# Ultraweak Continuity Implies Norm Continuity

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.

• This is valid in a more general context. Let $X,Y$ be Banach spaces, $S:Y^*\to X^*$ be linear. $S$ is weak$^*$-to-weak$^*$ continuous iff $\exists$ bounded $T:X\to Y$ such that $S=T^*$. Apply this to $X=Y=$trace class operators, and keep in mind that $X^*=B(H)$ and ultraweek topology is the weak$^*$ topology on $B(H)$. Mar 3 at 23:37
• @OnurOktay Yes this is very helpful and it is similar to what I was trying to do, It will be very good to keep this mind going foward thank you. Mar 3 at 23:44
• @OnurOktay However, I think this is not applicable in this specific case because we only know that $T|_{K(H)}:K(H) \rightarrow K(H)$ and we do not necessarily have $T|_{L^1(B(H))}:L^1(B(H)) \rightarrow L^1(B(H))$, only $T|_{L^1(B(H))}:L^1(B(H)) \rightarrow K(H)$ . Is there a way around this? Mar 3 at 23:46
• @TomásPacheco That’s not what Onur meant. What Onur’s comment suggested is that, more generally, if a linear map $T: Y^\ast \to X^\ast$ is weak$^\ast$-to-weak$^\ast$ continuous, without any other assumptions whatsoever, then it is automatically bounded, which can be proved via uniform boundedness principle. Mar 4 at 0:01
• @DavidGao Ah! and thats because if $S: X \rightarrow Y$ is bounded and such that $S^* = T$ then $T$ is bounded right? Mar 4 at 0:04

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 $$\|T\| \leq \|T|_{K(H)}\|$$.

• Thank you so much! I have a question, is it the case that $\|Tx_\lambda\| \rightarrow \|Tx\|$ ? If yes why does ultraweak convergence imply it? I am quite new into the ultraweak topology, do you recommend a reference where I can learn it more? Thank you so much for the reply Mar 3 at 23:38
• @TomásPacheco No, it is not true in general that $\|Tx_\lambda\| \to \|Tx\|$, but it is true that $\|Tx\| \leq \lim \sup \|Tx_\lambda\|$, which is enough to prove the claim. Mar 3 at 23:51
• @TomásPacheco I’m not sure if there’s any reference specifically on ultraweak topology, but most introductory textbooks on operator algebras cover this. See Chapter 4 of Murphy’s $C^\ast$-algebras and operator theory, and Chapters 2 and 3 of Conway’s A course in operator theory, for example. Mar 3 at 23:55
• Thank you so much for the book recomendations, it was what I was looking for. Regarding the inequality, I am having troubles seeing where it comes from, do you have a hint? Thank you again Mar 4 at 0:01
• @TomásPacheco Use the definition of ultraweak convergence to prove the following: if $x_\lambda \to x$ ultraweakly and $x_\lambda$ are all in the unit ball, then $x$ is in the unit ball as well. Hint: $\|x\| = \sup_{\varphi \in (B(H)_\ast)_1} |\varphi(x)| = \sup_{\varphi \in (B(H)_\ast)_1} \lim_\lambda |\varphi(x_\lambda)|$. Mar 4 at 0:06

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. $$\square$$

• I was quite close to proving this on my own, so $S$ isn't needed to prove that $T$ is bounded. I think I got it except one tiny thing. I am probably wrong but when you say that $T_y$ has to be an evaluation at some $x \in X$, isn't this the case only if $X$ is reflexive? And in case I haven't said it enough, thank you so much. Mar 4 at 0:43
• @TomásPacheco $T$, by assumption, is weak$^\ast$-to-weak$^\ast$ continuous. $T_y$ is $T$ composed with evaluation at $y$, which is weak$^\ast$ continuous, so $T_y$ is weak$^\ast$ continuous as well. The only weak$^\ast$ continuous linear functionals on $X^\ast$ are evaluation at some $x \in X$. This does not require any reflexivity assumptions. (It would require reflexivity if $T_y$ is just bounded, but here I’m saying it’s weak$^\ast$ continuous, which is a stronger condition.) Mar 4 at 0:48
• Ah, this is very subtle but I get it now, and allow me to say one last thank you so much. Mar 4 at 0:52