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 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 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 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.)