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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.

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Could you provide us with some motivation why you want to do this? Please check the FAQ if MO is appropriate for your question. – David Roberts Nov 13 '12 at 4:43
it's quite complicated to explain, i need it to investigate a class of partial differential equations – Ohad Asor Nov 13 '12 at 4:46
i'm not a student, this is for real research – Ohad Asor Nov 13 '12 at 4:46
up vote 3 down vote accepted

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.

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Thanks! Please note that I'm not talking about the convolution of the Fourier transforms, but of the functions themselves. Fourier transform doesn't appear in the question at all. – Ohad Asor Nov 13 '12 at 16:31
Ohad, where is Alexandre talking about convolution of Fourier transforms? He is talking about the Fourier transform of a convolution of functions being the pointwise product of the Fourier transforms. – Yemon Choi Nov 13 '12 at 22:06
OK I reread the answer and got it. – Ohad Asor Nov 14 '12 at 3:28
regarding compactness and finite measure: look at the set which is the union of all closed intervals [4n,4n+1] where n is natural. isn't it both compact and infinite measure set? – Ohad Asor Nov 14 '12 at 4:39
On my opinion, this set is NOT compact. I was taught (and teach my students) that the set on a line is compact if and only if it is closed and bounded, am I wrong? – Alexandre Eremenko Nov 14 '12 at 16:59


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.

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Thanks! I was not aware of Titchmarsh convolution theorem So I guess the answer whether there are such $f,g,h$ depends whether $C$ is belongs to their support or not. Is that what you meant? – Ohad Asor Nov 13 '12 at 7:26
I wrote the question mistakenly, I meant C is given. I edited the question to fix it. – Ohad Asor Nov 13 '12 at 7:44

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