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Consider a probability measure $\mu$ on, let's say, $\mathbb R$.

Is there a necessary and sufficient condition so that $\mu$ has compact support $Supp(\mu)$ ?

I agree this question is too vague, and it may tempting to answer it by quoting the definition of compactness for the support, so let me be more precise.

Imagine you know the Fourrier transform $F_\mu$ of $\mu$ (which contains all the information concerning $\mu$), that is $$ F_\mu(t)=\int e^{itx}\mu(dx),\qquad t\in\mathbb R, $$ or its Cauchy-Stieltjes transform $C_\mu$ (similar story), i.e. $$ C_\mu(z)=\int \frac{\mu(dx)}{z-x},\qquad z\in {\mathbb C }\setminus Supp(\mu), $$ is there any (necessary and) sufficient condition on $F_\mu$ or $C_\mu$ to force $Supp(\mu)$ to be a compact set ?

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Perhaps is relevant? – Qiaochu Yuan May 5 '13 at 5:15
up vote 8 down vote accepted

The Paley-Wiener theorem as it is presented in W.Rudin's "Functional analysis" (Theorem 7.23), will it be satisfactory for you? In your situation: $\mu$ has compact support if and only if $F_\mu$ can be extended as an entire function to $\mathbb C$, and for some $C>0$ and $r>0$ $$ |F_\mu(z)|\le C\cdot e^{r\cdot |{\rm Im} z|},\qquad z\in {\mathbb C}. $$

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Yes, it does ! By any chance do you know if there is a similar characterization involving $C_\mu$ ? – Adrien Hardy May 5 '13 at 9:22
No, actually, I don't know... – Sergei Akbarov May 5 '13 at 10:58

The answer in the second case is even easier. In order to avoid circularity, define the Stieltjes transform of a measure only on the complex plane minus the real axis . Then the measure has support in a compact interval if and only if this transform has an analytic continuous over the real axis outside of this interval. This has suitable generalisations to bounded measures (i.e., with not necessarily compact support) and even to distributions where it is a kingpin of the theory of distributions as boundary values of analytic functions (Köthe, Tillmann et. al).

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