Is there an "elementary" proof that $\alpha$-stable random variables only exist for $0 < \alpha \le 2$? By elementary I mean without using Fourier transforms. I'd be happiest with either a direct probabilistic argument, or a geometric argument, e.g., referring to embeddings of $L\_p$ spaces (as long as it doesn't depend on results which are themselves proved via Fourier analysis).


Note that what I'm interested in here is the fact that $\alpha$-stable random *don't* exist for $\alpha > 2$, not the fact that they *do* exist for $\alpha \le 2$.


**Edit:** To clarify, for the purposes of this question I define an $\alpha$-stable random variable as follows. Given $\alpha > 0$, $X$ is $\alpha$-stable if, whenever $X\_1, \ldots, X\_n$ are independent copies of $X$, there is a real number $d=d(n)$ such that $X\_1 + \cdots + X\_n$ has the same distribution as $n^{1/\alpha} X + d$. And I'm quite happy to simplify to the situation when $d=0$.



Nonexistence of stable random variables
I believe there to be a proof avoiding Fourier analysis in the literature given by Feller (1971) as part of Theorem VI.1.1.


The argument can be subdivided into two parts. The first part (which was already mentioned in the comments) states, that if an $\alpha$-stable random vector has a finite non-zero variance, then $\alpha=2$. The second part is showing that variance is finite when $\alpha>2$, which is done by subdividing the integral $E(X^2)$ into a suitable series. Thus, $Var(X)=0$ when $\alpha>2$, which completes the proof.


