# Weak convergence implies norm convergence for trace class operators?

It is known that weak convergence implies norm convergence in $\ell^1(\mathbb{N})$, see e.g. here.

Because of the typical analogies of the Schatten ideals $C_p \subset B(H)$ (where $H$ is a Hilbert space), it seems natural to ask whether weak convergence in $C_1$ (the ideal of trace-class operators) already implies convergence with respect to the trace-class norm. (This question has been asked on math.stackexchange by another user but didn't receive an answer, see here).

It is known that weak convergence $T_n \rightarrow T$ implies strong convergence for a sequence in $C_1$ if one additionally knows that $\|T_n\|_1 \rightarrow \|T\|_1$, see here.

It seems to me that completely dropping the restriction $\|T_n\|_1 \rightarrow \|T\|$ might be a little too much to ask, even though it is possible in $\ell^1(\mathbb{N})$.

However, if the answer to the above question is negative, then one can ask if one can do better:

• For example, is it sufficient to just have a uniform bound on $\|T_n\|$ (the operator norm) or $\|T_n\|_1$ (the trace class norm)?
• Is it sufficient to know that $\mathrm{tr}(T_n) \rightarrow \mathrm{tr}(T)$ (this was claimed by user Folkmar Bornemann in this post)?

(Of course, all additionally to knowing that $T_n\rightarrow T$ in the weak topology.)

It's still false, however. Let $T_n: v \mapsto \langle v, e_1\rangle e_n$ where $(e_n)$ is an orthonormal basis of $H$. Then $\|T_n\|_1 = 1$ for all $n$, but the sequence converges weakly to zero. For any $A \in B(H)$, the sequence ${\rm Tr}(AT_n)$ reads off the entries of the first row of $A$, which has to lie in $l^2$ and hence goes to zero.

This example also falsifies the two suggested weakenings --- $\|T_n\|$ and $\|T_n\|_1$ are both constantly 1, and ${\rm Tr}(T_n)$ is constantly zero (after the first term).

We can assume $T=0$. Let $H$ be a HIlbert space with orthonormal basis $(e_1,e_2,\dots)$. For $j\in{\mathbb N}$ let $T_j:H\to H$ be given by $T_j e_k=\delta_{j,k}e_k$ (Kronecker-delta). Then $T_j$ is trace class of trace norm one, but the sequence $T_j$ tends to zero strongly (and hence weakly).

Also, insisting $tr(T_j)\to 0$ does not help for you can replace $T_j$ with $S_j$ where $S_je_k=e_k$ if $k=j$, $S_je_k=-e_k$ if $k=j+1$ and zero otherwise.

• That's not correct --- your sequence $(T_j)$ converges to zero weak operator, but not weakly. Its pairing with the identity operator doesn't go to zero. Jan 21, 2016 at 13:08
• What does that mean? What is the pairingwith the identity?
– user1688
Jan 21, 2016 at 13:13
• ${\rm Tr}(IT_j) = 1 \not\to 0$. The map $B \mapsto {\rm Tr}(AB)$ is a bounded linear functional on $TC(H)$, for any $A \in B(H)$. Jan 21, 2016 at 13:21