# K(X,Y) is a linear subspace of B(X,Y)

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?

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

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.
• "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? Jul 12, 2021 at 12:22
• 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 Jul 12, 2021 at 12:31
• 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 Jul 12, 2021 at 13:48