Let $T\in \ell^1$, $Tx = (\lambda_1x_1,\dots,\lambda_nx_n,\dots)$. Want to show that if $T$ is compact, then $\lambda_n\to0$.
I know for $p\in(1,\infty]$, canonical basis $e_n \rightharpoonup 0$ (so $T(e_n)\rightharpoonup 0$), and $T(e_n)\in \overline{T(B(0,1))}$, which is compact, so we can get $T(e_n)\to 0$, and $\lambda_n\to 0$.
But what should I do with this problem when $p=1$, there is no weak convergence to $0$. Can someone help me with this? Thanks
Best Answer
Suppose that $T\colon\ell^1 \to \ell^1$ is compact. Then the sequence $\left(Te_n\right)_{n\geqslant 1}$ admits a subsequence $\left(Te_{n_k}\right)_{k\geqslant 1}$ which converges to some $v$ (strongly) in $\ell^1$. Look at $\left\lVert Te_{n_{k+1}}-Te_{n_k}\right\rVert_1$ to conclude that $\lambda_{n_k}\to 0$.
Now apply the previous result to $\left(Te_{N_j}\right)_{j\geqslant 1}$ for a fixed sequence $N_j\uparrow \infty$ instead of $\left(Te_n\right)_{n\geqslant 1}$ to see that each subsequence of $\left(\lambda_n\right)_{n\geqslant 1}$ admit a further subsequence with converges to $0$.