0
$\begingroup$

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.

Thanks!

$\endgroup$
2
$\begingroup$

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 $\delta+|\Im t|<b$.)

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.