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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)?

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up vote 4 down vote accepted

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.

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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.

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Thanks for the correction. – Denny Leung Dec 15 '12 at 1:40

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