Suppose $v \in \mathbb{R}^n$ then I need to prove
$\|v\| = \sup\{\langle v,u\rangle , u \in \mathbb{R}^n$ and $ \|u\| = 1\}$
I know that in the set on $S = \{\langle v,u\rangle \|u\| =1$} if $v \ne 0$ $u = \dfrac{v}{\|v\|}$
gives $\langle v, u\rangle = \|v\|$ but how to show this is the least upper bound for this set ?
From the Cauchy Schwarz inequality we get $$\langle v, u\rangle\leq \|v\|\cdot 1 =\|v\|.$$ Therefore $$\sup\{\langle v, u\rangle : \|u\|=1\}\leq \|v\|$$ but for $u=\frac{v}{\|v\|}$ we get $$\langle v, u\rangle=\frac{\|v\|^2}{\|v\|}= \|v\|$$ so $$\sup\{\langle v, u\rangle : \|u\|=1\}= \|v\|.$$