7
$\begingroup$

If $E$ is a (complex) Banach and $E^*$ is it's dual, is every bounded $T:E^*\rightarrow E^*$ an adjoint? I am mostly interested in when $T$ is an automorphism.

If not, is $\{T: E^*\rightarrow E^* \ | \ \mbox{T is an adjoint}\}$ dense in $B(E^*)$?

Thanks in advance!

$\endgroup$
  • $\begingroup$ Can you define what means to be an adjoint map? $\endgroup$ – alexp9 Jan 19 at 21:30
  • $\begingroup$ @Rhcpy99 Presumbaly $T$ is an adjoint if $T=S^*$ for some $S\in B(E)$. $\endgroup$ – Aweygan Jan 19 at 21:33
  • $\begingroup$ @Aweygan Yes, that's what I meant. Sorry for the confusion $\endgroup$ – RandomWalker Jan 19 at 21:34
  • $\begingroup$ What is easy: if $E$ is reflexive, then every $T\in B(E^*)$ is an adjoint. $\endgroup$ – amsmath Jan 19 at 21:55
8
$\begingroup$

A linear map $T\in B(E^*)$ is of the form $T=S^*$ for some $S\in B(E)$ if an only if $T$ is weak$^*$-to-weak$^*$ continuous.

Indeed, the reverse implication is easy (take a weak$^*$-convergent net $(x_\gamma^*)$, and show that $(Tx_\gamma^*)$ is weak$^*$-convergent). For the forward implication, for each $x\in X$, the map $E^*\to\mathbb C$ given by $x^*\mapsto \langle Tx^*,x\rangle$ is a weak$^*$-continuous linear functional, whence there is some $Sx\in E$ such that $\langle Tx^*,x\rangle=\langle x^*,Sx\rangle$ for all $x^*\in E^*$. Showing that the map $S:x\mapsto Sx$ is in $B(E)$ and $S^*=T$ isn't terribly difficult (the former follows from the closed graph theorem, and the latter is by construction).

As for the question about density, I'm not aware of any general results. If $E$ is reflexive, then $\{T: E^*\rightarrow E^* \ \mid \ T\mbox{ is an adjoint}\}=B(E^*)$, but there are some weird Banach spaces, and I imagine counterexamples exist.

EDIT Note that the adjoint map $B(E)\to B(E^*)$, $T\mapsto T^*$ is an isometry, whence the image is closed. Thus to show that the image of $B(E)$ under the adjoint map is not dense, it suffices to show that there is some element of $ B(E^*)$ which is not an adjoint. An example of such an operator is provided the comments following the answer on this question.

$\endgroup$
  • $\begingroup$ Extremely thorough and helpful. Thank you very much. $\endgroup$ – RandomWalker Jan 19 at 22:12
  • $\begingroup$ You're welcome. Glad to help! $\endgroup$ – Aweygan Jan 19 at 22:14
  • $\begingroup$ Follow up: is the subspace of adjoints complemented in $B(E^*)$? $\endgroup$ – RandomWalker Jan 19 at 22:14
  • $\begingroup$ Interesting question. I can't say anything concrete, but I doubt it. $\endgroup$ – Aweygan Jan 19 at 22:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.