Consider $T\colon\ell^2\to\ell^2$ an operator such that $$T((x_n))=(2^{-n}x_n); \forall x=(x_n)\in \ell^2 $$ Does anyone know how to prove that it is compact?
I understand that a linear operator $T:E_1\to E_2$ is considered compact if for every bounded sequence $(a_n)$ in $E_1$; $(Ta_n)$ has a converging semisequence in $E_2$.
So I understand that I have to find a converging semisequence in $\ell^2$ and show that $(X_n)$ is bounded. I got some hints but Im having trouble in understanding the whole picture here.
Thank you for your help. I do not understand how to sum up the hints that I got in the previous question or how to comment there that I dont fully understand the answers. I think that in this question I added some data that is important for focusing on what Im trying to figure out. can someone give me a push on that? Thanks, Jeremy