# Show compactness/ noncompactness of an operator by approximation

I have to show whether the following operator is compact or not: $$T\colon\ell^2\to\ell^2: (x_n)_{n\in\mathbb{N}}\mapsto\left(\frac{x_n+x_{n+1}}{2}\right)$$

My idea was to approximate $T$ by operators with finite rank but I cannot find such a sequence of operators...

Do you have an idea how to choose such a sequence?

EDIT: Concept of rank was confused with range.

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@David: almost: $T(e_n) = (e_{n-1} + e_n)/2$. –  Martin Feb 17 '13 at 12:34
@Martin Blech. Thanks –  David Mitra Feb 17 '13 at 12:38

The operator is not compact. As @Martin suggested $e_{4n} →0$ weakly but $∥Te_{4n} ∥=const$.

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$\lim_{n\to\infty} T_n = T$ in what sense? Stronly, yes. Not in norm: $\lVert Te_{2n} - T_{n}e_{2n}\rVert = \frac{1}{\sqrt{2}}$. This is not enough to deduce compactness of $T$. –  Martin Feb 17 '13 at 12:03
I have no idea what you want to use Banach-Saks for. The operator is not compact: $e_{4n} \to 0$ weakly but $\lVert Te_{4n} \rVert = \text{const.}$, for example. –  Martin Feb 17 '13 at 12:16
I don't follow your argument: you show that for every bounded sequence you can produce a norm-convergent sequence (not a subsequence!) and of course every bounded operator sends null-sequences to null-sequences. This shows nothing. An operator is compact if and only if it sends weakly convergent sequences to norm-convergent sequences and I showed you a weakly convergent sequence whose image is not norm convergent. –  Martin Feb 17 '13 at 12:28
Let me see if I got it: $T$ is NOT compact, because if it were compact, for every bounded sequence the range-sequence would have a convergent subsequence. But: Choose $(e_n)_{n\in\mathbb{N}}=((1,0,0,...),(0,1,0,0,...),...)$. This is a bounded sequence because $\lVert e_n\rVert_{\ell^2}=1<\infty$. But $\lVert Te_n-Te_m\rVert_{\ell^2}=\frac{1}{\sqrt{2}}$ and so $(Te_n)$ cannot have a convergent subsequence, because this convergent subsequence would be automatically a Cauchysequence what is, as shown, impossible. Is that right? –  math12 Feb 17 '13 at 13:01
@math12: yes, that's exactly right. –  Martin Feb 17 '13 at 13:28