# Is weakly continuous operator weakly compact?

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?

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.