Proof explanation - Existence of second derivative of characteristic function at 0 implies finite second moment

Question

Existence of second derivative of characteristic function of r.v. X at 0 implies that the second moment of X is finite.

There is a proof of that statement in Chung's "A Course in Probability Theory" (2001 edition, Theorem 6.4.1 - even more general, but it involves doing that for k = 2) which has one transition I can't wrap my mind around. It goes as follows:

$$ \begin{split} \varphi^{(2)}_X(0) &= \lim_{h \to 0} \frac{\varphi(h)-2\varphi(0)+\varphi(-h)}{h^2} \\ &= \lim_{h \to 0} \int_{-\infty}^{\infty} \frac{e^{ihx}-2+e^{-ihx}}{h^2}\mu_X(\mathrm{d}x) = -2 \lim_{h \to \infty} \int_{-\infty}^{\infty}\frac{1 - \cos(hx)}{h^2}\mu_X(\mathrm{d}x) \end{split} $$

On the other hand, by Fatou's Lemma:

$$ \begin{split} \mathbb{E}[X^2] = \int_{-\infty}^{\infty} x^2\mu_X(\mathrm{d}x) &= 2 \int_{-\infty}^{\infty} \lim_{h \to 0} \frac{1 - \cos(hx)}{h^2}\mu_X(\mathrm{d}x) \\ &\leq 2 \liminf_{h \to 0} \int_{-\infty}^{\infty} \frac{1 - \cos(hx)}{h^2}\mu_X(\mathrm{d}x) = -\varphi^{(2)}_X(0) < \infty \end{split} $$ QED.

Now, I want to know why the following part in second line is true:

$$ \int_{-\infty}^{\infty} x^2\mu_X(\mathrm{d}x) = 2 \int_{-\infty}^{\infty} \lim_{h \to 0} \frac{1 - \cos(hx)}{h^2}\mu_X(\mathrm{d}x) $$ ?

Answer 1

This follows from the fact that $$\lim_{h \to 0} \frac{1 - \cos(hx)}{h^2}=\frac{x^2}2$$ which can be seen by Taylor's expansion.

Answer 2

Yes, I agree with Davide Giraudo. Indeed the limit is easily verified: $$\cos(\alpha) = 1-\dfrac{\alpha^2}{2!} + \dfrac{\alpha^4}{4!} - \dfrac{\alpha^6}{6!} +...$$ $$1-\cos(hx) = \dfrac{h^2x^2}{2} - \dfrac{h^4 x^4}{4!} + \dfrac{h^6x^6}{6!} +...$$ $$\dfrac{1-\cos(hx)}{h^2} = \dfrac{x^2}{2} - \dfrac{h^2 x^4}{4!} + \dfrac{h^4x^6}{6!} +...$$ and as you can see the limit for $h\to 0 $ is clearly what expected: $$\lim_{h\to 0} \dfrac{1-\cos(hx)}{h^2} = \dfrac{x^2}{2}$$