Nuclear operators on Banach function spaces

Abstract

Let X be a Banach space and E be a perfect Banach function space over a finite measure space $(\Omega ,\Sigma ,\lambda )$ such that $L^\infty \subset E\subset L^1$. Let $E'$ denote the Köthe dual of E and $\tau (E,E')$ stand for the natural Mackey topology on E. It is shown that every nuclear operator $T:E\rightarrow X$ between the locally convex space $(E,\tau (E,E'))$ and a Banach space X is Bochner representable. In particular, we obtain that a linear operator $T:L^\infty \rightarrow X$ between the locally convex space $(L^\infty ,\tau (L^\infty ,L^1))$ and a Banach space X is nuclear if and only if its representing measure $m_T:\Sigma \rightarrow X$ has the Radon-Nikodym property and $|m_T|(\Omega )=\Vert T\Vert _{nuc}$ (= the nuclear norm of T). As an application, it is shown that some natural kernel operators on $L^\infty$ are nuclear. Moreover, it is shown that every nuclear operator $T:L^\infty \rightarrow X$ admits a factorization through some Orlicz space $L^\varphi$, that is, $T=S\circ i_\infty$, where $S:L^\varphi \rightarrow X$ is a Bochner representable and compact operator and $i_\infty :L^\infty \rightarrow L^\varphi$ is the inclusion map.

1 Introduction and preliminaries

We assume that $(X,\Vert \cdot \Vert _X)$ is a real Banach space. For terminology concerning Riesz spaces and function spaces, we refer the reader to [9, 13, 27].

We assume that $(\Omega ,\Sigma ,\lambda )$ is a finite measure space. Let $L^0$ denote the corresponding space of $\lambda$-equivalence classes of all $\Sigma$-measurable real functions on $\Omega$. Then $L^0$ is a super Dedekind complete Riesz space, equipped with the topology ${\mathcal {T}}_o$ of convergence in measure. By ${\mathcal {S}}(\Sigma )$ we denote the space of all real $\Sigma$-simple functions $s=\sum ^n_{i=1} c_i\mathbb {1}_{A_i}$, where the sets $A_i\in \Sigma$ are pairwise disjoint.

Let $(E,\Vert \cdot \Vert _E)$ be a Banach function space, where E is an order ideal of $L^0$ such that $L^\infty \subset E\subset L^1$, and $\Vert \cdot \Vert _E$ is a Riesz norm on E. By ${\mathcal {T}}_E$ we denote the $\Vert \cdot \Vert _E$-norm topology on E. By $E'$ we denote the Köthe dual of E, that is,

$$\begin{aligned} E':=\bigg \{v\in L^0:\int _\Omega |u(\omega )\,v(\omega )|\,d\lambda <\infty \ \text{ for } \text{ all } u\in E\bigg \}. \end{aligned}$$

The associated norm $\Vert \cdot \Vert _{E'}$ on $E'$ is defined for $v\in E'$ by

$$\begin{aligned} \Vert v\Vert _{E'}=\sup \left\{ \int _\Omega |u(\omega )\,v(\omega )|\,d\lambda :u\in E,\; \Vert u\Vert _E\le 1\right\} . \end{aligned}$$

We will assume that E is perfect, that is, $E=E''$ and $\Vert u\Vert _E=\Vert u\Vert _{E''}$. The order continuous dual $E^\sim _n$ of E separates the points of E and $E^\sim _n$ can be identified with $E'$ through the Riesz isomorphism $E'\ni v\mapsto F_v\in E^\sim _n$, where

$$\begin{aligned} F_v(u)=\int _\Omega u(\omega )v(\omega )\,d\lambda \ \text{ for } u\in E \ \text{ and } \ \Vert F_v\Vert =\Vert v\Vert _{E'} \end{aligned}$$

(see [13, Theorem 6.1.1]). The Mackey topology $\tau (E,E')$ is a locally convex-solid Hausdorff topology with the Lebesgue property (see [9, Corollary 82H]). Then $\tau (E,E')\subset {\mathcal {T}}_E$ and $\tau (E,E')={\mathcal {T}}_E$ if the norm $\Vert \cdot \Vert _E$ is order continuous.

The most important classes of Banach function spaces are Lebesgue spaces $L^p$ $(1\le p\le \infty )$ and Orlicz spaces $L^\varphi$ (see [19]).

Now we present a characterization of $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous linear operators $T:E\rightarrow X$ (see [17, Proposition 2.2]).

Proposition 1.1

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

• (i) T is $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous.

• (ii) $\Vert T(u_n)\Vert _X\rightarrow 0$ if $\,u_n(\omega )\rightarrow 0$ $\lambda$-a.e. and $|u_n(\omega )|\le |u(\omega )|$ $\lambda$-a.e. for some $u\in E$ and all $n\in {\mathbb {N}}$.

• (iii) For each $u\in E$, $\Vert T(u\mathbb {1}_{A_n})\Vert _X\rightarrow 0$ whenever $\lambda (A_n)\rightarrow 0$.

For terminology and basic facts concerning vector measure, we refer the reader to [4, 6, 7, 22]. For a finitely additive measure $m:\Sigma \rightarrow X$, by |m|(A) we denote the variation of m on $A\in \Sigma$. A measure $m:\Sigma \rightarrow X$ is said to be $\lambda$ -continuous if $\Vert m(A_n)\Vert _X\rightarrow 0$ whenever $\lambda (A_n)\rightarrow 0$.

Let $L^1(X)$ denote the Banach space of $\lambda$-equivalence classes of all X-valued Bochner integrable functions g defined on $\Omega$, equipped with the norm $\Vert g\Vert _1:=\int _\Omega \Vert g(\omega )\Vert _X d\lambda .$

Recall that a $\lambda$-continuous measure $m:\Sigma \rightarrow X$ of finite variation is said to have the Radon-Nikodym property with respect to $\lambda$ if there exists a function $g\in L^1(X)$ such that $m(A)=\int _A g(\omega )\,d\lambda$ for all $A\in \Sigma$. Then we write $m=g\lambda$ and a function g is called the density of m with respect to $\lambda$.

Assume that $m:\Sigma \rightarrow X$ is a $\lambda$-continuous measure. Following [6, § 13] for $A\in \Sigma$, we put

$$\begin{aligned} |m|_{E'}(A):=\sup \left\{ \sum ^n_{i=1} |c_i|\ \Vert m(A_i)\Vert _X\right\} , \end{aligned}$$

where the supremum is taken for all functions $s=\sum ^n_{i=1} c_i\mathbb {1}_{A_i}\in \ {\mathcal {S}}(\Sigma )$ such that $A_i\subset A$ for $1\le i\le n$ and $\Vert s\Vert _E\le 1$. The set function $|m|_{E'}$ will be called a $E'$ -variation of the measure m.

If, in particular, $E=L^\infty$, then $|m|_{L^1}(A)=|m|(A)$ for $A\in \Sigma$.

Let $L^0(X)$ stand for the linear space of $\lambda$-equivalence classes of all strongly $\Sigma$-measurable functions $g:\Omega \rightarrow X$. Let

$$\begin{aligned} E(X)=\left\{ g\in L^0(X):\Vert g(\cdot )\Vert _X\in E\right\} . \end{aligned}$$

Then E(X) equipped with the norm $\Vert g\Vert _{E(X)}:=\Vert \cdot \Vert g(\cdot )\Vert _X\Vert _{E}$ is a Banach space, called a Köthe-Bochner space (see [14]).

Definition 1.1

A bounded linear operator $T:E\rightarrow X$ is said to be Bochner representable, if there exists $g\in E'(X)$ such that

$$\begin{aligned} T(u)=\int _\Omega u(\omega )g(\omega )\,d\lambda \ \ \text{ for } \ \ u\in E. \end{aligned}$$

The concept of nuclear operators between Banach spaces in due to Ruston [21]. Grothendieck carried over the concept of nuclear operators to locally convex spaces [10, 11] (see also [26, p. 289], [18, 23, Chap. 3, § 7], [4, Chap. 6], [5, 22]).

Following [23, Chap. 3, § 7] (see also [2, Chap. 4], [12, 17.3, p. 379]), we have

Definition 1.2

A linear operator $T:E\rightarrow X$ is said to be $\tau (E,E')$ -nuclear if there exist a sequence $(v_n)$ in $E'$ such that the family $\{F_{v_n}:n\in {\mathbb {N}}\}$ is $\tau (E,E')$-equicontinuous, a bounded sequence $(x_n)$ in X and a sequence $(\alpha _n)\in \ell ^1$ such that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _\Omega u\, v_n \,d\lambda \right) x_n \ \ \text{ for } \ u\in E. \end{aligned}$$

(1.1)

Let

$$\begin{aligned} \Vert T\Vert _{nuc}:=\inf \left\{ \sum ^\infty _{n=1}|\alpha _n|\ \Vert v_n\Vert _{E'}\, \Vert x_n\Vert _X\right\} , \end{aligned}$$

where the infimum is taken over all sequences $(v_n)$ in $E'$, $(x_n)$ in X and $(\alpha _n)\in \ell ^1$ such that T admits a representation (1.1).

It is known that a $\tau (E,E')$-nuclear operator $T:E\rightarrow X$ is $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous and $\tau (E,E')$-compact, that is, T(V) is relatively norm compact in X for some $\tau (E,E')$-neighborhood V of 0 in E (see [23, Chap. 3, § 7, Corollary 1], [12, Corollary 4, p. 379]).

In this paper we study $\tau (E,E')$-nuclear operators $T:E\rightarrow X$. In Section 2 it is shown that every $\tau (E,E')$-nuclear operator $T:E\rightarrow X$ is Bochner representable (see Theorem 2.3 below). In particular, we obtain that a linear operator $T:L^\infty \rightarrow X$ is $\tau (L^\infty ,L^1)$-nuclear if and only if its representing measure $m_T:\Sigma \rightarrow X$ has the Radon-Nikodym property and $|m_T|(\Omega )=\Vert T\Vert _{nuc}$ (see Theorem 2.5 below). As an application, we obtain that some natural kernel operators on $L^\infty$ are $\tau (L^\infty ,L^1)$-nuclear (see Proposition 2.9 below). In Section 3 it is shown that every $\tau (L^\infty ,L^1)$-nuclear operator $T:L^\infty \rightarrow X$ admits a factorization through some Orlicz space $L^\varphi$, that is, $T=S\circ i_\infty$, where $S:L^\varphi \rightarrow X$ is a Bochner representable, compact operator and $i_\infty :L^\infty \rightarrow L^\varphi$ denotes the inclusion map (see Corollary 3.5).

2 Nuclear operators on Banach function spaces

Assume that $T:E\rightarrow X$ is a linear operator. Then the measure $m_T:\Sigma \rightarrow X$ defined by

$$\begin{aligned} m_T(A):=T(\mathbb {1}_A) \ \ \text{ for } \ A\in \Sigma \end{aligned}$$

is called a representing measure of T.

If, in particular, T is $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous, then using Proposition 1.1 we obtain that $m_T$ is countably additive. Since $m_T(A)=0$ if $\lambda (A)=0$, by the Pettis theorem $m_T$ is $\lambda$-continuous, that is, $m_T\ll \lambda$.

The following lemma will be useful.

Lemma 2.1

Let $T:E\rightarrow X$ be a $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous linear operator. If $|m_T|_{E'}(\Omega )<\infty$ and $m_T$ has the Radon-Nikodym property with respect to $\lambda$ with a density $g\in L^1(X)$, then $g\in E'(X)$ and

$$\begin{aligned} \Vert \mathbb {1}_A g\Vert _{E'(X)}=|m_T|_{E'}(A) \ \ \text{ for } \text{ all } \ A\in \Sigma , \end{aligned}$$

and

$$\begin{aligned} T(u)=\int _\Omega u(\omega )\,g(\omega )\,d\lambda \ \ \text{ for } \text{ all } \ u\in E. \end{aligned}$$

Proof

First we shall show that for $A\in \Sigma$,

$$\begin{aligned} |m_T|_{E'}(A)=\sup \left\{ \int _A|s(\omega )| \ \Vert g(\omega )\Vert _X \,d\lambda : s\in {\mathcal {S}}(\Sigma ), \ \Vert s\Vert _E\le 1\right\} . \end{aligned}$$

(2.1)

Note that $|m_T|(A)=\int _A\Vert g(\omega )\Vert _X\,d\lambda$. For $s=\sum ^k_{i=1}c_i\mathbb {1}_{A_i} \in {\mathcal {S}}(\Sigma )$, we have

$$\begin{aligned} \sum ^k_{i=1}c_i\, m_T(A\cap A_i)=\sum ^k_{i=1} c_i\int _{A\cap A_i} g(\omega )\,d\lambda = \int _A s(\omega )\, g(\omega )\, d\lambda . \end{aligned}$$

We now show that for $s=\sum ^k_{i=1} c_i\mathbb {1}_{A_i}\in {\mathcal {S}}(\Sigma )$ and $\Vert s\Vert _E\le 1$, we have

$$\begin{aligned} \sum ^k_{i=1}|c_i|\int _{A\cap A_i}\Vert g(\omega )\Vert _X \,d\lambda =\sum ^k_{i=1}|c_i|\ |m_T| (A\cap A_i)\le |m_T|_{E'}(A). \end{aligned}$$

Indeed, let $\varepsilon >0$ be given. Then for each $1\le i\le k$, there exists a $\Sigma$-partition $(A_{i,j})^{j_i}_{j=1}$ of $A\cap A_i$ such that

$$\begin{aligned} |m_T|(A\cap A_i)\le \sum ^{j_i}_{j=1}\big \Vert m_T(A_{ij})\big \Vert _X+\frac{\varepsilon }{k|c_i|}. \end{aligned}$$

Hence

$$\begin{aligned} \sum ^k_{i=1}|c_i|\,|m_T|(A\cap A_i)\le \sum ^k_{i=1}\left( \sum ^{j_i}_{j=1} \Vert c_i \,m_T(A_{ij})\Vert _X\right) +\varepsilon \le |m_T|_{E'}(A)+\varepsilon , \end{aligned}$$

because $\sum ^k_{i=1}\big (\sum ^{j_i}_{j=1}c_i\, \mathbb {1}_{A_{ij}}\big )=\sum ^k_{i=1} c_i\,\mathbb {1}_{A_i}.$

Then we have

$$\begin{aligned} \begin{array}{l} \displaystyle \sum ^k_{i=1}|c_i|\;\Vert m_T(A\cap A_i)\Vert _X \le \sum ^k_{i=1}|c_i|\;|m_T|(A\cap A_i)\\ \quad \displaystyle =\sum ^k_{i=1}|c_i|\int _{A\cap A_i}\Vert g(\omega )\Vert _X\,d\lambda =\int _A\left( \sum ^k_{i=1}|c_i| \,\mathbb {1}_{A_i}(\omega )\right) \Vert g(\omega )\Vert _X \,d\lambda \\ \quad \displaystyle =\int _A|s(\omega )|\ \Vert g(\omega )\Vert _X\, d\lambda \le |m_T|_{E'}(A). \end{array} \end{aligned}$$

Taking supremum on the left side, we get

$$\begin{aligned} |m_T|_{E'}(A)=\sup \left\{ \int _A|s(\omega )|\ \Vert g(\omega )\Vert _X \,d\lambda :s\in {\mathcal {S}}(\Sigma ),\ \Vert s\Vert _E\le 1\right\} . \end{aligned}$$

We shall now show that $g\in E'(X)$, that is, $\Vert g(\cdot )\Vert _X\in E'$.

Indeed, let $u\in E$. Then there exists a sequence $(s_n)$ in ${\mathcal {S}}(\Sigma )$ such that $0\le s_n(\omega )\uparrow |u(\omega )|$ $\lambda$-a.e. (see [13, Corollary I.6]). Choose $c>0$ such that $\Vert cu\Vert _E\le 1$. Then by the Fatou lemma,

$$\begin{aligned} \int _\Omega c\,|u(\omega )|\ \Vert g(\omega )\Vert _X \,d\lambda \le \sup _n\int _\Omega c \,s_n(\omega )\Vert g(\omega )\Vert _X \, d\lambda \le |m_T|_{E'}(\Omega ), \end{aligned}$$

and this means that $\Vert g(\cdot )\Vert _X\in E'$, that is, $g\in E'(X)$. Moreover, for $A\in \Sigma$ and $u\in E$ with $\Vert u\Vert _E\le 1$, using (2.1) we get

$$\begin{aligned} \int _\Omega |u(\omega )|\ \Vert \mathbb {1}_A(\omega )\,g(\omega )\Vert _X\, d\lambda \le \sup _n\int _\Omega s_n(\omega ) \Vert \mathbb {1}_A(\omega )\,g(\omega )\Vert _X\, d\lambda \le |m_T|_{E'}(A) \end{aligned}$$

and it follows that $\Vert \mathbb {1}_A g\Vert _{E'(X)}\le |m_T|_{E'}(A)$. In view of (2.1) we get $|m_T|_{E'}(A)\le \Vert \mathbb {1}_A g\Vert _{E'(X)}$. Hence $|m_T|_{E'}(A)=\Vert \mathbb {1}_A g\Vert _{E'(X)}$.

Note that for $s=\sum ^k_{i=1} c_i\mathbb {1}_{A_i}\in {\mathcal {S}}(\Sigma )$, we have

$$\begin{aligned} T(s)=\sum ^k_{i=1}c_i\, m_T(A_i)=\int _\Omega s(\omega )\,g(\omega )\,d\lambda . \end{aligned}$$

Let $u\in E$ be given. Then there exists a sequence $(s_n)$ in ${\mathcal {S}}(\Sigma )$ such that $|s_n(\omega )-u(\omega )|\rightarrow 0$ $\lambda$-a.e. and $|s_n(\omega )|\le |u(\omega )|$ $\lambda$-a.e. for all $n\in {\mathbb {N}}$. Then $|s_n(\omega )-u(\omega )|$ $\Vert g(\omega )\Vert _X\le 2|u(\omega )|$ $\Vert g(\omega )\Vert _X$ $\lambda$-a.e., where $u\,\Vert g(\cdot )\Vert _X\in L^1$. Hence by the Lebesgue dominated convergence theorem,

$$\begin{aligned} \bigg \Vert \int _\Omega s_n(\omega )\,g(\omega )\, d\lambda -\int _\Omega u(\omega )\,g(\omega )\, d\lambda \,\bigg \Vert _X\le \int _\Omega |s_n(\omega )-u(\omega )| \ \Vert g(\omega )\Vert _X\, d\lambda \rightarrow 0. \end{aligned}$$

On the other hand, in view of Proposition 1.1 we have

$$\begin{aligned} \Vert T(s_n)-T(u)\Vert _X=\Vert T(s_n-u)\Vert _X\rightarrow 0. \end{aligned}$$

Hence $T(u)=\int _\Omega u(\omega ) g(\omega )\,d\lambda$. $\square$

As a consequence of Lemma 2.1, we have

Proposition 2.2

Assume that $T:E\rightarrow X$ is a Bochner representable operator, that is, there exists $g\in E'(X)$ such that

$$\begin{aligned} T(u)=\int _\Omega u(\omega )\,g(\omega )\,d\lambda \ \ for \ all \ u\in E. \end{aligned}$$

Then the following statements hold:

• (i) T is $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous.

• (ii) For every $A\in \Sigma$, $|m_T|_{E'}(A)=\Vert \mathbb {1}_A g\Vert _{E'(X)}.$

Proof

• (i) Assume that $u_n(\omega )\rightarrow 0$ $\lambda$-a.e. and $|u_n(\omega )|\le |u(\omega )|$ $\lambda$-a.e. for some $u\in E$ and all $n\in {\mathbb {N}}$. Since for $n\in {\mathbb {N}}$,

  $$\begin{aligned} \Vert T(u_n)\Vert _X\le \int _\Omega |u_n(\omega )|\ \Vert g(\omega )\Vert _X\,d\lambda , \end{aligned}$$

  where $u\,\Vert g(\cdot )\Vert _X\in L^1$, by the Lebesgue dominated convergence theorem, we get $\Vert T(u_n)\Vert _X\rightarrow 0$. Hence in view of Proposition 1.1T is $(\tau (E,E'),\Vert \cdot \Vert _X)$-continuous.

• (ii) Assume that $s=\sum ^k_{i=1} c_i\mathbb {1}_{A_i}\in {\mathcal {S}}(\Sigma )$ and $\Vert s\Vert _E\le 1$. Then

  $$\begin{aligned} \begin{array}{rl} \displaystyle \sum ^k_{i=1}|c_i|\ \Vert m_T(A_i)\Vert _X &{} \displaystyle =\sum ^k_{i=1}|c_i|\int _{A_i} \Vert g(\omega )\Vert _X\,d\lambda =\int _\Omega |s(\omega )|\ \Vert g(\omega )\Vert _X\,d\lambda \\ &{} \displaystyle \le \Vert s\Vert _E\ \Vert g\Vert _{E'(X)}\le \Vert g\Vert _{E'(X)}. \end{array} \end{aligned}$$

Hence $|m_T|_{E'}(\Omega )\le \Vert g\Vert _{E'(X)}$. Using (i) and Lemma 2.1, we get $|m_T|_{e'}(A)=\Vert \mathbb {1}_A g\Vert _{E'(X)}$ for all $A\in \Sigma$. $\square$

The following result shows a relationship between $\tau (E,E')$-nuclear operators and Bochner representable operators $T:E\rightarrow X$.

Theorem 2.3

Assume that $T:E\rightarrow X$ is a $\tau (E,E')$-nuclear operator. Then T is Bochner representable and $|m_T|_{E'}(\Omega )\le \Vert T\Vert _{nuc}$.

Proof

Let $\varepsilon >0$ be given. Then there exists a bounded sequence $(v_n)$ in $E'$, a bounded sequence $(x_n)$ in X and a sequence $(\alpha _n)\in \ell ^1$ such that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _\Omega u\,v_n\, d\lambda \right) x_n \ \ \text{ for } \ u\in E \end{aligned}$$

and

$$\begin{aligned} \sum ^\infty _{n=1}|\alpha _n| \ \Vert v_n\Vert _{E'}\ \Vert x_n\Vert _X\le \Vert T\Vert _{nuc}+\varepsilon . \end{aligned}$$

(2.2)

Hence we have

$$\begin{aligned} m_T(A)=\sum ^\infty _{n=1}\alpha _n\left( \int _A v_n\, d\lambda \right) x_n. \end{aligned}$$

(2.3)

We shall now show that $|m_T|_{E'}(\Omega )\le \Vert T\Vert _{nuc}$. Indeed, let $s=\sum ^k_{i=1}c_i\mathbb {1}_{A_i}\in {\mathcal {S}}(\Sigma )$ with $\Vert s\Vert _E\le 1$. Then using (2.2) we get

$$\begin{aligned} \displaystyle \sum ^k_{i=1}|c_i|\ \Vert m_T(A_i)\Vert _X&\displaystyle =&\sum ^k_{i=1}\Vert c_i\sum ^\infty _{n=1}\alpha _n\left( \int _{A_i} v_n(\omega )\,d\lambda \right) x_n\Vert _X\\&\displaystyle \le&\sum ^\infty _{n=1}|\alpha _n|\left( \sum ^k_{i=1}|c_i|\int _{A_i}|v_n(\omega )|\,d\lambda \right) \Vert x_n\Vert _X\\&\displaystyle =&\sum ^\infty _{n=1}|\alpha _n|\ \Vert x_n\Vert _X\int _\Omega |s(\omega )| \ |v_n(\omega )|\,d\lambda \\&\displaystyle \le&\sum ^\infty _{n=1}|\alpha _n|\ \Vert x_n\Vert _X \ \Vert v_n\Vert _{E'}\le \Vert T\Vert _{nuc}+\varepsilon . \end{aligned}$$

Hence we get

$$\begin{aligned} |m_T|_{E'}(\Omega )\le \Vert T\Vert _{nuc}. \end{aligned}$$

(2.4)

For $n\in {\mathbb {N}}$, let $g_n:=\sum ^n_{i=1}\alpha _i v_i\otimes x_i$. Then for $n,k\in {\mathbb {N}}$ with $n>k$, we have

$$\begin{aligned} \begin{array}{rl} \displaystyle \int _\Omega \Vert g_n(\omega ) &{}\!\!\!\displaystyle -\,g_k(\omega )\Vert _X \,d\lambda =\int _\Omega \left\| \sum ^n_{i=k+1} \alpha _i \,v_i(\omega )x_i\right\| _X\, d\lambda \\ &{}\!\!\!\displaystyle \le \int _\Omega \left( \sum ^n_{i=k+1}|\alpha _i|\ |v_i(\omega )|\ \Vert x_i\Vert _X\right) \,d\lambda = \sum ^n_{i=k+1}|\alpha _i| \ \Vert x_i\Vert _X\int _\Omega |v_i|\,d\lambda \\ &{}\!\!\!\displaystyle \le \sum ^n_{i=k+1}|\alpha _i|\ \Vert x_i\Vert _X\ \Vert \mathbb {1}_\Omega \Vert _E\ \Vert v_i\Vert _{E'}\le \sup _{j\in {\mathbb {N}}}\Vert x_j\Vert _X \sup _{j\in {\mathbb {N}}}\Vert v_j\Vert _{E'}\Vert \mathbb {1}_\Omega \Vert _E \sum ^n_{i=k+1}|\alpha _i|. \end{array} \end{aligned}$$

This follows that $(g_n)$ is a Cauchy sequence in the Banach space $L^1(X)$, so there exists $g\in L^1(X)$ such that $\Vert g_n-g\Vert _1\rightarrow 0$. One can easily show that

$$\begin{aligned} \left( \int _A v\,d\lambda \right) x=\int _A v(\omega )\,x\,d\lambda \ \text{ for } \ v\in E', \ x\in X, \ A\in \Sigma . \end{aligned}$$

Hence for $A\in \Sigma$, we have

$$\begin{aligned} \begin{array}{l} \displaystyle \bigg \Vert \sum ^n_{i=1}\alpha _i\left( \int _A v_i \,d\lambda \right) x_i-\int _A g\,d\lambda \, \bigg \Vert _X=\bigg \Vert \int _A\sum ^n_{i=1}\alpha _i\, v_i(\omega )\,x_i \,d\lambda -\int _A g(\omega ) \,d\lambda \,\bigg \Vert _X\\ \quad \displaystyle \le \int _A\bigg \Vert \sum ^n_{i=1}\alpha _i\, v_i(\omega )\,x_i-g(\omega )\bigg \Vert _X\,d\lambda = \int _A\Vert g_n(\omega )-g(\omega )\Vert _X\,d\lambda \rightarrow 0. \end{array} \end{aligned}$$

Then in view of (2.3) for $A\in \Sigma$, we get

$$\begin{aligned} m_T(A)=\int _A g(\omega )\, d\lambda . \end{aligned}$$

Using Lemma 2.1 and (2.4) we see that $g\in E'(X)$ with $\Vert g\Vert _{E'(X)}=|m_T|_{E'}(\Omega )\le \Vert T\Vert _{nuc}$ and $T(u)=\int _\Omega u(\omega )\, g(\omega )\,d\lambda$ for all $u\in E$. $\square$

Now we shall study $\tau (L^\infty ,L^1)$-nuclear operators $T:L^\infty \rightarrow X$.

Making use of the Dunford–Pettis theorem (see [3, Theorem, p. 93]) we have

Theorem 2.4

For a subset H of $L^1$ the following statements are equivalent:

• (i) H is relatively weakly compact.

• (ii) $\sup _{v\in H}\Vert v\Vert _1<\infty$ and H is uniformly integrable.

• (iii) $\{F_v:v\in H\}$ is $\tau (L^\infty ,L^1)$-equicontinuous.

Note that in view of Theorem 2.4 and Definition 1.2, a linear operator T :  $L^\infty \rightarrow X$ is $\tau (L^\infty ,L^1)$ -nuclear if there exist a bounded uniformly integrable sequence $(v_n)$ in $L^1$, a bounded sequence $(x_n)$ in X and a sequence $(\alpha _n)\in \ell ^1$ such that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _\Omega u\, v_n\, d\lambda \right) x_n \ \text{ for } \text{ all } \ u\in L^\infty \end{aligned}$$

(2.5)

and then

$$\begin{aligned} \Vert T\Vert _{nuc}=\inf \left\{ \sum ^\infty _{n=1}|\alpha _n|\ \Vert v_n\Vert _1\ \Vert x_n\Vert _X\right\} , \end{aligned}$$

where the infimum is taken over all sequences $(v_n)$ in $L^1$, $(x_n)$ in X and $(\alpha _n)\in \ell ^1$ such that T admits a representation (2.5).

Now we can state our main result.

Theorem 2.5

For a linear operator $T:L^\infty \rightarrow X$ the following statements are equivalent:

• (i) $m_T$ has the Radon-Nikodym property with respect to $\lambda$.

• (ii) T is Bochner representable.

• (iii) T is a $\tau (L^\infty ,L^1)$-nuclear operator.

In this case $|m_T|(\Omega )=\Vert T\Vert _{nuc}$.

Proof

(i)$\Rightarrow$(ii) Assume that (i) holds with the density $g\in L^1(X)$. Note that for $s\in {\mathcal {S}}(\Sigma )$, we have

$$\begin{aligned} \int _\Omega s\,dm_T=\int _\Omega s(\omega )\,g(\omega )\,d\lambda . \end{aligned}$$

Let $u\in L^\infty$. Choose a sequence $(s_n)$ in ${\mathcal {S}}(\Sigma )$ such that $\Vert u-s_n\Vert _\infty \rightarrow 0$. Then we have

$$\begin{aligned} T(u)=\int _\Omega u\,dm_T=\lim \int _\Omega s_n \,dm_T=\lim \int _\Omega s_n(\omega )\,g(\omega )\, d\lambda =\int _\Omega u(\omega )\,g(\omega )\,d\lambda . \end{aligned}$$

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

(ii)$\Rightarrow$(iii) Assume that (ii) holds, that is, there exists $g\in L^1(X)$ such that

$$\begin{aligned} T(u)=\int _\Omega u(\omega )\,g(\omega )\,d\lambda \ \ \text{ for } \text{ all } \ u\in L^\infty . \end{aligned}$$

Let $L^1\hat{\otimes } X$ denote the projective tensor product of $L^1$ and X, equipped with the norm $\pi$ defined for $w\in L^1\hat{\otimes } X$, by

$$\begin{aligned} \pi (w):=\inf \left\{ \sum ^\infty _{n=1}|\alpha _n|\ \Vert v_n\Vert _1\ \Vert x_n\Vert _X\right\} , \end{aligned}$$

where the infimum is taken over all sequences $(v_n)$ in $L^1$, $(x_n)$ in X with $\lim \Vert v_n\Vert _1=0=\lim \Vert x_n\Vert _X$ and $(\alpha _n)\in \ell ^1$ such that $w=\sum ^\infty _{n=1}\alpha _n v_n\otimes \,x_n$ (see [22, Proposition 2.8, pp. 21–22]). It is known that $L^1\hat{\otimes }\, X$ is isometrically isomorphic to $L^1(X)$ through the isometry J, where

$$\begin{aligned} J(v\otimes x):=v(\cdot )\,x \ \text{ for } \ v\in L^1, \ x\in X \end{aligned}$$

(see [4, Example 10, p. 228], [22, Example 2.19, p. 29], [5, Theorem 1.1.10, p. 14]). Let $\varepsilon >0$ be given. Then there exist sequence $(v_n)$ in $L^1$ and $(x_n)$ in X with $\lim _n\Vert v_n\Vert _1=0=\lim \Vert x_n\Vert _X$ and $(\alpha _n)\in \ell ^1$ such that

$$\begin{aligned} J^{-1}(g)=\sum ^\infty _{n=1}\alpha _n\, v_n\otimes \, x_n \ \text{ in } \ \big (L^1\hat{\otimes }\, X,\pi \big ) \end{aligned}$$

and

$$\begin{aligned} \sum ^\infty _{n=1}|\alpha _n|\ \Vert v_n\Vert _1 \ \Vert x_n\Vert _X\le \pi (J^{-1}(g))+\varepsilon = \Vert g\Vert _1+\varepsilon . \end{aligned}$$

(2.6)

Hence

$$\begin{aligned} g=J\left( \sum ^\infty _{n=1}\alpha _n v_n\otimes \, x_n\right) =\sum ^\infty _{n=1} \alpha _n \,v_n(\cdot )\,x_n \ \text{ in } \ \big (L^1(X),\ \Vert \cdot \Vert _1\big ). \end{aligned}$$

Note that

$$\begin{aligned} \begin{array}{l} \displaystyle \bigg \Vert \int _\Omega u(\omega )\,g(\omega )\,d\lambda -\sum ^n_{i=1}\alpha _i\left( \int _\Omega u(\omega ) \,v_i(\omega ) \,d\lambda \right) x_i\bigg \Vert _X\\ \quad \displaystyle \le \int _\Omega |u(\omega )|\ \bigg \Vert g(\omega )-\sum ^n_{i=1}\alpha _i\, v_i(\omega )\,x_i\,\bigg \Vert _X \,d\lambda \\ \quad \displaystyle \le \Vert u\Vert _{\infty }\int _\Omega \bigg \Vert g(\omega )-\sum ^n_{i=1}\alpha _i\, v_i(\omega )\,x_i\,\bigg \Vert _X\, d\lambda = \Vert u\Vert _{\infty }\cdot \bigg \Vert g-\sum ^n_{i=1}\alpha _i \,v_i\otimes x_i\bigg \Vert _{1} \end{array} \end{aligned}$$

Hence

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _\Omega u\, v_n\, d\lambda \right) x_n \ \text{ for } \ u\in L^\infty . \end{aligned}$$

Since $\lim \Vert v_n\Vert _1=0$, the set $\{v_n:n\in {\mathbb {N}}\}$ is uniformly integrable and it follows that T is $\tau (L^\infty ,L^1)$-nuclear. In view of (2.6) we get

$$\begin{aligned} \Vert T\Vert _{nuc}\le \Vert g\Vert _{1}=|m_T|(\Omega ). \end{aligned}$$

(2.7)

(iii)$\Rightarrow$(ii) Assume that (iii) holds. Then by Theorem 2.3T is Bochner representable and

$$\begin{aligned} |m_T|(\Omega )=\Vert g\Vert _1\le \Vert T\Vert _{nuc}. \end{aligned}$$

(2.8)

Thus (i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iii) hold and using (2.7) and (2.8), we get $|m_T|(\Omega )=\Vert T\Vert _{nuc}$. $\square$

Let $ba_\lambda (\Sigma )$ denote the Banach space of all bounded finitely additive real measures $\mu$ on $\Sigma$ such that $\mu (A)=0$ if $\lambda (A)=0$, equipped with norm $\Vert \mu \Vert :=|\mu |(\Omega )$. The Banach dual $(L^\infty )^*$ of $L^\infty$ can be identified with $ba_\lambda (\Sigma )$ through the integration mapping $ba_\lambda (\Sigma )\ni \mu \mapsto F_\mu \in (L^\infty )^*$, where $F_\mu (u)=\int _\Omega u d\mu$ for all $u\in L^\infty$ and $|\mu |(\Omega )=\Vert F_\mu \Vert$.

Recall (see [26, p. 279]) that a linear operator $T:L^\infty \rightarrow X$ is said to be nuclear if there exist a bounded sequence $(\mu _n)$ in $ba_\lambda (\Sigma )$, a bounded sequence $(x_n)$ in X and a sequence $(\alpha _n)\in \ell ^1$ such that

$$\begin{aligned} T(u)=\sum ^\infty _{n=1}\alpha _n\left( \int _\Omega u\, d\mu _n\right) x_n \ \text{ for } \text{ all } \ u\in L^\infty . \end{aligned}$$

It is well known that a bounded linear operator $T:L^\infty \rightarrow X$ is $(\tau (L^\infty ,L^1),$ $\Vert \cdot \Vert _X)$-continuous if and only if its representing measure $m_T:\Sigma \rightarrow X$ is $\lambda$-continuous (see [17, Proposition 3.1]). Hence due to Swartz (see [24, Theorem 1 and Theorem 3]) we have

Theorem 2.6

Assume that $T:L^\infty \rightarrow X$ is a $(\tau (L^\infty ,L^1),\Vert \cdot \Vert _X)$-continuous linear operator. Then the following statements are equivalent:

• (i) T is nuclear.

• (ii) $m_T$ has the Radon-Nikodym property with respect to $\lambda$.

Combining Theorem 2.5 and Theorem 2.6, we get

Corollary 2.7

For a linear operator $T:L^\infty \rightarrow X$ the following statements are equivalent:

• (i) T is $\tau (L^\infty ,L^1)$-nuclear.

• (ii) T is $(\tau (L^\infty ,L^1),\Vert \cdot \Vert _X)$-continuous and nuclear.

As an application of Theorem 2.5 we show that some natural kernel operators on $L^\infty$ are $\tau (L^\infty ,L^1)$-nuclear.

Assume that $K_1$ and $K_2$ are compact Hausdorff spaces and $k(\cdot ,\cdot )\in C(K_2\times K_1)$. Let ${\mathcal {B}}o$ be the $\sigma$-algebra of Borel sets in $K_1$ and $\lambda :{\mathcal {B}}o\rightarrow [0,\infty )$ be a countably additive measure.

We will need the following lemma.

Lemma 2.8

For every $u\in L^\infty (\lambda )$, the mapping $\Psi _u:K_1\ni s\mapsto u(s)\,k(\cdot ,s)\in C(K_2)$ is continuous.

Proof

Let $s_0\in K_1$ and $\varepsilon >0$ be given. Then for every $t\in K_2$ there exist a neighborhood $V_t$ of t and a neighborhood $W_t$ of $s_0$ such that

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

Hence there exist $t_1,\dots ,t_n\in K_2$ such that $K_2=\bigcup ^n_{i=1} V_{t_i}$. Let us put $W:=\bigcap ^n_{i=1} W_{t_i}$. For $t\in K_2$, 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 $|k(t,s)-k(t,s_0|\le \frac{\varepsilon }{\Vert u\Vert _\infty }$. Hence

$$\begin{aligned} \Vert \Psi _u(s)-\Psi _u(s_0)\Vert _\infty =\sup _{t\in K_2}|k(t,s)-k(t,s_0)|\;\Vert u\Vert _\infty \le \varepsilon . \end{aligned}$$

This means that $\Psi _u$ is continuous. $\square$

Note that the Banach space $C(K_1,C(K_2))$ can be embedded in the Banach space $L^1(C(K_2))$ such that with each function from $C(K_1,C(K_2))$ is associated its $\lambda$-equivalence class in $L^1(C(K_2))$.

In view of Lemma 2.8 we can define a kernel operator $T_k:L^\infty (\lambda )\rightarrow C(K_2)$ by

$$\begin{aligned} T_k(u):=\int _{K_1}u(s)\,k(\cdot ,s)\,d\lambda (s) \ \ \text{ for } \text{ all } \ u\in L^\infty (\lambda ). \end{aligned}$$

For $t\in K_2$, let $\delta _t(w):=w(t)$ for all $w\in C(K_2)$. Then $\delta _t\in C(K_2)^*$ and according to the Hille’s theorem (see [DU, Theorem 6, p. 47]), we have

$$\begin{aligned} T_k(u)(t)=\delta _t(T_k(u))=\int _{K_1}u(s)\,k(t,s)\,d\lambda (s) \ \ \text{ for } \text{ all } \ u\in L^\infty (\lambda ), \ t\in K_2. \end{aligned}$$

Then

$$\begin{aligned} m_{T_k}(A)=T_k(\mathbb {1}_A)=\int _A k(\cdot ,s)\,d\lambda (s) \ \ \text{ for } \text{ all } \ A\in {\mathcal {B}}o, \end{aligned}$$

where the mapping $K_1\ni s\mapsto k(\cdot ,s)\in C(K_2)$ belongs to $L^1(C(K_2))$.

Hence for $A\in {\mathcal {B}}o$,

$$\begin{aligned} |m_{T_k}|(A)=\int _A \Vert k(\cdot ,s)\Vert _\infty \,d\lambda (s)=\int _A\sup _{t\in K_2} |k(t,s)|\,d\lambda (s). \end{aligned}$$

Hence as a consequence of Theorem 2.5 we have

Proposition 2.9

The kernel operator $T_k:L^\infty (\lambda )\rightarrow C(K_2)$ is $\tau (L^\infty (\lambda ),L^1(\lambda ))$-nuclear and

$$\begin{aligned} \Vert T_k\Vert _{nuc}=\int _{K_1}\sup _{t\in K_2} |k(t,s)|\,d\lambda (s). \end{aligned}$$

3 Application of the theory of Orlicz spaces to vector measures

First we recall terminology and basic facts concerning Orlicz spaces and Orlicz-Bochner spaces (see [19] for more details). By a Young function we mean here a convex continuous mapping $\varphi :[0,\infty )\rightarrow [0,\infty )$ that vanishes only at 0 and $\varphi (t)/t$ $\rightarrow 0$ as $t\rightarrow 0$ and $\varphi (t)/t\rightarrow \infty$ as $t\rightarrow \infty$. By $\varphi ^*$ we denote the complementary function of $\varphi$ in the sense of Young, that is, $\varphi ^*(t)=\sup \{ts-\varphi (s):s\ge 0\}$ for $t\ge 0$. Note that $\varphi ^{**}=\varphi$.

The corresponding Orlicz space $L^\varphi$ is an ideal of $L^0$ defined by

$$\begin{aligned} L^\varphi :=\bigg \{u\in L^0:\int _\Omega \varphi \,(\alpha \,|u\,(\omega )|\,)\,d\lambda <\infty \ \ \text{ for } \text{ some } \ \alpha >0\bigg \}, \end{aligned}$$

and equipped with the topology ${\mathcal {T}}_\varphi$, defined by two equivalent norms:

$$\begin{aligned} \begin{array}{l} \displaystyle \Vert u\Vert _\varphi :=\inf \bigg \{\alpha >0:\int _\Omega \varphi \,(|u(\omega )|\,/\,\alpha )\, d\lambda \le 1\bigg \},\\ \displaystyle \Vert u\Vert ^0_\varphi :=\sup \bigg \{\int _\Omega |u(\omega )\,v(\omega )|\, d\lambda :v\in L^{\varphi ^*},\int _\Omega \varphi ^* (|v(\omega )|)\,d\lambda \le 1\bigg \}, \end{array} \end{aligned}$$

called the Luxemburg norm and the Orlicz norm. Then we have:

$\Vert u\Vert _\varphi \le 1$ if and only if $\int _\Omega \varphi (|u(\omega )|)\,d\lambda \le 1$,

$\Vert u_n\Vert _\varphi \rightarrow 0$ if and only if $\int _\Omega \varphi (\alpha |u_n(\omega )|)\,d\lambda \rightarrow 0$ for every $\alpha >0$,

$$\begin{aligned} \begin{array}{rl} E^\varphi := &{}\! \displaystyle \bigg \{u\in L^\varphi :\int _\Omega \varphi (\alpha \,|\,u(\omega )|)\,d\lambda <\infty \ \ \text{ for } \text{ all } \ \alpha >0\bigg \}\\ = &{}\! \displaystyle \big \{u\in L^\varphi :\Vert \mathbb {1}_{A_n} u\Vert _\varphi \rightarrow 0 \ \ \text{ if } \ \lambda (A_n)\rightarrow 0\big \}, \end{array} \end{aligned}$$

$(L^\varphi )'=L^{\varphi ^*}.$

The Orlicz-Bochner space $L^\varphi (X)$ is defined by

$$\begin{aligned} \begin{array}{rl} L^\varphi (X) \,&{}\! \displaystyle :=\bigg \{g\in L^0(X):\int _\Omega \varphi (\alpha \,\Vert \,g(\omega )\Vert _X)\, d\lambda <\infty \ \ \text{ for } \text{ some } \ \alpha >0\bigg \} \\ &{}\!\!\!\displaystyle \,=\big \{g\in L^0(X):\Vert g(\cdot )\Vert _X\in L^\varphi \big \}. \end{array} \end{aligned}$$

For $g\in L^\varphi (X)$, let

$$\begin{aligned} \Vert g\Vert _\varphi :=\Vert \ \Vert g(\cdot )\Vert _X \ \Vert _\varphi \ \text{ and } \ \Vert g\Vert ^0_\varphi :=\Vert \ \Vert g(\cdot )\Vert _X \ \Vert ^0_\varphi . \end{aligned}$$

Then

$$\begin{aligned} E^\varphi (X):=\left\{ g\in L^0(X):\Vert g(\cdot )\Vert _X\in E^\varphi \right\} . \end{aligned}$$

Assume that $\varphi$ is Young function. If $m:\Sigma \rightarrow X$ is a $\lambda$-continuous measure, we write $|m|_{\varphi ^*}$ instead of $|m|_{(L^\varphi )'}$.

Lemma 3.1

Assume that $m:\Sigma \rightarrow X$ is a $\lambda$-continuous measure and $\varphi$ is Young function. Then for $A\in \Sigma$, we have

$$\begin{aligned} |m|(A)\le \left( \varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) \right) ^{-1}|m|_{\varphi ^*}(A). \end{aligned}$$

Proof

Let $A\in \Sigma$ and $\varepsilon >0$ be given. Then there is a $\Sigma$-partition $(A_i)^n_{i=1}$ of A such that

$$\begin{aligned} |m|(A)\le \sum ^n_{i=1}\Vert m(A_i)\Vert _X+\left( \varphi ^{-1}\left( \frac{1}{\lambda (A)} \right) \right) ^{-1}\varepsilon . \end{aligned}$$

It is known that $\Vert \mathbb {1}_A\Vert _\varphi =\left( \varphi ^{-1}\left( \frac{1}{\lambda (A)} \right) \right) ^{-1}$ (see [RR, Chap. 3.4, Corollary 7, p. 79]). Hence

$$\begin{aligned} \bigg \Vert \sum ^n_{i=1}\varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) \mathbb {1}_{A_i}\bigg \Vert _\varphi = \; \bigg \Vert \varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) \mathbb {1}_{A}\bigg \Vert _\varphi =1 \end{aligned}$$

and

$$\begin{aligned} \varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) |m|(A)\le \sum ^n_{i=1} \varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) \Vert m(A_i)\Vert _X+\varepsilon . \end{aligned}$$

It follows that $\varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) |m|(A)\le |m|_{\varphi ^*}(A)$, so

$$\begin{aligned} |m|(A)\le \varphi ^{-1}\left( \frac{1}{\lambda (A)}\right) ^{-1}|m|_{\varphi ^*}(A). \end{aligned}$$

$\square$

By $\gamma [{\mathcal {T}}_\varphi ,{\mathcal {T}}_0\big |_{L^\varphi }]$ (in brief, $\gamma _\varphi$) we denote the natural mixed topology on $L^\varphi$ (in the sense of Wiweger), that is, $\gamma _\varphi$ is the finest linear topology that agrees with ${\mathcal {T}}_0$ on ${\mathcal {T}}_\varphi$-bounded sets in $L^\varphi$ (see [1, 25] for more details). Then $\gamma _\varphi$ is a locally convex-solid Hausdorff topology and $\gamma _\varphi$ and ${\mathcal {T}}_\varphi$ have the same bounded sets. This means that $(L^\varphi ,\gamma _\varphi )$ is a generalized DF-space (see [20]) and it follows that $(L^\varphi ,\gamma _\varphi )$ is quasinormable (see [20, p. 422]). Hence as a consequence of the Grothendieck’s classical result (see [20, p. 429]), we have

Proposition 3.2

For a linear operator $S:L^\varphi \rightarrow X$ the following statements are equivalent:

• (i) S is $(\gamma _\varphi ,\Vert \cdot \Vert _X)$-continuous and compact.

• (ii) S is $\gamma _\varphi$-compact, that is, there exists a $\gamma _\varphi$-neighborhood V of $\,0$ in $L^\varphi$ such that T(V) is relatively norm compact in X.

We say that a Young function $\varphi$ increases essentially more rapidly than another $\psi$ (in symbols, $\psi \ll \varphi$) if for arbitrary $c>0$, $\psi (ct)/\varphi (t)\rightarrow 0$ as $t\rightarrow 0$ and $t\rightarrow \infty$. Note that $L^\varphi \subset E^\psi$ if $\psi \ll \varphi$.

The following result will be useful (see [16, Theorem 2.1]).

Theorem 3.3

Let $\varphi$ be a Young function. Then

• (i) $\gamma _\varphi$ is generated by the family of norms $\{\Vert \cdot \Vert _\psi \big |_{L^\varphi }:\psi \ll \varphi \}$.

• (ii) $(L^\varphi ,\gamma _\varphi )^*=\{F_v:v\in E^{\varphi ^*}\}$, where $F_v(u)=\int _\Omega u(\omega )\, v(\omega )\,d\lambda$ for all $u\in L^\varphi .$

According to [15, Corollary 1.6] we have the following identity:

$$\begin{aligned} L^1=\bigcup _\varphi E^\varphi , \text{ where } \varphi \text{ runs } \text{ over } \text{ the } \text{ family } \text{ of } \text{ all } \text{ Young } \text{ functions. } \end{aligned}$$

(3.1)

Now we can state the main result in this section.

Theorem 3.4

Assume that a measure $m:\Sigma \rightarrow X$ has the Radon-Nikodym property with the density $g\in L^1(X)$. Then there exists a Young function $\varphi$ such that $g\in E^{\varphi ^*}(X)$ and the following statements hold:

• (i) The operator $S_g:L^\varphi \rightarrow X$ defined by

  $$\begin{aligned} S_g(u):=\int _\Omega u(\omega )\, g(\omega )\, d\lambda \ \ \text{ for } \text{ all } \ u\in L^\varphi \end{aligned}$$

  is $(\gamma _\varphi ,\Vert \cdot \Vert _X)$-continuous and compact.

• (ii) $S_g$ is $\gamma _\varphi$-compact, that is, there exists a Young function $\psi$ with $\psi \ll \varphi$ such that $\{\int _\Omega u(\omega )g(\omega )\,d\lambda : u\in L^\varphi ,\Vert u\Vert _\psi \le 1\}$ is a relatively norm compact subset of X.

• (iii) $|m|_{\varphi ^*}(\Omega )<\infty$ and $|m|_{\varphi ^*}(A)=\Vert \mathbb {1}_A g\Vert ^0_{\varphi ^*}$ for every $A\in \Sigma$.

• (iv) $|m|_{\varphi ^*}$ is $\lambda$-continuous, that is, $|m|_{\varphi ^*}(A_n)\rightarrow 0$ if $\lambda (A_n)\rightarrow 0$.

Proof

In view of (3.1) there exists a Young function $\psi$ such that $g\in E^\psi (X)$, that is, $\Vert g(\cdot )\Vert _X\in E^\psi$. Hence $|u| \ \Vert g(\cdot )\Vert _X\in L^1$ for every $u\in L^{\psi ^*}$. Let us put $\varphi =\psi ^*$. Then $\varphi ^*=\psi ^{**}=\psi$ and $(L^\varphi )'=L^{\varphi ^*}=L^\psi$ and $L^{\psi ^*}=L^\varphi$, and so $g\in E^{\varphi ^*}(X)$. Then for $u\in L^\varphi$, we have

$$\begin{aligned} \Vert S_g(u)\Vert _X\le \int _\Omega |u(\omega )| \ \Vert g(\omega )\Vert _X \,d\lambda =F_{\Vert g(\cdot )\Vert _X}(|u|), \end{aligned}$$

where $\Vert g(\cdot )\Vert _X\in E^{\varphi ^*}$. The inequality shows that $S_g$ is $(|\sigma |(L^\varphi ,E^{\varphi ^*}),\Vert \cdot \Vert _X)$-continuous, where $|\sigma |$ denotes the absolute weak topology. Since $(|\sigma |(L^\varphi ,E^{\varphi ^*})$ is the coarsest locally convex-solid topology on $L^\varphi$ with dual $E^{\varphi ^*}$, and by Theorem 3.3(ii) $\gamma _\varphi$ is such a topology, it follows that $S_g$ is $(\gamma _\varphi ,\Vert \cdot \Vert _X)$-continuous.

To show that $S_g$ is compact, choose a sequence $(f_n)$ of X-valued simple functions on $\Omega$ such that $\Vert f_n(\omega )-g(\omega )\Vert _X\rightarrow 0$ and $\Vert f_n(\omega )\Vert _X\le \Vert g(\omega )\Vert _X$ $\lambda$-a.e. and for all $n\in {\mathbb {N}}$ (see [7, Theorem 6, p. 4]). Hence for every $\alpha >0$, we have that $\varphi ^*(\alpha \Vert f_n(\omega )-g(\omega )\Vert _X)\rightarrow 0$ $\lambda$-a.e. and $\varphi ^*(\alpha \Vert f_n(\omega )-g(\omega )\Vert _X) \le \varphi ^*(2\alpha \Vert g(\omega )\Vert _X)$ $\lambda$-a.e. and for all $n\in {\mathbb {N}}$, where $\int _\Omega \varphi ^*(2\alpha \Vert g(\omega )\Vert _X)\,d\lambda <\infty$. By the Lebesgue dominated convergence theorem, $\int _\Omega \varphi ^*(\alpha \Vert f_n(\omega )-g(\omega )\Vert _X)\,d\lambda \rightarrow 0$ and this means that $\Vert f_n-g\Vert ^0_{\varphi ^*}\rightarrow 0$.

For each $n\in {\mathbb {N}}$, let $S_n:L^\varphi \rightarrow X$ be a linear operator defined by

$$\begin{aligned} S_n(u):=\int _\Omega u(\omega )f_n(\omega )\,d\lambda \ \ \text{ for } \text{ all } \ u\in L^\varphi . \end{aligned}$$

Note that the range of each $S_n$ is contained in the span of finite set of values of $f_n$. Therefore $S_n$ is compact and by the Hölder’s inequality for each $u\in L^\varphi$, we have

$$\begin{aligned} \begin{array}{l} \displaystyle \Vert (S_n-S_g)(u)\Vert _X=\bigg \Vert \int _\Omega u(\omega )(f_n(\omega )-g(\omega ))\, d\lambda \,\bigg \Vert _X \\ \quad \displaystyle \le \int _\Omega |u(\omega )|\ \Vert f_n(\omega )-g(\omega )\Vert _X\, d\lambda \le \Vert u\Vert _\varphi \ \Vert f_n-g\Vert ^0_{\varphi ^*}. \end{array} \end{aligned}$$

It follows that $\Vert S_n-S_g\Vert \rightarrow 0$, so $S_g$ is a compact operator.

(ii) In view of (i) and Proposition 3.2$S_g$ is $\gamma _\varphi$-compact. Using Theorem 3.3 we obtain that (ii) holds.

(iii) Let $s=\sum ^k_{i=1} c_i\mathbb {1}_{A_i}\in {\mathcal {S}}(\Sigma )$ and $\Vert s\Vert _\varphi \le 1$. Then

$$\begin{aligned} \sum ^k_{i=1}|c_i| \ \Vert m(A_i)\Vert _X=\sum ^k_{i=1}|c_i|\int _{A_i} \Vert g(\omega )\Vert _X \,d\lambda = \int _\Omega |s(\omega )| \ \Vert g(\omega )\Vert _X \,d\lambda \le \Vert g\Vert ^0_{\varphi ^*} \end{aligned}$$

and hence $|m|_{\varphi ^*}(\Omega )\le \Vert g\Vert ^0_{\varphi ^*}$. Note that $m=m_{S_g}$. Hence using Lemma 2.1 we have that $|m_{\varphi ^*}|(A)=\Vert \mathbb {1}_A g\Vert ^0_{\varphi ^*}$ for $A\in \Sigma$.

(iv) This follows from (iii) because $g\in E^{\varphi ^*}(X)$. $\square$

Let $i_\infty :L^\infty \rightarrow L^\varphi$ denotes the inclusion map. Note that $i_\infty$ is $(\sigma (L^\infty ,L^1),$ $\sigma (L^\varphi ,L^{\varphi ^*}))$-continuous, and it follows that $i_\infty$ is $(\tau (L^\infty ,L^1),\tau (L^\varphi ,L^{\varphi ^*}))$-continuous (see [8, Theorem 8.6.1]). Since $\gamma _\varphi \subset \tau (L^\varphi ,L^{\varphi ^*})$, we obtain that $i_\infty$ is $(\tau (L^\infty ,L^1),\gamma _\varphi )$-continuous.

As a consequence of Theorem 3.4 and Theorem 2.5, we can show that every $\tau (L^\infty ,L^1)$-nuclear operator $T:L^\infty \rightarrow X$ admits a factorization through some Orlicz space $L^\varphi$.

Corollary 3.5

Assume that $T:L^\infty \rightarrow X$ is a $\tau (L^\infty ,L^1)$-nuclear operator. Then there exists a Young function $\varphi$ such that $T=S\circ i_\infty$, where

• (i) $S:L^\varphi \rightarrow X$ is a Bochner representable and $\gamma _\varphi$-compact linear operator.

• (ii) $|m_S|_{\varphi ^*}(A_n)\rightarrow 0$ if $\lambda (A_n)\rightarrow 0$.

Proof

• (i) In view of Theorem 2.5 the representing measure $m_T:\Sigma \rightarrow X$ has the Radon-Nikodym property with respect to $\lambda$, that is, $m_T=g\lambda$, where $g\in L^1(X)$. Hence according to Theorem 3.4 there exists a Young function $\varphi$ such that $g\in E^{\varphi ^*}(X)$ and an operator $S:L^\varphi \rightarrow X$ defined by

  $$\begin{aligned} S(u):=\int _\Omega u(\omega )\,g(\omega )\,d\lambda \ \text{ for } \text{ all } \ u\in L^\varphi , \end{aligned}$$

  is $\gamma _\varphi$-compact. Note that for $u\in L^\infty$, $T(u)=\int _\Omega u\,dm_T=\int _\Omega u(\omega )g(\omega )\,d\lambda =S(u)$.

• (ii) Since $m_S(A)=\int _A g(\omega )\,d\lambda$ for all $A\in \Sigma$, using Theorem 3.4 we have that $|m_S|_{\varphi ^*}(A_n)\rightarrow 0$ if $\lambda (A_n)\rightarrow 0$.

$\square$