# Question about compact operator in a Hilbert space.

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?

• I'm assuming $A$ should be $T$ in your post. Commented Jul 21, 2022 at 6:09

$$\text{Id}:\ell^2\to \ell^2$$ 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.

• 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 ? Commented Jul 19, 2022 at 3:31