In Durrett's probability theory, we want to show that if $\lim_{t \downarrow 0} (\varphi(t) - 1)/t^2 = c > -\infty$, then $EX = 0$ and $E|X|^2 = -2c < \infty$.
I know I have to use this theorem to get $E|X|^2 < \infty$:
Theorem 3.3.21 If $\limsup_{t \downarrow 0} \{ \varphi(t) - 2 \varphi(0) + \varphi(-t) \} / t^2 > -\infty$, then $E|X|^2 < \infty$
I'm having trouble satisfying the condition of the above theorem for $\lim_{t \downarrow 0} (\varphi(t) - 1)/t^2 = c > -\infty$. I can rewrite the limit as $$\limsup_{t \downarrow 0} { \varphi(t) - 2 \varphi(0) + \varphi(-t) \over t^2}= \underbrace{\limsup_{t \downarrow 0} {\varphi(t) - 1 \over t^2}}_{=c} + \underbrace{\limsup_{t \downarrow 0}{\varphi(-t) -1 \over t^2}}_{?}$$ I believe we need to use the fact that $\varphi(-t) = \overline{\varphi(t)}$. Is $$\limsup_{t \downarrow 0} {\varphi(t) - 1 \over t^2} = \limsup_{t \downarrow 0}{\overline{\varphi(t)} -1 \over t^2}$$
$${\overline{\varphi(t)} -1 \over t^2}$$ is the complex conugate of $$ {\varphi(t) - 1 \over t^2} $$ But here it is assumed that $c$ is a real number so $c$ is its own complex conjugate.