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

  • $\begingroup$ Thanks for the input. I feel blind sometimes. $\endgroup$
    – Rebellos
    Nov 25 '18 at 11:36
  • 2
    $\begingroup$ I guess that is the normal thing for people doing mathematics. We wander through dark rooms and when we finally find the switch to turn on the light we get the impression that everything was clear from the beginning. $\endgroup$ Nov 25 '18 at 11:38
  • $\begingroup$ Haha, well said ! $\endgroup$
    – Rebellos
    Nov 25 '18 at 11:38
  • 1
    $\begingroup$ In fact that was a more pessimistic version of a quote of Andrew Wiles: "Perhaps I can best describe my experience of doing mathematics in terms of a journey through a dark unexplored mansion. You enter the first room of the mansion and it’s completely dark. You stumble around bumping into the furniture, but gradually you learn where each piece of furniture is. Finally, after six months or so, you find the light switch, you turn it on, and suddenly it’s all illuminated. You can see exactly where you were. $\endgroup$ Nov 25 '18 at 11:41
  • 1
    $\begingroup$ Then you move into the next room and spend another six months in the dark. So each of these breakthroughs, while sometimes they’re momentary, sometimes over a period of a day or two, they are the culmination of—and couldn’t exist without—the many months of stumbling around in the dark that proceed them." $\endgroup$ Nov 25 '18 at 11:42

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


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