Let $X : \Omega \to \mathbb{R}$ be a random variable, and consider its characteristic function:
$$\phi_X(t) = \mathbb{E}[e^{itX}]$$
Suppose that $\phi_X(u) = 1$. How can I rigorously prove that $e^{iuX} \equiv 1$ a.s.? (More context.)
We have the obvious inequality:
$$ |\mathbb{E}[e^{itX}]| \le \mathbb{E}[|e^{itX}|] =1$$
so this is quite an intuitive result, but it's not immediately clear how to prove it. This seems especially difficult because we do not have the notion of continuity.
One idea I am considering is differentiating $\mathbb{E}[e^{itX}]$ under the integral. Since $1$ is a maximum for $|\mathbb{E}[e^{itX}]|$ by the above, the derivative at $t = u$ must be zero. But I am having some difficulty with making this work.
$E(1-\cos (tX))=\Re (1-Ee^{itX})=0$. Since $1-\cos (tX)\geq 0$, this implies $1-\cos (tX)=0$ a.s. But $\cos (tX)=1$ implies $\sin (tX)=0$ so $e^{itX}=1$ a.s.