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

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?

• While $\ell^1\subset\ell^2$, it may not be true that $A(\ell^1)\subset\ell^1$. You either need to (a) assume $\ell^1$ is an invariant subspace of $A$, or (b) redefine $\|\cdot\|_{alt}$ to be $$\|A\|_{alt}=\sup_{u\in\ell^1(\mathbb N)} \frac{\|Au\|_{\ell^2}}{\|u\|_{\ell^1}}.$$ Jun 6 '18 at 4:13
• fixed, thank you Jun 6 '18 at 4:41

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
• Thanks for complementing my comment. Do you have any input for the opposite situation? Namely, when $\|A\|_{\text{alt}}$ is finite but $A$ is unbounded as an operator on $\ell^2$? Jun 6 '18 at 4:31
• @Aweygan I think I have an answer to your question. Let $A$ be a Hamel basis for $l^{1}$ and $A \cup B$ be a Hamel basis for $l^{2}$ with $B \subset l^{2}\setminus l^{1}$ and $||b||=1$ for all $b \in B$. Since $l^{1}$ has inifnite codimension in $l^{2}$ there exists a sequence of distinct points $\{b_n\}$ in $B$. Let $T=0$ on $A$ and $Tb_n=n$. Let $Tb=0$ for $b \in B\setminus \{b_1,b_2,...\}$. Extend $T$ to $l^{2}$ by linearity. Then $T=0$ on $l^{1}$ but $T$ is not bounded. Jun 6 '18 at 8:25