# Showing that $\|x \| = \sup_{y \neq 0} \frac{|\langle x,y \rangle|}{\|y\|}$

Exercise :

Let $$H$$ be an inner product space and $$x \in H$$. Show that : $$\|x \| = \sup_{y \neq 0} \frac{|\langle x,y \rangle|}{\|y\|}$$

Attempt :

If $$x=0$$ then the equality follows imidiatelly as an equality with respect to $$0$$. Let it now be that $$x \neq 0$$. For $$y \in H$$ with $$y \neq 0$$, by the Cauchy-Schwarz inequality, it is : $$\langle x,y \rangle^2 \leq \langle x,x \rangle \cdot \langle y,y \rangle \Leftrightarrow |\langle x,x \rangle|\leq \langle x,x \rangle^{1/2} \cdot \langle y,y \rangle^{1/2} = \|x\|\cdot\|y\|$$

Thus, it also holds that :

$$\|x\| \geq \sup_{y \neq 0} \frac{|\langle x,y \rangle|}{\|y\|}$$

How would I show the other direction of the inequality, though, to prove that it must be equal to it ?

I thought about expressing $$\|x\|$$ as

$$\|x\| = \bigg\langle x,\frac{x}{\|x\|}\bigg\rangle$$

but can't see anything obvious.

Any tips will be greatly appreciated.

In the case $$x\neq 0$$ we have $$\Vert x \Vert = \frac{1}{\Vert x \Vert} \Vert x \Vert^2 = \frac{1}{\Vert x \Vert} \langle x, x \rangle = \frac{\vert \langle x, x \rangle \vert}{\Vert x \Vert} \leq \sup_{y\neq 0} \frac{\vert \langle x, y \rangle \vert}{\Vert y \Vert}$$
For another approach, note that, by the Riesz theorem, for each $$x\in H$$ there is a functional $$A_x:y\mapsto \langle y,x\rangle$$ such that $$\|A_x\|=\|x\|.$$
Therefore, $$\sup_{y \neq 0} \frac{|\langle x,y \rangle|}{\|y\|}= \|\overline {A_x}\|=\|A_x\|=\|x\|.$$