# Compact operator on a hilbert space. [duplicate]

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A compact operator is completely continuous.

I came across this question in a book, it is an exercise, I can't prove it. Can someone please help me.

If $X$ and $Y$ are Banach spaces, we have to prove that a compact linear operator is completely continuous. A mapping $T \colon X \to Y$ is called completely continuous, if it maps a weakly convergent sequence in $X$ to a strongly convergent sequence in $Y$ , i.e., $x_n\underset{n\to +\infty}\rightharpoonup x$ implies $\lVert Tx_n- Tx\rVert_Y\to 0$.

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## marked as duplicate by t.b., Davide Giraudo, Rudy the Reindeer, robjohn♦, anonJul 10 '12 at 12:15

So... johnathan's homework can be found verbatim in carla's book? :) which book? – t.b. Jul 10 '12 at 9:35
A study guide to functional analysis by R. K. Chopra. – carla Jul 10 '12 at 9:39
@t.b. ping on behalf of carla. – Rudy the Reindeer Jul 10 '12 at 10:43

Suppose not. Then there is some sequence $(x_n)$ in $X$ with $x_n \rightharpoonup x$ but $Tx_n \not\to Tx$. That is, there is some $\epsilon > 0$ such that $\|Tx_{n_j} - Tx\| \ge \epsilon$ for some subsequence $(x_{n_j})$. As $(x_{n_j})$ is bounded and $T$ is compact, some subsequence converges in $Y$, say $Tx_{n_{j_k}} \to y$. As $T$ is norm-to-norm continuous, it is weak-to-weak continuous, hence $Tx_{n_{j_k}} \rightharpoonup Tx$. Weak limits are unique, so $Tx = y$, that is $Tx_{n_{j_k}} \to Tx$, contradicting the choice of $(x_{n_j})$.
Hence $T$ is completely continuous.