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There is a natural correspondence between Fourier and Laurent series: If $f$ is a function on the circle whose Fourier coefficients $\{ a_n \}$ satisfy an exponential decrease $a_n = O(r^{|n|})$, $0<r<1$ as $|n| \to \infty$, then the associated Laurent series $g(z)=\sum_{n \in \mathbb{Z}}a_nz^n$ defines a holomorphic function on the annulus $r<|z|<1/r$ which is an extension of $f$ on the circle.

Is there any analogue statement for this fact for Fourier transform?

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For the sake of clearity, let $f$ be a (complex-valued)function on the real line which is in the class $L^1$ and the Fourier transform $\hat{f}(\xi)=O\left(e^{-a|\xi|}\right)$ for some $a>0$.

Can we deduce a holomorphic extension of $f$ on a strip containing the real line?

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  • $\begingroup$ Now you have slower than exponential decay. The corresponding condition would be $\lvert\hat{f}(\omega)\rvert \leqslant C e^{-\delta \lvert \omega\rvert}$ (for $\lvert \omega\rvert > R_0$), and that would guarantee that $f$ is (after redefining on a null set) the restriction of a holomorphic function on a strip. $\endgroup$ – Daniel Fischer Jan 12 '17 at 15:41
  • $\begingroup$ @DanielFischer I was wrong. The Fourier transform should have exponential decay. I'll correct it. $\endgroup$ – HyJu Jan 12 '17 at 15:46
  • $\begingroup$ @DanielFischer Anyway it is remarkable that there is such a nice analogue! $\endgroup$ – HyJu Jan 12 '17 at 15:47
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If there is an $a > 0$ such that $\hat{f}(\xi) e^{a\lvert\xi\rvert}$ is bounded, then the integral

$$F(z) = \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} \hat{f}(\xi) e^{i \xi z}\,d\xi$$

converges at least for $z \in \mathbb{C}$ with $\lvert\operatorname{Im} z\rvert < a$. By the dominated convergence theorem, $F$ is a continuous function on the strip, and we see that $F$ is holomorphic by Morera's theorem, or differentiation under the integral. By the Fourier inversion formula, we have

$$F(x) = f(x)$$

for almost all $x\in \mathbb{R}$. The result generalises to higher dimensions with essentially the same argument. If you use the form of the Fourier transform with kernel $e^{-2\pi i x\xi}$, then the strip would be $\lvert\operatorname{Im} z\rvert < \frac{a}{2\pi}$, but nothing essential changes.

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  • $\begingroup$ This is interesting since the variable $z$ is moved from the base to the exponent, which is inevitable modification to saturate the exponential decreasing condition. Thanks! $\endgroup$ – HyJu Jan 13 '17 at 13:57

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