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Let $X$ be a dual Banach space, i.e. $X=(X_*)^*$ for some Banach space $X_*$. Consider the weak* topology of $B(X)$, i.e. the topology of pointwise convergence on $X$ endowed with the $\sigma(X,X_*)$-topology.

Consider the set $B_{w^*}(X)$ of $w^*$-continuous bounded operators on $X$. Is it a closed subset of $B(X)$ for the weak* topology of $B(X)$ ?

Zouba
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    Interestingly enough for the more general case of $B(X^,Y^)$ (i.e. for two different dual Banach spaces) the answer to this question in general is no, cf. this counterexample by Jochen Glück given on mathoverflow. Of course this does not settle the special case $X=Y$ OP asked about but it might give some further intuition for this problem---after all the answer given below is flawed (as elaborated on in the comments below). – Frederik vom Ende Jan 05 '20 at 17:07

2 Answers2

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By this answer, you are looking at the set $$ B_{w^{\ast}}(X) = \{T^{\ast} : T \in B(X_{\ast})\} $$ which is convex.

Now if $\sigma$ denotes the weak-$\ast$-topology on $B(X)$, then $(B(X),\sigma)$ is a topological vector space whose topology is generated by the family of semi-norms given by $$ p_{f,y}(T) := |T(f)(y)| $$ for $f\in X, y\in X_{\ast}$. Hence, by a consequence of the Hahn-Banach separation theorem (see this answer), $B_{w^{\ast}}(X)$ is closed with respect to $\sigma$ iff it is closed in the norm topology of $B(X)$.

But this last fact is easy to prove:

If $X_{\ast}$ is a Banach space, then $B_{w^{\ast}}(X)$ is closed in $B(X)$ with respect to the norm topology.

Proof: If $\{T_n\} \subset B(X_{\ast})$ and $S\in B(X)$ such that $\|T_n^{\ast} - S\|_{B(X)} \to 0$, then we wish to show that $S$ is weak-$\ast$-continuous. To see this, choose a net $f_{\alpha} \in X$ such that $f_{\alpha}(y) \to f(y)$ for all $y\in X_{\ast}$. By the principle of uniform boundedness, $\exists M > 0$ such that $$ \sup_{\alpha} \|f_{\alpha}\|, \|f\| < M $$ (Here we need the fact that $X_{\ast}$ is a Banach space). Then for any $\epsilon > 0$ and fixed $y\in X_{\ast}$, choose $N \in \mathbb{N}$ such that $$ \|T_N^{\ast} - S\|_{B(X)} < \frac{\epsilon}{3M\|y\|} $$ and $\beta$ such that $$ \|f_{\alpha}(T_N(y)) - f(T_N(y))| < \epsilon/3 $$ for all $\alpha \geq \beta$. Then, for all $\alpha\geq \beta$, we have \begin{equation*} \begin{split} |S(f_{\alpha})(y) - S(f)(y)| &\leq |S(f_{\alpha}(y)) - T_N^{\ast}(f_{\alpha})(y)| + |T_N^{\ast}(f_{\alpha})(y) - T_N^{\ast}(f)(y)| + |T_N^{\ast}(f)(y) - S(f)(y)| \\ &< \epsilon \end{split} \end{equation*} and so $S(f_{\alpha}) \to S(f)$ pointwise as required.

Community
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Prahlad Vaidyanathan
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    It seems to me that there is a problem in the proof. A weak*-convergent net need not be bounded in general. The net $(f_\alpha)$ in not necessarily bounded. – Zouba Feb 02 '18 at 22:50
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    A second problem is the following, a closed convex subset is not necessarilly weak* closed. – Zouba Feb 03 '18 at 07:02
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    The fatal flaw in this answer is that it is for the weak topology that closure of convex sets agrees with the norm topology, and not the weak-* topology (as soon as we have a space that is not reflexive). The weak-* operator topology suggested in the question is still different. – Robert Furber Feb 09 '20 at 07:21
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    However, the highlighted statement in the answer (closure in the norm topology) is correct, although a flawed proof is given. As Zouba has pointed out, weak-* convergent nets need not be bounded. What we need is a version of the Banach-Dieudonné theorem or Krein-Šmulian theorem: For a normed space $X_$, it is a Banach space iff every linear functional $\phi : X_ \rightarrow \mathbb{R}$ that is weak-* continuous on the unit ball of $X$ is weak-* continuous on all of $X$. This allows us to restrict to bounded nets in $X$. – Robert Furber Feb 09 '20 at 07:25
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    Trying to use the uniform boundedness principle can't be made to work because there are normed spaces $X_$ that are barrelled, i.e.* satisfy the uniform boundedness principle, but are not complete. If $E_$ is an incomplete normed space, then $B_{w^}(E)$ is not closed even in the norm topology of $B(E)$ (hint: construct a sequence that leaves $B_{w^}(E)$ using rank-one operators and a Cauchy nonconvergent sequence in $E_$). – Robert Furber Feb 09 '20 at 07:51
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Over on the MathOverflow version of this question, I have answered it with a counterexample, $B_{w^*}(\ell^1)$ is not weak-* closed in $B(\ell^1)$.

Robert Furber
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