I am trying to prove that the characteristic function is uniformly continuous.
I understand how to get to this bound:
And I would like to find the $\delta$ as a function of $\epsilon$ but I am used to having $|f(x) -f(c)|$ instead of $|f(x+h) - f(x)|$ when doing uniform continuous proofs. Also $X$ is a random variable and I am unsure how to proceed.
Could I see how one would terminate this proof by finding the correct $\delta$?
If $E(X)$ is finite, the inequality $|e^{ihx}-1|\le |hx|$ gets you uniform continuity right away: $$|\varphi(t+h)-\varphi(t)|\le\int|hx|dF_X(x)=|h|E(|X|)\;.$$
If $X$ is not integrable, you've already found an upper bound that is free of $t$, so it suffices to show that $$\lim_{h\to0}\int|e^{ihx}-1|dF_X(x)=0\;,\tag1$$ since (1) then implies that for every $\varepsilon>0$ you can find $\delta>0$ such that $\int|e^{ihx}-1|dF_X(x)<\varepsilon$ whenever $|h|<\delta$. To show (1), you can use the 'sequence' approach: argue that $\int|e^{ih_nx}-1|dF_X(x)\to0$ whenever $h_n$ is a sequence tending to zero; this follows from the bounded convergence theorem.