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Let $f \in \mathcal{C}^{\infty}(\mathbb{R})$ be a compactly supported function ($supp(f)\Subset\mathbb{R})$.

I am wondering about the existence of a $g \in L^p(\mathbb{R})$, for some $p$, such that $$(f \ast g) \mathbb{1}_{Supp(f)}=f.$$

What if $f \in \mathcal{C}^{0}(\mathbb{R})$?


It is a well-known fact that no algebra of functions possesses an identity for the convolution, but the restriction to the support of $f$ may overcome this issue. It is quite trivial that $g$ cannot be more regular than $f$; moreover, the Holder inequality gives some bounds for the norms of $g$ and $\widehat{g}$.

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  • $\begingroup$ Not clear what you are asking. $\endgroup$ – Yiorgos S. Smyrlis Feb 21 '14 at 12:08
  • $\begingroup$ It looks perfectly clear to me. Given a compact supported $C^{\infty}$ function $f$, prove or disprove the existence of a $g\in L^p$ such that the identity holds. $\endgroup$ – Jack D'Aurizio Feb 21 '14 at 12:14
  • $\begingroup$ It is worth noticing that if such a $g$ exists (even if I do not believe so) it can be defined in any way outside of $\operatorname{Supp}(f)-\operatorname{Supp}(f)$, since values of $g$ in this set never play in the LHS. $\endgroup$ – Jack D'Aurizio Feb 21 '14 at 15:30
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If we define $S$ as $S=\operatorname{Supp}f$, since $f=f\cdot\mathbb{1}_S$, $$\widehat{f}=\widehat{f}*\widehat{\mathbb{1}}_S,\tag{1}$$ so the Fourier transform of your identity gives: $$\left(\widehat{f}\cdot(1-\widehat{g})\right)*\widehat{\mathbb{1}}_S=0.\tag{2}$$ By the Riemann-Lebesgue theorem we know that $g\in L^p$ implies $\widehat{g}=o(1)$, so, if the convolution with $\widehat{\mathbb{1}}_S$ is an injective map, there are no solutions to $(2)$. Anyway, we can assume that a $g$ satisfying the given identity is supported on the set $S-S$, that is symmetric around zero (see my previous comments). Hence we have that $f*g$ is supported on $S+S-S$ and: $$ \widehat{f},\widehat{g}\in C^{\omega}.\tag{3}$$ (Following Tao, we know much more: $\widehat{f}$ and $\widehat{g}$ are entire functions with at most exponential growth) By the inversion formula: $$ \forall x\in S\qquad \int e^{i x\xi}\,\widehat{f}(\xi)\,\left(1-\widehat{g}(\xi)\right)\,d\xi = 0,\tag{4}$$ so, by assuming that the interior of $S$ is non-empty, we should have that $\widehat{f}\cdot(1-\widehat{g})$ is almost-everywhere zero. Since $\widehat{f}\in C^{\omega}$ implies $\widehat{f}\neq 0$ almost everywhere, $$ \widehat{g}=1\;\text{a.e.}\tag{5}$$ follows, but $(5)$ contradicts the Riemann-Lebesgue theorem.

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  • $\begingroup$ A little side note: You might want to watch constants in the fourier transform and convolution theorem. These will not affect your result, so it's not a major issue. Using $$\hat f(\xi) := (2\pi)^{-\frac n2} \int_{\mathbb R^n} e^{-ix\xi} f(x)\ {\rm d}x$$ You get $$\widehat{f\ast g} = (2\pi)^{\frac n2} \hat f \hat g$$ for example. If working on $\mathbb T^n$, the constants change, of course... $\endgroup$ – AlexR Feb 23 '14 at 16:31
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A set of related questions is discussed in Hans G. Feichtinger and Walter Schachermayer
Local nonfactorization of functions on locally compact groups Arch. Math. (Basel), Vol.49 (1987) p.72--78 http://univie.ac.at/nuhag-php/bibtex/open_files/fesc87_lnfact.pdf

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  • $\begingroup$ Hi @hgfei, welcome to SE! This is a useful piece of information, but it is not an answer as such. Here at SE, the Answer section is only used to full answers to OP questions. This post would go as a comment of the question. $\endgroup$ – Andrea May 5 '16 at 18:11

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