Let $H$ be a separable infinite-dimensional real Hilbert space. We consider operators in $H.$ 


Nuclear norm of a nuclear operator is the sum of its singular values.
A nuclear, positive and self-adjoint operator is called S-operator.


Does the following criterion hold true?


A sequence $A\_n$ of S-operators converges in nuclear norm to S-operator $A$ if, and only if, $A\_n$ weakly converges to $A$ and there exists an orthobasis $e\_k$ such that the series $\Sigma\_{k=1}^{\infty}(A\_n e\_k, e\_k)$ converges uniformly in $n=1, 2\ldots$


I can prove the necessity only. But does the sufficiency hold true as well?



Convergence of nuclear operators
Looks correct. Split $H=H\_n\oplus H^n$ where $H\_n=span(e\_1,\dots,e\_n)$. Let $P\_n$ and $P^n$ be the corresponding projections. The conditions are $P\_nA\_kP\_n\to P\_nAP\_n$, $\operatorname{Tr}(P^nA\_kP^n)\le\varepsilon\_n$ with $\varepsilon\_n\to 0$ as $n\to\infty$.


Now just observe that $A\_k-A+\delta n^{-1} P\_n+\delta(P\_nA\_kP\_n+P\_nAP\_n)+(1+\delta^{-1})(P^nA\_kP^n+P^nAP^n)$ is an $S$-operator for large $k$ and fixed $n,\delta$. However, its trace is $\le |\operatorname{Tr}P\_n(A\_k-A)P\_n|+\delta(1+2\varepsilon\_1)+(3+\delta^{-1})\varepsilon\_n$, which can be made arbitrarily small by choosing $\delta, n, k$ in this order. Also, it is an $S$-perturbation of $A\_k-A$ of small trace, so we are done. 


