)|<\infty$ for all $\Phi \in C_b(X)'$.

• (iii) $\sum ^\infty _{n=1}\big |\int _X u_n\,d\mu \big |<\infty$ for all $\mu \in M(X)$.

Proof

(i)$\Leftrightarrow$(ii) It is well known (see [10, Chap. 5, Theorem 6, p. 44]).

(ii)$\Rightarrow$(iii) This follows from Theorem 1.1 because $\beta \subset \tau _u$.

(iii)$\Rightarrow$(i) Assume that (iii) holds. Then, for $\varepsilon _i=\pm 1$, $M\in {{\mathcal {F}}}({\mathbb {N}})$ and $\mu \in M(X)$, we have

$$\begin{aligned} \begin{aligned} \bigg |\int _X\bigg (\sum _{i\in M}\varepsilon _i \;u_i\bigg ){\text {d}}\mu \bigg |&=\bigg |\sum _{i\in M} \int _X\varepsilon _i\; u_i\,{\text {d}}\mu \bigg |\le \sum _{i\in M}\bigg |\int _X u_i\,{\text {d}}\mu \bigg |\\&\le \sum ^\infty _{n=1}\bigg |\int _X u_n\,{\text {d}}\mu \bigg |<\infty . \end{aligned} \end{aligned}$$

This means that $\{\sum _{i\in M}\varepsilon _i\; u_i:\varepsilon _i=\pm 1,$ $M\in {{\mathcal {F}}}({\mathbb {N}})\}$ is $\sigma (C_b(X),M(X))$-bounded, and hence it is $\beta$-bounded. It follows that $\sup \{\Vert \sum _{i\in M}\varepsilon _i\; u_i\Vert _\infty :\varepsilon _i=\pm 1$, $M\in {{\mathcal {F}}}({\mathbb {N}})\}<\infty$ because $\tau _u$ and $\beta$ have the same bounded sets. $\square$

The following theorem characterizes $\beta$-absolutely summing operators $T:C_b(X)$ $\rightarrow E$ (see [9, Proposition 2, p. 162], [22, Proposition 3.1] if X is compact).

Theorem 3.2

Let $T:C_b(X)\rightarrow E$ be a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator. Then the following statements are equivalent:

• (i) There exists $c>0$ such that for any finite set $\{u_1,\ldots ,u_n\}$ in $C_b(X)$,

  $$\begin{aligned} \sum ^n_{i=1}\Vert T(u_i)\Vert _E\le c\,\sup \left\{ \sum ^n_{i=1}\bigg |\int _X u_i\,d\mu \bigg |:\mu \in M(X),\,|\mu |(X)\le 1\right\} . \end{aligned}$$

• (ii) $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E<\infty \,$ if $\;\sum ^\infty _{n=1}|\int _X u_n\,d\mu |<\infty$ for every $\mu \in M(X)$.

• (iii) $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E<\infty \,$ if $\sum ^\infty _{n=1}u_n$ is unconditionally $\beta$-convergent.

• (iv) T is $\beta$-absolutely summing.

Proof

(i)$\Rightarrow$(ii) Assume that (i) holds. Let $(u_n)$ be a sequence in $C_b(X)$ such that $\sum ^\infty _{n=1}|\int _X u_n\,{\text {d}}\mu |<\infty$ for every $\mu \in M(X)$. Then, by Lemma 3.1, we have $\sum ^\infty _{n=1}|\Phi (u_n)|<\infty$ for all $\Phi \in C_b(X)'$. Hence, by [27, 1.2.3, pp. 23–24], we get

$$\begin{aligned} \Vert (u_n)\Vert ^w_1:=\sup \left\{ \sum ^\infty _{n=1}|\Phi (u_n)|:\Phi \in C_b(X)', \Vert \Phi \Vert '\le 1\right\} <\infty . \end{aligned}$$

Hence, for every $n\in {\mathbb {N}}$, we have

$$\begin{aligned} \sum ^n_{i=1}\Vert T(u_i)\Vert _E\le c\,\sup \left\{ \sum ^n_{i=1}|\Phi (u_i)|: \Phi \in C_b(X)',\Vert \Phi \Vert '\le 1\right\} \le c\,\Vert (u_n)\Vert ^w_1, \end{aligned}$$

and it follows that $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E<\infty$, as desired.

(ii)$\Rightarrow$(iii) Assume that (ii) holds and the series $\sum ^\infty _{n=1}u_n$ is unconditionally $\beta$-convergent in $C_b(X)$. Then $\sum ^\infty _{n=1}|\int _X u_n\,{\text {d}}\mu |<\infty$ for every $\mu \in M(X)$ and it follows that $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E<\infty$.

(iii)$\Rightarrow$(iv) Assume that (iii) holds and $(u_n)\in \ell ^1_s(C_b(X),\beta )$. Then a net $(s_M)_{M\in {{\mathcal {F}}}({\mathbb {N}})}$ is a $\beta$-Cauchy sequence, where $s_M:=\sum _{i\in M} u_i$ for $M\in {{\mathcal {F}}}({\mathbb {N}})$. Let $\sigma$ be a permutation of ${\mathbb {N}}$. Let $w\in {{\mathcal {W}}}_1$ and $\varepsilon >0$ be given. Then, there exists $M\in {{\mathcal {F}}}({\mathbb {N}})$ such that $p_w(\sum _{j\in L} u_j)\le \varepsilon$ for every $L\in {{\mathcal {F}}}({\mathbb {N}})$ with $L\cap M=\emptyset$. Choose $k\in {\mathbb {N}}$ such that $M\subset \{\sigma (i):1\le i\le k\}$. Then for $n,m\in {\mathbb {N}}$ with $m>n>k$, we have $p_w\big (\sum ^m_{i=n}u_{\sigma (i)}\big )\le \varepsilon$. This means that the partial sums $\sum ^n_{i=1}u_{\sigma (i)}$ form a $\beta$-Cauchy sequence in $C_b(X)$. Since the space $(C_b(X),\beta )$ is complete, we obtain that the series $\sum ^\infty _{n=1}u_n$ is unconditionally $\beta$-convergent in $C_b(X)$. Hence, we get $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E<\infty$.

(iv)$\Rightarrow$(i) Assume that (iv) holds. Let $w\in {{\mathcal {W}}}_1$. Then in view of [27, Theorem 2.1.2] there exists $c_w>0$ such that $\pi _E((T(v_n)))= \sum ^\infty _{n=1}\Vert T(v_n)\Vert _E\le c_w$ whenever $(v_n)\in \ell ^1_w(C_b(X),\beta )$ with ${{\mathcal {E}}}_{w,1}((v_n))\le 1$. Hence for $(v_n)\in \ell ^1_w(C_b(X),\beta )$, we have

$$\begin{aligned} \pi _E((T(v_n)))=\sum ^\infty _{n=1}\Vert T(v_n)\Vert _E\le c_w\,{{\mathcal {E}}}_{w,1}((v_n)). \end{aligned}$$

Let $u_i\in C_b(X)$ for $i=1,\ldots ,n$. Define $v_i=u_i$ for $i=1,\ldots ,n$ and $v_i=0$ for $i>n$. Then

$$\begin{aligned} \sum ^n_{i=1}\Vert T(u_i)\Vert _E\le c_w \sup \left\{ \sum ^n_{i=1}\bigg |\int _X u_i\,d\mu \bigg |:\mu \in U_w(1)^0\right\} . \end{aligned}$$

(3.1)

Note that $B_\infty (1):=\{u\in C_b(X):\Vert u\Vert _\infty \le 1\}\subset U_w(1)$. Hence, $U_w(1)^0\subset B_\infty (1)^0$, where the polars are taken with respect to the pairing $\langle C_b(X),M(X)\rangle$. In view of Theorem 1.1 for $\mu \in M(X)$, we have

$$\begin{aligned} |\mu |(X)=\sup \left\{ \bigg |\int _X u\,{\text {d}}\mu \bigg |:u\in C_b(X), \Vert u\Vert _\infty \le 1 \right\} . \end{aligned}$$

It follows that $B_\infty (1)^0=\{\mu \in M(X):|\mu |(X)\le 1\}$. By (3.1) we get

$$\begin{aligned} \sum ^n_{i=1}\Vert T(u_i)\Vert _E\le c_w \sup \left\{ \sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}\mu \bigg |:\mu \in M(X), |\mu |\,(X)\le 1\right\} . \end{aligned}$$

Thus (i) holds. $\square$

We show that dominated operators and $\beta$-absolutely summing operators on $C_b(X)$ coincide (see [27, 2.3.4, Proposition, p. 41]).

We will need the following lemma.

Lemma 3.3

Assume that $\mu \in M(X)$. Then for $O\in {{\mathcal {T}}}$, we have

$$\begin{aligned} |\mu |(O)=\sup \bigg \{\bigg |\int _X u\,d\mu \bigg |:u\in C_b(X), \Vert u\Vert _\infty =1 \ \text{ and } \ \mathrm{supp}\,u\subset O\bigg \}. \end{aligned}$$

(3.2)

Proof

For $u\in C_b(X)$ with $\Vert u\Vert _\infty =1$ and $\mathrm{supp}\,u\subset O$, we have

$$\begin{aligned} \bigg |\int _O u\,{\text {d}}\mu \bigg |\le \Vert u\Vert _\infty \,|\mu |(O)\le |\mu |(O). \end{aligned}$$

Now let $\varepsilon >0$ be given. Then there exists a ${{\mathcal {B}}}o$-partition $(A_i)^n_{i=1}$ of O such that

$$\begin{aligned} |\mu |(O)-\frac{\varepsilon }{3}\le \bigg |\sum ^n_{i=1}\mu (A_i)\bigg |. \end{aligned}$$

For $i=1,\ldots ,n$ choose $K_i\in {{\mathcal {K}}}$ with $K_i\subset A_i$ such that $|\mu |(A_i{\setminus }K_i)\le \frac{\varepsilon }{3n}$ for $i=1,\ldots ,n$. Choose pairwise disjoint $O_i\in {{\mathcal {T}}}$ with $K_i\subset O_i$ for $i=1,\ldots ,n$ such that $|\mu |(O_i{\setminus }K_i)\le \frac{\varepsilon }{3n}$. For $i=1,\ldots ,n$ choose $u_i\in C_b(X)$ with $0\le u_i\le \mathbb {1}_X$, $u_i\big |_{K_i}\equiv 1$ and $u_i\big |_{X{\setminus }(O_i\cap O)}\equiv 0$. Let $u:=\sum ^n_{i=1}u_i$. Then $\Vert u\Vert _\infty =1$ with $\mathrm{supp}\,u\subset O$ and

$$\begin{aligned} \int _O u\,{\text {d}}\mu =\sum ^n_{i=1}\int _O u_i\,{\text {d}}\mu =\sum ^n_{i=1}\int _{O_i\cap O}u_i\,{\text {d}}\mu . \end{aligned}$$

Then

$$\begin{aligned} \begin{array}{rl} |\mu |(O)-\frac{\varepsilon }{3} &{} \!\!\!\displaystyle \le \bigg |\sum ^n_{i=1}\mu (A_i)-\sum ^n_{i=1}\mu (K_i)\bigg |\\ &{}\!\!\! \displaystyle +\;\bigg |\sum ^n_{i=1}\int _{K_i}u_i\,{\text {d}}\mu -\sum ^n_{i=1}\int _{O_i\cap O}u_i\,{\text {d}}\mu \bigg |+ \bigg |\int _O u\,{\text {d}}\mu \bigg |\\ &{}\!\!\! \displaystyle \le \sum ^n_{i=1}|\mu |\,(A_i{\setminus }K_i)+\sum ^n_{i=1} |\mu |\,((O_i\cap O){\setminus }K_i)+\bigg |\int _O u\,{\text {d}}\mu \bigg |\\ &{}\!\!\! \displaystyle \le \frac{\varepsilon }{3}+\frac{\varepsilon }{3}+\bigg |\int _O u\,{\text {d}}\mu \bigg |, \end{array} \end{aligned}$$

that is, $|\mu |(O)\le |\int _O u\,{\text {d}}\mu |+\varepsilon$. Thus (3.2) holds. $\square$

Now we can state our main result (see [27, 2.3.4, Proposition, p. 41]).

Corollary 3.4

Assume that $T:C_b(X)\rightarrow E$ is a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator and ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ is its representing measure. Then the following statements are equivalent:

• (i) $|{\hat{m}}|(X)<\infty$.

• (ii) T is dominated.

• (iii) T is $\beta$-absolutely summing.

• (iv) T is absolutely summing.

In this case, $\Vert T\Vert _{as}=|{\hat{m}}|(X)$.

Proof

(i)$\Leftrightarrow$(ii) This follows from [25, Theorem 3.1].

(ii)$\Rightarrow$(iii) Assume that (ii) holds. Then T is dominated by $|{{\hat{m}}}|$, so

$$\begin{aligned} \Vert T(u)\Vert _E \le \int _X |u|\,{\text {d}}\,|{\hat{m}}|\ \ \text{ for } \ \ u\in C_b(X). \end{aligned}$$

Let $u_1,\ldots ,u_n\in C_b(X)$. Then we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \sum ^n_{i=1}\Vert T(u_i)\Vert _E &{}\!\!\! \displaystyle \le \sum ^n_{i=1}\int _X|u_i|\,{\text {d}}\,|{\hat{m}}|\le \int _X\bigg (\sum ^n_{i=1}|u_i|\bigg )\,{\text {d}}\,|{\hat{m}}|\\ &{}\!\!\! \displaystyle \le \sup _{t\in X}\bigg (\sum ^n_{i=1}|u_i(t)|\bigg )|{\hat{m}}|(X)= \sup _{t\in X}\bigg (\sum ^n_{i=1}\bigg |\int _X u_i \,{\text {d}}\delta _t\bigg |\bigg )|{\hat{m}}|(X)\\ &{}\!\!\!\displaystyle \le \sup \bigg \{\sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}\mu \bigg |:\mu \in M(X), \ |\mu |(X)\le 1\bigg \}\,|{\hat{m}}|(X). \end{array} \end{aligned}$$

In view of Theorem 3.2T is $\beta$-absolutely summing and $\Vert T\Vert _{\text {as}}\le |{\hat{m}}|(X)$.

(iii)$\Rightarrow$(i) Assume that (iii) holds. Then in view of Theorem 3.2, there exists $c>0$ such that for every $u_1,\ldots ,u_n\in C_b(X)$, we have

$$\begin{aligned} \sum ^n_{i=1}\Vert T(u_i)\Vert _E\le c\,\sup \bigg \{\sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}\mu \bigg |:\mu \in M(X),|\mu |(X)\le 1\bigg \}. \end{aligned}$$

Let $(u_n)$ be a sequence in $C_b(X)$ such that $\sup _n\Vert u_n\Vert _\infty =a<\infty$ and $\mathrm{supp}\,u_n\cap \mathrm{supp}\,u_k=\emptyset$ if $n\ne k$. Then, for $\mu \in M(X)$ with $|\mu |(X)\le 1$, we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}\mu \bigg | &{}\!\!\!\displaystyle \le \sum ^n_{i=1} \Vert u_i\Vert _\infty \,|\mu |(\mathrm{supp}\,u_i)\le a\sum ^n_{i=1}|\mu |\,(\mathrm{supp}\,u_i)\\ &{} \!\!\!\displaystyle =\,a \,|\mu |\,\bigg (\bigcup ^n_{i=1}\mathrm{supp}\,u_i\bigg )\le a\,|\mu |(X)\le a. \end{array} \end{aligned}$$

Then $\sum ^\infty _{n=1}\Vert T(u_n)\Vert _E\le ca<\infty$, so $\Vert T(u_n)\Vert _E\rightarrow 0$ and according to Theorem 2.2T is weakly compact. Hence by Theorem 2.3$m:=j_E\circ {\hat{m}}:{{\mathcal {B}}}o\rightarrow E$ is a Radon measure and

$$\begin{aligned} T(u)=\int _X u\,{\text {d}}m \ \ \text{ for } \ u\in C_b(X). \end{aligned}$$

Now, we shall show that $|m|(X)=|{{\hat{m}}}|(X)<\infty$. In fact, let $(A_i)^n_{i=1}$ be a ${{\mathcal {B}}}o$-partition of X and $\varepsilon >0$ be given. Choose $e'_1,\ldots ,e'_n\in B_{E'}$ such that $\Vert m\Vert (A_i)\le |m_{e'_i}|(A_i)+ \frac{\varepsilon }{4n}$ for $i=1,\ldots ,n$. Hence

$$\begin{aligned} \sum ^n_{i=1}\Vert m(A_i)\Vert _E\le \sum ^n_{i=1}\Vert m\Vert (A_i)\le \sum ^n_{i=1}|m_{e'_i}|(A_i) +\frac{\varepsilon }{4}. \end{aligned}$$

(3.3)

For each $i=1,\ldots ,n$ one can choose $K_i\in {{\mathcal {K}}}$ with $K_i\subset A_i$ such that $|m_{e'}|(A_i{\setminus }K_i)\le \frac{\varepsilon }{4n}$. Hence $|m_{e'}|(A_i)\le |m_{e'_i}|(K_i)+\frac{\varepsilon }{4n}$ for $i=1,\ldots ,n$. Then we can choose pairwise disjoint open sets $O_i$ with $K_i\subset O_i$ for $i=1,\ldots ,n$. According to Lemma 3.3 for each $i=1,\ldots ,n$ there exists $u_i\in C_b(X)$ with $\Vert u_i\Vert _\infty =1$ and $\mathrm{supp}\,u_i\subset O_i$ such that

$$\begin{aligned} |m_{e'_i}|(O_i)\le \bigg |\int _X u_i\,{\text {d}}m_{e'_i}\bigg |+\frac{\varepsilon }{2n}. \end{aligned}$$

(3.4)

Hence, by (2.2) and Lemma 3.3, we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}m_{e'_i}\bigg | &{}\!\!\! \displaystyle = \sum ^n_{i=1} |e'_i(T(u_i))| \le \sum ^n_{i=1}\Vert T(u_i)\Vert _E \\ &{}\!\!\! \displaystyle \le c\, \sup \bigg \{\sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}\mu \bigg |:\mu \in M(X), |\mu |\,(X)\le 1\bigg \}\\ &{}\!\!\! \displaystyle \le c \;\sup \bigg \{\sum ^n_{i=1}|\mu |\,(O_i):\mu \in M(X), |\mu |\,(X)\le 1\bigg \}\le c. \end{array} \end{aligned}$$

Hence using (3.3) and (3.4), we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \sum ^n_{i=1}\Vert m(A_i)\Vert _E &{}\!\!\! \displaystyle \le \sum ^n_{i=1}|m_{e'_i}|\,(A_i)+ \frac{\varepsilon }{4}\le \sum ^n_{i=1}\bigg (|m_{e'_i}|\,(K_i)+\frac{\varepsilon }{4n}\bigg )+\frac{\varepsilon }{4} \\ &{}\!\!\! \displaystyle \le \sum ^n_{i=1}|m_{e'_i}|\,(O_i)+\frac{\varepsilon }{2}\le \sum ^n_{i=1}\bigg |\int _X u_i\,{\text {d}}m_{e'_i}\bigg |+\frac{\varepsilon }{2}+\frac{\varepsilon }{2}\le c+\varepsilon . \\ \end{array} \end{aligned}$$

It follows that $\sum ^n_{i=1}\Vert m(A_i)\Vert _E\le c$, so $|m|(X)\le c$. Thus, $|{\hat{m}}|(X)\le c$ and hence $|{\hat{m}}|(X)\le \Vert T\Vert _{\text {as}}$.

(iii)$\Leftrightarrow$(iv) This follows from Lemma 1.3 and Theorem 3.2. $\square$

Let $\varphi \in L^1(\mu )$, where $\mu \in M^+(X)$. We define the multiplication operator $M_\varphi :C_b(X)\rightarrow L^1(\mu )$ by $M_\varphi (u):=\varphi \,u$ for $u\in C_b(X)$. For $A\in {{\mathcal {B}}}o$, let $m_\varphi (A):=\varphi \mathbb {1}_A$.

Proposition 3.5

Assume that $\varphi \in L^1(\mu )$, where $\mu \in M^+(X)$. Then the following statements hold:

• (i) $|m_\varphi |(A)=\int _A|\varphi |\,d\mu$ for $A\in {{\mathcal {B}}}o$ and $|m_\varphi |\in M^+(X)$.

• (ii) $\Vert M_\varphi (u)\Vert _1=\int _X|u|\,d\,|m_\varphi |$ for $u\in C_b(X)$, that is, $M_\varphi$ is dominated by $|m_\varphi |$.

• (iii) $m_\varphi :{{\mathcal {B}}}o\rightarrow L^1(\mu )$ is a Radon measure and

  $$\begin{aligned} M_\varphi (u)=\int _X u\,dm_\varphi \ \ for \ u\in C_b(X). \end{aligned}$$

• (iv) $M_\varphi$ is $\beta$-absolutely summing.

Proof

(i) Let $A\in {{\mathcal {B}}}o$ and $(A_i)^n_{i=1}$ be a finite ${{\mathcal {B}}}o$-partition of A. Then

$$\begin{aligned} \sum ^n_{i=1}\Vert m_\varphi (A_i)\Vert _1=\sum ^n_{i=1}\int _X|\varphi |\,\mathbb {1}_{A_i} {\text {d}}\mu =\int _A|\varphi |\,{\text {d}}\mu . \end{aligned}$$

Hence, $|m_\varphi |(A)=\int _A|\varphi |\,{\text {d}}\mu$ and it follows that $|m_\varphi |$ is countably additive. Since $|m_\varphi |\ll \mu$ and $\mu \in M^+(X)$, we obtain that $|m_\varphi |\in M^+(X)$.

(ii) From (i) it follows that $|\varphi |=\frac{{\text {d}}|m_\varphi |}{{\text {d}}\mu }$ (= the Radon–Nikodym derivative of $|m_\varphi |$ with respect to $\mu$). Since $C_b(X)\subset L^1(\mu )$, in view of [7, Theorem C.8, p. 380] for $u\in C_b(X)$, we get

$$\begin{aligned} \Vert M_\varphi (u)\Vert _1=\int _X |\varphi \,u|\,{\text {d}}\mu =\int _X |u|\,{\text {d}}|m_\varphi |. \end{aligned}$$

(iii) Since $\Vert m_\varphi \Vert (A)\le |m_\varphi |(A)$ for $A\in {{\mathcal {B}}}o$ and $|m_\varphi |\in M^+(X)$, we obtain that $m_\varphi$ is a Radon measure. Note that for $s\in {{\mathcal {S}}}({{\mathcal {B}}}o)$, $\int _X s\,{\text {d}}m_\varphi =\varphi \,s$.

Let $u\in C_b(X)$ and choose a sequence $(s_n)$ in ${{\mathcal {S}}}({{\mathcal {B}}}o)$ such that $\Vert u-s_n\Vert _\infty \rightarrow 0$. Hence

$$\begin{aligned} \Vert M_\varphi (u)-\varphi s_n\Vert _1=\int _X|\varphi u-\varphi s_n|\,{\text {d}}\mu \le \int _X|\varphi |\,{\text {d}}\mu \, \Vert u-s_n\Vert _\infty . \end{aligned}$$

This means that $M_\varphi (u)=\int _X u\,{\text {d}}m_\varphi$.

(iv) In view of (ii) and Proposition 1.7$M_\varphi$ is $(\beta ,\Vert \cdot \Vert _1)$-continuous. Hence, by Corollary 3.4$M_\varphi$ is $\beta$-absolutely summing. $\square$

The next result shows that every $\beta$-absolutely summing operator $T:C_b(X)\rightarrow E$ admits a factorization through $L^1$-space (see [9, Corollary 7, pp. 164–165], [11, Corollary 2.5], [43, Theorem 1.8] if X is compact).

Corollary 3.6

Let $T:C_b(X)\rightarrow E$ be a $\beta$-absolutely summing operator and ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ be its representing measure. Then, $m:=j_E\circ {{\hat{m}}}:{{\mathcal {B}}}o\rightarrow E$ is a Radon measure with $|m|\in M^+(X)$ and the following statements hold:

• (i) The inclusion map $I:C_b(X)\rightarrow L^1(|m|)$ is a $\beta$-absolutely summing operator with $\Vert I\Vert _{as}=|m|(X)$.

• (ii) The integration operator $S:L^1(|m|)\rightarrow E$ defined by

  $$\begin{aligned} S(v):=\int _Xv\,dm\ \ \text{ for } \text{ all }\ \ v\in L^1(|m|) \end{aligned}$$

  is bounded with $\Vert S\Vert \le 1$ and $T=S\circ I$.

Proof

In view of Theorem 2.3$m:=j_E\circ {{\hat{m}}}:{{\mathcal {B}}}o\rightarrow E$ is a Radon measure with $|m|\in M^+(X)$.

(i) Since $|m|\in M^+(X)$ in view of Proposition 3.5, I is $\beta$-absolutely summing and $\Vert I\Vert _{\text {as}}=\int _X\mathbb {1}_X\,{\text {d}}|{{\hat{m}}}|=|{\hat{m}}|(X)=|m|(X)$.

(ii) In view of Theorem 2.3 we have that $T(u)=\int _X u\,{\text {d}}m$ for $u\in C_b(X)$.

Thus, we get $T=S\circ I$, where by (2.4) $\Vert S\Vert \le 1$. $\square$

4 Compact operators

The tensor product $ca({{\mathcal {B}}}o)\otimes E$ consists of all measures $m:{{\mathcal {B}}}o\rightarrow E$ of the form $m=\sum ^n_{i=1}(\mu _i\otimes e_i)$, where $\mu _i\in ca({{\mathcal {B}}}o)$ and $e_i\in E$ for $i=1,\ldots ,n$. Then $m(A)=\sum ^n_{i=1}\mu _i(A) e_i$ for $A\in {{\mathcal {B}}}o$.

Now, we can state a characterization of $\beta$-compact operators $T\!:\!C_b(X)\!\!\rightarrow \!E$ in terms of their representing measures ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ (see [9, Theorem 18, p. 161], [34, Theorem 5.27] if X is compact).

Theorem 4.1

Let $T:C_b(X)\rightarrow E$ be a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator and ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ be its representing measure. Then the following statements are equivalent:

• (i) T is $\beta$-compact.

• (ii) ${\hat{m}}$ has a relatively norm compact range in $E''$.

Proof

(i)$\Rightarrow$(ii) Assume that (i) holds. Then $T'':C_b(X)''\rightarrow E''$ is compact and hence ${\hat{T}}:=T''\circ \pi :B({{\mathcal {B}}}o)\rightarrow E''$ is compact. Since

$$\begin{aligned} \big \{{\hat{m}}(A):A\in {{\mathcal {B}}}o\big \}=\big \{{\hat{T}}(\mathbb {1}_A):A\in {{\mathcal {B}}}o\big \}\subset \big \{{\hat{T}}(v):v\in B({{\mathcal {B}}}o),\Vert v\Vert _\infty \le 1\big \}, \end{aligned}$$

we obtain that ${\hat{m}}({{\mathcal {B}}}o)$ is relatively norm compact in $E''$.

(ii)$\Rightarrow$(i) Assume that (ii) holds. Since ${\hat{m}}({{\mathcal {B}}}o)$ is weakly compact, the corresponding integration operator ${\hat{T}}:B({{\mathcal {B}}}o)\rightarrow E''$ is weakly compact (see [19, Theorem 7]). Then, in view of (2.5), T is weakly compact, and by Theorem 2.3$m:=j_E\circ {\hat{m}}:{{\mathcal {B}}}o\rightarrow E$ is countably additive and $m({{\mathcal {B}}}o)$ is relatively norm compact in E. According to the proof of [34, Theorem 5.18], there exists a sequence $(m_k)$ in $ca({{\mathcal {B}}}o)\otimes E$ such that $\Vert m-m_k\Vert \rightarrow 0$.

For each $k\in {\mathbb {N}}$, let $T_k:C_b(X)\rightarrow E$ be the finite rank operator defined by $T_k(u):=\int _X u\,{\text {d}}m_k$. For $u\in C_b(X)$, we have

$$\begin{aligned} \Vert T_k(u)-T(u)\Vert _E=\bigg \Vert \int _Xu\,{\text {d}}(m_k-m)\bigg \Vert _E\le \Vert u\Vert _\infty \, \Vert m_k-m\Vert (X), \end{aligned}$$

and it follows that $\Vert T_k-T\Vert \rightarrow 0$. Hence, T is a compact operator and using Proposition 1.5 we have that T is $\beta$-compact. $\square$

5 Nuclear operators

We state our main result that characterizes $\beta$-nuclear operators $T:C_b(X)\rightarrow E$ in terms of their representing measures (see [9, Theorem 4, p. 179], [34, Proposition 5.30], [43, Proposition 1.2] if X is a compact Hausdorff space).

Let $(\Omega ,\Sigma ,\mu )$ be a finite measure space. Recall that a bounded linear operator $S:L^1(\mu )\rightarrow E$ is said to be representable if there exists an essentially bounded $\mu$-Bochner integrable function $f:\Omega \rightarrow E$ such that $S(v)=\int _\Omega v(\omega )f(\omega )\,{\text {d}}\mu$ for all $v\in L^1(\mu )$.

Theorem 5.1

Let $T:C_b(X)\rightarrow E$ be a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator and ${\hat{m}}:{{\mathcal {B}}}o\rightarrow E''$ be its representing measure. Then the following statements are equivalent:

• (i) T is $\beta$-nuclear.

• (ii) $|{\hat{m}}|(X)<\infty$ and m has a |m|-Bochner integrable derivative.

• (iii) $|{\hat{m}}|(X)<\infty$ and there exists a representable operator $S:L^1(|m|)\rightarrow E$ such that $T=S\circ I$, where $I:C_b(X)\rightarrow L^1(|m|)$ denotes the inclusion map.

In this case, $\Vert T\Vert _{\beta \text{- }{\text {nuc}}}=|{\hat{m}}|(X)=|m|(X)$.

Proof

(i)$\Rightarrow$(ii) This follows from [26, Theorem 3.1].

(ii)$\Rightarrow$(i) Assume that (ii) holds, that is, $|{\hat{m}}|(X)<\infty$ and there exists a function $f\in L^1(|m|,E)$ such that $m(A)=\int _Af(t)\,{\text {d}}|m|$ for $A\in {{\mathcal {B}}}o$. Then, $|m|(X)=\Vert f\Vert _1$. Hence, we easily obtain that

$$\begin{aligned} T(u)=\int _Xu(t)f(t)\,{\text {d}}|m|\ \ \text{ for }\ \ u\in C_b(X). \end{aligned}$$

Let $L^1(|m|){\hat{\otimes }}E$ denote the projective tensor product of $L^1(|m|)$ and E, equipped with the norm $\gamma$ defined for $w\in L^1(|m|){\hat{\otimes }}E$ by

$$\begin{aligned} \gamma (w):=\inf \Big \{\sum _{n=1}^\infty |\lambda _n|\Vert v_n\Vert _1\Vert e_n\Vert _E\Big \}, \end{aligned}$$

where the infimum is taken over all sequences $(v_n)$ in $L^1(|m|)$ and $(e_n)$ in E with $\lim _n\Vert v_n\Vert _1=0=\lim _n\Vert e_n\Vert _E$ and $(\lambda _n)\in \ell ^1$ such that $w=\sum _{n=1}^\infty \lambda _n(v_n\otimes e_n)$ (see [34, Proposition 2.8, pp. 21–22]). It is known that $L^1(|m|){\hat{\otimes }}E$ is isometrically isomorphic to the Banach space $(L^1(|m|, E),\Vert \cdot \Vert _1)$ throughout the isometry J, where

$$\begin{aligned} J(v\otimes e):=v(\cdot )\otimes e\ \ \text{ for }\ \ v\in L^1(|m|),\ e\in E, \end{aligned}$$

(see [9, Example 10, p. 228], [34, Example 2.19, p. 29]).

Let $\varepsilon >0$ be given. Then, there exist sequences $(v_n)$ in $L^1(|m|)$ and $(e_n)$ in $\,E\,$ with $\lim _n\Vert v_n\Vert _1=0=\lim _n\Vert e_n\Vert _E$ and $(\lambda _n)\in \ell ^1$ such that

$$\begin{aligned} J^{-1}(f)=\sum _{n=1}^\infty \lambda _n(v_n\otimes e_n)\ \ \text{ in }\ \ (L^1(|m|){\hat{\otimes }}E, \gamma ) \end{aligned}$$

and

$$\begin{aligned} \sum _{n=1}^\infty |\lambda _n|\Vert v_n\Vert _1\Vert e_n\Vert _E\le \gamma (J^{-1}(f))+\varepsilon =\Vert f\Vert _1+\varepsilon . \end{aligned}$$

(5.1)

Hence

$$\begin{aligned} f=J\Big (\sum _{n=1}^\infty \lambda _n(v_n\otimes e_n)\Big )=\sum _{n=1}^\infty \lambda _n(v_n\otimes e_n)\ \ \text{ in }\ \ (L^1(|m|,E), \Vert \cdot \Vert _1) \end{aligned}$$

and we obtain that

$$\begin{aligned} T(u)=\sum _{n=1}^\infty \lambda _n\Big (\int _Xu\,v_n\,{\text {d}}|m|\Big )e_n\ \ \text{ for }\ \ u\in C_b(X). \end{aligned}$$

For $n\in {\mathbb {N}}$, let

$$\begin{aligned} \mu _n(A):=\int _Av_n\,{\text {d}}|m|\ \ \text{ for }\ \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Note that $\mu _n\in M(X)$ and $|\mu _n|(X)=\Vert v_n\Vert _1$. Then we have $\sup _n|\mu _n|(X)=\sup _n\Vert v_n\Vert _1<\infty$. To show that the family $\{\mu _n: n\in {\mathbb {N}}\}$ is uniformly tight, let $\varepsilon >0$ be given. Since $\Vert v_n\Vert _1\rightarrow 0$, we can choose $n_\varepsilon \in {\mathbb {N}}\,$ such that $|\mu _n|(X)\le \varepsilon$ for $n>n_\varepsilon$. For $n=1,\ldots ,n_\varepsilon$ choose $K_n\in {{\mathcal {K}}}$ such that $|\mu _n|(X{\setminus }K_n)\le \varepsilon$. Denote $K:=\bigcup _{n=1}^{n_\varepsilon } K_n$. Then, $|\mu _n|(X{\setminus }K)\le \varepsilon$ for every $n\in {\mathbb {N}}$, as desired.

Clearly for $n\in {\mathbb {N}}$, we have (see [7, Theorem C.8]),

$$\begin{aligned} \int _Xu\,v_n\,{\text {d}}|m|=\int _Xu\,{\text {d}}\mu _n\ \ \text{ for }\ \ u\in C_b(X). \end{aligned}$$

Hence, we have

$$\begin{aligned} T(u)=\sum _{n=1}^\infty \lambda _n\Big (\int _X u\,{\text {d}}\mu _n\Big )e_n\ \ \text{ for }\ \ u\in C_b(X), \end{aligned}$$

and this means that T is $\beta$-nuclear. By (5.1) we get

$$\begin{aligned} \Vert T\Vert _{\beta \text{- }{\text {nuc}}}\le \Vert f\Vert _1=|m|(X). \end{aligned}$$

(5.2)

(ii)$\Rightarrow$(iii) Assume that (ii) holds, that is, $|{\hat{m}}|(X)<\infty$ and there exists a |m|-Bochner integrable function $f:X\rightarrow E$ such that $m(A)=\int _Af(t)\,{\text {d}}|m|$ for $A\in {{\mathcal {B}}}o$. Let

$$\begin{aligned} S(v):=\int _Xv\,{\text {d}}m\ \ \text{ for } \text{ all }\ \ v\in L^1(|m|). \end{aligned}$$

Then, $S(u)=T(u)$ for $u\in C_b(X)$ and $m(A)=S(\mathbb {1}_A)$ for $A\in {{\mathcal {B}}}o$. Hence, by [9, Lemma 4, p. 62] f is essentially bounded and

$$\begin{aligned} S(v)=\int _Xv(t)f(t)\,{\text {d}}|m|\ \ \text{ for } \text{ all }\ \ v\in L^1(|m|). \end{aligned}$$

(iii)$\Rightarrow$(ii) This is obvious.

Thus, (i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii) hold. Moreover, if T is $\beta$-nuclear and $\varepsilon >0$ is given, then there exist a uniformly bounded and uniformly tight sequence $(\mu _n)$ in M(X), a bounded sequence $(e_n)$ in E and a sequence $(\lambda _n)\in \ell ^1$ such that

$$\begin{aligned} T(u)=\sum _{n=1}^\infty \lambda _n\Big (\int _X u\,{\text {d}}\mu _n\Big )e_n\ \ \text{ for }\ \ u\in C_b(X) \end{aligned}$$

and

$$\begin{aligned} \sum _{n=1}^\infty |\lambda _n||\mu _n|(X)\Vert e_n\Vert _E\le \Vert T\Vert _{\beta \text{- }{\text {nuc}}}+\varepsilon . \end{aligned}$$

(5.3)

Following the proof of [26, Theorem 3.1], we have

$$\begin{aligned} m(A)=\sum _{i=1}^\infty \lambda _n\,\mu _n(A)\,e_n\ \ \text{ for }\ \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Now, if $\Pi$ is a finite ${{\mathcal {B}}}o$-partition of X, then

$$\begin{aligned} \begin{aligned} \sum _{A\in \Pi }\Vert m(A)\Vert _E&=\sum _{A\in \Pi }\Big \Vert \sum _{n=1}^\infty \lambda _n\,\mu _n(A)\,e_n\Big \Vert _E\le \sum _{A\in \Pi }\sum _{n=1}^\infty |\lambda _n||\mu _n(A)|\Vert e_n\Vert _E\\&=\sum _{n=1}^\infty |\lambda _n|\Big (\sum _{A\in \Pi }|\mu _n(A)|\Big )\Vert e_n\Vert _E\le \sum _{n=1}^\infty |\lambda _n||\mu _n|(X)\Vert e_n\Vert _E. \end{aligned} \end{aligned}$$

Thus, in view of (5.3), we get

$$\begin{aligned} |m|(X)\le \sum _{n=1}^\infty |\lambda _n||\mu _n|(X)\Vert e_n\Vert _E\le \Vert T\Vert _{\beta \text{- }{\text {nuc}}}+\varepsilon . \end{aligned}$$

Hence using (5.2) we have $\Vert T\Vert _{\beta \text{- }{\text {nuc}}}=|m|(X)=|{\hat{m}}|(X)$. Thus the proof is complete. $\square$

In view of Theorem 5.1 and Corollary 3.4, we get (see [9, Corollary 5, p. 174]).

Corollary 5.2

Assume that $T:C_b(X)\rightarrow E$ is a $(\beta ,\Vert \cdot \Vert _E)$-continuous linear operator.

• (i) If the operator T is $\beta$-nuclear, then T is $\beta$-absolutely summing and $\Vert T\Vert _{as}=\Vert T\Vert _{\beta \text{- }nuc}$.

• (ii) If E has the Radon–Nikodym property, then T is $\beta$-absolutely summing if and only if T is $\beta$-nuclear.

As a consequence of Corollarys 3.4 and 5.2, we have

Corollary 5.3

Let $T:C_b(X)\rightarrow E$ be a dominated operator. If E has the Radon–Nikodym property, then T is $\beta$-compact.

Remark 5.4

If X is a compact Hausdorff space, the related result to Corollary 5.3 was obtained in the different way by Uhl [44, Theorem 1].

Remark 5.5

A relationship between vector measures $m:\Sigma \rightarrow E$ with a $\mu$-Bochner integrable derivatives (with respect to a finite measure $\mu$) and the nuclearity of the corresponding integration operators $T_m:L^\infty (\mu )\rightarrow E$ has been studied by Swartz [42] and Popa [30].

6 Nuclearity of kernel operators

It is well known that if K is a compact Hausdorff space, $\mu \in M^+(K)$ and $k(\cdot ,\cdot )\in C(K\times K)$, then the corresponding kernel operator $T:C(K)\rightarrow C(K)$ between Banach spaces, defined by

$$\begin{aligned} T(u)(t)=\int _Xu(s)k(t,s)\,{\text {d}}\mu (s)\ \ \text{ for }\ \ u\in C(K), t\in K, \end{aligned}$$

is nuclear (see [16, Theorem V.22, p. 99] if $X=[a,b]$).

Now as an application of Theorem 5.1, we extend this result to the setting, where X is a k-space and the kernel operator $T:C_b(X)\rightarrow C(K)$ is acting from the space $(C_b(X),\beta )$ to a Banach space $(C(K),\Vert \cdot \Vert _\infty )$, where K is a compact Hausdorff space.

From now on we assume that $\mu \in M^+(X)$ and $k(\cdot ,\cdot )\in C_b(K\times X)$ with $\sup _{t\in K}|k(t,s)|\ge c$ for every $s\in X$ and some $c>0$.

We start with the following lemma.

Lemma 6.1

For every $v\in B({{\mathcal {B}}}o)$, the mapping $\Psi _v:X\ni s\mapsto v(s)k(\cdot ,s)\in C(K)$ is $({{\mathcal {T}}},\Vert \cdot \Vert _\infty )$-continuous.

Proof

Let $s_0\in X$ and $\varepsilon >0$ be given. Then for every $t\in K$ there exist a neighborhood $V_t$ of t and a neighborhood $W_t$ of $s_0$ such that

$$\begin{aligned} |k(z,s)-k(t,s_0)|\le \frac{\varepsilon }{\Vert v\Vert _\infty }\ \ \text{ for } \text{ all }\ \ z\in V_t, s\in W_t. \end{aligned}$$

Hence there exist $t_1,\ldots ,t_n\in K$ such that $K=\bigcup _{i=1}^nV_{t_i}$. Let us put $W:=\bigcap _{i=1}^nW_{t_i}$. Let $t\in K$ and choose $i_0$ with $1\le i_0\le n$ such that $t\in V_{t_{i_0}}$. Then for $s\in W$, we have $|k(t,s)-k(t,s_0)|\le \frac{\varepsilon }{\Vert v\Vert _\infty }$. Hence

$$\begin{aligned} \Vert \Psi _v(s)-\Psi _v(s_0)\Vert _\infty \le \Vert v\Vert _\infty \,\sup _{t\in K}|k(t,s)-k(t,s_0)|\le \varepsilon . \end{aligned}$$

This means that $\Psi _v$ is $({{\mathcal {T}}},\Vert \cdot \Vert _\infty )$-continuous. $\square$

Let $L^1(\mu , C(K))$ stand for the Banach space of $\mu$-Bochner integrable functions on X with values in C(K). In view of [23, Theorem 5.1] we have

$$\begin{aligned} C_b(X,C(K))\subset L^1(\mu , C(K)). \end{aligned}$$

Hence, in view of Lemma 6.1, we can define the kernel operator $S:B({{\mathcal {B}}}o)\rightarrow C(K)$ by

$$\begin{aligned} S(v):=\int _X\Psi _v(s)\,{\text {d}}\mu (s)=\int _Xv(s)k(\cdot ,s)\,{\text {d}}\mu (s)\ \ \text{ for } \text{ all }\ \ v\in B({{\mathcal {B}}}o). \end{aligned}$$

For $t\in K$, let $\phi _t(w):=w(t)$ for $w\in C(K)$. Then $\phi _t\in C(K)'$ and using Hille’s theorem (see [13, §1, Section J, Theorem 36]), we get

$$\begin{aligned} S(v)(t)=\int _Xv(s)k(t,s)\,{\text {d}}\mu (s)\ \ \text{ for } \text{ all }\ \ v\in B({{\mathcal {B}}}o), t\in K. \end{aligned}$$

Then for $v\in B({{\mathcal {B}}}o)$,

$$\begin{aligned} \begin{array}{rl} \Vert S(v)\Vert _\infty &{}\!\!\! \displaystyle =\sup _{t\in K}|S(v)(t)|\le \sup _{t\in K}\int _X|v(s)||k(t,s)|\,{\text {d}}\mu (s)\\ &{}\!\!\! \displaystyle \le \int _X|v(s)|\sup _{t\in K}|k(t,s)|\,{\text {d}}\mu (s)\le \Vert v\Vert _\infty \sup _{t\in K,\, s\in X}|k(t,s)|\,\mu (X), \end{array} \end{aligned}$$

that is, S is a $(\Vert \cdot \Vert _\infty ,\Vert \cdot \Vert _\infty )$-bounded operator.

Define a measure $m_k:{{\mathcal {B}}}o\rightarrow C(K)$ by

$$\begin{aligned} m_k(A):=S(\mathbb {1}_A)=\int _Ak(\cdot ,s)\,{\text {d}}\mu (s)\ \ \text{ for }\ \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Then,

$$\begin{aligned} S(v)=\int v\,{\text {d}}m_k\ \ \text{ for } \text{ all }\ \ v\in B({{\mathcal {B}}}o) \end{aligned}$$

and for $A\in {{\mathcal {B}}}o$, $t\in K$, we have

$$\begin{aligned} m_k(A)(t)=\int _Ak(t,s)\,{\text {d}}\mu (s). \end{aligned}$$

Proposition 6.2

The measure $m_k$ has the following properties:

• (i) $m_k$ is of bounded variation and for every $A\in {{\mathcal {B}}}o$,

  $$\begin{aligned} |m_k|(A)=\int _A\sup _{t\in K}|k(t,s)|\,{\text {d}}\mu (s). \end{aligned}$$

• (ii) $|m_k|\in M^+(X)$ and $m_k$ is a Radon measure and

  $$\begin{aligned} m_k(A)=\int _A\frac{k(\cdot ,s)}{\sup _{t\in K}|k(t,s)|}\,{\text {d}}|m_k|(s)\ \ \text{ for } \text{ all }\ \ A\in {{\mathcal {B}}}o, \end{aligned}$$

  where the function $X\ni s\mapsto \frac{k(\cdot ,s)}{\sup _{t\in K}|k(t,s)|}\in C(K)$ belongs to $L^1(|m_k|,C(K))$.

Proof

(i) See [9, Theorem 4, p. 46].

(ii) Note that $|m_k|(A)\le \mu (A)\, \sup \{|k(t,s)|:t\in K, s\in X\}$ for $A\in {{\mathcal {B}}}o$. Then $|m_k|\in M^+(X)$ and hence $m_k$ is a Radon measure. From (i) it follows that $\sup _{t\in K}|k(t,\cdot )|=\frac{{\text {d}}|m_k|}{{\text {d}}\mu }$ (= the Radon–Nikodym derivative of $|m_k|$ with respect to $\mu$). Since $|m_k|\in M^+(X)$, using [23, Theorem 5.1] we get $C_b(X,C(K))\subset L^1(|m_k|,C(K))$.

Let $v(s):=\frac{1}{\sup _{t\in K}|k(t,s)|}$ for $s\in X$. Then $v\in B({{\mathcal {B}}}o)$ and by Lemma 6.1 the function

$$\begin{aligned} X\ni s\mapsto \frac{k(\cdot ,s)}{\sup _{t\in K}|k(t,s)|}\in C(K) \end{aligned}$$

belongs to $L^1(|m_k|,C(K))$. Hence we can define the measure $m_0:{{\mathcal {B}}}o\rightarrow C(K)$ by

$$\begin{aligned} m_0(A):=\int _A\frac{k(\cdot ,s)}{\sup _{t\in K}|k(t,s)|}\,{\text {d}}|m_k|(s)\ \ \text{ for }\ \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Using Hille’s theorem and [7, Theorem C.8] for $A\in {{\mathcal {B}}}o$ and each $\tau \in K$, we get

$$\begin{aligned} \begin{array}{rl} m_0(A)(\tau )&{}\!\!\! \displaystyle =\phi _\tau (m_0(A))=\phi _\tau \left( \int _X\frac{\mathbb {1}_A(s)k(\cdot ,s)}{\sup _{t\in K}|k(t,s)|}\,{\text {d}}|m_k|(s)\right) \\ &{}\!\!\! \displaystyle =\int _X\frac{\mathbb {1}_A(s)k(\tau ,s)}{\sup _{t\in K}|k(t,s)|}\,{\text {d}}|m_k|(s)\\ &{}\!\!\! \displaystyle =\int _X\frac{\mathbb {1}_A(s)k(\tau ,s)}{\sup _{t\in K}|k(t,s)|}\sup _{t\in K}|k(t,s)|\,{\text {d}}\mu (s)\\ &{}\!\!\! \displaystyle =\int _Ak(\tau ,s)\,{\text {d}}\mu (s)=m_k(A)(\tau ). \end{array} \end{aligned}$$

Thus $m_k(A)=m_0(A):=\int _A\frac{k(\cdot ,s)}{\sup _{t\in K}|k(t,s)|}\,{\text {d}}|m_k|(s)$ for every $A\in {{\mathcal {B}}}o$. $\square$

Define the kernel operator $T:C_b(X)\rightarrow C(K)$ by

$$\begin{aligned} T(u):=\int _Xu(s)k(\cdot ,s)\,{\text {d}}\mu (s)\ \ \text{ for } \text{ all }\ \ u\in C_b(X). \end{aligned}$$

Let us consider the mapping $\lambda :K\ni t\mapsto \mu _t\in M(X)$, where for $t\in K$,

$$\begin{aligned} \mu _t(A):=\int _Ak(t,s)\,{\text {d}}\mu (s)\ \ \text{ for } \text{ all }\ \ A\in {{\mathcal {B}}}o. \end{aligned}$$

Then

$$\begin{aligned} T(u)(t)=\int _Xu(s)\,{\text {d}}\mu _t(s)\ \ \text{ for } \text{ all }\ \ u\in C_b(X), t\in K, \end{aligned}$$

that is, T is a kernel operator in the sense of Sentilles (see [39, 40]) with the kernel $\lambda$ and $T(u)(t)=\lambda (u)(t)$ for $u\in C_b(X)$, $t\in K$.

Now, we are ready to state our desire result.

Theorem 6.3

The kernel operator $T:C_b(X)\rightarrow C(K)$ is $\beta$-nuclear and

$$\begin{aligned} \Vert T\Vert _{\beta \text{- }nuc}=\int _X\sup _{t\in K}|k(t,s)|\,{\text {d}}\mu (s). \end{aligned}$$

Proof

For every $u\in C_b(X)$, using Proposition 6.2, we get

$$\begin{aligned} \begin{array}{rl} \Vert T(u)\Vert _\infty &{}\!\!\! \displaystyle =\sup _{t\in K}|T(u)(t)|=\sup _{t\in K}\Big |\int _Xu(s)k(t,s)\,{\text {d}}\mu (s)\Big |\\ &{}\!\!\! \displaystyle \le \int _X|u(s)|\sup _{t\in K}|k(t,s)|\,{\text {d}}\mu (s)=\int _X|u(s)|\,{\text {d}}|m_k|(s). \end{array} \end{aligned}$$

Hence, T is dominated, and by Proposition 1.7T is $(\beta ,\Vert \cdot \Vert _\infty )$-continuous and weakly compact. In view of Theorem 2.3,

$$\begin{aligned} T(u)=\int _Xu\,{\text {d}}m\ \ \text{ for } \text{ all }\ \ u\in C_b(X), \end{aligned}$$

where $m:=j_{C(K)}\circ {{\hat{m}}}$ and ${{\hat{m}}}$ is the representing measure of T.

On the other hand,

$$\begin{aligned} T(u)=S(u)=\int _Xu\,{\text {d}}m_k\ \ \text{ for } \text{ all }\ \ u\in C_b(X), \end{aligned}$$

and since $m_k$ and m are Radon measures, we derive that $m_k=m$. In view of Proposition 6.2 and Theorem 5.1, we obtain that T is a $\beta$-nuclear operator and

$$\begin{aligned} \Vert T\Vert _{\beta \text{- }{\text {nuc}}}=|m|(X)=|m_k|(X)=\int _X\sup _{t\in K}|k(t,s)|\,{\text {d}}\mu (s). \end{aligned}$$

$\square$