# Compact Operators: Weak Convergence [duplicate]

Problem

Given Banach spaces $X$ and $Y$.

Consider a compact operator $C\in\mathcal{C}(X,Y)$.

Then weak convergence is turned into strong convergence: $$x_n\rightharpoonup x\implies Cx_n\to Cx$$ I'd like to try proving it but would need some hints.

Attempt

Denote the sup-norm by: $$F:\Omega\to E:\quad\|F\|_\Omega:=\sup_{\omega\in\Omega}\|F(\omega)\|_E$$

Weak convergence is preserved under continuous operators: $$\|l(Cx_n-Cx)\|=\|(C'l)(x_n-x)\|\to0$$ By uniform boundedness principle weak convergence implies boundedness: $$x_n\rightharpoonup x:\quad\|l(x)\|_\mathbb{N}<\infty\implies\|x\|_\mathbb{N}<\infty$$ Hence one can exploit compactness of the operator: $$(x_n)_{n\in\mathbb{N}}\text{ bounded}\implies C(x_n)_{n\in\mathbb{N}}\text{ precompact}$$ And one obtains strongly convergent subsequences.

Should I combine these now and how?

## marked as duplicate by AlexR, user147263, ncmathsadist, apnorton, Daniel W. FarlowFeb 11 '15 at 1:37

As a general principle, if we have a unique possible limit, together with some form of sequential compactness, we get convergence. This is a corollary of the following: $x_n \to x$ if and only if for every subsequence $(x_{n_k})$, there is a further subsequence $x_{n_{k_j}} \to x$ (To prove the less obvious implication, suppose it doesn't converge and consider a sequence bounded away from $x$ by some $\epsilon > 0$).
We can apply this to the given problem as follows: $Tx_n \rightharpoonup Tx$ by continuity. Thus if any subsequence has a strong limit, it certainly is $Tx$. But compactness guarantees every subsequence has a subsequence that converges to something: that something is $Tx$ by uniqueness, and so by our above equivalence with convergence, we have $Tx_n \to Tx$.
• Ah ok so one derives a contradiction by: $x_n\nrightarrow x:\quad\|x_{n_k}-x\|\geq\varepsilon,\|x_{n_{k_j}}-x\|\to0$ – C-Star-W-Star Feb 10 '15 at 19:12