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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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The orthogonal of $V_{\varepsilon}$ has a finite dimension. Take $\{x_n\}$ a sequence which converges weakly to $0$ and bounded by $1$. Fix an orthonormal basis $\{e_1^{\varepsilon},\dots,e_{N_{\varepsilon}}^{\varepsilon}$ of the orthogonal of $V_{\varepsilon}$. Write $x_n=\underbrace{x_n^{\varepsilon}}_{\in V_{\varepsilon}} +\sum_{j=1}^{N_{\varepsilon}}\alpha_{n,j}^{\varepsilon}e_j^{\varepsilon}$. Then use the weak converges with the linear functional $L_J\colon x\mapsto \langle x,e_j^{\varepsilon}\rangle$ to see that $\alpha_{n,j}^{\varepsilon}\to 0$. – Davide Giraudo Jul 10 '12 at 9:58
With this proof, I think we need that $X$ is separable. – Andrea Jul 10 '12 at 12:03
Where is the separability needed? – Davide Giraudo Jul 10 '12 at 12:12
If $T \colon X \to Y$ is sequentially continuous from weak topology to strong topology and $X$ is separable and reflexive, then $T$ is compact. – Andrea Jul 10 '12 at 12:18
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, $Tx_{n_k}\to Tx$. – Davide Giraudo Jul 10 '12 at 12:24
up vote 5 down vote accepted

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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Simpler than what I had. We notice that we don't need $T$ to be injective. – Davide Giraudo Jul 10 '12 at 10:57
Good answer! :) – Andrea Jul 10 '12 at 12:03

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