Is weakly continuous operator weakly compact?

Question

Suppose we are given an operator $T\colon X^*\to Y$ and both $X$ and $Y$ are Banach spaces. Assume that operator $T$ is continuous for the weak$^*$ topology of $X^*$ and weak topology of $Y$. Does it imply that operator $T$ is weakly compact?

Answer 1

Let $S^*$ be a unit ball of $X^*$. As $X$ is a Banach space, then so is its dual. Hence, from Banach-Alaoglu theorem, we deduce that $S^*$ is weakly$^*$ sequentially compact. Now, since $T$ is continuous from $X^*$ endowed with weak$^*$ topology to $X$ with weak topology, we deduce that $T(S^*)$ is weakly compact.