Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathcal{A}$ be a $W^*$-algebra. Is the map $a \mapsto a^2$, or more generally the map $a \mapsto a^k$, ultraweakly continuous? (Of course, products are not jointly ultraweakly continuous in general, but products of something with itself are a special case!) If not, does it become so when restricted to the normal/self-adjoint/positive elements of $\mathcal{A}$?

share|cite|improve this question
up vote 0 down vote accepted

The map $a \mapsto a^2$ on a von Neumann algebra is not always ultraweakly continuous, even on the positive elements:

Claim. The sequence of positive elements $a_1$, $a_2$, ... in $B(\ell^2(\mathbb{N}))$ given by

$$a_n := \left|n\right\rangle\!\!\left\langle1\right| + \left|1\right\rangle\!\!\left\langle n\right| + 1$$

converges ultraweakly to 1, but $a_1^2$, $a_2^2$, ... converges ultraweakly to $\left|1\right\rangle\!\!\left\langle1\right|+1 \neq 1^2$.

Proof.

  1. $\left|n\right\rangle \!\!\left\langle n\right| \rightarrow 0$ ultraweakly. Write $s_n := \left|1\right\rangle\left\langle1\right| + \cdots + \left|n\right\rangle\left\langle n\right|$. Then $s_1 < s_2 < \ldots$ is an ascending sequence with supremum 1. Thus for every normal state $\omega$ on $B(\ell^2(\mathbb{N}))$, the ascending sequence $\omega(s_1) \leq \omega(s_2) \leq \ldots$ has supremum $\omega(1)=1$. Thus $\sum_{n=1}^\infty \omega(\left|n \right\rangle\!\!\left\langle n\right|)=1$ and so $\omega(\left| n \right\rangle\!\!\left\langle n \right|) \rightarrow 0$. Thus $\left|n\right\rangle \!\!\left\langle n\right| \rightarrow 0$ ultraweakly.
  2. $\left|1\right\rangle \!\!\left\langle n\right| \rightarrow 0$ ultraweakly. Let $\omega$ be a normal state. By Cauchy-Schwarz for positive linear functionals, we get $$ \left| \omega( \left|1\right\rangle \!\!\left\langle n\right|) \right|^2 = \left| \omega(1^* \cdot \left|1\right\rangle \!\!\left\langle n\right|) \right|^2 \leq \omega(1^*1) \ \omega(\ \left|n\right\rangle \!\!\left\langle 1\right|\left|1\right\rangle \!\!\left\langle n\right|\ ) = \omega(\left|n\right\rangle \!\!\left\langle n\right|).$$ Thus $\left| \omega( \left|1\right\rangle \!\!\left\langle n\right|) \right|^2 \rightarrow 0$. Thus $\left| \omega( \left|1\right\rangle \!\!\left\langle n\right|) \right| \rightarrow 0$. Hence $ \left|1\right\rangle \!\!\left\langle n\right| \rightarrow 0 $ ultraweakly.
  3. $\left|n\right\rangle \!\!\left\langle 1\right| \rightarrow 0$ ultraweakly. Indeed, note that $a \mapsto a^*$ is ultraweakly continuous.
  4. $a_n\rightarrow 1$ ultraweakly. Follows from the previous two points.
  5. $a^2_n \rightarrow \left|1\right\rangle\!\!\left\langle1 \right| + 1$ ultraweakly. Follows from $a^2_n = \left|n\right\rangle\!\!\left\langle n\right| + \left|1\right\rangle\!\!\left\langle 1\right| + 2\left|n\right\rangle\!\!\left\langle1\right| + 2\left|1\right\rangle\!\!\left\langle n\right| + 1$ and the first three points.
  6. $a_n$ is positive. The square of the self-adjoint element $\left|n\right\rangle\!\!\left\langle1\right| + \left|1\right\rangle\!\!\left\langle n\right|$ is a projection and thus by the C$^*$-identity its norm is 1. Thus $a_n$ is positive, because by Gel'fand $\|x\|+x \geq 0$ for any self-adjoint $x$.
share|cite|improve this answer

If your function were continuous on selfadjoints, then every bounded ultraweak-convergent net would be ultrastrong convergent. Because if $a_j\to0$ weakly for a net of selfadjoints, the continuity implies that $a_j^2\to0$ weakly and so $a_j\to0$ strongly. For an arbitrary convergent net, since the adjoint is weakly continuous one deduces that the real and imaginary parts converge weakly, and so strongly by the hypothesis.

Of course, the ultraweak and ultrastrong are well-known to be different: for instance, a ultrastrong limit of unitaries is a unitary, while the unitaries are ultraweakly dense in the unit ball.

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.