Which smooth compactly supported functions are convolutions? If $f,g$ are smooth functions with support in the interval $[-r,r]$ for some $r>0$, then their convolution $f*g$ is smooth with support in $[-2r,2r]$. My question is about the converse: Given smooth $h$ with support in $[-2r,2r]$, can I always write it as $h=f*g$ with $f,g$ as above? (By Fourier transform, one can formulate this problem also as a decomposition of entire functions of exponential type $2r$ into a product of entire functions of exponential type $r$ with additional restrictions on the real line.)
 A: The general issue of whether test functions are convolutions of two others, or finite sums, and/or limitations, is the subject of (at least) two classic papers: 
Pierre Cartier, ‘Vecteurs diffe ́rentiables dans les repre ́sentations unitaires des groupes de Lie’, expose ́ 454 of Seminaire Bourbaki 1974/1975. Lecture Notes in Mathematics 514, Springer, 1976.
Jacques Dixmier and Paul Malliavin, ‘Factorisations de fonctions et de vecteurs inde ́finiment diffe ́ren- tiables’, Bull. Sci. Math. 102 (1978), 305–330.
On Lie groups in general, the answer is that a test function can be written as a finite sum of convolutions of pairs of test functions. At some point, the sum must contain an indefinite number of summands, but for many applications this is irrelevant.
A: After some more searching I found the solution in the literature. In the paper
L. Ehrenpreis, "Solution of some problems of division. IV", Amer. J. Math. 82 (1960), 522-588
Ehrenpreis posed the question if any $h\in C_c^\infty({\mathbb R}^n)$ can be represented as a convolution $f*g$ of two functions $f,g\in C_c^\infty({\mathbb R}^n)$, this question is therefore known as the "Ehrenpreis factorization problem". 
For $n\geq2$ the answer is no (shown 1978-1980 by several authors, cited in the paper below), but for $n=1$ such a factorization is always possible. This has been proven much later in 
R. S. Yulmukhametov, "Solution of the Ehrenpreis factorization problem", Sb. Math. 190 (1999) 597, doi:10.1070/SM1999v190n04ABEH000400
via the complex analysis approach, i.e. by factoring the entire Fourier transform of $h$ (and in particular its zeros) in an appropriate manner. In his Theorem 10, Yulmukhametov also answers the sharpened version, including the support conditions supp $h\subset[-2r,2r]$, supp $f$, supp $g\subset[-r,r]$, affirmatively. This is precisely the question posted here.
