I have the following proof in my notes:

T and S are compact operators. We need to proof that a linear combination is still a compact operator Let $\{x_n\}$ be a bounded sequence in $X$, then because $T$ is compact there exists a convergent subsequence $\{Tx_{n_k}\}$ . Because $S$ in compact, there exists a convergent subsequence $\{Sx_{n_{k_j}}\}$. Hence, $\alpha Tx_{n_{k_j}} + \beta Sx_{n_{k_j}}$ is convergent.

I'm lost after getting $\{Tx_{n_k}\}$ . In particular I don't understand why they are taking a subsequence of a subsequence here $\{Sx_{n_{k_j}}\}$, since I don't know if $\{Tx_{n_k}\}$ or $\{Sx_{n_k}\}$ are bounded in order to extract a subsequence of that subsequence. I only know that $\{x_n\}$ is bounded.

I would do it like this: I can say for sure is that there exist convergent subsequences $\{Tx_{n_k}\}$ and $\{Sx_{n_k}\}$. Then I would say that the linear combination $\alpha Tx_{n_k} + \beta Sx_{n_k}$ is convergent so I have a convergent subsequence of $\{(\alpha T + \beta S)x_n\}$ , so $\alpha T + \beta S$ is compact. Isn't this correct?

  • $\begingroup$ What is $T$, what is $S$?, got some hypotheses you’d like to share? Limited is a strange term, do you mean bounded? $\endgroup$ Commented Jul 12, 2021 at 11:54
  • $\begingroup$ @CharlieFrohman Sorry I meant bounded. I fixed it and added the hypotheses $\endgroup$ Commented Jul 12, 2021 at 11:55

1 Answer 1


No, your argument is not correct. You can only say that $Tx_{n_k}$ is convergent for some $(n_k)$ and $Sx_{m_k}$ is convergent for some $(m_k)$. You do not know that you get the same subsequence for $T$ and $S$.

After getting $x_{n_k}$ you look at the new bounded sequence $(x_{n_k})$ and use compactness of $S$ to get a further subsequence $(n_{k_j})$ with $(Sx_{n_{k_j}})$ convergent. Then note that $(Tx_{n_{k_j}})$ is subsequence of $(Tx_{n_k})$ so it is also convergent. Now you see that $\alpha Tx_{n_{k_j}}+\beta Sx_{n_{k_j}}$ is convergent.

  • $\begingroup$ "note that $(Tx_{n_{k_j}})$ is subsequence of $(Tx_{n_k})$ so it is also convergent". Don't I need to know if the set I am extracting the subsequence from is limited first in order to apply Bolzano-Weierstrass? How to know if $(Tx_{n_k})$ is limited? $\endgroup$ Commented Jul 12, 2021 at 12:22
  • $\begingroup$ Convegent sequneces are bounded (limited in your langauge) but I don't see why you want to prove that $T(x_{n_k})$ is bounded. I am not applying compactness of $T$ at this stage. I am just using the fact that any subsequnce of a convergent sequence is convergent. @J.C.VegaO $\endgroup$ Commented Jul 12, 2021 at 12:31
  • $\begingroup$ Bolzano-Weistrass's theorem states that to each bounded sequence has a convergent subsequence. If the sequence is $Tx_{n_{k}}$ . I thought we had to use that theorem to extract the convergent subsequence$Tx_{n_{k_j}}$ , for which I first would have to make sure $Tx_{n_{k}}$ is bounded $\endgroup$ Commented Jul 12, 2021 at 13:48
  • $\begingroup$ I guess I can't use that since the sequence is not necessarily in an Euclidean space, can I? $\endgroup$ Commented Jul 12, 2021 at 13:57

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