Is trace functional strongly continuous?

Question

Trace functional is defined on space of trace class operators as $Tr:B_1(\mathcal H)\to \mathbb C$ $$Tr(A)=\sum_{\alpha\in I}<Ae_\alpha,e_\alpha>$$ where $\{e_\alpha\}_{\alpha\in I}$ is an orthonormal basis for the Hilber space $\mathcal H$. I was wondering if this functional is continuous with respect to the strong topology on $B_1(\mathcal H)$ inherited from $B(\mathcal H)$?

Answer 1

Hint: there exist positive numbers $a_{nk}$ such that $\sum_k (a_{nk})^{2} \to 0$ as $n \to \infty$ but $\sum_k a_{nk}$ does not tend to $0$ as $n \to \infty$. On $\ell^{2}$ define $T_n(x)= \sum\limits_{k=1}^{\infty} a_{nk} \langle x, e_k \rangle e_k$ where $(e_n)$ is the usual orthonormal basis. Then trace of $T_n$ (which is $\sum_k a_{nk}$) does not tend to $0$ but $\|T_n\| \to 0$.

Answer 2

Your question is somewhat ambiguous because there is no such thing as "strong topology on $\mathcal B^1(\mathcal H)$ inherited from $\mathcal B(\mathcal H)$". If you mean $\mathcal B^1(\mathcal H)$ together with the operator norm $\|\cdot\|$ (which turns $\mathcal B(\mathcal H)$ into a Banach space) then the answer to your question is negative (cf. Kavi's answer and the respective comments).

If you mean (which given your wording makes more sense to me) the weak topology on the Banach space $\mathcal B^1(\mathcal H)$, i.e. $$ A_i\overset{i\to\infty}\to 0\text{ weakly if }|\operatorname{tr}(A_iB)|\overset{i\to\infty}\to 0\text{ for all }B\in\mathcal B(\mathcal H) . $$ Here one uses that the dual space $(\mathcal B^1(\mathcal H))'\simeq\mathcal B(\mathcal H)$ contains precisely the functionals $A\mapsto \operatorname{tr}(AB)$ for some $B\in\mathcal B(\mathcal H)$. Note that the strong topology (on spaces of operators on $\mathcal H$) is something very different.

For the question itself, the trace is trivially weakly continuous because it is obviously is of trace-form. Indeed, if $A_i\to 0$ weakly, then choosing $B=\operatorname{id}_{\mathcal H}$ in the above definition shows $|\operatorname{tr}(A_i)|\to 0$.