# Finding the operator norm of $\phi \in ((\ell^{2}, \|\cdot \|_{2}))^{*}$

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 $$\ell^{2}$$?

## 2 Answers

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$$\lVert\phi\rVert=\sqrt{\sum_{n=1}^\infty\frac1{n^2}}.$$

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

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