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Let $\ell^2$ denote the space of square summable sequences of complex numbers. Let $L:\ell^2\to\ell^2$ be a linear operator with $\Vert L\Vert=1$ such that for all $x\in\ell^2\setminus\{0\}$, $\Vert L(x)\Vert_2<\Vert x\Vert_2$. Give an example of a compact operator with these properties or show that no such compact operator exists.

I thought of some bounded operators with the above properties and none of them were compact, so I feel like no such compact operator exists. Since $\Vert L\Vert=\sup_{\Vert x\Vert_2=1}\Vert L(x)\Vert_2=1$, there exists a sequence $x_n\in\ell^2$ such that $\Vert x_n\Vert_2=1$ for all $n$ and $\Vert L(x_n)\Vert_2\to 1$. My idea was to show that no subsequence of $L(x_n)$ converges (as that tended to be the case in the examples I thought of) which would show that the image of the closed unit ball is not contained in any compact set and hence, $L$ is not compact. But I'm not sure whether that is actually true, and if it is true, I'm not sure how to prove it.

Anonymous
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    The eigenvalues of compact operators can only accumulate at 0. Guess no. – JRen Jan 16 '19 at 23:40
  • @T.Bongers I don't get the part where you assume that $x_n$ can pass to a subsequence that converges. Why is that so? – BigbearZzz Jan 16 '19 at 23:46
  • @JRen Oh, that's right, I totally forgot that property. For any compact operator, that operator only has finitely many eigenvalues outside a ball centered at the origin of the complex plane. Thus, the sequence $x_n$ can't have the property that $\Vert L(x_n)\Vert_2\to 1$ if $L$ were compact. Thanks! – Anonymous Jan 16 '19 at 23:48
  • @JRen On second thought, looking at the eigenvalues doesn't appear to work. The mapping $L(x_1,x_2,...)=(0,0,\frac{1}{2}x_2,\frac{2}{3}x_3,\frac{3}{4}x_4,...)$ is a (not compact) bounded linear operator with the above properties which has no nonzero eigenvalues. – Anonymous Jan 17 '19 at 00:29
  • @Anonymous I'm not generating to all the bounded linear operator. – JRen Jan 17 '19 at 17:13

2 Answers2

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Let $T$ be a compact operator with $\|T\|=1$. Then there is a sequence $(x_n)$ in $H$ with $\|x_n\|=1$ with $\|Tx_n\|\to1$ as $n\to\infty$. As $T$ is compact, there is a subsequence $(x_{n_k})$ of $(x_n)$ such that $Tx_{n_k}\to y$ for some $y\in H$. But as the image closed unit ball $B$ under $T$ is compact (see Douglas, Banach Algebra Techniques in Operator Theory, Corollary 5.4), it follows that $y\in T(B)$, i.e., there is some $x\in H$ with $\|x\|\leq 1$ and $y=Tx$. But then $\|x\|=1$, and $1=\|y\|=\|Tx\|$.

Aweygan
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    https://math.stackexchange.com/questions/404516/image-of-closed-unit-ball-under-a-compact-operator – Tsemo Aristide Jan 16 '19 at 23:53
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    The image of the closed unit ball under a compact operator is relatively compact. Why is it compact? – Kavi Rama Murthy Jan 16 '19 at 23:56
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    @KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced. – Aweygan Jan 16 '19 at 23:58
  • @KaviRamaMurthy According to Tsemo's link, it looks like this is true in any reflexive Banach space (and hence for Hilbert spaces), which is quite interesting. – Anonymous Jan 17 '19 at 00:00
  • @Aweygan I really like this solution! I had no idea the image of the closed unit ball was compact under a compact operator for Hilbert spaces. That's good to know. I'm also going to add my own answer using a suggestion from one of the comments. Thank you! – Anonymous Jan 17 '19 at 00:02
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    Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity. – Kavi Rama Murthy Jan 17 '19 at 00:36
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    @Anonymous I would be very interested to see if the approach hinted at by JRen does work. It's fairly easy if you assume the operator is normal, but in general I can't see it. – Aweygan Jan 17 '19 at 00:58
  • @Aweygan I was actually about to write up the solution using the things JRen noted about eigenvalues, but found that it doesn't quite work. At first glance, it looks like a bounded linear operator with the properties stated in my question would eigenvalues arbitrarily close to the unit circle in the complex plane, but as seen in my most recent comment, that's not actually the case. If it were the case, it would contradict the fact that a compact operator can only have finitely many eigenvalues outside any ball centered at the origin in the complex plane. – Anonymous Jan 17 '19 at 01:12
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The same argument without touching sequences: As $T$ is weakly continuous and the unit ball of $\ell^2$ is weakly compact its impage $T(B)$ is weakly compact hence weakly closed and thus closed in $\ell^2$. At the same time it is relatively compact and thus compact. Hence the continuous map $y\mapsto \|y\|$ realizes its supremum on $T(B)$.

Jochen
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