$\newcommand{\R}{\mathbb{R}}$ This is only possible when $g=1$. Indeed, otherwise, by Hoeffding's Lemma 3.2, page 217, one of the following two alternatives must take place:

(i) $g$ is the characteristic function of a degenerate distribution, so that $g(t)=e^{itx}$ for some real $x\ne0$ and all real $t$, and hence $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$ is false;

(ii) there are positive real numbers $b$ and $c$ such that $1-|g(t)|\ge ct^2$ for $|t|\le b$, which again precludes $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$, because $|1-g(t)|\ge1-|g(t)|$.

$\newcommand{\R}{\mathbb{R}}$ This is only possible when $g=1$ identically. Indeed, otherwise, by Hoeffding's Lemma 3.2, page 217, one of the following two alternatives must take place:

(i) $g$ is the characteristic function of a degenerate distribution, so that $g(t)=e^{itx}$ for some real $x\ne0$ and all real $t$, and hence $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$ is false;

(ii) there are positive real numbers $b$ and $c$ such that $1-|g(t)|\ge ct^2$ for $|t|\le b$, which again precludes $\frac{1-g(t)}{t^2}\underset{t\to0}\longrightarrow0$, because $|1-g(t)|\ge1-|g(t)|$.