# Extend a function by convolution

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 $\widehat{g}$.

• Not clear what you are asking. – Yiorgos S. Smyrlis Feb 21 '14 at 12:08
• 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. – Jack D'Aurizio Feb 21 '14 at 12:14
• 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. – Jack D'Aurizio Feb 21 '14 at 15:30

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.
• 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... – AlexR Feb 23 '14 at 16:31