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.)
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$\begingroup$ Do you allow $f$ and $g$ to be complex-valued (even if $h$ is real)? $\endgroup$– Noah SteinCommented Feb 6, 2014 at 15:20
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$\begingroup$ I was thinking of real $f,g,h$, but also a complex factorization would be interesting. This would represent $h$ as $h=f*g-f'*g'$ with real $f,g,f',g'$ (real and imaginary parts of the complex factors), and thus be close to the solution of the "real" question. $\endgroup$– Gandalf LechnerCommented Feb 6, 2014 at 15:54
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$\begingroup$ If you try with Fourier transform the restriction to real $f,g$ would not be very natural. By the way, doesn't Paley-Wiener-Schwartz say that an entire function is the Fourier transform of a smooth function with support in $[-t,t]$ if and only if it is bounded by $c_n (1+|z|)^{-n} \exp(r |\mathrm{Im}(z)|)$ for every $n\in\mathbb N$? $\endgroup$– Jochen WengenrothCommented Feb 6, 2014 at 16:56
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$\begingroup$ Yes, that's right (with $t=r$ ;-)), but I can't see how it solves the problem. For example, if you were asking for even more, namely representing $h$ as a convolution square, $h=f*f$ (i.e. requiring also $f=g$), you would be led to seek square roots of the Fourier transform. Whereas the growth of the root works as it should (taking $r/2$ instead of $r$), analyticity of the root is not so clear because the Fourier transform of $h$ will have zeros. $\endgroup$– Gandalf LechnerCommented Feb 6, 2014 at 17:01
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1$\begingroup$ I think $\hat{h}$ will always have infinitely many zeros. For if it had only finitely many, dividing by a polynomial $p$ would give an entire functions without zeros, which can be represented as $e^k$ with some other entire function $k$, i.e. $\hat{h}(z)=p(z)\,e^{k(z)}$. Now, since $\hat{h}$ is of exponential type and $p$ is a polynomial, $k$ must be bounded by a polynomial of order 1, and thus $k(z)=\alpha+\beta z$. One then sees that the Fourier transform of such a function exists only as a distribution. $\endgroup$– Gandalf LechnerCommented Feb 6, 2014 at 21:00
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2 Answers
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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.
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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.
answered Feb 6, 2014 at 23:57
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2$\begingroup$ I think that a representation of $h$ as a series of convolutions is quite a different question which can be tackled with linear functional analysis because it asks for the surjectivity of the linear operator $\mathscr D([-r,r]) \tilde{\otimes}\mathscr D([-r,r]) = \mathscr D([-r,r]^2) \to \mathscr D([-2r,2r])$, $\phi \mapsto (x\mapsto \int \phi(x-y,y)dy$. $\endgroup$ Commented Feb 7, 2014 at 7:25
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$\begingroup$ Thanks for bringing up the smooth vectors point of view and Dixmier/Malliavin, Paul. That gives us a representation of $h$ as a finite sum of convolutions $f_n*g_n$. I am not quite sure if one is still free to choose the supports of these $f_n,g_n$ in $[-r,r]$, though. As Jochen points out, the original question is somewhat different. Given that already Dixmier/Malliavin is a quite non-trivial result, I now doubt that a factorization $h=f*g$ without linear combinations is always possible. $\endgroup$ Commented Feb 7, 2014 at 9:21