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Let $\;H\;$ be a Hilbert space and consider $\;T_n:H\rightarrow H \;$ a sequence of compact operators. It is known that if $\; \vert \vert T_n - T \vert \vert \to 0\;$ as $\;n\to \infty\;$ then $\;T\;$ is also compact.

Substituting the above convergence by $\;T_nx \to Tx\; \forall x\in H\;$ the statement isn't valid.

I'm trying to find a counterexample to confirm this but I lack good ideas. Any help would be valuable.

Thanks in advance!!

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Define $T_n : \ell^2 \to \ell^2$ by $T_nx = (x_1, x_2, ..., x_n,0,..)$, then we have that $\dim \operatorname{ran} T_n < \infty$ and hence it is compact. We also have ($x \in \ell^2$) $$\|T_nx - Tx\|_2^2 = \sum_{i=n}^\infty x_i^2 \to 0$$

But $T_n$ converges to the identity operator, which is not compact since $\dim \ell^2 = \infty$.

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  • $\begingroup$ Thanks a lot for your quick reply. I guess $\;l^2\;$ space is always appropriate for counterexamples! :) $\endgroup$ – kaithkolesidou Jun 5 '17 at 11:25

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