Is left shift operator compact? Let $T:l^2 \to l^2$ be $T(a_1,a_2,...)=(a_2,a_3,...)$. Is this linear operator compact. If yes, how to prove it? If no, please give an example.
I want to show that if $(a_2,a_3,...)$ has a cluster point or not. I think it is sufficient to show that if $(a_1,a_2,...)\in l^2$ , does it have a cluster point?
 A: If $T$ were compact then the image of any bounded sequence would contain a convergent subsequence. However, you can take the sequence $x_n=(0,\dots,0,1,0,0,\dots)$ with a $1$ in the $n$th position.
A: Some quick explanations already, but here's a nice quick proof:
It is well known that $T^*$ is the right-shift operator, which is an isometry.  We see that $TT^* = I$ is not compact, which means that neither $T$ nor $T^*$ can be compact (since the compact operators are absorbing under composition).
A: No. If $T$ were compact, then every $ 0 \neq \lambda \in \sigma(T)$ would be an eigenvalue.
On the other hand, $\sigma(T)$ is the unit disk in $\mathbb C$, but we will not need the full strength of this.
Note that $T(1, \lambda, \dots)=(\lambda, \lambda^2, \dots)$ for any $|\lambda|<1$, so that
$$\{\lambda \in \mathbb C \mid |\lambda|<1\}\subset \sigma(T),$$
and taking it closure will give the closed unit desk. Consider $\lambda=1$.
If this were an eigenvalue, then  $(a_2-a_1,a_3-a_2, \dots)=0$. But this in turn implies that $a_1=a_2=a_3, \dots$, whch is in $\ell^1$ if and only if $a_1=0$.
