A question was recently asked by a new user, SomeoneHAHA, and then deleted by the user, after receiving an answer. I think the question and the answer (QA) to it may be of interest to some users. Therefore, I am resurrecting the QA. Here is the question, in a bit brushed-up form:

Let $(X_t)_{t\in\mathbb R}$ be a normalized wide-sense stationary stochastic process in $\mathbb R$, so that $g(t):=\text{Cov}(s,s+t)$ does not depend on $s$ and $g(0)=1$. By Bochner's theorem, $g$ must then be the characteristic function of a probability distribution.

The question is: Is it possible that $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$?

A question was recently asked by a new user, SomeoneHAHA, and then deleted by the user, after receiving an answer. I think the question and the answer (QA) to it may be of interest to some users. Therefore, I am resurrecting the QA. Here is the question, in a bit brushed-up form:

Let $(X_t)_{t\in\mathbb R}$ be a normalized wide-sense stationary stochastic process in $\mathbb R$, so that $g(t):=\text{Cov}(X_s,X_{s+t})$ does not depend on $s$ and $g(0)=1$. By Bochner's theorem, $g$ must then be the characteristic function of a probability distribution.

The question is: Is it possible that $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$?