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I have the suspect that the following statement is true, but I don't how to prove it. Any suggestion? Thanks to all!

Let $X$, $Y$ be Hilbert spaces and let $T \colon X \to Y$ be a linear continuous injective map. Suppose that for every $\epsilon > 0$ there exists a closed vector subspace $V_\epsilon \subseteq X$ of finite codimension such that $\Vert Tv \Vert_Y \leq \epsilon \Vert v \Vert_X$ for all $v \in V_\epsilon$. Then $T$ is compact.

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    I'm not sure we need separability (just reflexivity). Indeed, we have to show that $T(B)$, where $B$ is the closed unit ball, is compact. Sequentially compact is enough, since we are in a metric space. To see that, take $\{y_n\}\in T(B)$, then $y_n=Tx_n$ for $x_n\in B$. We extract a weakly converging subsequence (to some $x$), and by what we showed, $Tx_{n_k}\to Tx$.2012-07-10

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Let $P_\epsilon$ be the orthogonal projection onto $V_\epsilon$ and $ T_n := T(\mathbb I - P_{1/n}) $ $T_n$ is a finite-rank operator and $ \lVert (T - T_n)v \rVert = \lVert T P_{1/n} v \rVert \leq \frac 1 n \lVert v \rVert $ and so $ \lVert T - T_n \rVert \leq \frac 1 n \to 0 $ We can conclude $T$ is compact because it is limit in the operator norm of a sequence of finite-rank operators.

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    Good answer! :)2012-07-10