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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?

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    $\begingroup$ Does "weak convergence" mean "convergence with respect to the weak operator topology"? $\endgroup$ – Jochen Glueck Jun 10 '18 at 17:43
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    $\begingroup$ Yes, it means that $(A_n x, y)$ converges to $(A x, y),$ for all $x, y \in $H$. $\endgroup$ – Alexander Kukush Jun 10 '18 at 18:57
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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.

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  • $\begingroup$ Alas, in the proposed answer I do not see why this long operator is positive for large k and fixed n, $\delta.$ And I cannot understand the origin of the term $\delta n^{-1} P_n$. $\endgroup$ – Alexander Kukush Jun 10 '18 at 20:23
  • $\begingroup$ @AlexanderKukush $\delta(P_nA_kP_n+P_nAP_n)+\delta^{-1}(P^nA_kP^n+P^nAP^n)$ dominates the "off-diagonal" part of $A_k-A$; $\delta n^{-1}P_n$ is just to dominate $-P_n(A_k-A)P_n$. Is that much clear? $\endgroup$ – fedja Jun 10 '18 at 20:34
  • $\begingroup$ @fedija Now I understand your solution thoroughly. Brilliant! You helped me a lot. As a result I have a criterion for weak convergence of Gaussian measures in H: Measures $N(m_n, S_n)$ converge weakly to $N(m, S)$ if, and only if, $m_n$ converges strongly to $m$ and $S_n$ converges to $S$ in nuclear norm. $\endgroup$ – Alexander Kukush Jun 11 '18 at 7:07

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