# The Fourier transform of holomorphic functions

Jordan Bell
November 4, 2014

For $f\in L^{1}(\mathbb{R})$, define

 $\widehat{f}(\xi)=\int_{-\infty}^{\infty}e^{-2\pi i\xi x}f(x)dx,\qquad\xi\in% \mathbb{R}.$

For $a>0$, write

 $S_{a}=\{z\in\mathbb{C}:|\mathrm{Im}\,z|

We define $\mathfrak{F}_{a}$ to be the set of functions $f$ that are holomorphic on $S_{a}$ and for which there is some $A>0$ such that

 $|f(x+iy)|\leq\frac{A}{1+x^{2}},\qquad x+iy\in S_{a}.$ (1)

For example, for $f(z)=e^{-\pi z^{2}}$,

 $|f(z)|=|e^{-\pi(x+iy)^{2}}|=|e^{-\pi x^{2}-2\pi ixy+\pi y^{2}}|=e^{-\pi x^{2}}% e^{\pi y^{2}},$

and for any $a>0$, $f\in S_{a}$.

The following is from Stein and Shakarchi.11 1 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 114, Theorem 2.1.

###### Theorem 1.

If $a>0$ and $f\in\mathfrak{F}_{a}$, then for any $0\leq b,

 $\widehat{f}(\xi)=e^{-2\pi|\xi|b}\int_{-\infty}^{\infty}e^{-2\pi i\xi x}f(x-i% \cdot\mathrm{sgn}\,\xi\cdot b)dx,\qquad\xi\in\mathbb{R}.$
###### Proof.

If $b=0$ then the claim is immediate. If $0, we define $g(z)=e^{-2\pi i\xi z}f(z)$. Because $f\in\mathfrak{F}_{a}$ there is some $A>0$ such that $f$ satisfies (1). We prove the claim separately for $\xi\geq 0$ and $\xi\leq 0$. For $\xi\geq 0$, with $R>0$,

 $\displaystyle\left|\int_{-R-ib}^{-R}g(z)dz\right|$ $\displaystyle\leq\int_{-R-ib}^{-R}|e^{-2\pi i\xi z}f(z)|dz$ $\displaystyle=\int_{-b}^{0}|e^{-2\pi i\xi(-R+iy)}f(-R+iy)|dy$ $\displaystyle=\int_{-b}^{0}e^{2\pi\xi y}|f(-R+iy)|dy$ $\displaystyle\leq\int_{-b}^{0}e^{2\pi\xi y}\frac{A}{1+R^{2}}dy$ $\displaystyle=O(R^{-2})$

and likewise

 $\left|\int_{R}^{R-ib}g(z)dz\right|=O(R^{-2}).$

$g$ is holomorphic on $S_{a}$, so by Cauchy’s integral theorem, taking $R\to\infty$,

 $\int_{-\infty}^{\infty}g(z)dz=\int_{-\infty-ib}^{\infty-ib}g(z)dz,$

i.e.,

 $\displaystyle\widehat{f}(\xi)$ $\displaystyle=\int_{-\infty-ib}^{-\infty-ib}e^{-2\pi i\xi z}f(z)dz$ $\displaystyle=\int_{-\infty}^{\infty}e^{-2\pi i\xi(x-ib)}f(x-ib)dx$ $\displaystyle=e^{-2\pi\xi b}\int_{-\infty}^{\infty}e^{-2\pi i\xi x}f(x-ib)dx.$

For $\xi\leq 0$, with $R>0$,

 $\displaystyle\left|\int_{-R+ib}^{-R}g(z)dz\right|$ $\displaystyle\leq\int_{-R}^{-R+ib}|e^{-2\pi i\xi z}f(z)|dz$ $\displaystyle=\int_{0}^{b}|e^{-2\pi i\xi(-R+iy)}f(-R+iy)|dy$ $\displaystyle=\int_{0}^{b}e^{2\pi\xi y}|f(-R+iy)|dy$ $\displaystyle\leq\int_{0}^{b}e^{2\pi\xi y}\frac{A}{1+R^{2}}dy$ $\displaystyle=O(R^{-2}),$

and likewise

 $\left|\int_{R}^{R+ib}g(z)dz\right|=O(R^{-2}).$

By Cauchy’s integral theorem, taking $R\to\infty$,

 $\int_{-\infty}^{\infty}g(z)dz=\int_{-\infty+ib}^{\infty+ib}g(z)dz,$

i.e.,

 $\displaystyle\widehat{f}(\xi)$ $\displaystyle=\int_{-\infty+ib}^{\infty+ib}e^{-2\pi i\xi z}f(z)dz$ $\displaystyle=\int_{-\infty}^{\infty}e^{-2\pi i\xi(x+ib)}f(x+ib)dx$ $\displaystyle=e^{2\pi\xi b}\int_{-\infty}^{\infty}e^{-2\pi i\xi x}f(x+ib)dx.$

###### Corollary 2.

If $a>0$ and $f\in\mathfrak{F}_{a}$, then for any $0\leq b there is some $B$ such that

 $|\widehat{f}(\xi)|\leq Be^{-2\pi b|\xi|},\qquad\xi\in\mathbb{R}.$
###### Proof.

Because $f\in\mathfrak{F}_{a}$ there is some $A>0$ such $f$ satisfies (1). Put

 $B=A\int_{-\infty}^{\infty}\frac{1}{1+x^{2}}dx=\pi A.$

By Theorem 1,

 $\displaystyle|\widehat{f}(\xi)|$ $\displaystyle\leq e^{-2\pi|\xi|b}\int_{-\infty}^{\infty}|e^{-2\pi i\xi x}f(x-i% \cdot\mathrm{sgn}\,\xi\cdot b)|dx$ $\displaystyle=e^{-2\pi|\xi|b}\int_{-\infty}^{\infty}|f(x-i\cdot\mathrm{sgn}\,% \xi\cdot b)|dx$ $\displaystyle\leq e^{-2\pi|\xi|b}\int_{-\infty}^{\infty}\frac{A}{1+x^{2}}dx$ $\displaystyle=e^{-2\pi|\xi|b}\cdot B.$

Define

 $\mathfrak{F}=\bigcup_{a>0}\mathfrak{F}_{a}.$

We now prove the Fourier inversion formula for functions belonging to $\mathfrak{F}$.22 2 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 115, Theorem 2.2.

###### Theorem 3.

If $f\in\mathfrak{F}$, then

 $f(x)=\int_{-\infty}^{\infty}e^{2\pi ix\xi}\widehat{f}(\xi)d\xi,\qquad x\in% \mathbb{R}.$
###### Proof.

Say $f\in\mathfrak{F}_{a}$, write

 $\int_{-\infty}^{\infty}e^{2\pi ix\xi}d\xi=\int_{0}^{\infty}e^{2\pi ix\xi}d\xi+% \int_{-\infty}^{0}e^{2\pi ix\xi}d\xi=I_{1}+I_{2},$

and take $0. First we handle $I_{1}$. By Theorem 1, for $\xi>0$,

 $\widehat{f}(\xi)=e^{-2\pi\xi b}\int_{-\infty}^{\infty}e^{-2\pi i\xi u}f(u-ib)% du=\int_{-\infty}^{\infty}e^{-2\pi i\xi(u-ib)}f(u-ib)du,$

with which, because $\xi b>0$,

 $\displaystyle\int_{0}^{\infty}e^{2\pi ix\xi}\widehat{f}(\xi)d\xi$ $\displaystyle=\int_{0}^{\infty}e^{2\pi ix\xi}\left(\int_{-\infty}^{\infty}e^{-% 2\pi i\xi(u-ib)}f(u-ib)du\right)d\xi$ $\displaystyle=\int_{-\infty}^{\infty}f(u-ib)\int_{0}^{\infty}e^{-2\pi i\xi(u-% ib-x)}d\xi du$ $\displaystyle=\int_{-\infty}^{\infty}f(u-ib)\frac{1}{2\pi i(u-ib-x)}du$ $\displaystyle=\frac{1}{2\pi i}\int_{L_{1}}\frac{f(\zeta)}{\zeta-x}d\zeta.$

where $L_{1}=\{u-ib:u\in\mathbb{R}\}$ traversed left to right. Now we handle $I_{2}$. By Theorem 1, for $\xi<0$,

 $\widehat{f}(\xi)=e^{2\pi\xi b}\int_{-\infty}^{\infty}e^{-2\pi i\xi u}f(u+ib)dx% =\int_{-\infty}^{\infty}e^{-2\pi i\xi(u+ib)}f(u+ib)dx,$

with which, because $\xi b<0$,

 $\displaystyle\int_{-\infty}^{0}e^{2\pi ix\xi}\widehat{f}(\xi)d\xi$ $\displaystyle=\int_{-\infty}^{0}e^{2\pi ix\xi}\left(\int_{-\infty}^{\infty}e^{% -2\pi i\xi(u+ib)}f(u+ib)du\right)d\xi$ $\displaystyle=\int_{-\infty}^{\infty}f(u+ib)\int_{-\infty}^{0}e^{-2\pi i\xi(u+% ib-x)}d\xi du$ $\displaystyle=\int_{-\infty}^{\infty}f(u+ib)\frac{-1}{2\pi i(u+ib-x)}du$ $\displaystyle=-\frac{1}{2\pi i}\int_{L_{2}}\frac{f(\zeta)}{\zeta-x}d\xi,$

where $L_{2}=\{u+ib:u\in\mathbb{R}\}$ traversed left to right. Thus

 $\int_{-\infty}^{\infty}e^{2\pi ix\xi}\widehat{f}(\xi)d\xi=\frac{1}{2\pi i}\int% _{L_{1}}\frac{f(\zeta)}{\zeta-x}d\zeta-\frac{1}{2\pi i}\int_{L_{2}}\frac{f(% \zeta)}{\zeta-x}d\xi.$ (2)

Let $\gamma-R$ be the rectangle starting at $-R-ib$, going to $R-ib$, going to $R+ib$, going to $-R+ib$, going to $-R-ib$. Because this rectangle and its interior are contained in $S_{a}$, on which $f$ is holomorphic, by the residue theorem we have, for $R>|x|$,

 $\int_{\gamma_{R}}\frac{f(\zeta)}{\zeta-x}d\zeta=2\pi i\cdot\mathrm{Res}_{\zeta% =x}\frac{f(\zeta)}{\zeta-x}=2\pi i\cdot f(x).$

We estimate the integrand on the vertical sides of $\gamma_{R}$. For the left side, taking $A$ such that $f$ satisfies (1),

 $\left|\int_{-R+ib}^{-R-ib}\frac{f(\zeta)}{\zeta-x}d\zeta\right|\leq\int_{-b}^{% b}\left|\frac{f(-R+iy)}{-R+iy-x}\right|dy\leq\int_{-b}^{b}\frac{A}{1+R^{2}}% \cdot\frac{1}{R-|x|}dy=O(R^{-3}).$

For the right side,

 $\left|\int_{R-ib}^{R+ib}\frac{f(\zeta)}{\zeta-x}d\zeta\right|\leq\int_{-b}^{b}% \left|\frac{f(R+iy)}{R+iy-x}\right|dy\leq\int_{-b}^{b}\frac{A}{1+R^{2}}\cdot% \frac{1}{R-|x|}dy=O(R^{-3}).$

Thus, taking $R\to\infty$ we get

 $\int_{L_{1}}\frac{f(\zeta)}{\zeta-x}d\zeta-\int_{L_{2}}\frac{f(\zeta)}{\zeta-x% }d\zeta=2\pi i\cdot f(x),$

which by (2) is

 $\int_{-\infty}^{\infty}e^{2\pi ix\xi}\widehat{f}(\xi)d\xi=f(x),$

proving the claim. ∎

We now prove the Poisson summation formula.33 3 Elias M. Stein and Rami Shakarchi, Complex Analysis, p. 118, Theorem 2.4.

###### Theorem 4.

If $f\in\mathfrak{F}$, then

 $\sum_{n\in\mathbb{Z}}f(n)=\sum_{n\in\mathbb{Z}}\widehat{f}(n).$
###### Proof.

Say $f\in\mathfrak{F}_{a}$, take $0, and for $N$ a positive integer let $\gamma_{N}$ be the rectangle starting at $-N-\frac{1}{2}-ib$, going to $N+\frac{1}{2}-ib$, going to $N+\frac{1}{2}+ib$, going to $-N-\frac{1}{2}+ib$, going to $-N-\frac{1}{2}-ib$. Because $f\in\mathfrak{F}_{a}$, $\frac{f(z)}{e^{2\pi iz}-1}$ is meromorphic on a region containing $\gamma_{N}$ and its interior, and has poles at $z=-N,\ldots,N$, with residues

 $\mathrm{Res}_{z=n}\frac{f(z)}{e^{2\pi iz}-1}=\frac{f(n)}{2\pi ie^{2\pi in}}=% \frac{f(n)}{2\pi i}.$

Thus the residue theorem gives us

 $\int_{\gamma_{N}}\frac{f(z)}{e^{2\pi iz}-1}dz=2\pi i\sum_{|n|\leq N}\frac{f(n)% }{2\pi i}=\sum_{|n|\leq N}f(n).$ (3)

For the left side of $\gamma_{N}$, with $z=-N-\frac{1}{2}+iy$, $-b\leq y\leq b$,

 $|e^{2\pi iz}-1|=|e^{-2\pi iN-\pi i-2\pi y}-1|=|-e^{-2\pi y}-1|\geq 1,$

so, taking $A>0$ such that $f$ satisfies (1),

 $\displaystyle\left|\int_{-N-\frac{1}{2}+ib}^{-N-\frac{1}{2}-ib}\frac{f(z)}{e^{% 2\pi iz}-1}dz\right|$ $\displaystyle\leq\int_{-b}^{b}\left|f\left(-N-\frac{1}{2}+iy\right)\right|dy$ $\displaystyle\leq\int_{-b}^{b}\frac{A}{1+\left(-N-\frac{1}{2}\right)^{2}}dy$ $\displaystyle=O(N^{-2}).$

Likewise,

 $\left|\int_{N+\frac{1}{2}-ib}^{N+\frac{1}{2}+ib}\frac{f(z)}{e^{2\pi iz}-1}dz% \right|=O(N^{-2}).$

Therefore, taking $N\to\infty$, (3) becomes

 $\int_{L_{1}}\frac{f(z)}{e^{2\pi iz}-1}dz-\int_{L_{2}}\frac{f(z)}{e^{2\pi iz}-1% }dz=\sum_{n\in\mathbb{Z}}f(n),$

where $L_{1}=\{x-ib:x\in\mathbb{R}\}$, traversed left to right, and $L_{2}=\{x+ib:x\in\mathbb{R}\}$, traversed left to right. Then, as $b>0$,

 $\displaystyle\sum_{n\in\mathbb{Z}}f(n)$ $\displaystyle=\int_{L_{1}}f(z)\frac{e^{-2\pi iz}}{1-e^{-2\pi iz}}dz+\int_{L_{2% }}f(z)\frac{1}{1-e^{2\pi iz}}dz$ $\displaystyle=\int_{L_{1}}f(z)e^{-2\pi iz}\sum_{n=0}^{\infty}(e^{-2\pi iz})^{n% }dz+\int_{L_{2}}f(z)\sum_{n=0}^{\infty}(e^{2\pi iz})^{n}$ $\displaystyle=\sum_{n=0}^{\infty}\int_{L_{1}}e^{-2\pi i(n+1)z}f(z)dz+\sum_{n=0% }^{\infty}\int_{L_{2}}e^{2\pi inz}f(z)dz$ $\displaystyle=\sum_{n=1}^{\infty}\int_{-\infty}^{\infty}e^{-2\pi in(x-ib)}f(x-% ib)dx+\sum_{n=0}^{\infty}\int_{-\infty}^{\infty}e^{2\pi in(x+ib)}f(x+ib)dx$ $\displaystyle=\sum_{n=1}^{\infty}e^{-2\pi nb}\int_{-\infty}^{\infty}e^{-2\pi inx% }f(x-ib)dx$ $\displaystyle+\sum_{n=0}^{\infty}e^{-2\pi nb}\int_{-\infty}^{\infty}e^{2\pi inx% }f(x+ib)dx.$

Using Theorem 1 this becomes

 $\sum_{n\in\mathbb{Z}}f(n)=\sum_{n=1}^{\infty}\widehat{f}(n)+\sum_{n=0}^{\infty% }\widehat{f}(-n)=\sum_{n\in\mathbb{Z}}\widehat{f}(n),$

proving the claim. ∎

Take as granted that

 $\int_{-\infty}^{\infty}e^{-2\pi i\xi x}e^{-\pi x^{2}}dx=e^{-\pi\xi^{2}},\qquad% \xi\in\mathbb{R}.$

For $t>0$ and $a\in\mathbb{R}$, with $y=t^{1/2}(x+a)$,

 $\displaystyle\int_{-\infty}^{\infty}e^{-2\pi i\xi x}e^{-\pi t(x+a)^{2}}dx$ $\displaystyle=\int_{-\infty}^{\infty}e^{-2\pi i\xi(t^{-1/2}y-a)}e^{-\pi y^{2}}% t^{-1/2}dy$ $\displaystyle=e^{2\pi i\xi a}t^{-1/2}\int_{-\infty}^{\infty}e^{-2\pi i\xi t^{-% 1/2}y}e^{-\pi y^{2}}dy$ $\displaystyle=e^{2\pi i\xi a}t^{-1/2}e^{-\pi\xi^{2}t^{-1}}.$

With $f(x)=e^{-\pi t(x+a)^{2}}$, this shows us that

 $\widehat{f}(\xi)=e^{2\pi i\xi a}t^{-1/2}e^{-\pi\xi^{2}t^{-1}},$

and applying the Poisson summaton gives

 $\sum_{n\in\mathbb{Z}}e^{-\pi t(n+a)^{2}}=\sum_{n\in\mathbb{Z}}e^{2\pi ina}t^{-% 1/2}e^{-\pi n^{2}t^{-1}}.$ (4)

Define

 $\vartheta(t)=\sum_{n\in\mathbb{Z}}e^{-\pi n^{2}t},\qquad t>0.$

Using (4) with $a=0$ gives

 $\vartheta(t)=t^{-1/2}\vartheta\left(\frac{1}{t}\right).$