# Continuity of the Characteristic Function of a RV

Defining the Characteristic Function $\quad \phi(t) := \mathbb{E} \left[ e^{itx} \right]$ for a random variable with distribution function $F(x)$ in order to show it is uniformly continuous I say

$$|\phi(t+u) - \phi(t)| = \left |\int e^{itx}(e^{iux} - 1) dF(x) \right| \le \\ \int 1 \cdot|e^{iux} -1|dF(x) \to 0 \quad as \quad u\to0$$

Now my question is, does the convergence I state in the last line follow directly, or do I need to be a little carful before I conclude it is true ? (i.e. can I directly use that $|e^{itu} -1| \to 0 ? )$

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## 2 Answers

There is a typo: it should be $$|\phi(t+u)-\phi(t)|=\left|\int (e^{i(t+u)x}-e^{itx})dF(x)\right|\leq\int |e^{iux}-1|dF(x).$$ Now take a sequence $u_n$ which converges to $0$ and put $f_n(x):=|e^{iu_nx}-1|$. Then $f_n(x)\to 0$ for each $x$ and $|f_n(x)|\leq 2$ which is integrable so we conclude by the dominated convergence theorem that $\lim_{n\to\infty}\int|e^{iux}-1|dF(x)=0$ and so $\phi$ is uniformly continuous on $\mathbb R$.

Note that if the random variable is integrable we don't need the dominated convergence theorem, since we can write $$|\phi(t+u)-\phi(t)|\leq\int |e^{iux}-1|dF(x)=2\int|\sin(ux/2)|dF(x)\leq |u|\int |x|dF(x).$$

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You probably meant "as $u \to 0$" in that last line, right? In any case, it follows by the dominated (or bounded) convergence theorem applied to the measure $dF$.

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 indeed, correcting that now, tks ! – Beltrame Feb 12 '12 at 22:15 cld you comment why I need to use dominated convergence? I m not sure I see why it doesn't follow directly ! – Beltrame Feb 12 '12 at 22:17