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Let $ \mathcal{H} $ and $ \mathcal{K} $ be Hilbert spaces, and let $ T: \mathcal{H} \to \mathcal{K} $ be a bounded linear operator. Show that if $ T $ is a compact operator, then $$ \lim_{n \to \infty} \| T(e_{n}) \|_{\mathcal{K}} = 0 $$ for every orthonormal sequence $ (e_{n})_{n \in \mathbb{N}} $ in $ \mathcal{H} $. Is the converse of this statement true?

Thanks.

  • 0
    I guess $H$ is a Hilbert space. What is $K$? Use the fact that the sequence $\{e_n\}$ is weakly convergent to $0$ and the fact that a compact operator transform a weakly convergent sequence in a convergent sequence (for the norm).2011-09-21
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    could you explain that why $\{e_{n}\}\to 0$ weakly?2011-09-21
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    This really reads like a homework question; so perhaps it would be good to consult the FAQ: http://meta.math.stackexchange.com/questions/1803/how-to-ask-a-homework-question2011-09-21
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    If the Hilbert space $H$ is separable, we can show that $\lim_{n\to\infty}\langle e_n,v\rangle=0$ for all $v$ with $v=\sum_{k=1}^N\alpha_ke_k$. We can conclude that $\lim_{n\to\infty}\langle e_n,v\rangle=0$ for all $v$, since the vectors of the form $\sum_{k=1}^N\alpha_ke_k$, $N\in\mathbb N,\alpha_k\in\mathbb C$ is dense in $H$.2011-09-21
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    @DavideGiraudo: Another way to see this is that by Bessel's inequality, $\sum_{n=1}^\infty |\langle e_n, v \rangle|^2 \le ||v||^2 < \infty$. Since the series converges, its terms must go to 0.2011-09-21
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    what about the converse of the theorem? I think it holds in Hilbert separable space because basically $\|Te_n\|\to 0$ holds even for the orthonormal countable basis, which exists for sure. But what about non separable Hilbert space? Is the converse still true?2011-09-23
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    @uforoboa: I initially had trouble with the non-separable case, until I obtained the solution explained below. After doing some more research, I discovered that Paul Halmos had proved the converse result in the same manner, by exploiting Hilbert space geometry. If I am not wrong, the result is true even if one does not assume $ T $ to be a bounded operator, just one that sends every orthonormal sequence to $ 0_{\mathcal{H}} $.2012-10-14
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    @MatthewDaws In fact, it's a verbatim of Conway's *A Course in Functional Analysis*, Chapter II, section 5, exercise 6.2014-12-20
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    @Vahid: Don’t forget to vote for the best answer. You don’t seem to have done that for the questions that you’ve asked so far.2014-12-23

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