Weakly compact operators between Banach spaces 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.
 A: 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.
A: 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.
