Let $X$ and $Y$ be complex Banach spaces and $B(X,Y)$ be the Banach space of all bounded operators. An operator $T\in B(X,Y)$ is weakly compact if $T(\{ x\in X;\; \| x\| \leq 1\})$ is relatively compact in the weak topology of $Y$. If $X$ or $Y$ is reflexive, then every operator in $B(X,Y)$ is weakly compact. I guess that the converse holds as well: if every operator in $B(X,Y)$ is weakly compact then either $X$ or $Y$ has to be reflexive. But I cannot find a satisfactory argument for this. I know about some characterizations of weakly compact operators: factorization through a reflexive Banach space or continuity with respect to the right topology in $X$. However it seems to me that there must be a simple argument which forces that either $X$ or $Y$ is reflexive if every operator in $B(X,Y)$ is weakly compact. I am asking if someone knows this simple argument or if there is a paper with an answer to my question.
2 Answers
The fact that each $T\in B(X,Y)$ is weakly compact does not imply $X$ or $Y$ reflexive. For example, every non-weakly compact operator $T:\ell_\infty\to Y$ is an isomorphism on a subspace isomorphic to $\ell_\infty$ (See Prop. 2.f.4 in Classical Banach spaces I, by Lindenstrauss ans Tzafriri).
Thus if $Y$ is a non-reflexive space containing no copy of $\ell_\infty$ (e.g. $\ell_1$, or a separable non-reflexive space) then every $T\in B(\ell_\infty,Y)$ is weakly compact.
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$\begingroup$ Thank you for the answer! My intuition was wrong. Now I see, why I couldn't find an answer to my question in books and papers. However, is there a known general result which says what type of spaces have to be $X$ and $Y$ if every operator in $B(X,Y)$ is weakly compact? $\endgroup$ Commented Dec 17, 2015 at 8:28
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1$\begingroup$ I do not know a general answer; only special cases. Denoting $W(X,Y)$ the weakly compact operators, if $X$ is separable and $L(X,C[0,1])=W(X,C[0,1])$ or $L(X,\ell_\infty)=W(X,\ell_\infty)$, then $X$ is reflexive. If $X$ contains no copies of $\ell_1$ and weakly convergent sequences in $Y$ are norm convergent (like in $\ell_1$) then $B(X,Y)=W(X,Y)$. $\endgroup$ Commented Dec 17, 2015 at 9:54
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$\begingroup$ Thank you once more for your answer and useful information. $\endgroup$ Commented Dec 17, 2015 at 10:15
I don't know if the following helps:
Let $B(X,Y) = W(X,Y)$. Then the following hold.
(i) If there exists a surjection $T\in B(X,Y)$ then $Y$ is reflexive.
(ii) If there exists an injection $T\in B(X,Y)$ with closed range then $X$ is reflexive.
Indeed, denote by $B_r$ and $K_r$ the open and closed ball around $0$, respectively. By the open mapping theorem, $TB_1$ is open. Thus, there exists $r > 0$ such that $K_r\subset TB_1\subset TK_1$, i.e., $K_r$ is weakly compact in $Y$, meaning that $Y$ is reflexive. If $T$ is an injection with closed range then a similar argument as above shows that $\operatorname{ran} T$ is reflexive. But as $T : X\to\operatorname{ran} T$ admits a bounded inverse, $X$ is reflexive.
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$\begingroup$ Thank you for those results. Any new information about weakly compact operators is welcome. Although that I am not studying these operators somehow they have appeared in my research work. I am especially interested in the question when $B(X,Y)=W(X,Y)$. It seems that there is no complete answer to this question in literature. $\endgroup$ Commented Dec 28, 2015 at 12:34