Timeline for Which smooth compactly supported functions are convolutions?
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| Feb 8, 2014 at 13:06 | vote | accept | Gandalf Lechner | ||
| Feb 8, 2014 at 13:06 | answer | added | Gandalf Lechner | timeline score: 17 | |
| Feb 7, 2014 at 1:03 | comment | added | Suvrit | @Gandalf: I was also thinking along the lines of delta functions, but could not make the argument rigorous --- but I think your justification sounds right. Thanks! | |
| Feb 6, 2014 at 23:57 | answer | added | paul garrett | timeline score: 4 | |
| Feb 6, 2014 at 22:57 | comment | added | Gandalf Lechner | @Suvrit: I am not sure if these are "very few". Intuitively speaking, if you pick $g$ close to a delta distribution, then $f*g$ will approximate $f$, so you will at least get all smooth functions supported in $[-r,r]$ "up to $\varepsilon$". | |
| Feb 6, 2014 at 22:48 | comment | added | Suvrit | Wouldn't very few functions admit such a decomopsition (i.e., into a conv of two functions) | |
| Feb 6, 2014 at 21:34 | history | edited | Gandalf Lechner |
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| Feb 6, 2014 at 21:04 | comment | added | Gandalf Lechner | Along the same lines as in the comment by Will Sawin, one can use the complex analysis point of view to show that not every $h\in C_c^\infty([-2r,2r])$ can be represented as a convolution square, i.e. $h=f*f$ with some $f\in C_c^\infty({\mathbb R})$. In fact, after Fourier transformation we would have a square of an entire function, and thus all zeros of $\hat{h}$ would need to be even. But this is not the case for all $h\in C_c^\infty([-2r,2r])$. However, the original question is not settled by this argument. | |
| Feb 6, 2014 at 21:00 | comment | added | Gandalf Lechner | 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. | |
| Feb 6, 2014 at 20:03 | comment | added | Will Sawin | Consider the factorization $\hat{h}=\hat{f}\hat{g}$, and let $s=\hat{f}/\hat{g}$. Then $s$ has zeroes and poles whenever $\hat{h}$ has a singularity of odd order. $s$ cannot grow too large - $s$ can only be exponentially large in regions where $\hat{h}$ undershoots its upper bound, and the same with $s^{-1}$. If $\hat{h}$ has only finitely many zeroes of odd order, this is no problem - one can just choose $s$ to be meromorphic. If $\hat{h}$ has infinitely many odd order zeroes, $s$ has an essential singularity, and I'm not sure what growth problems that can cause. | |
| Feb 6, 2014 at 17:05 | comment | added | Gandalf Lechner | For completeness, I add another closely related question. If you consider the linear span $K_r$ of all $f*g$ with $f,g\in C_c^\infty([-r,r])$, do we then have $K_r=C_c^\infty([-2r,2r])$? It seems clear that $K_r$ is a dense subspace, but is it all of $C_c^\infty([-2r,2r])$? | |
| Feb 6, 2014 at 17:01 | comment | added | Gandalf Lechner | 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. | |
| Feb 6, 2014 at 16:56 | comment | added | Jochen Wengenroth | 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 $n\in\mathbb N$? | |
| Feb 6, 2014 at 15:54 | comment | added | Gandalf Lechner | 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. | |
| Feb 6, 2014 at 15:20 | comment | added | Noah Stein | Do you allow $f$ and $g$ to be complex-valued (even if $h$ is real)? | |
| Feb 6, 2014 at 9:22 | history | edited | Gandalf Lechner |
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| Feb 5, 2014 at 19:53 | review | First posts | |||
| Feb 5, 2014 at 19:54 | |||||
| Feb 5, 2014 at 19:34 | history | asked | Gandalf Lechner | CC BY-SA 3.0 |