Exercise about compact operator.

Question

In $X=\ell^p$, $p\in[1,\infty]$ we consider: $$ T(x_1,x_2,x_3,\ldots)=(0,x_1,0,x_3,\ldots) $$ Prove that $T$ isn't a compact operator and that $T^2$ is a compact operator. I think I solved the second point of the exercise because $T^2 x= 0$ always considering $T^2=T(T(x))$ (but I'm not sure this is $T^2$...). Anyway I'd like to find a sequence $\{x_n\}$ such that $x_n\rightharpoonup x$ but $Tx_n\nrightarrow Tx$.

Answer 1

Consider the subspace $U=\{(x_1,0,x_3,0,\dots)\}\subset\ell^p$. The restriction $T_{|_U}$ is the right shift operator on $U$. But we know that this is not a compact operator (e.g. looking at the spectrum).

Answer 2

Take the action of $T$ two distinct unit vectors $e_i$ (by vector I mean infinite dimensional) which is obviously bounded. Then use the equivalent formulation of a compact operator in terms of the properties of $(Tx_n)$.

Answer 3

Take $\boldsymbol{x}_n=\boldsymbol{e}_{2n+1},\,n\in\mathbb N$ - where $\boldsymbol{e}_{n}$is the element with all zeros except at position $n$, where it has an $1$. Then $\boldsymbol{x}_{n}$ is a bounded sequence in $\ell^p$, as $\|\boldsymbol{x}_{n}\|_p=1$, and $$T\boldsymbol{x}_n=T\boldsymbol{e}_{2n+1}=\boldsymbol{e}_{2n+2}$$ does not have a converging subsequence, since $$ \|\boldsymbol{e}_{m}-\boldsymbol{e}_{n}\|=2^{1/p}, $$ wherever $m\ne n$.