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Let $(a_n)\in l_\infty$ and $ T:l_2 \to l_2$ given by $T((x_n)) = (a_nx_n)$. Show that $T \in \mathcal{L}(l_2,l_2)$ $\|T\| = \|(a_n)\|_{\infty}$

It's easy to prove that $T$ is continuous since $T$ is linear and: $$\|T((x_n))\|_2 = \sqrt{\sum_{n=1}^{\infty}{|a_nx_n|^2}} \leq \sqrt{\sum_{n=1}^{\infty}{\|(a_n)\|_{\infty}^2|x_n|^2}} = \|(a_n)\|_{\infty}\sqrt{\sum_{n=1}^{\infty}{|x_n|^2}} = \|(a_n)\|_{\infty} \|(x_n)\|_2 $$

Thus $\|T\| \leq \|(a_n)\|_{\infty}$.

I'm having trouble with the other inequality.

Any hints?

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  • $\begingroup$ $T e_m$ where $m$ is the index such that $|a_m|$ is maximum, and if there is no such index (if $m= \infty$) you have to prove $\sup_m \|Te_m\| \ge \|a\|_\infty-\epsilon$ for every $\epsilon > 0$ (what Martin wrote) $\endgroup$ – reuns Jun 3 '16 at 21:56
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Let $\varepsilon>0$. Then there exists $m$ such that $\|a\|_\infty\leq|a_m|+\varepsilon$. If $x\in\ell^2$ has $x_m=1$ and $x_n=0$ otherwise, then $$ \|T\|\geq\|Tx\|=|a_m|\geq\|a\|_\infty-\varepsilon.$$

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