I found, by assuming that exists a random variable $X$ that accepts $\exp(-t^4)$ as characteristic function , $E[X^2] = 0$, which means that $P[X=0] = 1$ implying the function is $0$. It's acceptable to use this fact to prove what I want, but I really don't know a formal way to say that, would someone help me with formality?
Thaks in advance.
The characteristic function is defined as: $$ \varphi(t) = \mathbb{E}[e^{itX}] \tag{1}$$ and assuming either $X\in L^2$ or $\varphi\in C^2$ we have: $$ \frac{d^2}{dt^2}\,\varphi(t) = \mathbb{E}[-X^2 e^{itX}]\tag{2} $$ by the linearity of the expected value, hence $\varphi''(0)=0$ implies $\mathbb{E}[X^2]=0$. As you already noticed, that leads to a contradiction since it implies that the distribution of $X$ is concentrated in a single point, $\mathbb{E}[X]$. A more striking contradiction is also given by: $$ -24 = \varphi^{IV}(0) = \mathbb{E}[X^4] \geq 0.\tag{3}$$
