Background Given a (translation bounded) positive definite measure $\gamma$ lets say on $\mathbb R^d$, its Fourier transform as a tempered distribution is a positive measure $\widehat{\gamma}$.

I am interested in the Lesbegue decomposition $$\widehat{\gamma}=\widehat{\gamma}_{pp}+\widehat{\gamma}_{ac}+\widehat{\gamma}_{sc}$$ at the level of $\gamma$. The decomposition $\widehat{\gamma}=\widehat{\gamma}_{pp}+\widehat{\gamma}_{c}$ is well understood, so the problem reduces simply to understanding the decomposition of the continuous component.

So, the main question becomes: Given a measure $\gamma$ such that $\widehat{\gamma}$ is continuous, how can we identify $\gamma_1, \gamma_2$ such that $\widehat{\gamma_1}$ is a.c. and $\widehat{\gamma_2}$ is s.c.? Note that $\gamma_1, \gamma_2$ might actually not be measures. This leads to:

The Question

If we fix a compact set $K \subset \mathbb R^d$, can we get a nice description of the set

$$ \{ \widehat{f} | f \in L^1(\mathbb R^d), \operatorname{supp}(f) \subset K \} \,?$$

Note that for the type of problem I am interested, convolving $f$ with a continuous compactly supported function is OK. So any extra restriction on $f$ which can be obtained by convolutions with a class of compactly supported continuous functions (for example $f$ is continuous) is OK. Also, the condition $ \operatorname{supp}(f) \subset K$ can be replaced by $f$ compactly supported.


1 Answer 1


The Fourier transform of a function with compact support is automatically holomorphic. This support can be deduced from growth conditions on the former, at least in the case when it is a euclidean ball and the function is square integrable. This is the celebrated Payley-Wiener theorem, upon which there is a comprehensive Wikipedia article for starters.

  • $\begingroup$ But the Fourier Transform of any compactly supported tempered distribution is an entire function too. This tells us that the space of holomorphic functions is much bigger then the set I described. $\endgroup$
    – Nick S
    Dec 5, 2013 at 19:20
  • $\begingroup$ As I said, you don't get all holomorphic functions---there are also growth conditions which depend on the support of the function. The Payley-Wiener theorem gives a precise relationship between these and the support in the case of balls. It is possible to refine these for convex sets, using indicator functions. There are precise versions also for the case of distributions and smooth functions of compact support but the example of the FT for functions on the line suggests that there will be no such equivalences in the case of integrable functions. $\endgroup$
    – User4891
    Dec 5, 2013 at 19:39

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