Assume a positive Borel measure $\mu$ on $\mathbb{R}$ is given and that there is $b>0$ such that $$\int_{\mathbb{R}}e^{b|x|}d\mu(x)<\infty.$$ Then it should hold that the Fourier transform $$\hat{\mu}(t)=\int_{\mathbb{R}}e^{itx}d\mu(x)$$ is an analytic function in the strip $\{t\in\mathbb{C} \mid |\Im t|<b\}$.

I think that there is a variant of Paley-Wiener theorem stating exactly this, though I can not find it anywhere.

I have looked in some books: first the Hormander's book and next the Reed&Simon's book (vol.2), but a variant of the theorem suitable to my case is not there.

Can anybody help me with that?

Recall that there is a theorem stating exactly what I need for Fourier transform of $\textbf{functions}$ from $L^{2}(\mathbb{R})$ [e.g., R&SII Thm. IX.13]. A more general theorem dealing with $\textbf{tempered distributions}$ (of course, the measure $\mu$ can be understood as a tempered distribution) also exists, but the one I found in [R&SII, Thm. IX.14] gives only the opposite implication to what I need.



If you're not finding exactly this it's because this is the trivial direction; the converse you mention is harder. For the result you ask about just differentiate under the integral!

Or to put it another way: Fix $t$ in that strip. The hypothesis turns out to be exactly what you need to use dominated convergence to show that $\lim_{h\to0}(\hat\mu(t+h)-\hat\mu(t))/h$ exists. (Restrict to $|h|<\delta$, where $\delta+|\Im t|<b$.)

  • $\begingroup$ Yes, you're right. It's indeed quite easy...thanks for giving a clue. $\endgroup$ – Twi Sep 7 '15 at 15:51

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