If $\|u\| \leq \|u+av\|$ for all $a \in F$, How can I show that $\langle u,v\rangle=0$?

Question

If $\|u\| \leq \|u+av\|$ for all $a \in F$,

How can I show that $\langle u,v\rangle=0$?

I know a standard solution uses $\operatorname{Re}$ and $t =$ something but was wondering if there was perhaps an easier solution? Thanks!

Answer 1

For any $a > 0$, \begin{align*} 2a \langle u, v \rangle &= \langle u , u \rangle - \langle u - av, u - av \rangle + a^2\langle v, v \rangle \\ &= \left\|{u}\right\|^2 - \left\|{u - av}\right\|^2 + a^2 \left\|v\right\|^2 \\ &\le a^2 \left\|v\right\|^2 \\ -2a \langle u, v \rangle &= \langle u , u \rangle - \langle u + av, u + av \rangle + a^2\langle v, v \rangle \\ &= \left\|{u}\right\|^2 - \left\|{u + av}\right\|^2 + a^2 \left\|v\right\|^2 \\ &\le a^2 \left\|v\right\|^2 \\ \end{align*}

Therefore, \begin{align*} \left| \langle u, v \rangle \right| = \frac{\left|2a \langle u, v \rangle\right| }{\left|2a\right|} \le \frac{a^2 \| v\|^2}{\left|2a\right|} = \frac{|a| \|v\|^2}{2} \end{align*}

Since $a$ can be arbitrarily small, we conclude $\langle u, v \rangle = 0$.

See also robjohn's answer. I suspect this is as simple as it gets.