Problem:
Prove that $e^{-d|z|^\alpha}$ is characteristic function of a stable distribution, if $d\geq0$ and $0<\alpha\leq2$.
A note on the definition of stable:
Note that a measure $\mu$ is called stable if there exists a number $a_n>0$ and a number $b_n\in\mathbb{R}$ for all $n=1,2,\ldots$ such that $$\hat{\mu}(z)^n=\hat{\mu}(a_nz)e^{i<z,b_n>}$$.
Result so far: I have shown that $e^{-d|z|^\alpha}$ is indeed stable.
$$\hat{\mu}(z)^n=(e^{-d|z|^\alpha})^n=e^{-dn|z|^\alpha}$$ $$\hat{\mu}(a_nz)=e^{-d|a_nz|^\alpha}e^{izb_n}=e^{izb_n-da_n^\alpha|z|^\alpha}$$ Now we need to find $a_n$ and $b_n$ such that $$-dn|z|^\alpha\overset{!}{=}izb_n-da_n^\alpha|z|^\alpha.$$ Therefore we denote $a_n:= n^{1/\alpha}$ and $b_n:=0$. It follows that $e^{-d|z|^\alpha}$ is stable.
However I have no clue how to show that $e^{-d|z|^\alpha}$ is indeed the characteristic function of a distribution if $d>0$ and $0<\alpha\leq2$. I tried to check the conditions for Polya's theorem (check on wikipedia) but I failed. It seems that the second derivative is not greater equal zero.
Remaining question:
How can I check/prove that $e^{-d|z|^\alpha}$ is indeed the characteristic function of a distribution if $d>0$ and $0<\alpha\leq2$?
Edit 1: Indeed it is possible to show above for $0<\alpha\leq1$ with Polya. However, the question remains for $1<\alpha\leq2$. Anyone know how to check positiv definitess here?