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Fix a sequence $x = (x_n)_{n \geq 0}$. Then define $M_x: \ell^2(\Bbb N)\to \ell^2(\Bbb N)$ by: $$M_xy = (x_ny_n)_{n \geq 0},$$if $y = (y_n)_{n \geq 1}$. Supposedly we have $\|M_x\| = \|x\|_\infty$. On one hand, we have: $$\|M_xy\|_2^2 = \sum_{n \geq 0}|x_ny_n|^2 \leq \sum_{n \geq 0}\|x\|_\infty^2|y_n|^2 = \|x\|^2_\infty \sum_{n \geq 0}|y_n|^2 = \|x\|_\infty^2\|y\|_2^2,$$ so that $\|M_xy\|_2 \leq \|x\|_\infty\|y\|_2$. From this we get $\|M_x\| \leq \|x\|_\infty$.

The exercise doesn't says if $x \in \ell^2(\Bbb N)$ or if $x \in \ell^{\infty}(\mathbb{N})$. I'm having trouble coming up with a sequence that shows the $\geq$ part. The first guesses are $x/\|x\|_2$ and $x/\|x\|_\infty$ but they don't seem to work.

Can someone help? Thanks.

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Note that $\ell^2(\mathbb N)\subseteq\ell^{\infty}(\mathbb N)$, so one may assume that $x\in \ell^{\infty}(\mathbb N)$ in any case. Fix $\varepsilon>0$ and pick some $n_0\in\mathbb N$ such that $|x_{n_0}|>\|x\|_{\infty}-\varepsilon$. Now define $y\in\ell^2(\mathbb N)$ as follows: \begin{align*} y_n\equiv\begin{cases}1&\text{if $n=n_0$,}\\0&\text{otherwise.}\end{cases} \end{align*} Clearly, $\|y\|_2=1$ and $$\|M_x(y)\|_2=\sqrt{\left|x_{n_0}\right|^2}=|x_{n_0}|>\|x\|_{\infty}-\varepsilon.$$ It follows that $\|M_x\|>\|x\|_{\infty}-\varepsilon$ for each $\varepsilon>0$, so...

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    $\begingroup$ Very neat. Thank you :) $\endgroup$ – Ivo Terek Jun 24 '15 at 17:41
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    $\begingroup$ My pleasure! :-) $\endgroup$ – triple_sec Jun 24 '15 at 17:47

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