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

Question

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.

Answer 1

In the case $x\neq 0$ we have $$ \Vert x \Vert = \frac{1}{\Vert x \Vert} \Vert x \Vert^2 = \frac{1}{\Vert x \Vert} \langle x, x \rangle = \frac{\vert \langle x, x \rangle \vert}{\Vert x \Vert} \leq \sup_{y\neq 0} \frac{\vert \langle x, y \rangle \vert}{\Vert y \Vert} $$

Answer 2

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, $\sup_{y \neq 0} \frac{|\langle x,y \rangle|}{\|y\|}= \|\overline {A_x}\|=\|A_x\|=\|x\|.$

Answer 3

In case it's useful, here I show that $\|x\| = \sup_{\|y\| = 1} |\langle x, y \rangle|$.

For any $x$, define the linear functional $f : H \to \mathbb{C}$ as $x \to \langle x, y \rangle$. Then we have

$\| f \| \\ = \sup_{\|y\| = 1} |f(y)| \\ = \sup_{\|y\| = 1} |\langle x, y \rangle| \\ \le \sup_{\|y\| = 1} (\|x\| \|y\|) \\ = \| x \|$

And likewise,

$\| x \|^2 \\ = \langle x, x \rangle^2 \\ = | f(x) | \\ \le \|f\|\|x\| \\ \implies \|x\| \le \|f\|$

So we have $\|x\| = \|f\| = \sup_{\|y\| = 1} |\langle x, y \rangle|$.

In the Riesz representation theorem, this is the proof that $\| f \| = \| x_0 \|$, where $f(x)$ is a bounded linear function in a Hilbert space $H$, and $x_0$ is such that $f(x) = \langle x, x_0 \rangle$.