Proving that the convolution of nonzero compactly supported measures is again nonzero

Question

This is a self-answering question, as allowed by MSE's policies. This is mostly due to me figuring it out while writing the question, so I might as well "share with the class", I thought.

The measures considered here are complex measures, I apologize for forgetting to specify that in the previous version of this question.

Given two Borel measures $\mu$ and $\nu$ on $\mathbb{R}$, one can define the convolution (product) of $\mu$ and $\nu$, denoted by $\mu \ast \nu$, as being the unique measure $\kappa$ such that, for all bounded Borel-measurable functions $h$ we have: $$\int_{\mathbb{R}} h(x) \,\mathrm{d}\kappa(x) = \iint_{\mathbb{R}^2} h(y+z)\mathrm{d}\mu(y)\mathrm{d}\nu(z)$$ One can verify that for measures of the form $f(x)\mathrm{d}x$ and $g(x)\mathrm{d}x$ this is exactly the usual convolution product of $L^1$ functions.

The question then is the following:

How can we show that the convolution of two nonzero compactly supported measures is again nonzero?

This would neatly make the algebra of compactly supported Borel measures equipped with addition and convolution into an integral domain, since the support of the convolution is closed and contained in the sum of the supports if the underlying measures, which is compact because both supports are compact.

This was claimed in the article titled "Fourier transform for mean periodic functions" by László Székelyhidi without proof nor references, meaning that it's probably a classic result, and so I was wondering how to approach this.

Given the counterexamples in Example of non-zero functions with identically zero convolution and If $f*g(x)=0$ then is $f$ identically zero?, it stands to reason that the fact that the supports are compact is the key, and looking at the linked posts maybe we can use Fourier transforms (in the distributional sense), but I can't see how to really start.

Feel free to edit and re-tag if/when appropriate or necessary.

Answer 1

Past me had the right idea with Fourier transforms, as thanks to the Paley-Wiener theorem we have a characterisation of the Fourier transforms of compactly supported distributions.

First of all, we will define the Fourier transform of a measure and show that it is the same as the distributional Fourier transform. Next, we'll provide a formula for the Fourier transform of the convolution. Finally, we'll use a part of the Paley-Wiener theorem to conclude.

The Fourier transform of a measure is defined as follows: $$\hat{\mu} : \xi \in \mathbb{R} \mapsto \int_{\mathbb{R}} e^{-ix\xi} \mathrm{d}\mu(x)$$ This is a well-defined function inducing a tempered distribution for compactly supported measures, and, for all Schwartz class functions $\phi$ we have, denoting temporarily by $\mathcal{F}$ instead of $\hat{\cdot}$ the usual Fourier transform: $$\begin{split}\langle \hat{\mu}, \varphi\rangle &= \int_{\mathbb{R}} \hat{\mu}(\xi) \phi(\xi)\mathrm{d}\xi\\ &= \int_{\mathbb{R}} \left(\int_{\mathbb{R}} e^{-ix\xi} \mathrm{d}\mu(x)\right) \phi(\xi) \mathrm{d}\xi\\ &= \int_{\mathbb{R}} \left(\int_{\mathbb{R}} e^{-ix\xi} \phi(\xi)\mathrm{d}\xi\right) \mathrm{d}\mu(x)\\ &= \langle \mu, \mathcal{F}\phi\rangle = \langle \mathcal{F}\mu, \phi\rangle \end{split}$$ Therefore $\hat{\mu} = \mathcal{F}\mu$.

Now, for two compactly supported measures $\mu$ and $\nu$, we get: $$\begin{split}\widehat{\mu \ast \nu}(\xi) &= \int_{\mathbb{R}} e^{-ix\xi}\mathrm{d}(\mu \ast \nu)(x) \\ &= \iint_{\mathbb{R}^2} e^{-i(y+z)\xi} \mathrm{d}\mu(y)\mathrm{d}\nu(z)\\ &= \left(\int_{\mathbb{R}} e^{-iy\xi}\mathrm{d}\mu(y)\right)\left(\int_{\mathbb{R}} e^{-iz\xi}\mathrm{d}\nu(z)\right)\\ &= \hat{\mu}(\xi)\hat{\nu}(\xi) \end{split}$$ Let's then suppose that $\mu$ and $\nu$ are nonzero. It follows, for example by looking at it from the tempered distributions' perspective, that $\hat{\mu}$ and $\hat{nu}$ are also nonzero.

However, by the Paley-Wiener theorem, $\hat{\mu}$ and $\hat{\nu}$ are restrictions of entire functions (we won't have to use the regularity at infinity given by the theorem, so this could be proved directly to be fair), therefore their product cannot be zero: indeed, if that was the case, then, since $\mathbb{C}$ is not totally disconnected thus not the union of disjoint totally disconnected sets, at least one of $\{\xi \in \mathbb{C} \mid \hat{\mu}(\xi) = 0\}$ or $\{\xi \in \mathbb{C} \mid \hat{\nu}(\xi) = 0\}$ would have accumulation points, hence at least one of $\hat{\mu}$ or $\hat{\nu}$ would be zero by the isolated zeroes principle.

This implies in turn that $\hat{\mu}\hat{\nu}$ is nonzero on $\mathbb{R}$ as the restriction of a nonzero entire function on a set which has accumulation points, and so, since we have showed that $\widehat{\mu \ast \nu} = \hat{\mu}\hat{\nu}$, we finally obtain that $\mu \ast \nu$ is nonzero, concluding the proof.

Answer 2

It can be proved in general that if $\mu$ and $\nu$ are (positive) Borel measures on $\mathbb{R}$ (or $\mathbb{R}^n$), then $$ \overline{\operatorname{supp}(\mu)+\operatorname{supp}(\nu)}= \operatorname{supp}(\mu*\nu) $$ In particular, if $x_0\in\operatorname{supp}(\mu)$ and $y_0\in\operatorname{supp}(\nu)$ then $x_0+y_0\in\operatorname{supp}(\mu*\nu)$; thus, for any open neighborhood $U$ of $0$, $\mu*\nu(x_0+y+0+U)>0$. Indeed, let $V$ be a symmetric open neighborhood of $0$ such that $V+V\subset U$. Then $\mu(x_0+V)\nu(y_0+V)>0$. Notice that $$\mathbb{1}_{\{x_0+V\}}(x)\mathbb{1}_{\{y_0+V\}}(y)\leq\mathbb{1}_{\{x_0+y_0+U\}}(x+y)$$ Integrating with respect to the product measure $\mu\otimes\nu$ $$0<\mu(x_0+V)\nu(y_0+V)\leq \int\mathbb{1}_{x_0+y_0+U}(x+y)\mu(dx)\nu(dy)=\mu*\nu(x_0+y_0+V)$$ One consequence of this is that $\mu*\nu$ is not the zero measure (unless one of the measures $\mu$ or $\nu$ is $0$, which in turn means that $\operatorname{supp}(\mu)+\operatorname{supp}(\nu)=\emptyset $).

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For convolutions of functions, say $f,g\in L_1(m)$ $$\operatorname{supp}(f*g)\subset \overline{\operatorname{supp}(f)+\operatorname{supp}(g)}\tag{1}\label{one}$$ but equality does not hold in general -unless $f,g\geq0$ for in such case, $\mu_f=f\cdot dm$ and $\mu_g\cdot dm$ are nonnegative Borel measures.

For an example in which equality in \eqref{one} does not hold, consider a pair of functions $f,g\in\mathcal{D}=\mathcal{C}^\infty_{00}(\mathbb{R})$. Let $\phi$ and $\psi$ the corresponding Fourier transforms of $f$ and $g$. Then $\phi,\psi\in\mathcal{S}$ (Schwartz space) and $\phi*\psi=0$.

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Edit: The OP added the assumption that measures are complex. In that general setting, there are no compactly supported complex Borel measures $\mu$ and $\nu$ such that $\mu*\nu=0$. To se this, notice that is the support if $\mu$ is a finite positive measure then $\hat{\mu}(z)=\int e^{izx}\,\mu(dx)$ is an entire function. Indeed, suppose $\operatorname{supp}(\mu)\subset(-a,a)$. $$\left|\frac{e^{ix(z+h)}-e^{izx}}{h}\right|\leq e^{|z|a}\frac{e^{|h||x|}-1}{|h|}\leq |h|ae^{|h||a|}$$ Hence, by dominated convergence $\hat{\mu}'(z)$ exists and $\hat{\mu}'(z)=\int ixe^{izx}\,\mu(dx)$ for all $z\in\mathbb{C}$. For a general real measure of finite variation $\mu=\mu_+-\mu_-$, we have that $\operatorname{supp}(\mu_+)\cup\operatorname{supp}(\mu_-) \subset\operatorname{supp}(|\mu|)$.

If $\mu$ has compact support (i.e., $|\mu|$ has compact support), then $\hat{\mu}(z)=\hat{\mu_+}(z)-\hat{\mu_-}(z)$ is entire. By looking at the real and complex part measures of a complex measure $\nu$, we conclude that $\hat{\nu}(z)$ is entire.

Now, if $\mu$ and $\nu$ are non-zero complex measures with compact support, then $\operatorname{supp}(\mu*\nu)\subset\overline{\operatorname{supp}(\mu)+\operatorname{supp}(\nu)}=\operatorname{supp}(\mu)+\operatorname{supp}(\nu)$ and so, $\operatorname{supp}(\mu*\nu)$ is also compact. Being $\hat{\mu}$ and $\hat{\nu}$ entire functions and not $0$, neither of them vanish inside any open set. Since $$\widehat{\mu*\nu}(z)=\hat{\mu}(z)\hat{\nu}(z)$$ is entire, .$\widehat{\mu*\nu}$ does not vanish in any open set. Hence $\mu*\nu\neq0$.