An operator on $\ell^2(\mathbb N)$ is restricted to $\ell^1(\mathbb N)$. What happens to the corresponding operator norms?

Question

Let $A:\ell^2(\mathbb N)\to \ell^2(\mathbb N)$ be a linear operator. We define the operator norm as usual:

$$\|A\|=\sup_{u\in\ell^2(\mathbb N)} \frac{\|Au\|_{\ell^2}}{\|u\|_{\ell^2}}.$$

Recall that $\ell^1(\mathbb N)\subset \ell^2(\mathbb N)$. We can define an alternative operator norm as follows:

$$\|A\|_{\mathrm{alt}}=\sup_{u\in\ell^1(\mathbb N)} \frac{\|Au\|_{\ell^2}}{\|u\|_{\ell^1}}.$$

Is there a connection between $\|A\|$ and $\|A\|_{\mathrm{alt}}$? In particular, is it possible for some choice of $A$ that $\|A\|$ is finite but $\|A\|_{\mathrm{alt}}$ is infinite, or vice versa?

Answer 1

Teeing off of Aweygan's great comment, if you assume that you want the $\ell^2$ norm instead and that $A$ has finite norm, then

\begin{align} \|A\|_{\text{alt}} & = \sup \frac{\|A u\|_2}{\|u\|_1}\\ & \le \sup \frac{\|A\|_2\|u\|_2}{\|u\|_1} \\ & \le \sup \frac{\|A\|_2\|u\|_1}{\|u\|_1} \end{align}