How can I prove that $\varphi(t)=e^{-|t|^{\alpha}}$ is not a characteristic function for $\alpha > 2$

Question

I am trying to prove that for $\alpha > 2$ this is not a characteristic function $$ \varphi(t)=e^{-|t|^{\alpha}} $$

Suppose $t > 0$. Differentiating twice gives us $$ (e^{-t^{\alpha}})^{(2)} = \alpha^2e^{-t\alpha}t^{2\alpha -2}-\alpha e^{-t^{\alpha}}(\alpha-1)t^{\alpha-2} $$

Now, $\mathbb{E}X^2=\frac{\varphi^{(2)}(0)}{i^2}$ = 0. Beacuse $0 \leq \text{Var}X \leq \mathbb{E}X^2 = 0$ we know that $X$ has a variance of $0$ which means X is constant and equal to $\mathbb{E}X$. The same happens when $t<0$. How can I proceed? Is it a contradiction already?

I could not find how does a characteristic function of a constant variable look like.

Answer 1

Theorem 4.1.1 in Lukacs's book "characteristic functions" states that

The only characteristic function which has the form $f(t)=1+o(t^2)$ as $t\to 0$ is the function $f(t)\equiv 1$.

A corollary of this theorem is that a non-constant characteristic cannot have a first and second derivative vanishing at zero. If it does, then by Taylor's theorem for sufficiently small $t$: $$f(t)=f(0)+f'(0)t+f''(0)t^2+o(t^2)=1+o(t^2)$$

(Recall that $f(0)=1$ for every characteristic function).