# Compact operator maps weakly convergent sequences into strongly convergent sequences

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.

• @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. – t.b. Oct 6 '11 at 18:19
• @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. – Beni Bogosel Oct 6 '11 at 18:21

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 $x_n \to x$.

• This surely is a very nice argument. Thank you. – Beni Bogosel Oct 6 '11 at 18:22
• You're welcome! This little lemma is simple yet useful to know, IMHO. – Giuseppe Negro Oct 6 '11 at 21:29
• One of the most beautiful and useful techniques in analysis and problems in sequences. I love it – Fardad Pouran May 8 '18 at 8:29
• I don't understand how this lemma can help us? The lemma speaks about every subsequence. While we found only one convergent subsequence? – Kamil Dec 15 '19 at 12:30
• @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 – Giuseppe Negro Dec 16 '19 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 $Tu_n\to Tu$.

• Why is it "immediate" that you get $Tu_n \rightharpoonup Tu$ from the weak convergence of the $u_n$? – maximumtag Mar 11 '14 at 21:56
• @maximumtag what is your definition of weak convergence? – math Mar 25 '14 at 10:23
• @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. – MathGuy Feb 19 at 20:52
• @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. – Sam Wong Apr 5 at 14:23
• @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. – Martin Argerami Apr 5 at 16:26