$(Fx)(t) = f(t)x(t)$ is unitary iff $|f(t)| = 1$ almost everywhere

Question

The problem comes from Naylor's Linear Operator Theory Section 5.19 Problem 4

Let $I = [a, b]$ be a bounded interval and define $F : L_2(I) → L_2(I)$ by $(F x)(t) = f(t)x(t)$, where $f ∈ L_∞(I)$. Show that $F$ is a unitary mapping if and only if $|f(t)| = 1$ almost everywhere.

I've already shown $F$ is unitary given $|f(t)| = 1$ a.e.. It's the forward direction that is giving me trouble. I know $\langle Fx,Fy\rangle=\langle x,y\rangle$ implies $\int_I f(t)^2x(t)y(t)dt= \int_I x(t)y(t)dt.$

I feel like I'm missing something super simple. Any help is appreciated.

Answer 1

Take $x=y=I_E$ where $E$ is any measurable set in $[a,b]$. You get $\int_E \phi (t)dt=0$ where $\phi (t)=1-|f(t)|^{2}$. This implies $\phi=0$ a.e.: take $E=\{x: \phi (t) >0\}$ to see that $\phi (t) \leq 0$ a.e. and then take $E=\{x: \phi (t) <0\}$ to see that $\phi (t) \geq 0$ a.e.