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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!
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.
