I found the following property of compact operators in a proof, and I can't prove it.

Prove that if $T \in \mathcal{L}(E,F)$ is compact, and if $u_n \rightharpoonup u$ (the sequence converges weakly to $u$ in $\sigma(E,E^*)$) then $Tu_n \to Tu$ strongly in $F$.

I was able to prove that $Tu_n$ has a convergent subsequence ($u_n \rightharpoonup u$ implies that $(u_n)$ is bounded in $E$. Then, because $T$ is compact it must follow that $(Tu_n)$ must contain some strong convergent subsequence), but didn't manage to finalize the proof.

Any reference or hints are welcome.

  • $\begingroup$ @all: please ignore my vote to close. I'm pretty sure this was discussed several times already, but I can't seem to find the thread at the moment. The thread I linked to is not a duplicate of this question. Beni, sorry about that. $\endgroup$
    – t.b.
    Commented Oct 6, 2011 at 18:19
  • $\begingroup$ @t.b.: It's ok. I was wondering why it was voted to close. I also searched the site before posting, and didn't find it. $\endgroup$ Commented Oct 6, 2011 at 18:21

2 Answers 2


Make use of the following topological lemma.

Lemma Let $X$ be a topological space and $\mathbf{x}=(x_n)_{n\in \mathbb{N}}$ be a sequence of elements of $X$. If every subsequence of $\mathbf{x}$ contains a subsequence convergent to $x$ then $x_n \to x$.

  • $\begingroup$ This surely is a very nice argument. Thank you. $\endgroup$ Commented Oct 6, 2011 at 18:22
  • 1
    $\begingroup$ You're welcome! This little lemma is simple yet useful to know, IMHO. $\endgroup$ Commented Oct 6, 2011 at 21:29
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    $\begingroup$ One of the most beautiful and useful techniques in analysis and problems in sequences. I love it $\endgroup$ Commented May 8, 2018 at 8:29
  • $\begingroup$ I don't understand how this lemma can help us? The lemma speaks about every subsequence. While we found only one convergent subsequence? $\endgroup$
    – Kamil
    Commented Dec 15, 2019 at 12:30
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    $\begingroup$ @Kamil: you have to apply this lemma to an arbitrary subsequence of $u_n$. Since $u_n$ is bounded, and since $T$ is compact, however you choose a subsequence of $u_n$ there is a sub-subsequence of $Tu_n$ that converges. By assumption, each and every one of these sub-subsequences converge to the same limit. This is, roughly speaking, how you should reason. Hope this helps $\endgroup$ Commented Dec 16, 2019 at 9:36

This is also true by the following reason:

Since you've already proved that there is a strongly convergent subsequence, let's say $ Tu_{n_k} \to u^* $ for $ k \to \infty $. Then by the weak convergence of $ u_n \rightharpoonup u $ you get immediately that $ Tu_n \rightharpoonup Tu $. Now since strong convergence implies weak convergence and from the uniqueness of the limit of a weak convergent sequence it must be true that

$ u^*= Tu $.

Therefore $ Tu $ is a limit point of the sequence $ (Tu_n)_{n \in \mathbb{N}} $. Now there's just one thing left to prove your statement

Claim: There's no other limit point, hence it must be the limit.

Proof: Suppose there's another limit point $ z $ of the sequence $ (Tu_n)_{n \in \mathbb{N}} $. Again there must be a subsequence $ (Tu_{n_m})_{m \in \mathbb{N}} $ converging to $ z $. Hence this subsequence $ u_{n_m} \rightharpoonup u$. Last step, use the same argumentation as above to conclude that $ z = Tu = u^*$.

Therefore $ Tu_n \to Tu $ as $ n\to \infty $.

To be precise, at this point you know that

$ Tu_{n_k} \to Tu $ and that the limit $ Tu $ is the only limit point of $ (Tu_n)_{n\in \mathbb{N}} $. Use now a contradiction argument to prove that $ Tu_n\to Tu $.

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    $\begingroup$ Why is it "immediate" that you get $Tu_n \rightharpoonup Tu$ from the weak convergence of the $u_n$? $\endgroup$
    – maximumtag
    Commented Mar 11, 2014 at 21:56
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    $\begingroup$ @maximumtag what is your definition of weak convergence? $\endgroup$
    – math
    Commented Mar 25, 2014 at 10:23
  • $\begingroup$ @maximumtag you can also use this fact: a linear operator is (norm to norm ) continuous if and only if it is weak to weak continuous. [ see Megginson "An introduction to Banach space theory", theorem 2.5.11]. In this context, a compact operator is of course continuous. $\endgroup$
    – MathGuy
    Commented Feb 19, 2020 at 20:52
  • $\begingroup$ @MartinArgerami Hi Martin, what you are talking about seems to be the weak-$*$ convergence unless the bracket means inner product, but our spaces may not be Hilbert spaces. $\endgroup$
    – Sam Wong
    Commented Apr 5, 2020 at 14:23
  • $\begingroup$ @SamWong: it's hard to know what I was talking about a year and a half ago since the question I was answering has been deleted. I'll delete that comment. $\endgroup$ Commented Apr 5, 2020 at 16:26

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