$\mathcal{V}$ is an inner product space over $\mathbb{F}$ of finite-dimension, and $\phi : \mathcal{V} \to \mathbb{F}$ is a linear functional. Let $\textbf{r}$ be the Riesz vector for $\phi$. Prove that for $\textbf{v} \in \mathcal{V}$, $$\textrm{max}\{|\phi(\textbf{v})| : \Vert \textbf{v} \Vert = 1\} = \Vert \textbf{r} \Vert$$
I know that $$\phi(\textbf{v}) = \langle \textbf{v}, \textbf{r} \rangle $$ And also (from the definition of the norm derived from the inner product): $$\phi(\textbf{r}) = \langle \textbf{r}, \textbf{r} \rangle = \Vert \textbf{r} \Vert^{2} $$ And, clearly when $\Vert \textbf{v} \Vert = 1$, then $\Vert \textbf{v} \Vert^{2} = 1$.
Plus, given an orthonormal basis for $\mathcal{V}$, $u_1...u_n$, the Riesz vector can be represented:
$$ \textbf{r} = \phi(u_1) \cdot u_1 \, + \, ... \, + \, \phi(u_n) \cdot u_n$$
But I can't seem to collect these observations into something that works to prove the statement. Any ideas?
$|\phi (v)| \leq \|v\| \|r\|$ be Cauchy -Schwraz inequlaity. This gives LHS $\leq $ RHS. For the reverse inequality just take $v= \frac {r} {\|r\|}$. (The case $r=0$ has to be treated separately).