I am reading Compact operator in a Hilbert space which is defined as
Let $H$ be a Hilbert space and $T$ be a bounded operator, we say $T$ is a compact operator if for any bounded sequence $x_n$ in $H$, The image $A(x_n)$ has a convergent subsequence.
I want to construct an example for a non-compact operator in $l^2$ for that I am thinking to define an identity map in $l^2$ but I am not getting which sequence to define in $l^2$ so that it is convergent but it has no convergent subsequence. Am I missing something here?
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$\begingroup$ I'm assuming $A$ should be $T$ in your post. $\endgroup$– Disintegrating By PartsCommented Jul 21, 2022 at 6:09
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$\text{Id}:\ell^2\to \ell^2$ is not compact operator as $\text{Id}({\overline{B(0, 1)}}) =\overline{B(0, 1)}$ is not compact.
A norm space is finite dimensional iff closed unit ball is compact. $\ell^2$ is not finite dimensional.
More specifically, choose $(e_n) _{n\in\Bbb{N}}$ where $e_n=(0, 0,\ldots, 1,0,\ldots) $ i.e $1$ in the $n$- th place and zeros everywhere.
Then $\|e_n\|_2=1 \space \forall n$ implies $(e_n) $ is a bounded sequence in $\ell^2$ space but This sequence doesn't have any convergent subsequence.
I am not getting which sequence to define in $ l_2 $ so that it is convergent but it has no convergent subsequence.
This is impossible as every subsequence of a convergent sequence is convergent.
answered Jul 18, 2022 at 18:43
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$\begingroup$ if I define a sequence $x_n$ in $l^2$ and a Operator $T \colon l^2 \to l^2$ such that $T(x_n)$=$n^2x_n$ where $x_n$=$\sqrt{\frac{1}{n}}$,is it correct example ? $\endgroup$ Commented Jul 19, 2022 at 3:31