Show that $0\leq\|f\|^2+2 r\int_X |fg|d\mu+r^2\|g\|^2$ for all $r\in\mathbb{R}$ implies that $\int_X |fg|d\mu\leq\|f\|\|g\|$.

Question

The Problem arises from Walter Rudin's "Principles of Mathematical Analysis":

My Question: How exactly does $\int_X |fg|d\mu\leq\|f\|\|g\|$ in the theorem follow from the highlighted texts? Surely we have $\|f\|^2+r^2\|g\|^2\geq2r\|f\|\|g\|$ for all $r\in\mathbb{R}$; but that does not seem to be of any help here. Any hint would be greatly appreciated.

Answer 1

It is basically from the algebraic fact regarding quadratic equation that:

if $ax^2+bx+c\ge 0$ ($a>0$), then the discriminant $ b^2-4ac\le 0\;. $

In your case, consider $x=\lambda$, $a=\|g\|^2$, $b=2\int_X |fg|\;d\mu$, and $c=\|f\|^2$.

Answer 2

You have an estimate of the form $0 \leq K(f, g, \lambda)$. To get what Rudin says, note that $0 \leq \inf_{\lambda \in \mathbb{R}}K(f, g, \lambda)$ and that this infimum is easily computed using calculus.