If $X$ has zero mean and finite variance $\sigma^{2}$, show that $|\Phi_{X}(t)-1|\leq 2t^{2}\sigma^{2}$

Question

As stated in the title, I want to show

If $X$ has zero mean and finite variance $\sigma^{2}$, show that $|\Phi_{X}(t)-1|\leq 2t^{2}\sigma^{2}$, where $\Phi_{X}$ is the characteristic function of $X$.

I have some attempts but got stuck in the end:

We can use the zero mean property to derive the following computation: $$|\Phi_{X}(t)-1|=\Big|\int(e^{itx}-1)d\alpha\Big| =\Big|\int(e^{itx}-1-itx)d\alpha\Big|,$$ then what I should do next? Hopefully we have $|e^{itx}-1-itx|\leq 2t^{2}x^{2}$, but I don't know how to show it..

Since the last inequality is somehow a complex analysis question, I will also put a tag of complex analysis.

Any idea? Thank you!

Answer 1

Actually $|e^{itx}-1-itx| \leq \frac {t^{2}x^{2}} 2$ so we have the stronger inequality $|\Phi_X(t)-1| \leq \frac {t^{2} \sigma^{2}} 2$.

[$|1-e^{ia}|=|\int_0^{a} ie^{it} dt| \leq |a|$. Next note that $\int_0^{a} (1-e^{ix}) dx=a-(e^{ia} -1)/i$. Can you finish the proof of the inequality from here?]