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Define $f: \ell^2(\mathbb{N}) \to \Bbb C$ by: $$f(x) = \sum_{n \geq 1}\frac{x_n}{n^2},$$where $x = (x_n)_{n \geq 1} \in \ell^2(\mathbb{N})$. It is pretty clear that $f$ is linear. Also, since $\ell^2(\mathbb{N}) \subset \ell^\infty(\mathbb{N})$ and $\|x\|_\infty \leq \|x\|_2$, we have: $$\begin{align}|f(x)| &= \left|\sum_{n \geq 1}\frac{x_n}{n^2}\right| \leq \sum_{n \geq 1}\frac{|x_n|}{n^2} \\ &\leq \sum_{n \geq 1}\frac{\|x\|_\infty}{n^2} = \left(\sum_{n \geq 1}\frac{1}{n^2}\right)\|x\|_\infty \\ &\leq \frac{\pi^2}{6}\|x\|_2. \end{align} $$So $f$ is bounded and $\|f\| \leq \pi^2/6$. I would like to compute what exactly is this norm. The professor said that the book from where he took the exercise says that the norm is $\pi^2/(3\sqrt{10})$, which is indeed possible, but there's no comments on it. He thinks that the norm is actually that $\pi^2/6$ we have up here, but I can't come up with convenient sequences to prove or refute this. If the domain were $\ell^\infty(\mathbb{N})$ instead of $\ell^2(\mathbb{N})$ we could take $(1,1,1,\cdots)$ and the norm would be $\pi^2/6$, but this sequence is not in $\ell^2(\mathbb{N})$. Help?

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By Cauchy Schwarz,

$$|f(x)| =\left| \sum \frac{x_n}{n^2}\right|\le \sqrt{\sum \frac{1}{n^4}} \|x\|_2 = \sqrt{\frac{\pi^4}{90}}\|x\|_2$$

(See here for the calculation). Note that equality holds when $x_n = \frac{1}{n^2}$.

(Indeed, if you think of $f\in \ell^2(\mathbb N)^*$, then $f(x) = \langle x, y\rangle$, where $y_n= \frac{1}{n^2}$. Thus $\|f\| = \|y\|$)

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    $\begingroup$ Oh boy, that was fast! Thanks. So fast I can't even accept it yet haha $\endgroup$ – Ivo Terek Jun 22 '15 at 0:51

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