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

Question

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.

Answer 1

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.