If X and Y are Banach spaces, and $A: X \to Y$, $B: X \to Y$ are both compact operators, then $A + B$ is compact.
A + B is compact if and only if for every bounded sequence $\lbrace x_n \rbrace$ in X, the sequence $\lbrace (A + B) x_n \rbrace$ has a convergent subsequence; certainly both $\lbrace A x_n \rbrace$ and $\lbrace B x_n \rbrace$ each have a convergent subsequence, and I've seen a few proofs that seem to think it straightforward that this implies $\lbrace (A + B) x_n \rbrace$ has one. This doesn't seem quite right to me.
The way it appears one should try to make a convergent subsequence is to intersect the indices of the convergent subsequences for A and B individually. So, for example, if $\lbrace A x_{n_j} \rbrace$ and $\lbrace B x_{n_k} \rbrace$, then the idea is to take the set of indices $\lbrace n_m \rbrace := \lbrace n_j \rbrace \cap \lbrace n_k \rbrace$ and say that $\lbrace (A + B) x_{n_m} \rbrace$ converges as a sum of two further subsequences of the convergent subsequences for A and B.
But what if that intersection is empty, for example if $\lbrace n_j \rbrace = 1, 3, 5, …$ and $\lbrace n_k \rbrace = 2, 4, 6, …$? Instinctively I would say "if they're disjoint, just pick a different pair", but how can one be sure that the intersection isn't empty for any chosen pair of convergent subsequences?
To be clear, I believe that A + B is compact, it's this form of proof of the fact that seems dubious to me.
Best Answer
Find a convergent subsequence for $A$, then take a sub-subsequence for $B$. You can't necessarily intersect two subsequences, but you can easily take a sub-subsequence.