The Fourier transform of holomorphic functions

For $f\in L^1(ℝ)$, define

$$\widehat{f}(\xi )={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i\xi x}f(x)𝑑x,\xi \in ℝ.$$

For $a>0$, write

We define ${𝔉}_a$ to be the set of functions $f$ that are holomorphic on $S_a$ and for which there is some $A>0$ such that

$$|f(x+iy)|\le \frac{A}{1+x^2},x+iy\in S_a.$$ (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.

Theorem 1.

If $a\mathrm{>}\mathrm{0}$ and $f\mathrm{\in }{\mathrm{F}}_a$, then for any ,

$$\widehat{f}(\xi )=e^{-2\pi |\xi |b}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i\xi x}f(x-i\cdot \mathrm{sgn}\xi \cdot b)𝑑x,\xi \in ℝ.$$

Proof.

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

$\left|{\displaystyle {\int }_{-R-ib}^{-R}}g(z)𝑑z\right|$ $\le {\displaystyle {\int }_{-R-ib}^{-R}}|e^{-2\pi i\xi z}f(z)|𝑑z$

$={\displaystyle {\int }_{-b}^0}|e^{-2\pi i\xi (-R+iy)}f(-R+iy)|𝑑y$

$={\displaystyle {\int }_{-b}^0}{e}^{2\pi \xi y}|f(-R+iy)|𝑑y$

$\le {\displaystyle {\int }_{-b}^0}{e}^{2\pi \xi y}{\displaystyle \frac{A}{1+R^2}}𝑑y$

$=O(R^{-2})$

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 $R\to \mathrm{\infty }$,

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}g(z)𝑑z={\int }_{-\mathrm{\infty }-ib}^{\mathrm{\infty }-ib}g(z)𝑑z,$$

i.e.,

$\widehat{f}(\xi )$ $={\displaystyle {\int }_{-\mathrm{\infty }-ib}^{-\mathrm{\infty }-ib}}{e}^{-2\pi i\xi z}f(z)𝑑z$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi (x-ib)}f(x-ib)𝑑x$

$=e^{-2\pi \xi b}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi x}f(x-ib)𝑑x.$

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

$\left|{\displaystyle {\int }_{-R+ib}^{-R}}g(z)𝑑z\right|$ $\le {\displaystyle {\int }_{-R}^{-R+ib}}|e^{-2\pi i\xi z}f(z)|𝑑z$

$={\displaystyle {\int }_0^b}|e^{-2\pi i\xi (-R+iy)}f(-R+iy)|𝑑y$

$={\displaystyle {\int }_0^b}{e}^{2\pi \xi y}|f(-R+iy)|𝑑y$

$\le {\displaystyle {\int }_0^b}{e}^{2\pi \xi y}{\displaystyle \frac{A}{1+R^2}}𝑑y$

$=O(R^{-2}),$

and likewise

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

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

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

i.e.,

$\widehat{f}(\xi )$ $={\displaystyle {\int }_{-\mathrm{\infty }+ib}^{\mathrm{\infty }+ib}}{e}^{-2\pi i\xi z}f(z)𝑑z$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi (x+ib)}f(x+ib)𝑑x$

$=e^{2\pi \xi b}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi x}f(x+ib)𝑑x.$

∎

Corollary 2.

If $a\mathrm{>}\mathrm{0}$ and $f\mathrm{\in }{\mathrm{F}}_a$, then for any there is some $B$ such that

$$|\widehat{f}(\xi )|\le B{e}^{-2\pi b|\xi |},\xi \in ℝ.$$

Proof.

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

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

By Theorem 1,

$|\widehat{f}(\xi )|$ $\le e^{-2\pi |\xi |b}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|e^{-2\pi i\xi x}f(x-i\cdot \mathrm{sgn}\xi \cdot b)|𝑑x$

$=e^{-2\pi |\xi |b}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}|f(x-i\cdot \mathrm{sgn}\xi \cdot b)|𝑑x$

$\le e^{-2\pi |\xi |b}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{\displaystyle \frac{A}{1+x^2}}𝑑x$

$=e^{-2\pi |\xi |b}\cdot B.$

∎

Define

$$𝔉=\bigcup _{a>0}{𝔉}_a.$$

We now prove the Fourier inversion formula for functions belonging to $𝔉$.

Theorem 3.

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

$$f(x)={\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2\pi ix\xi }\widehat{f}(\xi )𝑑\xi ,x\in ℝ.$$

Proof.

Say $f\in {𝔉}_a$, write

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2\pi ix\xi }𝑑\xi ={\int }_0^{\mathrm{\infty }}{e}^{2\pi ix\xi }𝑑\xi +{\int }_{-\mathrm{\infty }}^0{e}^{2\pi ix\xi }𝑑\xi =I_1+I_2,$$

and take . First we handle $I_1$. By Theorem 1, for $\xi >0$,

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

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

${\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{2\pi ix\xi }\widehat{f}(\xi )𝑑\xi$ $={\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{2\pi ix\xi }\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi (u-ib)}f(u-ib)𝑑u\right)𝑑\xi$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(u-ib){\displaystyle {\int }_0^{\mathrm{\infty }}}{e}^{-2\pi i\xi (u-ib-x)}𝑑\xi 𝑑u$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(u-ib){\displaystyle \frac{1}{2\pi i(u-ib-x)}}𝑑u$

$={\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{L_1}}{\displaystyle \frac{f(\zeta )}{\zeta -x}}𝑑\zeta .$

where $L_1=\{u-ib:u\in ℝ\}$ traversed left to right. Now we handle $I_2$. By Theorem 1, for ,

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

with which, because ,

${\displaystyle {\int }_{-\mathrm{\infty }}^0}{e}^{2\pi ix\xi }\widehat{f}(\xi )𝑑\xi$ $={\displaystyle {\int }_{-\mathrm{\infty }}^0}{e}^{2\pi ix\xi }\left({\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi (u+ib)}f(u+ib)𝑑u\right)𝑑\xi$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(u+ib){\displaystyle {\int }_{-\mathrm{\infty }}^0}{e}^{-2\pi i\xi (u+ib-x)}𝑑\xi 𝑑u$

$={\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}f(u+ib){\displaystyle \frac{-1}{2\pi i(u+ib-x)}}𝑑u$

$=-{\displaystyle \frac{1}{2\pi i}}{\displaystyle {\int }_{L_2}}{\displaystyle \frac{f(\zeta )}{\zeta -x}}𝑑\xi ,$

where $L_2=\{u+ib:u\in ℝ\}$ traversed left to right. Thus

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{2\pi ix\xi }\widehat{f}(\xi )𝑑\xi =\frac{1}{2\pi i}{\int }_{L_1}\frac{f(\zeta )}{\zeta -x}𝑑\zeta -\frac{1}{2\pi i}{\int }_{L_2}\frac{f(\zeta )}{\zeta -x}𝑑\xi .$$ (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),

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

For the right side,

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

Thus, taking $R\to \mathrm{\infty }$ 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

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

proving the claim. ∎

We now prove the Poisson summation formula.

Theorem 4.

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

$$\sum _{n\in \mathbb{Z}}f(n)=\sum _{n\in \mathbb{Z}}\widehat{f}(n).$$

Proof.

Say $f\in {𝔉}_a$, take , 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 $f\in {𝔉}_a$, $\frac{f(z)}{e^{2\pi iz}-1}$ is meromorphic on a region containing ${\gamma }_N$ and its interior, and has poles at $z=-N,\mathrm{\dots },N$, with residues

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

Thus the residue theorem gives us

$${\int }_{{\gamma }_N}\frac{f(z)}{e^{2\pi iz}-1}𝑑z=2\pi i\sum _{|n|\le N}\frac{f(n)}{2\pi i}=\sum _{|n|\le N}f(n).$$ (3)

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

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

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

$\left|{\displaystyle {\int }_{-N-\frac{1}{2}+ib}^{-N-\frac{1}{2}-ib}}{\displaystyle \frac{f(z)}{e^{2\pi iz}-1}}𝑑z\right|$ $\le {\displaystyle {\int }_{-b}^b}\left|f\left(-N-{\displaystyle \frac{1}{2}}+iy\right)\right|𝑑y$

$\le {\displaystyle {\int }_{-b}^b}{\displaystyle \frac{A}{1+{\left(-N-\frac{1}{2}\right)}^2}}𝑑y$

$=O(N^{-2}).$

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 $N\to \mathrm{\infty }$, (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 $L_1=\{x-ib:x\in ℝ\}$, traversed left to right, and $L_2=\{x+ib:x\in ℝ\}$, traversed left to right. Then, as $b>0$,

${\displaystyle \sum _{n\in \mathbb{Z}}}f(n)$ $={\displaystyle {\int }_{L_1}}f(z){\displaystyle \frac{e^{-2\pi iz}}{1-e^{-2\pi iz}}}𝑑z+{\displaystyle {\int }_{L_2}}f(z){\displaystyle \frac{1}{1-e^{2\pi iz}}}𝑑z$

$={\displaystyle {\int }_{L_1}}f(z)e^{-2\pi iz}{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(e^{-2\pi iz})}^{n}dz+{\displaystyle {\int }_{L_2}}f(z){\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{(e^{2\pi iz})}^n$

$={\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle {\int }_{L_1}}{e}^{-2\pi i(n+1)z}f(z)𝑑z+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle {\int }_{L_2}}{e}^{2\pi inz}f(z)𝑑z$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi in(x-ib)}f(x-ib)𝑑x+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{2\pi in(x+ib)}f(x+ib)𝑑x$

$={\displaystyle \sum _{n=1}^{\mathrm{\infty }}}{e}^{-2\pi nb}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi inx}f(x-ib)𝑑x$

$+{\displaystyle \sum _{n=0}^{\mathrm{\infty }}}{e}^{-2\pi nb}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{2\pi inx}f(x+ib)𝑑x.$

Using Theorem 1 this becomes

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

proving the claim. ∎

Take as granted that

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-2\pi i\xi x}{e}^{-\pi x^2}𝑑x=e^{-\pi {\xi }^2},\xi \in ℝ.$$

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

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

$=e^{2\pi i\xi a}{t}^{-1/2}{\displaystyle {\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}}{e}^{-2\pi i\xi t^{-1/2}y}{e}^{-\pi y^2}𝑑y$

$=e^{2\pi i\xi a}{t}^{-1/2}{e}^{-\pi {\xi }^2{t}^{-1}}.$

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

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

Using (4) with $a=0$ gives

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