Is the trace map continuous on the trace class?

Question

$\DeclareMathOperator{\tr}{tr}$ Let $A$ be trace-class, i.e., $\tr{|A|}<\infty$ where $|A|=\sqrt{A^*A}$. Then is $A\mapsto \tr{A}$ continuous wrt the uniform topology, SOT or maybe WOT? I know that $|| A || \le \tr{|A|}$, but I'm curious about other topologies.

Answer 1

In finite-dimensions all of these topologies (trace norm, operator norm, SOT, WOT) coincide on the trace class (which itself coincides with the space of all linear operators on the corresponding Hilbert space) so in this case the trace, unsurprisingly, is continuous.

As soon as one gets to infinite-dimensions the map $\operatorname{tr}:(\mathcal B^1(\mathcal H),\tau)\to\mathbb F$ is not continuous for $\tau=$ norm, SOT, WOT.

To see this we will construct a sequence of trace-class operators which converge to 0 in the operator norm but the corresponding traces do not converge. Moreover if $\operatorname{tr}$ is not continuous in operator norm it cannot be continuous with respect to any weaker topology on the domain (such as SOT or WOT, cf. here).

Let $(e_k)_{k\in\mathbb N}$ be an orthonormal basis of your underlying (separable, infinite-dimensional) Hilbert space $\mathcal H$ and define $T_n:\mathcal H\to\mathcal H$ for any $n\in\mathbb N$ via $T_n(e_k):=\frac1ne_k$ for all $k=1,\ldots,n$ and $T_n(e_k)=0$ for all $k>n$ as well as its linear extension onto all of $\mathcal H$. One readily verifies that each $T_n$

• is of finite-rank, hence trace-class.
• has operator norm $\frac1n$ (as no "matrix element" is larger than $\frac1n$).
• has trace $\operatorname{tr}(T_n)=\sum_{k=1}^n\frac1n=1$.

As by the second point $\|T_n\|=\frac1n\overset{n\to\infty}\to0$ but $\operatorname{tr}(T_n)=1\overset{n\to\infty}{\not\to}0=\operatorname{tr}(0)$.