Given the operator $T:l^1\to c_0$ with $T(x=(x_k))=\left(\sum\limits_{k=1}^\infty x_k,\sum\limits_{k=2}^\infty x_k,…,\sum\limits_{k=n}^\infty x_k,…\right)$.
I have to decide if this is a compact operator. I think it is, and to prove it I have the sequence of operators:
$T_n(x=(x_k))=\left(\sum\limits_{k=1}^\infty x_k,\sum\limits_{k=2}^\infty x_k,…,\sum\limits_{k=n}^\infty x_k,0,0,…\right)$
whose image is finite dimensional. So if I prove that $T_n\to T$, I have that $T$ is compact.
I get $||T_n(x)-T(x)||\leq \sum\limits_{k=n+1}^\infty|x_k|$, but I want to find $||T_n(x)-T(x)||\leq \varepsilon_n||x||$ where $\varepsilon_n\to0$. If I prove that, I have that $||T_n-T||\leq\varepsilon_n\to 0$ and I conclude my proof.
Can somebody help me to find that $\varepsilon_n$? Thank you
Best Answer
In fact it's pretty clear that $||T_n-T||=1$ (for example because the standard isomorphism of $\ell_1^*$ and $\ell_\infty$ is an isometry.) That doesn't prove that $T$ is not compact, but it's not; for example $||Te_n-Te_m||=1$ for $n\ne m$, so the sequence $(Te_n)$ has no convergent subsequence.