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