Suppose there is a random variable $X$ whose characteristic function is $f(t) = E[e^{itX}]$. I tried this: $$ f^{(2k)}(t) = \lim_{\Delta t \to 0} i^{2k-1}E[e^{itX}X^{2k-1}\frac{e^{i\Delta tX}-1}{\Delta t}] $$ I know that from Fatou's lemma: \begin{align*} \lim_{\Delta t \to 0} E|e^{itX}X^{2k-1}\frac{e^{i\Delta tX}-1}{\Delta t}| &=\lim_{\Delta t \to 0} E|X^{2k-1}\frac{e^{i\frac{\Delta t}{2}X}-e^{-i\frac{\Delta t}{2}X}}{\Delta t}|\\ &=\lim_{\Delta t \to 0} E|X^{2k-1}\frac{2\sin(\frac{\Delta t}{2}X)}{\Delta t}|\\ &\geq E\liminf_{\Delta t \to 0}|X^{2k-1}\frac{2\sin(\frac{\Delta t}{2}X)}{\Delta t}|\\ &=E[X^{2k}] \end{align*} But I can only get $|f^{(2k)}(t)| \leq \liminf_{\Delta t \to 0} E|e^{itX}X^{2k-1}\frac{e^{i\Delta tX}-1}{\Delta t}|$, then I cannot proceed anymore.
My professor proved the case when $k=1$ and $t=0$: \begin{align*} -f''(0) &= -\lim_{h \to 0} \frac{f(h)-2f(0)+f(-h)}{h^2}\\ &= 2\lim_{h \to 0} E[\frac{1-\cos (hX)}{h^2}]\\ &\geq 2E[\liminf_{h\to 0} \frac{1-\cos (hX) }{h^2}]\\ &=E[X^2] \end{align*} Thus $E[X^2]$ exists. Then by induction, I can prove the general case when $f^{(2k)}(0) $ exists, $E[X^{2k}]$ exists. But is there anyway to edit my computation in the beginning part to derive the answer? Thank you!