# Strong limit of compact operators

$X$ and $U$ are Banach spaces. A linear map $\textbf{C} : X \rightarrow U$ is called compact if the image $\textbf{C}B$ of the unit ball $B$ in X is precompact in $U$. A subset S of a complete metric space is called precompact if its closure is compact

I cannot come up with an counter to prove that the strong limit of a sequence of compact operators need not to be compact.

Since all closed subsets of a compact space are closed, every linear map to a compact space is compact. Hence $U$ should not be compact at least..

• I think that would be hard since it was an exercise we had. If it (or I) did not do a typo. – Johan Dec 3 '12 at 11:57
• I just realize that by "strong limit", you mean pointwise limit (not limit in norm as I believed). – Davide Giraudo Dec 3 '12 at 12:01

Let $\{e_k:k\in\mathbb{N}\}$ be some orthnormal system of Hilbert space $H$. Consider orthogonal projections $$P_n:H\to H:x\mapsto \sum\limits_{k=1}^n\langle x,e_k\rangle e_k$$ They are of finite rank, hence compact. Thier strong limit is projection $P$ on $\mathrm{span}\{e_k:k\in\mathbb{N}\}$. You can show that $P$ is not compact by noticing that $P(B_H)$ is a unit ball of infinite dimensional subspace $\mathrm{span}\{e_k:k\in\mathbb{N}\}$.