$x$ is orthogonal to $y$ iff $\|x+ay\| \geq \|x\|$ where $x \in C$

Question

Show that in an inner product space, $x$ is orthogonal to $y$ iff $\|x+ay\| \geq \|x\|$ where $x \in C$.

Proof:

LHS: If $x$ is orthogonal to $y$, then $\langle x,y\rangle =0$. Let $a \in \mathbb C$.

$\|x+ay\|^2 = \langle x+ay,x+ay\rangle =\langle x,x\rangle +\langle ay,ay\rangle +\langle x,ay\rangle +\langle ay,x\rangle $ $\geq \langle x,x\rangle =\|x\|^2$

Hence $\|x+ay\| \geq \|x\|$.

RHS: If $\|x+ay\| \geq \|x\|$ for $a \in \mathbb C$, then

$\|x+ay\|^2 = \langle x+ay,x+ay\rangle =\langle x,x\rangle +\langle ay,ay\rangle +\langle x,ay\rangle +\langle ay,x\rangle = 1 + \bar{a}\langle x,y\rangle +a\langle y,x\rangle =1 + \bar{a}\bar{\langle y,x\rangle }+a\langle y,x\rangle $

But I don't know how to continue. Can someone guide me to finish this proof.

Answer 1

You have shown that, for any $a\in\mathbb C$, $$ \|x\|\leq\|x+ay\|=\|x\|+2\,\text{Re}\,a\langle y,x\rangle, $$ or $$ 2\,\text{Re}\,a\langle y,x\rangle\geq0. $$ In particular this would hold for $a_0=-\overline{\langle y,x\rangle}$, so $$ -2|\langle y,x\rangle|^2=2\,\text{Re}\,a_0\langle y,x\rangle\geq0. $$ The only way this can happen is when $\langle y,x\rangle =0$.