What additional conditions are required for convergence in operator norm to imply convergence in trace norm Let $\mathcal{H},\mathcal{K} $ be a separable Hilbert spaces, and $T_n: \mathcal{H}\mapsto \mathcal{K}$ a sequence of compact operators  satisfying $\|T_n - T\|_{op} \to 0$, with $\|\cdot\|_{op}$ denoting the standard operator norm. Evidently this is not enough to imply that $\|T_n - T\|_1 \to 0$, where $\|\cdot\|_1$ is the nuclear (trace) norm. See e.g. this post: Convergence in operator and trace-class norm
Is it known what (sharp?) additional conditions on top of $\|T_n - T\|_{op} \to 0$ imply that $\|T_n - T\|_1 \to 0$? Are there weaker additional conditions under which $\|T_n - T\|_{op} \to 0$ imply that $\|T_n \|_1 \to \| T\|_1$? What if $\|\cdot\|_{op}$ is replaced with $\|\cdot\|_2$, the Frobenius norm?
 A: I still think there is not satisfactory answer to your question, as vague as it is (very open to be proven wrong, though).
Let us restrict a bit, to see what happens in some cases. We can take $T=0$ with no loss, but I will also assume (big restriction) that all the $T_n$ are simultaneously unitarily diagonalizable (equivalently, that they are normal and mutually commuting). Basically, we are considering the analog problem but in $c_0$.
So we have
$$
T_n=(t_{n1},t_{n2},\ldots).
$$
We have
$$
\|T_n\|_{\rm op}=\max\{|t_{nk}|:\ k\},\qquad \|T_n\|_1=\sum_k|t_{nk}|,\qquad \|T_n\|_2=\sum_k|t_{nk}|^2.
$$
Let us restrict even more, for the sake of simplicity, and also assume that $|t_{n1}|\geq|t_{n2}|\geq\cdots$ for all $n$.
Then
$$
\|T_n\|_{\rm op}=|t_{n1}|,\qquad \|T_n\|_1=\sum_k|t_{nk}|.
$$
The question becomes, in this setting, that we assume that $t_{n1}\to0$, and what conditions will guarantee that
$$\tag1
\sum_k|t_{nk}|\xrightarrow[n\to\infty]{}0.
$$
One example that might be enlightening is to consider
$$
T_n=\big(\overbrace{\tfrac1n,\ldots,\tfrac1n}^{n^2 \text{ times}},0,\ldots\big).
$$
Then $$\|T_n\|_{\rm op}=\frac1n,\qquad\text{ while } \qquad \|T_n\|_1=n$$
for all $n$.
Other than talking specifically about the series (that is, prescribing directly that $\|T_n\|_1\to0$), I cannot see what condition would guarantee $(1)$.
