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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 Stein
    Feb 6 '14 at 15:20
  • $\begingroup$ 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. $\endgroup$ Feb 6 '14 at 15:54
  • $\begingroup$ 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 $n\in\mathbb N$? $\endgroup$ Feb 6 '14 at 16:56
  • $\begingroup$ 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. $\endgroup$ Feb 6 '14 at 17:01
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    $\begingroup$ 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. $\endgroup$ Feb 6 '14 at 21:00
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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.

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    $\begingroup$ 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 $\mathscr D([-r,r]) \tilde{\otimes}\mathscr D([-r,r]) = \mathscr D([-r,r]^2) \to \mathscr D([-2r,2r])$, $\phi \mapsto (x\mapsto \int \phi(x-y,y)dy$. $\endgroup$ Feb 7 '14 at 7:25
  • $\begingroup$ 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. $\endgroup$ Feb 7 '14 at 9:21

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