# If $\phi$ is a characteristic function, then $1-|\phi(2t)|\leq 8\{1-|\phi(t)|\}$

Question

If $$\phi$$ is a characteristic function, show that $$\text{Re}\{1-\phi(t)\}\geq \frac{1}{4}\text{Re}(1-\phi(2t))$$ and deduce that $$1-|\phi(2t)|\leq 8\{1-|\phi(t)|\}$$.

My attempt

I have managed to show the first part, but unable to deduce the second part.

Here is a proof of the first part. It is immediate that $$\text{Re}\{1-\phi(t)\}=E(1-\cos tX)$$ and $$\text{Re}(1-\phi(2t))=E(1-\cos 2tX)$$, but then $$4E(1-\cos tX)-E(1-\cos2tX)=2E(1-\cos tX)^2\geq 0$$ from which the first claim follows.

My problem

I have not been able to use the first part to deduce the second claim. I tried squaring and both sides and using the fact that $$|\text{Re}\; z|\leq |z|$$ (so in particular $$1-|\phi(2t)|\leq 1-|\text{Re}\;\phi(2t)|$$) but did not get too far.

Any help is appreciated.

Hints: fix $$t$$ and choose $$\theta$$ such that $$\phi(t)=|\phi(t)|e^{it\theta}$$. Note that $$\psi (s)=e^{is\theta}\phi(s)$$ is also a characteristic function. Apply first part to $$\psi$$. Use the fact that $$2\cos(2t\theta) \leq 2$$ to complete the proof.