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

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.

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

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 $$Se_n=\frac1n\,e_n$$.

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.

• How can we say $S^{-1}$ is continuous? – Anupam Oct 24 '18 at 2:52
• I read 'one-one' as bijective? Did you mean injective ? – daw Oct 24 '18 at 6:06
• Yeah I meant injective. – Anupam Oct 24 '18 at 7:47
• 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). – Anupam Oct 24 '18 at 8:00