Fourier transform division theorem in $\mathbb R^n$

Question

It is known that if $f \in L^1(\mathbb R)$, $\widehat f(\xi) \neq 0$ for any $\xi \in \mathbb R$, then for any $h \in L^1(\mathbb R)$ such that $\widehat h$ is compactly supported there exists $g \in L^1(\mathbb R)$ with $\widehat g = \frac{\widehat h}{\widehat f}$. This theorem can be found in many books. In several articles the authors also use the multidimensional version but I can't find any proof. My question is whether this theorem can be found in some book in the case of $\mathbb R^n$ (for references)?

Answer 1

I don't have a reference for you, but I can sketch a very simple elegant proof. Of course all this is closely related to Wiener's Tauberian Theorem. A few years ago I found the proof of WTT that I'd been looking for all my life! Was all excited, then of course it turned out I'd been scooped again.

But "my" proof is the "right" proof, although it's not actually mine, and it's nearly as well known as it should be. So my vote is you give the proof instead of citing a reference, just to make the "right" argument a little better known.

CBAID = Commutative Banach Algebra with Idenity.

If only $L^1(R^n)$ had an identity. Then WTT would be trivial from the basic stuff about CBAIDs. Instead you see proofs of WTT that use techniques that remind you of the proofs of the basic CBAID theorem (summing geometric series, etc). All my life I felt it should instead be possible to get WTT directly from the CBAID stuff, instead of repeating similar arguments.

It is possible to get WTT from the basic results on CBAIDs! And the result you cite is of course the main step.

Ok, end of explanation of why the following is fascinating, and on to the following:

Say $f$ and $h$ are as you say. Let $B$ be, say, a large closed ball such that $\hat h$ is supported in a compact subset of the interior of $B$.

Let $I$ be the space of all $g\in L^1(R^n)$ such that $\hat g$ vanishes identically on $B$. Then $I$ is a closed ideal in $L^1$. So the quotient space $$A=L^1/I$$is a CBA. The big deal is that $A$ actually has an identity! Indeed, if $e\in L^1$ is such that $\hat e=1$ everywhere on $I$ then it's easy to check that $e+I$ is an identity for $A$.

So $A$ is a CBAID. And it's easy to verify that the maximal ideal space of $A$ "is" $I$, in the sense that the maps $g+I\mapsto\hat g(\xi)$ for $\xi\in B$ give all the complex homomorphisms of $A$.

So $\phi(f+I)\ne0$ for every complex homomorphism $\phi$ of $A$. Hence $f+I$ is invertible in $A$ (!). This says that there exists $g\in L^1$ with $\hat f\hat g=1$ everywhere on $B$. And now since $\hat h$ is supported in $B$ it follows that $$\hat g\hat h = \hat h/\hat f.$$

Oops, my "$g$" is not your $g$. Sorry...