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Let $T:l^2 \to l^2$ be $T(a_1,a_2,...)=(a_2,a_3,...)$. Is this linear operator compact. If yes, how to prove it? If no, please give an example.

I want to show that if $(a_2,a_3,...)$ has a cluster point or not. I think it is sufficient to show that if $(a_1,a_2,...)\in l^2$ , does it have a cluster point?

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    $\begingroup$ If $B$ denotes the unit ball of $\ell^2$, what is $T(B)$? Is it (pre)compact? $\endgroup$ – Aweygan Nov 2 '17 at 0:00
  • $\begingroup$ @NYRAHHH: You are not going to find a single point with a cluster point in any normed space. You have to consider sets or sequences of elements of $\ell^2$ to make sense using the definition of compact operator. $\endgroup$ – Jonas Meyer Nov 2 '17 at 0:02
  • $\begingroup$ Look at @Aweygan's suggestion for a straightforward hint that works directly from the definition. $\endgroup$ – copper.hat Nov 2 '17 at 0:09
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    $\begingroup$ You wrote down the left shift operator $\endgroup$ – Andres Mejia Nov 2 '17 at 0:12
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If $T$ were compact then the image of any bounded sequence would contain a convergent subsequence. However, you can take the sequence $x_n=(0,\dots,0,1,0,0,\dots)$ with a $1$ in the $n$th position.

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  • $\begingroup$ This is nice as well +1 $\endgroup$ – Andres Mejia Nov 2 '17 at 0:24
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Some quick explanations already, but here's a nice quick proof:

It is well known that $T^*$ is the right-shift operator, which is an isometry. We see that $TT^* = I$ is not compact, which means that neither $T$ nor $T^*$ can be compact (since the compact operators are absorbing under composition).

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  • $\begingroup$ Why do we need to mention that the left-shift is an isometry, or that it's adjoint to $T$? Is it not just enough to note that if we let $S$ be the left-shift then $TS=I$, which is not compact, so $T$ isn't either? Either way nice solution, +1 $\endgroup$ – Alex Mathers Nov 2 '17 at 0:18
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    $\begingroup$ @AlexMathers I only mention it because it makes the solution more intuitive to me; $T^*$ is an isometry means that $T^{**}T^* = I$, and so on. In particular, this generalizes to the statement that neither an isometry nor its adjoint can be compact (in an infinite dimensional Hilbert space). $\endgroup$ – Omnomnomnom Nov 2 '17 at 0:19
  • $\begingroup$ +1, this is nice, I tried to think of something clever using the adjoint, but didn't manage. $\endgroup$ – Andres Mejia Nov 2 '17 at 0:24
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No. If $T$ were compact, then every $ 0 \neq \lambda \in \sigma(T)$ would be an eigenvalue.

On the other hand, $\sigma(T)$ is the unit disk in $\mathbb C$, but we will not need the full strength of this. Note that $T(1, \lambda, \dots)=(\lambda, \lambda^2, \dots)$ for any $|\lambda|<1$, so that $$\{\lambda \in \mathbb C \mid |\lambda|<1\}\subset \sigma(T),$$

and taking it closure will give the closed unit desk. Consider $\lambda=1$.

If this were an eigenvalue, then $(a_2-a_1,a_3-a_2, \dots)=0$. But this in turn implies that $a_1=a_2=a_3, \dots$, whch is in $\ell^1$ if and only if $a_1=0$.

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  • $\begingroup$ Well well well! $\endgroup$ – Alex Mathers Nov 2 '17 at 0:38
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    $\begingroup$ oh my, it was sheaf keef all along! Hope all is well, it looks like you got around to that functional we spoke about. $\endgroup$ – Andres Mejia Nov 2 '17 at 0:43

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