# Compact operator $L:\ell^2\to\ell^2$ with $\Vert L\Vert=1$ such that $\Vert L(x)\Vert<\Vert x\Vert$ for all $x$

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.

• 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 0:29
• @Anonymous I'm not generating to all the bounded linear operator. – JRen Jan 17 '19 at 17:13

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 $$1=\|y\|=\|Tx\|$$.
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 $$T(B)$$.