Show the following are equivalent, for $f, g \in L^2[a,b]$
a. $|\langle f, g \rangle| = \|f\|_2 \|g\|_2$
b. $\exists \alpha, \beta \in \mathbb{R} $ s.t. $\alpha f = \beta g $
where $\langle \cdot , \cdot \rangle$ : inner product.
Try
(b $\Rightarrow$ a) If $\alpha=0$, things are trivial. Otherwise, note $f = \frac{\beta}{\alpha} g$
$|\langle f, g \rangle| = |\frac{\beta}{\alpha}| \langle g, g \rangle = |\frac{\beta}{\alpha}| \|g\|_2 \cdot\|g\|_2 = \|f\|_2 \|g\|_2 $
But I'm stuck at (a $\Rightarrow$ b).
One can consider the function $f:\mathbb{R}\rightarrow \mathbb{R},x\mapsto ||f-xg||^2$. Since we have $||f-xg||^2=\langle f-xg,f-xg\rangle=||f||^2-2x\langle f,g\rangle + x^2||g||^2$, this gives rise to a quadratic function, which has a real root, because the discriminant is $\left(2\langle f,g\rangle\right)^2-4||u||^2||v||^2=4\langle f,g\rangle^2-4\langle f,g\rangle^2=0$. Hence there is some $x\in \mathbb{R}$, s.t. $f-xg=0$, or equivalently $f=xg$.