Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

1 Answer 1

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


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.