Question about a weak*-norm continuity of a linear operator

Question

Let $X$ and $Y$ be infinite dimensional normed linear spaces and let $S:Y^*\to X^*$ be a one-one linear operator. I want to show that $S$ can not be weak*-norm continuous.

My idea is to choose a sequence $(y_n^*)$ in $Y^*$ which weak* converges to some $y^*$ in $Y^*$ (for this Banach-Alaoglu theorem will have to be used) such that $S(y^*_n)\not\to S(y^*)$ in norm. But I am not getting how infinite dimension of $X$ and $Y$ (and of course $X^*,Y^*$) will help. Any suggestion will be appreciated.

Answer 1

I don't think the result is true as stated.

Take $X=Y=H$, an infinite-dimensional separable Hilbert space. Then $X^*=Y^*=H$, and the weak$^*$-topology agrees with the weak topology. Now any injective compact operator contradicts the assertion, as compact operators map weak (hence weak$^*$)-convergent sequences into norm-convergent sequences.

As an example, fix an orthonormal basis $\{e_n\}$ and let $S$ be the linear operator induced by $Se_n=\frac1n\,e_n$.

Answer 2

The claim is true for bijective $S$, I have no idea how to extend to the general case.

Assume that $S:Y^*\to X^*$ is weak-star-to-strong continuous. By the Banach-Alaoglu theorem, $S$ maps bounded sets to relatively compact sets, hence it is compact.

If $S$ is bijective, then it is continuously invertible and compact, hence the space is finite-dimensional, a contradiction.