TITLE: Banach spaces $X$ for which the subspace of adjoint operators has finite codimension in $X^*$
QUESTION [8 upvotes]: For a Banach space $X$, let $A \subseteq \mathcal{B}(X^*)$ denote the closed subspace of those operators $T : X^* \to X^*$ such that $T = S^*$ for some $S \in \mathcal{B}(X)$, that is, $T$ is the adjoint of some bounded linear operator on $X$. It can be shown that this condition is equivalent to $T$ being weak*-continuous. Certainly $A=\mathcal{B}(X^*)$ for a Hilbert space. Do there exist Banach spaces $X$ for which $A$ is of finite positive codimension in $\mathcal{B}(X^*)$, i.e. the quotient space $\mathcal{B}(X^*)/A$ is finite-dimensional (but nontrivial)?

REPLY [2 votes]: The answer is no.
As noted in Denny Leung's answer, if $X$ is reflexive then $A=B(X^\ast)$.
A dual version of Denny Leung's subsequent argument for a certain non-reflexive case actually gives the complete answer in the nonreflexive case. Indeed, suppose $X$ is nonreflexive, fix $x^{\ast\ast}\in X^{\ast\ast}\setminus X$ and let $(x_n^\ast)$ be a linearly independent sequence in $X^\ast$. Since $x^{\ast\ast}$ is not weak$^\ast$-continuous, the sequence $(x^{\ast\ast}\otimes x_n^\ast)$ is linearly independent over $A$, hence $B(X^\ast)/A$ is infinite dimensional.