It is known that weak convergence implies norm convergence in $\ell^1(\mathbb{N})$, see e.g. [here](http://www.math.ucla.edu/~nickcook/l1weakstrong.pdf).


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](https://math.stackexchange.com/questions/1578728/weak-convergence-norm-convergence-for-trace-class-operators)).




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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](http://www.ams.org/journals/proc/1981-083-01/S0002-9939-1981-0619978-4/S0002-9939-1981-0619978-4.pdf).


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](https://mathoverflow.net/questions/72747/convergence-of-fredholm-determinants) post)?


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



Weak convergence implies norm convergence for trace class operators?
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).


