I'm taking a course in Real Analysis and have come across the following question which I initially thought looked quite simple:
Let $X$ be an inner product space over $\mathbb{K}$ ($\mathbb{K} = \mathbb{C}$ or $\mathbb{R}$) Show that x and y are orthogonal if and only if $||\lambda x + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$ for all $\mu, \lambda \in \mathbb{K}$.
I have proven the statement in one direction (starting with orthogonality and ending up with $||\lambda x + \mu y||^2 = ||\lambda x||^2 + ||\mu y||^2$). However, I seem to be getting stuck trying to go in the other direction. The following is what I have so far:
$||\lambda x + \mu y||^2 = ||\lambda x||^2 + 2\bar{\lambda}\mu Re<x,y> + ||\mu y||^2$
Then we must have 2$\bar{\lambda}\mu Re<x,y> = 0$ for the desired equality to hold, and so $Re<x,y> = 0$.
However, for orthogonality I want $<x,y> = 0$ and can't see how I can make this next step.
Any help would be greatly appreciated.
We can assume
$$\| \lambda x + \mu y\|^2 = \|\lambda x\|^2+\|\mu y\|^2 $$
for all $\lambda, \mu \in \mathbb C$. Taking $\lambda = \mu =1$ gives
$$\|x+y\|^2 = \|x\|^2+\|y\|^2+\langle x, y \rangle + \langle y, x \rangle $$
$$ =\|x\|^2+\|y\|^2 +\langle x, y \rangle + \overline{\langle x, y \rangle} = \|x\|^2+\|y\|^2+2 \operatorname{Re}\langle x, y \rangle$$
so that $\operatorname{Re}\langle x, y \rangle = 0$.
Now take $\lambda = i$, $\mu=1$. We obtain
$$\|ix+y\|^2 = \|ix\|^2+ \|y\|^2+\langle ix, y \rangle + \langle y,ix \rangle$$
$$ =\|ix\|^2+\|y\|^2-i \langle x,y\rangle + i \langle y, x\rangle$$
$$=\|ix\|^2+\|y\|^2-i \left(\langle x,y\rangle - \overline{\langle x,y \rangle}\right) $$
$$ =\|ix\|^2+\|y\|^2- 2i \operatorname{Im}\langle x,y\rangle,$$
so that $\operatorname{Im}\langle x,y\rangle = 0.$