 Strict stability implies self-decomposability

Let $X$ have a strictly stable distribution of index $\alpha$, and let $X^{\prime }$ be an independent copy. For integers $1\le k<n$, split a sum of $n$ independent copies into blocks of sizes $k$ and $n-k$. Strict stability of each block and of the full sum givesFor any $0<c<1$, choose rationals $k_m/n_m\to c^{\alpha }$. On the fixed probability space carrying $X,X^{\prime }$, the left side converges almost surely to $cX+(1-c^{\alpha }{)}^{1/\alpha }{X}^{\prime }$. Thus it converges in distribution; since its probability distribution is constant, the limit also has the law of $X$. This proves a self-decomposable distribution with the displayed remainder. The proof does not require any finite moments.