If $E$ is Banach and $E^*$ is its dual, is every $T:E^*\rightarrow E^*$ an adjoint?

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 $$B(E^*)$$?

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

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.
• Follow up: is the subspace of adjoints complemented in $B(E^*)$? – RandomWalker Jan 19 '19 at 22:14