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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.

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  • $\begingroup$ $S$ is any one one linear operator $\endgroup$ – Anupam Oct 23 '18 at 9:38
  • $\begingroup$ @daw The identity map is norm-weak$^*$ continuous, not the other way around. $\endgroup$ – Theo Bendit Oct 23 '18 at 9:49
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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 $Se_n=\frac1n\,e_n$.

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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.

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  • $\begingroup$ How can we say $S^{-1}$ is continuous? $\endgroup$ – Anupam Oct 24 '18 at 2:52
  • $\begingroup$ I read 'one-one' as bijective? Did you mean injective ? $\endgroup$ – daw Oct 24 '18 at 6:06
  • $\begingroup$ Yeah I meant injective. $\endgroup$ – Anupam Oct 24 '18 at 7:47
  • $\begingroup$ In the example how does compactness of the operator implies weak*-norm continuity? Please see An introduction to Banach Space Theory by R E Megginson exercise 3.44 (a) and (b). $\endgroup$ – Anupam Oct 24 '18 at 8:00

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