Question
Show that the only symmetric $\alpha$ stable distributions with $0<\alpha\leq 2$ are $\phi_{\alpha}(t)=\exp(-|t|^\alpha)$ and their scaled versions (where $\phi$ denotes the characteristic function).
For this question we say that $X$ has a symmetric $\alpha$ stable law if $\phi_{X}(t/n^{1/\alpha})^n=\phi_{X}(t)$ for all $t$ and $n\geq 1$ and if $\phi_{X}$ is real.
It is shown in Durrett that $\phi_{\alpha}$ is a characteristic function for $0<\alpha\leq 2$ and by the definition given above it is clearly a symmetric $\alpha$ stable law. Indeeed, $\phi_{\alpha}$ is real and moreover $$ \left(e^{-|ct/n^{1/\alpha}|^\alpha}\right)^n=e^{-|ct|^\alpha} $$ for any $c\in\mathbb{R}$ which implies that the scaled versions of $\phi_{\alpha}$ are symmetric $\alpha$ stable laws for $0<\alpha\leq 2$.
My problem is how to deduce the reverse inclusion i.e. supposing that $\phi_{X}(t/n^{1/\alpha})^n=\phi_{X}(t)$ for all $t$ and $n\geq 1$ , $\phi_{X}$ is real and $0<\alpha\leq 2$, how to deduce that $\log\phi_{X}(t)=-|ct|^\alpha$ for some $c$?. Any help/hints are appreciated.
