Is the following true?


Let $X$ and $Y$ be separable Banach spaces and consider their dual spaces $X^\*$ and $Y^\*$ equipped with weak\* topology. Suppose that a linear map $T:X^\*\to Y^\*$ is sequentially continuous. Then, is it true that $T$ is continuous?


I suspect this may be true using the following two facts:


(i) Any closed ball in $X^\*$ (and in $Y^\*$) is metrizable as $X$ and $Y$ are separable.


(ii) A convex set in $X^\*$ (and in $Y^\*$) is weak\* closed iff it is sequentially closed (by Krein-Smulian theorem).


From (i), we have that $T$, when restricted to a closed ball, is continuous. But I am not able to go further from this.


Any help or hint would be appreciated. Thank you.



Sequential Continuity in dual spaces of separable Banach Spaces
This is true. To show it, in the following I will use $\langle \mbox{-}, \mbox{-} \rangle$ for the pairing between a space and its dual (with the vectors from the space on the left, the dual on the right).


For each $y \in Y$, if we define $f(y) = \langle y, \mbox{-}\rangle \circ T$, then $f(y)$ is a sequentially continuous linear functional on $X^\*$. Its kernel is therefore a sequentially closed convex set, so is weak-\* closed in $X^\*$ (by, according to taste, Krein-Šmulian or Banach-Dieudonne or Grothendieck's completeness theorem), and therefore $f(y)$ is actually weak-\* continuous. Therefore there exists a $g(y) \in X$ such that $\langle g(y), \mbox{-}\rangle = f(y)$. It is then not difficult to show that $g(y)$ is a linear map $Y \rightarrow X$ such that $\langle g(y), \phi \rangle = \langle y, T(\phi) \rangle$ for all $y \in Y$ and $\phi \in X^\*$. This implies that $T$ is continuous from the weak-\* of $X^\*$ to the weak-\* of $Y^\*$ (not hard to prove directly, but a reference is Schaefer's *Topological Vector Spaces* IV.2.1).


Note, however, that the analogous statement for incomplete normed spaces is false -- any element $x$ of the completion of a normed space $X$ that is not part of $X$ defines a linear functional that is weak-\* continuous on the ball of $X^\*$ but not on $X^\*$.


