KU Math 800: Lecture 2/13/2026
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Given a map $f:\RR \to \RR$, along with some regularity conditions at $\pm \infty$, we can define its Fourier transform as
\[\hat{f}(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i x \xi} \dd{x}, \qquad \xi\in \RR.\]We also have Fourier inversion defined by
\[f(x) = \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x \xi} \dd{\xi}.\][example=%counter% (Exercise 1 from Ch. 2)]
\[\int_{-\infty}^\infty e^{-\pi x^2} e^{-2\pi i x\xi} \dd{x} = e^{-\pi \xi^2}.\][/example]
[example=%counter% (Exercise 3 from Ch. 3)]
\[\int_{-\infty}^\infty \frac{1}{\cosh(\pi x)} e^{-2\pi i x \xi} \dd{x} = \frac{1}{\cosh(\pi \xi)}.\][/example]
Let $a \in \RR_{> 0}$. We define $\mathcal F_a$ to be the class of functions satisfying
- $f$ is holomorphic in the strip $\Im(z) < a$
- There exists a constant $A > 0$ such that $|f(x + iy)| \le \frac{A}{1 + x^2}$ for all $x \in \RR$ and $|y| < a$.
[example=%counter%]
- $f(z) = e^{-\pi z^2} \in \mathcal F_a$ for all $a > 0$
- $f(z) = \frac{1}{\pi} \cdot \frac{c}{c^2 + z^2} \in \mathcal F_a$ if $0 < a < c$
- $f(z) = \frac{1}{\cosh(\pi z)}\in \mathcal F_a$ if $|a| < \frac{1}{2}$
[/example]
[remark=%counter%] If $f \in \mathcal F_a$, then $f^{(n)} \in \mathcal{F}_b$ for $0 < b < a$ by Cauchy’s Integral Formula. [/remark]
[theorem=%counter% (Theorem 2.1)] If $f \in \mathcal F_a$, then
\[|\hat{f}(\xi)| \le Be^{-2\pi b |\xi|}\]for some constant $b$ and any $0 \le b < a$. [/theorem]
[remark=%counter%] If $f \in \mathcal{F}$, then $\hat{f}$ has rapid decay at infinity. [/remark]
[proof=Proof of Theorem] Notice that the $b = 0$ case is easy. Suppose that $0 < b < a$. Then
\[\hat{f}(\xi) = \int_{-\infty}^\infty f(x) e^{-2\pi i x\xi} \dd{x}, \quad \xi\in \RR.\]As a side note, notice that the sign of $\xi$ matters.
Case 1: If $\xi > 0$, then let
\[g(z) = f(z) e^{-2\pi i z \xi}.\]Then $g$ is holomorphic. Then we will define the rectangular contour that goes
\[-R \;\;\xrightarrow{I}\;\; R \;\;\xrightarrow{II}\;\; R - ib \;\;\xrightarrow{III}\;\; -R - ib \;\;\xrightarrow{IV}\;\; -R.\]The idea here is that $\hat{f}(\xi)$ will show up in the contour integral. We claim that
\[\lim_{R\to \infty} \left|\int_{II}\right| + \left|\int_{IV}\right| = 0.\]Since
\[\begin{align*} \left|\int_{IV} g(z) \dd{z}\right| &= \left|\int_{-R -ib}^{-R} g(z) \dd{z}\right| \\ &\le \int_0^b |f(-R - it)|\left|e^{-2\pi i(-R - it)\xi}\right| \dd{t} \\ &\le \int_{0}^b \frac{A}{1 + R^2} \cdot e^{-2\pi t \xi} \dd{t} = O(R^{-2}). \end{align*}\]So then $\lim_{R\to \infty} \int_{IV} = 0$. Similarly $\int_{II} \to 0$. Cauchy’s integration then tells us that
\[\hat{f}(\xi) = \lim_{R \to \infty}\int_{I} g(z) \dd{z} = \int_{-\infty}^\infty f(x - ib) e^{-2\pi i (x - ib)\xi} \dd{x}.\]Thus,
\[\begin{align*} |\hat{f}(\xi)| &\le \int_{-\infty}^\infty |f(x - ib)|\cdot e^{-2\pi b \xi} \dd{x} \\ &= \int_{-\infty}^\infty \frac{A}{1 + x^2} e^{-2\pi b\xi} \dd{x} \le B e^{-2\pi b \xi}. \end{align*}\]Thus, we have proved Case 1. A similar argument holds by choosing the contour
\[-R \;\;\xrightarrow{I}\;\; R \;\;\xrightarrow{II}\;\; R + ib \;\;\xrightarrow{III}\;\; -R + ib \;\;\xrightarrow{IV}\;\; -R\]for Case 2 where $\xi < 0$. [/proof]
[lemma=%counter% (Lemma 2.3)] Let $A > 0$ and $B \in \RR$. Then
\[\int_{0}^\infty e^{-(A + iB)\xi} \dd{\xi} = \frac{1}{A + iB}.\][/lemma]
[theorem=%counter% (Theorem 2.2)] If $f \in \mathcal{F}_a$ for some $a$, then
\[f(x) = \int_{-\infty}^\infty \hat{f}(\xi) e^{2\pi i x \xi} \dd{\xi}\]for all $x \in \RR$. [/theorem]
[theorem=%counter% (Poisson Summation Formula)] If $f \in \mathcal{F}_a$ for some $a$, then
\[\sum_{n\in \ZZ} f(n) = \sum_{n\in \ZZ}\hat{f}(n).\][/theorem]
[example=%counter%] Suppose $f(z) = \frac{1}{\cosh(\pi x)}$. Then $\hat{f}(\xi) = f(\xi)$ and so PSF
\[\sum_{n=-\infty}^\infty \frac{e^{-2\pi i a n}}{ \cosh(\pi n/t)} = \sum_{n=-\infty}^{n} \frac{t}{\cosh(\pi(n + a)t)}.\][/example]