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If $X$ is a reflexive Banach space, is it true that the natural embedding $\Lambda$ of $X$ into its double dual is a homeomorphism? Here we equip $X$ with the weak topology, and $X^{**}$ with the weak-* topology, i.e. $\sigma(X^{**},X^*)$. I think that this is true. To start with I have shown that it is at least continuous by using nets. I also wanted to show that the inverse is continuous (but I could not figure this out using nets), but here's how far I got: So assume that the net $f_a \rightarrow f$ weakly (here the $f_a$ lie in $X^{**}$). We want to show $\Lambda^{-1}(f_a) \rightarrow \Lambda^{-1}(f)$ weakly. Note that $f_a \rightarrow f$ weakly is equivalent to $f_a(x) \rightarrow f(x)$ for all $x \in X^{*}$. And that the conclusion is equivalent to $x(\Lambda^{-1}(f_a)) \rightarrow x(\Lambda^{-1}(f))$ weakly for all $x \in X^*$. This is where I'm getting a bit confused: Is $x(\Lambda^{-1}(f_a))=f_a(x)$?

Vercingetorix
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Yes, but don't use nets, as that is very cumbersome. Since $X$ is isometric to $X^{**}$ under the canonical embedding, that means they are isomorphic under the weak topologies. But since $X^{**}$ is reflexive, its weak and weak-* topologies coincide.

Ben W
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  • Why must they be isomorphic under the weak topologies if $X$ is isometric to $X^{**}$ under the canonical embedding? – Vercingetorix Jan 06 '19 at 16:01
  • The weak topologies are generated by the respective norm topologies. Since the latter are identical, so are the former. – Ben W Jan 06 '19 at 16:04
  • See here for more info: https://math.stackexchange.com/questions/900903/if-t-x-to-y-is-norm-norm-continuous-then-it-is-weak-weak-continuous – Ben W Jan 06 '19 at 16:06