Suppose $X,X_1,X_2,\ldots$ are independent and identically distributed random variables with a symmetric distribution $F$. We say $F$ is symmetric $\alpha$ stable where $0<\alpha\le 2$, if $$X_1+X_2+\cdots+X_n \stackrel{d}{=} n^{1/\alpha}X, \ \ \mbox{for all} \ n\in\mathbb{N}$$
I wish to show that the characteristic function of $F$ is given by $\psi(t)=e^{-c|t|^{\alpha}}$.
Breiman's probability book solve the general version of this problem (consider stable only, not symmetric). However, it uses infinite divisible distributions and lot of calculations.
So I was wondering if there is a short argument for the symmetric case.
My try: Note that using the equal in distribution relation
$$\psi(t)=\psi\left(\frac{t}{n^{1/\alpha}}\right)^n$$ Enough to take $t>0$ as $\psi(t)=\psi(-t)$. Now taking limit we have $$\psi(t)=\lim_{n\to\infty}\psi\left(\frac{t}{n^{1/\alpha}}\right)^n=\exp\left(-\lim_{x\to 0^+}\frac{1-\psi(tx^{1/\alpha})}{x}\right)=\exp\left(-t^{\alpha}\lim_{z\to 0^+}\frac{1-\psi(z)}{z^\alpha}\right)$$
So it is enough to show that $\displaystyle \lim_{z\to 0^+}\frac{1-\psi(z)}{z^\alpha}$ exists and is non zero. However, I don't have any clue how to proceed from here. Any help/suggestions.
By applying the functional equation $$ \psi(t)=\psi\left(t\cdot n^{-1/\alpha}\right)^n $$ twice, it follows that for any natural numbers $n$ and $m$ we have that $$ \psi\left(\frac{n^{1/\alpha}}{m^{1/\alpha}}\right)=\psi(1)^{n/m}. $$ When combined with $\psi(t)=\psi(-t)$, it follows that $\psi(t)=e^{-c|t|^{\alpha}}$ whenever $t^{\alpha}$ is a rational number, where $e^{-c}=\psi(1)$.
The characteristic function of an arbitrary random variable is continuous, by the bounded convergence theorem. Since we have established that $\psi(t)=e^{-c|t|^{\alpha}}$ on a dense set, it follows that $\psi(t)=e^{-c|t|^{\alpha}}$ holds for all real values of $t$.