# A compact operator is completely continuous.

I have a question. 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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Since it's a homework question I will just give some steps.

1. By linearity, we can assume that $x=0$.
2. We have to show that for each subsequence of $\{Tx_n\}$, we can extract a further subsequence which converges to $0$ in norm in $Y$.
3. A weakly converging sequence is bounded.
4. $T$ maps bounded sets to sets with a compact closure.

Once the second steps is shown, we can conclude. Indeed, assume that $Tx_n$ doesn't converge to $0$. Then we are able to find $\delta>0$ and $A$ an infinite subset of the natural numbers such that $\lVert Tx_k\rVert_Y\geq\delta$ for each element of $A$. We can consider it as a subsequence, and we can't extract a further subsequence which converges to $0$, a contradiction.

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How do we show the second step? – johnathan Jul 9 '12 at 21:52
Could you please give some details. Thank you. – johnathan Jul 9 '12 at 21:55
Both, actually. – johnathan Jul 9 '12 at 22:01
It fact the step 2.,3. and 4. are linked (you have to use 3. and 4. to show the assertion of 2.). – Davide Giraudo Jul 9 '12 at 22:06