Show that $\|v\|=\sup \{|\langle v,w \rangle|, \|w\|\leq 1\}$

Question

I need some help.

Let $(H,\langle\cdot,\cdot\rangle)$ a Hilbert space, and for every $v\in H$, defines a continuous function $\langle v,\cdot \rangle:H\to \mathbb{R}$.

Show that for every $v\in H$:

$$\|v\|=\sup \{|\langle v,w \rangle|, \|w\|\leq 1\}$$

I already prove the $\geq$ inequality, but i have troubles with the other side. I suspect I should take $w=\frac{v}{\|v\|}$, but I don't know how I can conclude from it. I would greatly appreciate your help.

Answer 1

To show that $x \leq \sup{A}$, it suffices to show that for some $a \in A$, we have that $x \leq A$. This is what you showed - taking $w := \frac{v}{\|v\|}$, we have that $\|w\| \leq 1$ and $|\langle v,w \rangle| = \|v\|$, so $\|v\| \leq \sup\{|\langle v,w \rangle| : \|w\| \leq 1\}$.