Ultraweak Continuity Implies Norm Continuity

Question

I was reading the section on Schur Multipliers in Ozawa's book "C*-algebras and Finite-Dimensional Approximations" and I am having troubles understanding the proof of Proposition D.6, namely the part about ultraweak continuity.

I think I can skip most of the context and I am stuck on a particular part, so I'll write it differently.

Let $H$ be a Hilbert (in the problem we have $H=\ell^2(\Gamma)$) and let $T: B(H) \rightarrow B(H)$ be an ultraweak continuous linear operator on $H$, moreover, we know that the restriction $T|_{K(H)}:K(H) \rightarrow K(H) $ is well defined and not only ultraweak continuous, but also norm continuous. To my understanding of the proof in the book, this implies that $T$ is also norm continuous on the whole $B(H)$, however I am not being able to prove this.

I tried using the fact that trace class operators (predual of $B(H)$) are compact, but did not reach any relevant conclusion, moreover I am almost sure the Uniform Boundness Theorem is required.

If this affirmation is false and more context is needed, it can be found in the mentioned book, Proposition D.6.

Answer 1

I’m adding another answer, as community wiki, to expand upon @OnurOktay ’s comment:

Claim: If $T: X^\ast \to Y^\ast$ is weak$^\ast$-to-weak$^\ast$ continuous, where $X$ and $Y$ are Banach spaces, then $T$ is bounded. In fact, there exists bounded $S: Y \to X$ s.t. $T = S^\ast$.

Proof: For each $y \in (Y)_1$, we let $T_y: X^\ast \to \mathbb{C}$ be defined by $T_y(\phi) = [T(\phi)](y)$. We observe that $T_y$ is bounded. Indeed, evaluation at $y$ is a weak$^\ast$ continuous linear functional on $Y^\ast$, so by weak$^\ast$-to-weak$^\ast$ continuity of $T$, $T_y$ is a weak$^\ast$ continuous linear functional on $X^\ast$, so it is given by evaluation at some $x \in X$ and thus in particular bounded.

Consider the collection of maps $\{T_y: y \in (Y)_1\}$. For each fixed $\phi \in X^\ast$, $|T_y(\phi)| = |[T(\phi)](y)| \leq \|T(\phi)\|$ since $\|y\| \leq 1$, i.e., $\sup_{y \in (Y)_1} |T_y(\phi)| < \infty$ for any fixed $\phi$. So by uniform boundedness principle, $\sup_{y \in (Y)_1, \phi \in (X^\ast)_1} |T_y(\phi)| < \infty$, but,

$$\sup_{y \in (Y)_1, \phi \in (X^\ast)_1} |T_y(\phi)| = \sup_{\phi \in (X^\ast)_1} \sup_{y \in (Y)_1} |[T(\phi)](y)| = \sup_{\phi \in (X^\ast)_1} \|T(\phi)\| = \|T\|$$

So $\|T\| < \infty$, i.e., $T$ is bounded.

Finally $S$ is simply defined as the adjoint of $T$ restricted to $Y \subset Y^{\ast\ast}$. As we have seen in the first paragraph of the proof, composing $T$ with evaluation at $y \in Y$ gives evaluation at some $S(y) \in X$. Since $S$ is a restriction of $T^\ast$ and $T$ is bounded, $S$ is bounded as well. That $T$ is the adjoint of $S$ follows easily from the definition. $\square$

Answer 2

By Kaplansky density theorem, for any $x \in B(H)$, there exists a net $x_\lambda \in K(H)$ with $x_\lambda \to x$ ultraweakly and $\|x_\lambda\| \leq \|x\|$ for all $\lambda$. Then $\|Tx_\lambda\| \leq \|T|_{K(H)}\|\|x_\lambda\| \leq \|T|_{K(H)}\|\|x\|$. By ultraweak continuity of $T$, we have $Tx_\lambda \to Tx$ ultraweakly, whence $\|Tx\| \leq \|T|_{K(H)}\|\|x\|$. Since $x \in B(H)$ is arbitrary, this means $T$ is bounded and in fact $\|T\| \leq \|T|_{K(H)}\|$.