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Let $X$ be a Banach space and let $A\in B(X^*)$. I want to know if there exists some operator $B\in B(X)$ such that $B^*=A$, that is if the "adjoint" operation in surjective. Thank you !

Djalal Ounadjela
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  • In general, no. If $X$ is reflexive, yes. Perhaps you can find an example of an operator on $l^1$ that is not the dual of any operator on $c_0$, even though $c_0^* = l^1$. – GEdgar Jul 21 '17 at 17:57

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$A = B^*$ for some $B$ if and only if $A$ is continuous in the weak-* topology. See e.g. the answer to this question. One of the comments there gives an example of an $A$ that is not continuous in the weak-* topology.

Robert Israel
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