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Let $\alpha \in \ell^\infty$ and $T_\alpha:\ell^p \rightarrow \ell^p$ $(1\leq p \leq \infty)$ given by \begin{equation} T_\alpha(x)=(\alpha_1x_1,\dots,\alpha_nx_n,\dots). \end{equation} Note that $||T_\alpha(x)||\leq ||\alpha|| \ ||x||$, thus $T_\alpha(x) \in \ell^p$.

Now, I want to prove that $T_\alpha$ is continuous and $||T_\alpha||=\alpha$.

So, let $(x_n)=(x_1^n,x_2^n,\dots)\rightarrow(0,0,\dots)\in \ell^p$. Then $||x_n||\rightarrow 0 $ and by the above inequallity $||T_\alpha(x)||\rightarrow 0$. Then we get that $T_\alpha$ is continuous.

Computing the norm, I went for this idea:

\begin{equation} \begin{split} ||\alpha||&=\sup_{||x||=1 \\\ \ x\neq 0} ||\alpha|| \ ||x|| \\ &= \sup_{||x||=1 \\\ \ x\neq 0} \sup_{i \in \mathbb{N}} |\alpha_i| \ ||x|| \\ &=\sup_{||x||=1 \\\ \ x\neq 0} \sup_{i \in \mathbb{N}} |\alpha_i| \ \big(\sum_{j=1}^\infty |x_i|^p \big)^{1/p} \\ &=\sup_{i \in \mathbb{N}}\sup_{||x||=1 \\\ \ x\neq 0} \big(\sum_{j=1}^\infty |\alpha_i|^p|x_i|^p \big)^{1/p} = \sup_{i \in \mathbb{N}} ||T_\alpha||=||T_\alpha||. \end{split} \end{equation}

Then, in both cases when $p$ is finite or $p=\infty$ we have that $||T_\alpha||=||\alpha||$. Is this proof correct? What I'm missing here? Thanks for sharing!

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Your transition from the third to the fourth line needs remediation, specifically with your indexing. Your double $\sup$'s are also a bit hard to follow. Let me offer your a suggestion.

Define $A,B\subseteq l^p$ by $$A=\{e_i\in l^p:i\in \mathbb{N}\}$$ $$B=\{x\in l^p:\|x\|_p=1 \}$$ Because $A \subseteq B$ and $$\{|\alpha_i|:i\in \mathbb{N}\}=\Big\{\|T_{\alpha}(x)\|_p:x\in A\Big\}$$ we can say that $$\|\alpha\|_{\infty}=\sup_{i\in \mathbb{N}}|\alpha_i|=\sup_{x\in A}\|T_{\alpha}(x)\|_p\leq \sup_{x\in B}\|T_{\alpha}(x)\|_p=\|T_{\alpha}\|_{\mathcal{L}(\ell^p,\ell^p)} $$

This proves $\|\alpha\|_{\infty}\leq\|T_{\alpha}\|_{\mathcal{L}(\ell^p,\ell^p)}$. You can use your inequality $\|T_{\alpha}(x)\|_p\leq \|x\|_p\|\alpha\|_{\infty}$ to show $\|\alpha\|_{\infty}\geq \|T_{\alpha}\|_{\mathcal{L}(\ell^p,\ell^p)}$.

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