Question

Hi,

Given C>0. Let $f,g,h$ be $L^2$ functions such that $f,g,h$ have a compact and finite-measure support, and $f*f(x)=g*g(x)=h*h(x)=f*g(x)=g*h(x)=f*h(x)=0$ (where $*$ is the convolution) for all $|x| > C$.

Does that imply $f(x)=g(x)=h(x)=0$ for all $x$? What is I won't require $f*f=g*g=h*h=0$? What if we had two functions $f,g$ instead of three $f,g,h$, and $f*f(x)=g*g(x)=f*g(x)=0$ for all $|x| > C$? Will it imply they both identically zero everywhere?

If not, please provide (or hint to) a counterexample.

Answer 1

I don't know why you say "compact and finite measure". Every compact set has finite measure. Fourier transform of a function with compact support is an entire function. Fourier transform of a convolution is the square of the Fourier transform. So EACH of the equations $f*f=0$ or $f*g=0$ already implies that $f=0$ and $g=0$, and so on. This answers your first question.

But in the second question you ask what follows from $f*f(z)=0$ for large $x$. Nothing follows. If you have two functions with compact support then the support of the convolution is also compact.

However, if $[a,b]$ is the convex hull of the support of $f$ and $[c,d]$ is the convex hull of the support of $g$, then the convex hull of the support of $f*g$ is exactly $[a+c,b+d]$. Not smaller! (Here I use the assumption that functions are in $L^2$, and that we are dealing with functions of ONE variable). This is Titchmarsh's theorem.

Answer 2

No:

First: L^1 is an algebra under convolution, but if $f$ has compact support and is in $L^2$ then it is in $L^1$.

Your functions have all compact support. Since $supp(f * g)\subseteq \pm supp(f) \pm supp(g)$ your condition with $C$ is always satisfied.