Banach spaces $X$ for which the subspace of adjoint operators has finite codimension in $X^*$ 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)?
 A: If $X$ is reflexive, then $A= B(X^*)$.  
If $X^{**}/X$ is infinite dimensional, let $(x_n^{**})$ be a sequence in $X^{**}$ linearly independent over $X$.  Fix a nonzero $x^*\in X^*$.  If a finite linear combination $\sum a_nx^{**}_n\otimes x^*\in A$, then $\sum a_nx^{**}_n\in X$ and hence $a_n =0$ for all $n$. Thus $(x^{**}_n\otimes x^*)$ is linearly independent over $A$.  This shows that $B(X^*)/A$ is infinite dimensional.
This leaves the quasi-reflexive spaces ($X^{**}/X$ finite dimensional).  I'm pretty sure that $B(X^*)/A$ is finite dimensional and nontrivial for $X=$ James Space.  
A: 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.
