A nice inequality about characteristic function

Question

I need help with this please.

$\psi$ is characteristic function of a probability measure, then:

$ |\psi(t)-\psi(s)|^2\leq 2|1-\psi(t-s)|$

Answer 1

Denote by $\mu$ the associated probability measure, i.e. $$\psi(t) = \int e^{ixt} \, \mu(dx).$$

Since

$$|\psi(t)-\psi(s)|^2 = \int (\sin(tx)-\sin(sx))^2 \, \mu(dx) + \int (\cos(tx)-\cos(sx))^2 \, \mu(dx)$$

it follows from

$$\sin u- \sin v = 2 \cos \frac{u+v}{2} \sin \frac{u-v}{2} \qquad \cos u - \cos v = - 2 \sin \frac{u+v}{2} \sin \frac{u-v}{2}$$

that

$$|\psi(t)-\psi(s)|^2 \leq 4 \int \sin^2 \frac{(t-s)x}{2} \, \mu(dx).$$

As

$$\sin^2(u) = \frac{1}{2} (1-\cos(2u))$$

this gives

$$|\psi(t)-\psi(s)|^2 \leq 2 \int (1-\cos((t-s)x)) \, \mu(dx) = 2 |1-\text{Re} \, \psi(t-s)|;$$

in particular,

$$|\psi(t)-\psi(s)|^2 \leq 2 |1-\psi(t-s)|.$$