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.


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.
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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 $T_n \rightarrow T_0$


$K$ 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 $T_n \rightarrow T_0$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.