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Let $c_0=\{x=(x_1,x_2,\dots): \lim_{n\rightarrow\infty}x_n=0\}$. Supplied with the supremum norm this is a Banach space. Now we can view, if $1\leq p<\infty$, $\ell^p$ as an abstract subspace of $c_0$. Suppose that $T: (\ell^p,||\cdot||_{\infty})\rightarrow(\ell^p,||\cdot||_{\infty})$ is a bounded linear map that maps this subspace into itself. Now we can view the restriction of $T$ to this subspace as a linear map $T_p: (\ell^p,||\cdot||_p)\rightarrow(\ell^p,||\cdot||_p)$.

I have to show that $T_p$ is also bounded. I have already shown that the inclusion map $\iota: \ell^p\rightarrow c_0$ is continuous (hence bounded). Now I thought that maybe $T$ is a composition of the inclusion map and the restriction $T_p$, and I can prove from that that $T_p$ must also be continuous? Is this the right thought? If not, can someone push me in the right direction?

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  • $\begingroup$ the notation $\ell^p$ is not used generally for subspaces of $c_0$ if not for spaces of bounded sequences under the $p$-norm. $\endgroup$ – Masacroso Nov 28 '17 at 10:25
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    $\begingroup$ The inclusion map is not enough, that train of thoughts only gives you that $T\colon (\ell^p, \|\cdot\|_p)\to (\ell^p, \|\cdot\|_\infty)$ is continuous. I think that here you need the theorems of functional analysis (open mapping, or closed graph, or uniform boundedness). Which theorem, and how to use it, I have no idea. $\endgroup$ – Giuseppe Negro Nov 28 '17 at 12:23
  • $\begingroup$ @GiuseppeNegro thanks for your comment. I will resume my search then! $\endgroup$ – user428890 Nov 29 '17 at 7:41
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We show that the graph of $T_p$ is closed, that will imply that $T_p$ is continuous by the closed graph theorem. Take two sequences $x_n\in \ell^p,Tx_n\in\ell^p$ converging in $\ell^p$, that is:

$$\begin{align}& x_n\to x\in\ell^p \qquad \mbox{in}\,\, \ell^p, \\ & Tx_n\to y\in\ell^p \qquad \mbox{in}\,\, \ell^p.\end{align}$$

We need to show that $y=Tx$. Since for all $x\in\ell^p$ we have $\|x\|_{\ell^\infty}\le \|x\|_{\ell^p}$, then $\|x_n-x\|_{\ell^\infty}\le \|x_n-x\|_{\ell^p}$ and $\|Tx_n-y\|_{\ell^\infty}\le \|Tx_n-y\|_{\ell^p}$, then we also have the convergence:

$$\begin{align}& x_n\to x\qquad \mbox{in}\,\, \ell^\infty, \\ & Tx_n\to y \qquad \mbox{in}\,\, \ell^\infty.\end{align}$$

By hypothesis $T$ is continuous with respect to the $\ell^\infty$-norm, then $y=Tx$ and we are done.

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  • $\begingroup$ Why didn't I think of that before? I was trying to use the closed graph theorem but I got stuck. Thank you! $\endgroup$ – user428890 Nov 30 '17 at 9:55

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