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.



Convolutional Equation
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.


