Proof

1. (1) Let $(x_n)$ and $(f_n)$ be two sequences in X and $Z'$ , respectively, such that $x_n\overset {\sigma (X,X')}{\longrightarrow } 0$ and $f_n\overset {\sigma (Z',Z)}{\longrightarrow } 0$ . Remark that $Tx_n\overset {\sigma (Y,Y')}{\longrightarrow } 0$ and $f_n \circ S\overset {\sigma (Y',Y)}{\longrightarrow } 0$ . Hence,

   • • if T is weak^* Dunford–Pettis, then $(f_n \circ S)\circ T(x_n)\longrightarrow 0$ .

   • • if S is weak^* Dunford–Pettis, then $f_n \circ S ( T(x_n))\longrightarrow 0$ .

   In both cases, we have $f_n \circ (S\circ T)(x_n)\longrightarrow 0$ , so $S\circ T$ is weak^* Dunford–Pettis.

2. (2) It is a consequence of the first property of the current Lemma 2.1 and [Reference El Kaddouri, H’michane, Bouras and Moussa13, Corollary 3.3] since $T=id_Y\circ T=T\circ id_X $ .

3. (3) It is a direct consequence of (1) and (2).

Proof Assume that the norm of $E'$ is not order-continuous and $F\neq \{0\}$ and consider $0 < y \in F^+$ . It follows from [Reference Meyer-Nieberg18, Theorem 2.4.14 and Proposition 2.3.11] the existence of a sublattice of E isomorphic to $\ell ^1$ , and hence also a positive projection $P : E \rightarrow \ell ^1$ . Consider the positive operator S defined by

$$ \begin{align*}\begin{array}{llll} S :& \ell^1& \rightarrow &F \\ & (\lambda_n)_n & \mapsto &= (\sum\limits_{n=1}^{+\infty}\lambda_n)y \end{array}\end{align*} $$

and put $T= S\circ P$ . According to Lemma 2.1, T is a positive weak^* Dunford–Pettis (because $ \ell ^1$ has DP*property). However, T is not M-weakly compact. Indeed, if $(e_n)_n$ is the canonical basis of $\ell ^1$ , then that $T(e_n)=y \nrightarrow 0$ .

Proof Assume that the norm of E is not order-continuous, and that there exists a lattice embedding $S : \ell ^\infty \rightarrow F$ from $\ell ^\infty $ to F. It follows from [Reference Aliprantis and Burkinshaw2, Theorem 4.14] that there exist $y \in E^+$ and a disjoint sequence $(x_n)\in [0, y]$ which does not converge to zero in norm. We may assume that $\Vert x_n\Vert = 1$ for all $n\in \mathbb {N}^*$ . Hence, by [Reference Aqzzouz, Elbour and H’michane3, Lemma 3.4], there exists a positive disjoint sequence $(g_n)$ of $E^+$ with $\Vert g_n\Vert = 1$ such that $g_n(x_m) =1$ if $n = m$ and $g_n(x_m) =0$ if $n \neq m$ for every $m,n\in \mathbb {N}^*$ . We define the positive operator P as follows:

$$ \begin{align*}\begin{array}{llll} P : & E & \rightarrow & \ell^\infty\\ & x &\mapsto & P(x) = (g_n(x))_n \end{array}\end{align*} $$

and put $T = S\circ P $ . Since $ \ell ^\infty $ has DP* property, according to Lemma 2.1, T is a positive weak^* Dunford–Pettis. However, it is not order weakly compact, because $(x_n)$ is an order bounded disjoint sequence in E, but $\Vert T(x_n)\Vert \nrightarrow 0$ .

Proof Let E be a separable and Dedekind $\sigma $ -complete Banach lattices. According to [Reference Aliprantis and Burkinshaw2, Exercise 6, p. 337], every operator from E to $\ell ^\infty $ is order weakly compact, and then, by corollary 2.3, E has an order-continuous norm.

Proof Let $(x_n)$ be a norm bounded disjoint sequence of $E^+$ . We have to show that $\lim \Vert Tx_n\Vert = 0$ . To this end, assume by way of contradiction that $\lim \Vert Tx_n\Vert \neq 0$ . Then, there exist some $\varepsilon> 0$ and a subsequence $ (y_n)$ of $(x_n)$ satisfying $\Vert Ty_n\Vert>\varepsilon $ for all $n\in \mathbb {N}^*$ . For each n, there exists a continuous linear functional $f_n$ on X such that $\Vert f_n\Vert =1$ and

(2.1) $$ \begin{align} f_n\left( Ty_n\right)=\Vert Ty_n\Vert>\varepsilon. \end{align} $$

Since X is a separable Banach space, then $(f_n)$ has a weak^*-convergent subsequence $\left (f_{\phi (n)}\right )$ (see [Reference Aliprantis and Burkinshaw2, Exercise 7, p. 167]). Let $f\in X'$ be its weak^*-limit, i.e., $f_{\phi (n)} \overset {\sigma (X',X)}{\longrightarrow } f$ . On the other hand, since $E'$ has an order-continuous norm and $(y_n)$ is disjoint, it follows from [Reference Meyer-Nieberg18, Corollary 2.4.3] that $y_n \overset {\sigma (E,E')}{\longrightarrow } 0$ and hence $f\left ( Ty_n\right ) \longrightarrow 0$ . Finally, since T is weak^* Dunford–Pettis, $ (f_{\phi (n)}- f) (Ty_{\phi (n)} ) \longrightarrow 0$ , and consequently we have $f_{\phi (n)} (Ty_{\phi (n) }) \longrightarrow 0$ , which contradicts (2.1). Therefore, $\lim \Vert Tx_n\Vert = 0$ must hold true.

Proof Let $T:E\rightarrow \mathcal {C}(K)$ be a weak^* Dunford–Pettis operator, and let $(x_n)_n$ be a bounded disjoint sequence in E. We have to prove that $\lim \limits _{n} \Vert T(x_n) \Vert =0$ . Assume that $\left (\Vert T(x_n) \Vert \right )$ does not converge to $0$ . In this situation, we can suppose that there exists a sequence $(t_n)\in K^{\mathbb {N}}$ such that $\vert T(x_n)(t_n) \vert \geq 1$ for every $n\in \mathbb {N}$ . Next, since K is strictly semi-compact, the sequence $(t_n)$ has a subsequence (that we denote always $(t_n)$ ) which converges to $t\in K$ . Hence, the sequence $(\delta _{t_n}-\delta _{t})$ converges to $0$ for the topology $\sigma \left (F',F \right )$ , where $F=\mathcal {C}(K)$ and $\delta _{t_n},\delta _{t}$ are the Dirac measures. Once again, as the norm of $E'$ is order-continuous, the sequence $(x_n)$ converges weakly to $0$ . Hence, $\delta _{t}(T(x_n))\longrightarrow 0$ , and since T is weak^* Dunford–Pettis, $\left (\delta _{t_n}-\delta _{t}\right )\left (T(x_n)\right )\longrightarrow 0$ . We deduce then that $\delta _{t_n}(T(x_n))\longrightarrow 0$ , which presents a contradiction, because $1\leq \vert T(x_n) (t_n)\vert $ for every $n\in \mathbb {N}$ .

Proof

• $(1) \Rightarrow (2)$ Let $S, T : E\rightarrow F$ be two operators such that $0 \leq S \leq T$ and T is weak^* Dunford–Pettis. So, T is M-weakly compact and hence it is clear that S is M-weakly compact.

• $(2) \Rightarrow (1)$ Obvious.

Proof By definition, there exists a lattice embedding $T : E_1 \rightarrow E$ with $E_1=l^\infty $ (resp. $E_1=c_0$ ). Denote by $(e_n)$ the canonical basis of $E_1$ and put $v_n=T(e_n)$ . Then $\left ( v_n \right )$ is a bounded disjoint sequence in E, and by [Reference Aqzzouz, Elbour and H’michane3, Lemma 3.4], there exists a positive bounded and disjoint sequence $\left (f_n \right )$ in $E'$ such that for all $m,n\in \mathbb {N}$ , we have

$$ \begin{align*}f_n(v_m)=\begin{cases} 1, \text{ if } m=n, \\ 0,\text{ if } m\neq n.\end{cases}\end{align*} $$

Now, the operator

$$ \begin{align*}\begin{array}[t]{llll} P:& E &\longrightarrow & l^\infty\\ & x & \mapsto & (f_n(x)) \end{array}\end{align*} $$

answers the question.

Proof

• • Step 1. In [Reference El Fahri, H’michane, El Kaddouri and Aboutafail12, Theorem 2.2], it has been proved that if every positive weak^* Dunford–Pettis operator $T:E\rightarrow F$ is weakly compact, then either the norm of $E^{\prime }$ is order-continuous or F is reflexive. We repeat here this proof. Assume that the norm of $E'$ is not order-continuous. As before E, there exists a positive projection $P : E \rightarrow \ell ^1$ . Consider the operator $S : \ell ^1 \rightarrow F$ defined by $S((\alpha _i)) =\sum \limits _{i=1}^{+\infty }\alpha _ix_i$ for each $(\alpha _i) \in \ell ^1$ . The composed operator $T = S\circ P $ is weak^* Dunford–Pettis and hence by our hypothesis T is weakly compact. So, the sequence $(x_n) = (T(e_n))$ has a subsequence which converges weakly to an element of F, where $(e_n)$ is the canonical basis of $ \ell ^1$ . So, by the Eberlein–Smulian’s theorem, F is reflexive.

• • Step 2. Suppose by the way of contradiction that $l^{\infty }$ is lattice embeddable in F and E is not reflexive. According to the Lozanovsky theorem [Reference Aliprantis and Burkinshaw2, Theorem 4.71], $l^1$ or $c_0$ is lattice embeddable in E. If $l^1$ is lattice embeddable in E, then the norm of $E'$ is not order-continuous, in this case, by Step 1 of the current proof, F is reflexive, which contradicts the fact that $l^{\infty }$ is lattice embeddable in F. So, assume that $c_0$ is lattice embeddable in E and apply Lemma 3.1. Then there exist positive operators $T_F:l^\infty \rightarrow F$ , $P_F : F \rightarrow l^\infty $ , $T_E:c_0\rightarrow E$ , and $P_E : E \rightarrow l^\infty $ so that $P_F\circ T_F$ is the identity operator of $l^\infty $ and $P_E\circ T_E$ is the canonical injection from $c_0$ into $l^\infty $ . Since $l^\infty $ has the Dunford–Pettis* property, the composite operator $S:=T_F\circ P_E$ is weak^* Dunford–Pettis, but it fails to be weakly compact because the composite $P_F\circ S\circ T_E=P_F\circ T_F\circ P_E\circ T_E$ is the canonical injection from $c_0$ to $l^\infty $ which is not weakly compact.

Proof The direct implication follows from the previous theorem, since if F is Dedekind $\sigma $ -complete, its norm is order-continuous if and only if $\ell ^\infty $ is not lattice embeddable in F (see [Reference Aliprantis and Burkinshaw2, Theorem 4.51 (Lozanovsky–Mekler–Meyer-Nieberg)]).

For the converse, if E or F is reflexive, the result is obvious, and if the norms of $E'$ and F are order-continuous, the result follows from Theorem 2.2 of [Reference El Fahri, H’michane, El Kaddouri and Aboutafail12].

Proof

• $(1)\Rightarrow (2)$ In fact, in this case, according to [Reference Aliprantis and Burkinshaw2, Theorem 5.24], every operator from E to F is weakly compact.

• $(2)\Rightarrow (4)$ Obvious.

• $(4)\Rightarrow (3)$ Obvious.

• $(3)\Rightarrow (1)$ According to [Reference Aliprantis and Burkinshaw2, Exercise 15, p. 254], $\ell ^1$ is lattice embeddable in E; hence, the norm of $E'$ is not order-continuous and consequently, from Corollary 3.2, F is reflexive.

Proof This follows from Theorems 3.2 and 2.7.

Proof By the preceding corollary, the norm of $\mathcal {C}(K)$ is order-continuous. Since its closed unit ball is an order interval, then it is weakly compact. Consequently, $\mathcal {C}(K)$ is reflexive, and therefore it has a finite dimension [Reference Aliprantis and Burkinshaw2, Exercise 14, p. 253].

Proof In this situation, each operator $T:E\rightarrow c_0$ is weak $^*$ Dunford–Pettis and hence weakly compact. As consequence, E has the positive Grothendieck property (see [Reference Witold21, p. 762]).

Proof

1. (1) Suppose that the norm of $E'$ is order-continuous and consider a weak^* Dunford–Pettis operator $T:E\rightarrow Y$ . Let $(x_n)$ be a bounded and disjoint sequence in E and $(f_n)$ a weakly convergent sequence in $Y'$ . Since the norm of $E'$ is order-continuous, according to [Reference Meyer-Nieberg18, Corollary 2.4.3], we have $x_n\overset {\sigma (E,E')}{\rightarrow } 0$ . Hence, if we put f the weak limit of $(f_n)$ , then $f\circ T(x_n)\rightarrow 0$ and $(f_n-f)\circ T (x_n)\rightarrow 0$ . Consequently, we have $f_n\circ T (x_n)\rightarrow 0$ .

2. (2) Suppose that the norm of $E'$ is not order-continuous. Then $(E')^{a}\subsetneq E'$ , and therefore there exists $f\in E'$ such that $f\notin (E')^{a}$ . Fix $y\in Y$ nonnull and consider the operator

   $$ \begin{align*}\begin{array}{llll} T : & E & \longleftrightarrow & Y\\ & x & \longmapsto & f(x).y. \end{array}\end{align*} $$

   Let $(x_n)$ be a null weak sequence in E and $(f_n)$ a null weak^* sequence in $Y'$ . Hence, $f_n\circ T (x_n)=f_n(y). f (x_n)\rightarrow 0$ . This proves that T is weak^* Dunford–Pettis. On the other hand, there exists $g\in Y'$ satisfying $g(y)=1$ , so $T'(g)=f \notin (E')^{a}$ . Now, according to [Reference Elbour, Afkir and Sabiri14, Proposition 2], T is not almost M-weakly compact.

Proof

1. (1) Suppose that the norm of F is order-continuous and consider a weak^* Dunford–Pettis operator $T:X\rightarrow F$ . Let $(x_n)$ be a weakly convergent sequence in X, x be its weak limit, and $(f_n)$ be a bounded and disjoint sequence in F. As above, since the norm of F is order-continuous and according to [Reference Meyer-Nieberg18, Corollary 2.4.3], we have $f_n\overset {\sigma (F',F)}{\longrightarrow } 0$ . Hence, $f_n\circ T(x)\rightarrow 0$ , $f_n\circ T (x_n-x)\rightarrow 0$ , and consequently $f_n\circ T (x_n)\rightarrow 0$ .

2. (2) Suppose that the norm of F is not order-continuous. Again, $F^{a}\subsetneq F$ , and there exists $y\in F$ such that $y\notin F^{a}$ . Fix $f\in X'$ nonnull and consider the operator

   $$ \begin{align*}\begin{array}{llll} T : & X & \longleftrightarrow & F\\ & x & \longmapsto & f(x).y. \end{array}\end{align*} $$

   T is weak^* Dunford–Pettis and $T(X) \nsubseteq F^{a}$ , so according to [Reference Elbour, Afkir and Sabiri14, Proposition 1], T is not almost L-weakly compact.