JOSÉ AGUAYO-GARRIDO

Weakly compact operators and the Dunford-Pettis
property on uniform spaces

Annales mathématiques Blaise Pascal, tome 5, n° 2 (1998), p. 1-6

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# Weakly Compact Operators and the Dunford-Pettis Property on Uniform Spaces

JOSÉ AGUAYO-GARRIDO †

July, 1998

**ABSTRACT.** Let $(X, \mathcal{U})$ be a Hausdorff uniform space and $C_b(X)$ the space of all bounded continuous real-valued functions on $X$. The subspace of $C_b(X)$, consisting of the all uniformly continuous functions with respect to $\mathcal{U}$, is denoted by $C_{ub}(X)$. In this paper we give a characterization of weakly compact operators and $\beta_u$-continuous defined from $C_{ub}(X)$ into a Banach space $E$, where $\beta_u$ is the finest locally convex topology agreeing with the pointwise topology on each uniformly equicontinuous and bounded subsets of $C_{ub}(X)$. We also show that $(C_{ub}(X), \beta_u)$ has the Strict Dunford-Pettis Property and the Dunford-Pettis Property, both under special conditions.

## 1. INTRODUCTION AND NOTATIONS

All uniform spaces $(X, \mathcal{U})$ are assumed to be Hausdorff uniform spaces. Basic references for the measures theory on topological spaces are in Varadarajan [7]. We will denote by $C_b(X)$ the space of all real-valued bounded continuous functions defined on $X$, and $C_{ub}(X)$, the subspace of $C_b(X)$, consists of those functions which are uniformly continuous. $\mathcal{H}$ will denote the collection of all uniformly equicontinuous and bounded (U.E.B.) subsets of $C_{ub}(X)$. $\beta_u$ will denote the finest locally convex topology agreeing with the pointwise topology on each $H \in \mathcal{H}$. A uniform measure on $X$ is defined to be a bounded linear functional on $C_{ub}(X)$ which is pointwise continuous on each $H \in \mathcal{H}$ (see [1], [2], [4]) and the space of all uniform measures will be denoted by $M_u(X)$. It is well known that the dual of $(C_{ub}(X), \beta_u)$ is $M_u(X)$.

Let $M(X)$ be the dual of $(C_{ub}(X), \|\cdot\|)$, where the $\|\cdot\|$ denotes the supremum norm. We denote by $\mathcal{H}$-top the locally convex topology on $M(X)$ of uniform convergence on the U.E.B. sets. Let $M_d(X)$ be the subspace of $M(X)$ generated by the Dirac measures. It was proved in [4] that the $\mathcal{H}$-top closure of $M_d(X)$ is the space $M_u(X)$.

We denote by $\beta_t$ the locally convex topology on $C_b(X)$ agreeing with the compact-open topology on each norm-bounded subset of $C_b(X)$. Sentilles [6] proved that the

<footnote>† 1991 Mathematics Subject Classification. Primary 46E10, 47B38. Secondary 46E05</footnote>

<footnote>This research is supported by Fondecyt N°1950546 and Proyecto No. 98.015.013-1.0, Dirección de Investigación, Universidad de Concepción.</footnote>
dual of $(C_b(X), \beta_t)$ is the space $M_t(X)$ of all tight measures on $X$. One of the
Sentilles's results that we will use here is the following: "a subset $A$ of $M_t(X)$ is
uniformly tight if, and only if, $A$ is $\beta_t$-equicontinuous. It is also known that $C_{ub}(X)$
is $\beta_t$-dense on $C_b(X)$ (see [1]).

The proof of the following technical lemma is a simple verification and it will be
omitted..

**Lemma 1.** The canonical mapping $\Phi : (C_{ub}(X), \beta_u) \to (C_{ub}(\hat{X}), \beta_u)$ defined by $f \to \hat{f}$ is an isomorphism, where $\hat{X}$ denotes the completion of $(X, \mathcal{U})$ and $\hat{f}$ is the unique uniform extension of $f$ to $\hat{X}$.

## 2. WEAKLY COMPACT OPERATORS

Let $E$ be a Banach space. In this section we will study $E$-valued linear norm-
continuous operators on $C_{ub}(X)$, in particular, weakly compact operators.

**Definition 1.** Let $T$ be a $E$-valued linear continuous operator on $C_{ub}(X)$. We shall say that $T$ is a tight additive operator if its restriction to the unit ball of $C_{ub}(X)$ is continuous for the compact-open convergence topology.

Note that $T$ is tight additive if, and only if, $T$ is $\beta_t$-continuous and so, by the density of $C_{ub}(X)$ in $C_b(X)$ via the topology $\beta_t$, $T$ has a unique continuous extension to $C_b(X)$.

From now on we will assume that $T: C_{ub}(X) \to E$ is a weakly compact operator, that is, $T$ transforms the unit ball of $C_{ub}(X)$ into a relatively weakly compact subset of $E$. By the denseness, $(C_{ub}(X), \beta_t)' = (C_b(X), \beta_t)' = M_t(X)$, the space of all tight measures (see [6]), and by the weakly compactness of $T$, $T''(C_{ub}(X)'' \subset E$. On the other hand, $T$ has a unique $\beta_t$-continuous extension $\tilde{T}$ to $C_b(X)$ and, by the latter,

$$ \tilde{T}''(C_b(X)'' ) = \tilde{T}''(M_t(X)') = T'''(C_{ub}(X)'' ) \subset E. $$

Then, $\tilde{T}$ is also a weakly compact operator.

The following theorem will give a characterization of tight operators and the proof is based on the well known result which says that $\beta_u$ and the norm topology have the same bounded sets (see [1]).

**Theorem 2.** Suppose that $\mathcal{U}$ is metrizable and $(X, \mathcal{U})$ is complete. Then, the following statements are equivalent:

(a) $T$ is $\beta_u$-continuous

(b) $T$ is tight additive
**Proof.** Since $(X, \mathcal{U})$ is metrizable and complete, we have that the pointwise topology and the compact-open topology coincide on each U.E.B. subset of $C_{ub}(X)$. (a) ⇒ (b) Since $T$ is $\beta_u$-continuous, we have that $\{x' \circ T : \|x'\| \le 1\} \subset M_u(X)$. On the other hand, since $T$ is also weakly compact operator,

$$T' (\{x' : \|x'\| \le 1\}) = \{x' \circ T : \|x'\| \le 1\}$$

is relatively $\sigma(M(X), M(X'))$-compact. By this fact and since $M(X)$ is an AL-space, the closure of solid hull of $\{x' \circ T : \|x'\| \le 1\}$ is $\sigma(M(X), M(X'))$-compact (see Cor. 8.8, p. 119, [5]). Since $\sigma(M(X), M(X'))$ is finer than $\sigma(M(X), C_{ub}(X))$, we have that $\{|x' \circ T| : \|x'\| \le 1\}$ is relatively $\sigma(M(X), C_{ub}(X))$-compact. Then, by [1], p. 239, $\{|x' \circ T| : \|x'\| \le 1\}$ is a tight set.

Now, take a net $(f_\alpha)_{\alpha \in I}$ in the unit ball of $C_{ub}(X)$ converging to 0 in the topology of compact-open convergence; hence by the above and the tightness of

$$\{|x' \circ T| (f_\alpha) : \|x'\| \le 1\},$$

we have $|x' \circ T(f_\alpha)| \le |x' \circ T| (f_\alpha) \to 0$ uniformly for $\|x'\| \le 1$. This argument shows that $\|Tf_\alpha\| \to 0$.

(b) ⇒ (a) By the tightness of $T$, its restriction to each U.E.B. set is continuous for the topology of compact-open convergence. But, this topology coincides with the topology of pointwise convergence on each U.E.B. set; therefore $T$ is $\beta_u$-continuous.

In the next theorem we shall use the following notations: If $(X, \mathcal{U})$ is a uniform space and $d$ is uniformly continuous pseudometrics (u.c.p.) on $X$, then $\hat{X}_d$ denotes the completion of the metric space which comes from $X$, $d$ and the corresponding projection, $\pi_d$.

**Theorem 3.** Let $T$ be a weakly compact $E$-valued operator defined on $C_{ub}(X)$. Then, the following statements are equivalent:

1. $T$ is $\beta_u$-continuous

2. $T|_H$ is pointwise continuous for each U.E.B. set $H$.

3. $\{x' \circ T : x' \in E'; \|x'\| \le 1\}$ is $\beta_u$-equicontinuous.

4. $\{|x' \circ T| : x' \in E'; \|x'\| \le 1\}$ is relatively $\sigma(M_u(X), C_{ub}(X))$-compact

5. For each u.c.p. $d$ on $X$, $\pi_d \circ T$ (natural definition) is a tight additive operator.
**Proof.** The equivalences (3) ⇔ (1) ⇔ (2) are clear. The equivalence (3) ⇔ (4) follows from [1, p. 228 and 241].

(1) ⇒ (5) Let $d$ be a $u.c.p.$ on $X$. Since $\pi_d$ is uniformly continuous, we have that $\pi_d \circ T : C_{ub}(X) \to E$, defined by $(\pi_d \circ T)(\hat{f}) = T(\hat{f} \circ \pi_d)$, is $\beta_u$-continuous and a weakly compact operator. Therefore, by Th. 2.2, $\pi_d \circ T$ is a tight additive operator.
(5) ⇒ (1) Let $H \in \mathcal{H}$; hence $d_H(x, y) = \sup \{|f(x) - f(y)| : f \in H\}$ is a $u.c.p.$ on $X$. Denote by $\pi_H$ the corresponding projection of $d_H$ and defined by $\hat{f}(\pi_H(x)) = f(x)$, for any $f \in H$. It is not difficult to see that the function $\hat{f}$ is well defined, it belongs to $C_{ub}(X_{d_H})$ and the $\hat{H} = \{\hat{f} : f \in H\}$ is a U.E.B. subset of $C_{ub}(X_{d_H})$.

Take a net $(f_\alpha)_{\alpha \in I}$ in $H$ such that $f_\alpha \to 0$ pointwise. It easily follows that $\hat{f}_\alpha \to 0$ pointwise and, from the hypothesis, $\|(\pi_d \circ T)(\hat{f}_\alpha)\| \to 0$. Therefore, since $\|Tf\| = \|(\pi_d \circ T)(\hat{f})\|$, for any $f \in H$, we get that $T$ is $\beta_u$-continuous.

### 3. DUNFORD-PETTIS AND STRICT DUNFORD-PETTIS PROPERTY

In this section we will analyze the Strict Dunford-Pettis and the Dunford-Pettis Property of the locally convex space $(C_{ub}(X), \beta_u)$. We begin with the definition of these properties which were given by Grothendieck in his well known paper "Sur les applications linéaires faiblement compact d'espace du type $C(K)$", Canad. J. Math. 5(1974), 183-201.

**Definition 2.** We shall say that a Hausdorff locally convex space $E$ has the Dunford-Pettis Property (resp. Strict Dunford-Pettis Property) if for any Banach space $F$ and every linear continuous and weakly compact operator $T: E \to F$, $T(C)$ is relatively compact (resp. $\{T(x_n)\}$ is Cauchy) in $F$ for any absolutely convex weakly compact subset $C$ (resp. weak Cauchy sequence $\{x_n\}$) in $E$.

**Theorem 4.** If $\mathcal{U}$ is metrizable, then $(C_{ub}(X), \beta_u)$ has the Dunford-Pettis and Strict Dunford-Pettis Properties.

**Proof.** First we shall assume that $(X, \mathcal{U})$ is a complete metrizable uniform space. Let $T$ be a weakly compact and $\beta_u$-continuous linear operator defined from $C_{ub}(X)$ into an arbitrary Banach space $F$. By Th. 2.2, $T$ is a tight operator and then $T$ admits a unique extension $\tilde{T}$ to $C_b(X)$ which is $\beta_t$-continuous.

We shall first prove that $(C_{ub}(X), \beta_u)$ has the Strict Dunford-Pettis Property. Let $\{f_n\}_{n \in \mathbb{N}}$ be a $\sigma(C_{ub}(X), M_u(X))$-Cauchy sequence in $C_{ub}(X)$. Since $M_u(X) =$
$M_t(X)$ and $\{f_n\}_{n \in \mathbb{N}}$ is in $C_b(X)$, we have that this sequence is $\sigma(C_b(X), M_t(X))$ - Cauchy. Now, by [3], we know that $(C_b(X), \beta_t)$ has the Strict Dunford-Pettis Property, therefore $\{Tf_n\}_{n \in \mathbb{N}} = \{\tilde{T}f_n\}_{n \in \mathbb{N}}$ is convergent in $F$.
If $\mathcal{U}$ is metrizable, then the conclusion follows from Lemma 1.1, since

$$T \circ \Phi : C_{ub}(\hat{X}) \to F$$

is $\beta_u$-continuous and weakly compact operator.

Now, to prove that $(C_{ub}(X), \beta_u)$ has the Dunford-Pettis Property, we again assume that $(X, \mathcal{U})$ is complete metrizable uniform space and follows the similar argument given above.

One of the open problems that we still face, is whether or not $(C_{ub}(X), \beta_u)$ has the Strict Dunford-Pettis Property. We already proved that the answer is yes if $\mathcal{U}$ is metrizable. In the next theorem we will assume that $(C_{ub}(X), \beta_u)$ has the Strict Dunford-Pettis Property and we will prove that it has the Dunford-Pettis Property under the condition that $X$ is $\sigma$-compact.

**Theorem 5.** If $(C_{ub}(X), \beta_u)$ possesses the Strict Dunford-Pettis Property and $X$ is $\sigma$-compact, then $(C_{ub}(X), \beta_u)$ has the Dunford-Pettis Property.

**Proof.** Let $\{K_n\}_{n \in \mathbb{N}}$ be an increasing sequence of compact subsets of $X$ such that $\bigcup_{n=1}^{\infty} K_n$ is dense on $X$. We will denote by $L_n$ the closed absolutely convex hull of $K_n$ in $M_u(X)$ ($X$ is a uniform subspace of $M_u(X)$). Since $K_n$ is a compact subset of $M_u(X)$ in the $\mathcal{H}$-top and $(M_u(X), \mathcal{H}$-top) is complete, we have that $L_n$ is a compact subset of $M_u(X)$. Moreover, $\{L_n\}_{n \in \mathbb{N}}$ is an increasing sequence.

We claim that $\bigcup_{n=1}^{\infty} L_n$ is $\mathcal{H}$-top dense in $M_u(X)$. In fact, take $\mu \in M_u(X)$ and a balanced neighborhood $V$ of $\mu$. Since $M_d(X)$ is $\mathcal{H}$-top dense in $M_u(X)$, $V$ contains some element $\nu = \sum_{i=1}^{p} \alpha_i \delta_{x_i}$ of $M_d(X)$, with $x_1, x_2, \dots, x_p \in X$. Suppose that $0 < \alpha = \sum_{i=1}^{p} |\alpha_i| \le 1$ (if $\alpha = 0$, $V \cap (\bigcup_{n=1}^{\infty} L_n) \ne \emptyset$ and we are done) and take neighborhoods $W_i$ of $\delta_{x_i}$, $i = 1, 2, \dots, p$, such that $\sum_{i=1}^{p} \alpha_i W_i \subset V$. Since $W_i \cap X$ is a neighborhood of $\delta_{x_i}$ in $X$, we get $\delta_{y_i} \in K_{n_i}$ such that $\delta_{y_i} \in W_i \cap X$. Thus, $\sum_{i=1}^{p} \alpha_i \delta_{y_i} = \sum_{i=1}^{p} \alpha_i W_i \subset V$ and $\sum_{i=1}^{p} \alpha_i \delta_{y_i} \in L_N$, where $N = \max\{n_i : i = 1, 2, \dots, p\}$. Therefore, $V \cap (\bigcup_{n=1}^{\infty} L_n) \ne \emptyset$.

Suppose now that $\alpha > 1$; hence $\alpha \sum_{i=1}^{p} \frac{\alpha_i}{\alpha} \delta_{x_i} \in V$ and so $\sum_{i=1}^{p} \frac{\alpha_i}{\alpha} \delta_{x_i} \in \frac{1}{\alpha}V \subset V$. Applying
the above argument to $\frac{1}{\alpha}V$, we get $V \cap (\bigcup_{n=1}^{\infty} L_n) \neq \emptyset$.

From this, we have that $(M_u(X), \mathcal{H}-\text{top})$ is a $\sigma$-compact space, which implies that $(M_u(X), \sigma(M_u(X), C_{ub}(X)))$ is also a $\sigma$-compact space. The conclusion of the theorem follows from [3], Th. 1.

## REFERENCES

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[2] M. Z. Frolik, Measures uniformes, C. R. Acad. Sci., Paris, 277 (1973),105-108.

[3] S. S. Khurana, Dunford-Pettis Property, J. Math. Anal. Appl., 65 (1978), 361-364.

[4] J. Pachl, Measures as functional of uniformly continuous functions, Pacific J. Math., 82 (1979), 515-521.

[5] H. H. Schaefer, Banach Lattices and Positives Operators, Springer-Verlag, Berlin, Heidelberg, New York, 1974.

[6] F. Sentilles, Bounded Continuous Functions on a Completely Regular Spaces, Trans. Amer. Math. Soc., V. 168(1972).

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