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$A,B$ are locally convex topological spaces, $f:A\to B$ is linear, continuous. I want to verify the adjoint map has dense image in $A^*$ with respect to the weak* topology if and only if $f$ is injective.

Attempted Proof: If $fx=0$ then $\langle fx,g\rangle=0$ for each $g\in A^*$. Hence $\langle x,f^*g\rangle=0$ so the denseness implies $\langle x,g\rangle=0$ for each $g\in A^*$. Thus $x=0$.

On the other hand, suppose the range is not dense. By Hahn Banach, there is some nonzero $g\in A^*$ such that $g(fx)=0$ for each $x\in A$. Then $\langle x,f^*g \rangle=0$ for all $x$. So $f^*g=0$. This is a contradiction since injectivity of $f^*$ means that $g=0$.

Question: Am I using the correct version of Hahn-Banach? And does my argument show that $f^*$ has dense image with respect to the weak topology?

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  • $\begingroup$ In your proof, are you using the fact that the dual space of a locally convex space separates points? $\endgroup$ – user193319 Jun 16 at 17:24
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HINT Your proof that the range of $f^*$ is weak$^*$-dense implies the injectivity of $f$ is correct. For the converse, you are on the right track but you applied the Hahn-Banach theorem on the wrong space. You should apply (the same version of) the Hahn-Banach theorem to $B^*$ with the weak$^*$ topology.

Remember you're trying to show that injectivity of $f$ implies the range of $f^*$ is dense, not that the inejctivity of $f^*$ implies the range of $f^*$ is dense.


If $f^*(B^*)$ is not weak$^*$-dense in $A^*$, then by the Hahn-Banach theorem there is some nonzero $x\in A$ (i.e. a nonzero weak$^*$ linear functional on $A^*$) such that $\langle fx,g\rangle=\langle x,f^*g\rangle=0$ for all $g\in B^*$. Thus $f(x)=0$, and therefore $f$ is not injective.

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  • $\begingroup$ Can you clarify? I am not sure how to apply the Hahn-Banach with the weak topology. $\endgroup$ – Iced Palmer Feb 5 '17 at 6:41
  • $\begingroup$ @IcedPalmer I edited to include what I had in mind. Does that help? $\endgroup$ – Aweygan Feb 5 '17 at 7:00

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