Example of a bounded operator on $\ell^2$ that does not preserve $\ell^1$?

Question

Does there exist a bounded operator $A:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$ such that for some $u\in\ell^1(\mathbb N)\subset\ell^2(\mathbb N)$, $Au\notin\ell^1(\mathbb N)$?

Answer 1

Following Jalex Stark's hint, let \begin{align*} v \doteq (1/1,1/2,1/3,\dots), \end{align*} so that $v \notin \ell^1$ and $v \in \ell^2$. Now define the bounded operator $A : \ell^2 \to \ell^2$ as, \begin{align*} Au \doteq \langle v,u \rangle v. \end{align*} Indeed, $A$ is bounded, since by Cauchy Schwartz, \begin{align*} \|A\| = \sup_{u \neq 0}\frac{\|Au\|}{\|u\|} = \sup_{u \neq 0}\frac{|\langle v,u\rangle|\|v\|}{\|u\|} \leq \sup_{u \neq 0} \|v\|^2 = \|v\|^2. \end{align*} Now take $u = (1/1^2,1/2^2,1/3^2,\dots) \in \ell^1$, and we will have \begin{align*} Au = \langle v,u \rangle v = \left(\sum_{n=1}^{\infty}1/n^3 \right)v \notin \ell^1. \end{align*}