For symmetric stable distributions, why is $\alpha \le 2$? I'm preparing a lecture on stable distributions, and I'm trying to find a simple explanation of the following fact.
Suppose we are trying to come up with stable distributions.  From the definition, it's clear that a distribution is stable iff its characteristic function $\phi$  satisfies $\phi(t)^n = e^{i t b_n} \phi(a_n t)$.  The normal distribution, with chf $\phi(t) = e^{-t^2/2}$ clearly satisfies this with $b_n = 0$, $a_n = \sqrt{n}$.  This suggests that we look for distributions with chfs of the form $\phi(t) = e^{-c |t|^\alpha}$.  For $0 \le \alpha \le 2$, this is indeed a chf, and there is a nice proof in Durrett's book, constructing it as a weak limit using Lévy's continuity theorem.  But:

For $\alpha > 2$,  is there a simple reason why $\phi(t) = e^{-c |t|^\alpha}$ cannot be a chf?

Breiman's Probability proves a general formula for the chf of a stable distribution, using a representation formula for infinitely divisible distributions, but it's  more work  than I want to do for this.
 A: If $\phi$ is a characteristic function, then,  for every real values of $s$ and $t$, $K(t,s)\geqslant0$ where $K(t,s)$ is the determinant 
$$
K(t,s)=\det\begin{pmatrix}\phi(0) & \phi(t) & \phi(t+s) \\ \phi(-t) & \phi(0) & \phi(s) \\ \phi(-t-s) & \phi(-s) & \phi(0)\end{pmatrix}.
$$ 
Using $\phi_\alpha(t)=\mathrm e^{-c|t|^\alpha}$ for every $t$, one gets, for every fixed $x$, $K_\alpha(t,xt)=c^2|t|^{2\alpha}k_\alpha(x)+o(|t|^{2\alpha})$ when $t\to0$, with 
$$
k_\alpha(x)=2x^\alpha(1+x)^\alpha+2x^\alpha+2(1+x)^\alpha−x^{2\alpha}−(1+x)^{2\alpha}−1.
$$
If $\alpha>2$, $k_\alpha(x)=−\alpha^2x^2+o(x^2)$ when $x\to0$ hence $k_\alpha(x)<0$ for some values of $x$ and $K_\alpha(t,tx)<0$ for some (small) values of $t$ and $x$. This proves that $\phi_\alpha$ is not a characteristic function.
First edit To prove that the condition that $K$ is nonnegative is necessary for $\phi$ to be a characteristic function, consider more generally the matrix $M=(M_{k,\ell})$ where $M_{k,\ell}=\mathrm E(\mathrm e^{\mathrm i(t_k-t_\ell)X})$ for some given real numbers $(t_k)$. Then, for every complex valued vector $v=(v_k)$,
$$
v^*Mv=\sum\limits_{k,\ell}M_{k,\ell}v_k\bar v_\ell=\mathrm E\left(\sum\limits_{k,\ell}Z_k\bar Z_\ell v_k\bar v_\ell\right)=\mathrm E\left(\left|\sum\limits_{k}Z_kv_k\right|^2\right),
$$
with $Z_k=\mathrm e^{\mathrm it_kX}$, hence $v^*Mv\ge0$ for every $v$. This means that $M$ represents a nonnegative form, and in particular, $\det M\geqslant0$.
Second edit Here is an alternative proof. I seem to remember that the second moment of $X$ with characteristic function $\phi$, be it finite or not, is
$$
\mathrm E(X^2)=\lim\limits_{t\to0}\ t^{-2}(2-\mathrm E(\mathrm e^{\mathrm itX})-\mathrm E(\mathrm e^{-\mathrm itX})).
$$
Assuming $\phi_\alpha$ is the characteristic function of $X_\alpha$ and using $\phi_\alpha(t)=1-c|t|^\alpha+o(|t|^\alpha)$ when $t\to0$, one gets $\mathrm E(X_\alpha^2)=0$ when $\alpha>2$, which is absurd.
A: Lemma. If $X$ is a random variable such that $\mathbb E\cos(Xt)=1+o(t^2)$ as $t\to 0$, then $X=0$ a.e.
Proof. Observe that $\lim_{n\to\infty}2n^2\bigl(1-\cos\frac{x}{n}\bigr)=x^2$ for all real $x$. Thus by Fatou's lemma,
$$
\mathbb EX^2\leq \liminf_{n\to\infty}\ 2n^2\bigl(1-\mathbb E\cos\frac{x}{n}\bigr)=0.
$$
Corollary. There does not exist a random variable $X$, an $\alpha>2$, and a constant $c>0$ such that
$
\mathbb Ee^{iXt}=e^{-c|t|^{\alpha}}$ for all $t\in\mathbb R$.
Proof.
Suppose such a random variable existed. Then
$$
\mathbb E\cos(Xt)=\Re\bigl(\mathbb Ee^{iXt}\bigr)=1+O(|t|^{\alpha})=1+o(t^2)
$$ as $t\to 0$. Applying the lemma yields $X=0$ a.e., contradicting the hypothesis.
