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Let $\phi(x):=\sum\limits_{n\in \mathbb N} \frac{x_{n}}{n}$ where $\phi \in ((\ell^{2}, \| \cdot \|_{2}))^{*}$

Compute: $\|\phi \|_{*}$

Let $x \in \ell^{2}$ and then $\vert \phi(x) \vert=\vert \langle x,(\frac{1}{n})_{n}\rangle\vert\leq\| x\|_{2}\bigl\|\bigl(\frac{1}{n}\bigr)_{n}\bigr\|_{2}\Rightarrow \|\phi \|_{*}\leq \bigl\|\bigl(\frac{1}{n}\bigr)_{n}\bigr\|_{2}<\infty$

But I have no idea, how to show the converse $\geq $

I have been told to use the fact that the Cauchy inequality is an equality if $x$ and $\bigl(\frac{1}{n}\bigr)_{n}$ are linearly dependent. But how can I construct a constant $x$ that is still in $\ell^{2}$?

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From your computations, $\lVert\phi\rVert\leqslant\sqrt{\sum_{n=1}^\infty\frac1{n^2}}$. On the other hand,$$\phi\left(\left(\frac1n\right)_{n\in\mathbb N}\right)=\sum_{n=1}^\infty\frac1{n^2}=\left\lVert\left(\frac1n\right)_{n\in\mathbb N}\right\rVert_2^2,$$and therefore$$\phi\left(\frac{\left(\frac1n\right)_{n\in\mathbb N}}{\left\lVert\left(\frac1n\right)_{n\in\mathbb N}\right\rVert_2}\right)=\left\lVert\left(\frac1n\right)_{n\in\mathbb N}\right\rVert_2=\sqrt{\sum_{n=1}^\infty\frac1{n^2}}.$$This proves that$$\lVert\phi\rVert=\sqrt{\sum_{n=1}^\infty\frac1{n^2}}.$$

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  • $\begingroup$ Could one use Cauchy-Schwarz for equality at all? $\endgroup$ – MinaThuma Jul 19 at 21:39
  • $\begingroup$ I do not see how. $\endgroup$ – José Carlos Santos Jul 19 at 21:41
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We have

$$|\phi(x) |=\left|\left\langle x,\left(\frac{1}{n}\right)_{n}\right\rangle\right|\le \| x\|_{2}\left\|\left(\frac{1}{n}\right)_{n}\right\|_{2}$$ and we know that equality in Cauchy-Schwarz holds if and only if $x$ and $\left(\frac{1}{n}\right)_{n}$ are colinear. Hence for $x = \left(\frac{1}{n}\right)_{n}$ we get

$$\left|\phi\left(\left(\frac{1}{n}\right)_{n}\right)\right| = \left\|\left(\frac{1}{n}\right)_{n}\right\|_2^2$$

We conclude $\|\phi\|_* = \left\|\left(\frac{1}{n}\right)_{n}\right\|_2$.

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