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maybe this question is too elementary but I could not find any results in the functional analysis textbooks I own:

For separable Hilbert spaces $H$, $K$, let $S_p(H,K)$ be the $p$th Schatten class of linear operators $T \colon H \to K$. I'm interested in conditions which imply that for a sequence $(T_n)_{n \in \mathbb{N}_0} \subseteq S_p(H,K)$ the implication

$$\lVert T_n \rVert_p \to \lVert T_0 \rVert_p \Rightarrow \lVert T_n - T_0 \rVert_p \to 0$$

is satisfied.

In finite dimension, a related result states that for $p>0$ and a sequence $(f_n) \subseteq L^p(\mathbb{R},\mu)$ which satisfies $f_n \to f_0$ almost surely, one has that $\lVert f_n \rVert_2 \to \lVert f_0 \rVert_p$ implies $\lVert f_n - f_0 \rVert_p \to 0$ (see for example Lemma 1.32 in Kallenberg's Foundations of modern probability). Unfortunately the proof does not seem to be transferable but gives as a heuristic that the implication holds in $S_p$ if convergence in the strong operator topology is assumed in addition.

UPDATE:

Thanks to everybody for the comments! Apparently my question is a bit too general (I asked it out of pure interest with no specific research goal in mind), so let me be more specific:

Does the above implication hold if we assume any or some of:

  1. $T_n \to T_0$ in the strong operator topology
  2. The singular values of $T_0$ and/or the $T_n$ have some property.
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    $\begingroup$ The other hypothesis is $T_n \to T_0$ in some sense. $\endgroup$ Dec 16, 2019 at 20:58
  • $\begingroup$ @mr_e_man You can not replace $f_0$ by $-f_0$, I should have given more details though. The full statement of the Lemma is: Let $(f_n)$ be a sequence in $L^p$ such that $f_n \to f_0$ almost everywhere. Then $\lVert f_n \rVert_p \to \lVert f \rVert_p$ implies $\lVert f_n - f_0 \rVert_p$. I have edited my question as well, so hopefully it's clear now. $\endgroup$ Dec 16, 2019 at 22:52
  • $\begingroup$ @GeraldEdgar How do you know that hypotheses of different forms (for example a specific form of the spectrum or the singular values) can be excluded a priori? $\endgroup$ Dec 16, 2019 at 22:57
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    $\begingroup$ If $x_n\to x$ weakly in a Hilbert space and $\|x_n\|\to\|x\|$, then $\|x_n-x\|\to 0$. This gives such a result for $p=2$. $\endgroup$ Dec 16, 2019 at 23:00
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    $\begingroup$ In uniformly convex spaces weak convergence of $x_n$ together with $\|x_n\|\to\|x\|$ implies convergence. I am not sure whether this helps. $\endgroup$
    – Christian
    Dec 17, 2019 at 5:37

1 Answer 1

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In the case $E$ evn :

if $\{T_n ;n\in \mathbb N\}$ compact and $\text{card}(\{n \in \mathbb N; ||T_n||=||T_0||\})\in\mathbb N$

then $||T_n|| \rightarrow ||T_0||$ imply $T_n \rightarrow T_0$

or

$K$ compact with $\forall n \in \mathbb N, T_n\in K$

if $\text{card}(\{a \in K; ||a||=||T_0||\})=1$ then $||T_n|| \rightarrow ||T_0||$ imply $T_n \rightarrow T_0$

Remark : if $\{T_n ;n\in \mathbb N\}$ isn't compact then we have not $T_n \rightarrow T_0$

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  • $\begingroup$ These conditions seem too restrictive $\endgroup$
    – Yemon Choi
    Dec 18, 2019 at 3:54
  • $\begingroup$ @YemonChoi : and now? $\endgroup$
    – Dattier
    Dec 18, 2019 at 4:24
  • $\begingroup$ @Dattier Thanks for your answer! I have to agree with Yemon Choi and think that these conditions are quite restrictive and also make no specific use of the Schatten-norms, so I'll keep the question open. $\endgroup$ Dec 19, 2019 at 8:03
  • $\begingroup$ @r_faszanatas well, it's your choice $\endgroup$
    – Dattier
    Dec 19, 2019 at 8:06

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