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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.

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    $\begingroup$ The eigenvalues of compact operators can only accumulate at 0. Guess no. $\endgroup$
    – JRen
    Commented Jan 16, 2019 at 23:40
  • $\begingroup$ @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? $\endgroup$
    – BigbearZzz
    Commented Jan 16, 2019 at 23:46
  • $\begingroup$ @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! $\endgroup$
    – Anonymous
    Commented Jan 16, 2019 at 23:48
  • $\begingroup$ @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. $\endgroup$
    – Anonymous
    Commented Jan 17, 2019 at 0:29
  • $\begingroup$ @Anonymous I'm not generating to all the bounded linear operator. $\endgroup$
    – JRen
    Commented Jan 17, 2019 at 17:13

2 Answers 2

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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 $1=\|y\|=\|Tx\|$.

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    $\begingroup$ math.stackexchange.com/questions/404516/… $\endgroup$ Commented Jan 16, 2019 at 23:53
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    $\begingroup$ The image of the closed unit ball under a compact operator is relatively compact. Why is it compact? $\endgroup$ Commented Jan 16, 2019 at 23:56
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    $\begingroup$ @KaviRamaMurthy It's actually compact in the Hilbert space setting. See the book I referenced. $\endgroup$
    – Aweygan
    Commented Jan 16, 2019 at 23:58
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    $\begingroup$ Thanks guys for educating me on the image of the closed unit ball under a compact operator under reflexivity. $\endgroup$ Commented Jan 17, 2019 at 0:36
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    $\begingroup$ @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. $\endgroup$
    – Aweygan
    Commented Jan 17, 2019 at 0:58
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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 $T(B)$.

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