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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?

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    $\begingroup$ 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 $$\|A\|_{alt}=\sup_{u\in\ell^1(\mathbb N)} \frac{\|Au\|_{\ell^2}}{\|u\|_{\ell^1}}.$$ $\endgroup$
    – Aweygan
    Jun 6 '18 at 4:13
  • $\begingroup$ fixed, thank you $\endgroup$
    – Darren Ong
    Jun 6 '18 at 4:41
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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}

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    $\begingroup$ 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 $\ell^2$? $\endgroup$
    – Aweygan
    Jun 6 '18 at 4:31
  • $\begingroup$ Hmm good question.. Let me think on this for a bit. $\endgroup$ Jun 6 '18 at 4:35
  • $\begingroup$ I keep chasing my tail on it. I can't quite seem to make it work. $\endgroup$ Jun 6 '18 at 5:49
  • $\begingroup$ @Aweygan Your question is very interesting. Why don't you post it as a new question? $\endgroup$ Jun 6 '18 at 6:00
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    $\begingroup$ @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. $\endgroup$ Jun 6 '18 at 8:25

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