The Fourier transform of holomorphic functions

Jordan Bell
November 4, 2014

For fL1(), define

f^(ξ)=-e-2πiξxf(x)𝑑x,ξ.

For a>0, write

Sa={z:|Imz|<a}.

We define 𝔉a to be the set of functions f that are holomorphic on Sa and for which there is some A>0 such that

|f(x+iy)|A1+x2,x+iySa. (1)

For example, for f(z)=e-πz2,

|f(z)|=|e-π(x+iy)2|=|e-πx2-2πixy+πy2|=e-πx2eπy2,

and for any a>0, fSa.

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 fFa, then for any 0b<a,

f^(ξ)=e-2π|ξ|b-e-2πiξxf(x-isgnξb)𝑑x,ξ.
Proof.

If b=0 then the claim is immediate. If 0<b<a, we define g(z)=e-2πiξzf(z). Because f𝔉a there is some A>0 such that f satisfies (1). We prove the claim separately for ξ0 and ξ0. For ξ0, with R>0,

|-R-ib-Rg(z)𝑑z| -R-ib-R|e-2πiξzf(z)|𝑑z
=-b0|e-2πiξ(-R+iy)f(-R+iy)|𝑑y
=-b0e2πξy|f(-R+iy)|𝑑y
-b0e2πξyA1+R2𝑑y
=O(R-2)

and likewise

|RR-ibg(z)𝑑z|=O(R-2).

g is holomorphic on Sa, so by Cauchy’s integral theorem, taking R,

-g(z)𝑑z=--ib-ibg(z)𝑑z,

i.e.,

f^(ξ) =--ib--ibe-2πiξzf(z)𝑑z
=-e-2πiξ(x-ib)f(x-ib)𝑑x
=e-2πξb-e-2πiξxf(x-ib)𝑑x.

For ξ0, with R>0,

|-R+ib-Rg(z)𝑑z| -R-R+ib|e-2πiξzf(z)|𝑑z
=0b|e-2πiξ(-R+iy)f(-R+iy)|𝑑y
=0be2πξy|f(-R+iy)|𝑑y
0be2πξyA1+R2𝑑y
=O(R-2),

and likewise

|RR+ibg(z)𝑑z|=O(R-2).

By Cauchy’s integral theorem, taking R,

-g(z)𝑑z=-+ib+ibg(z)𝑑z,

i.e.,

f^(ξ) =-+ib+ibe-2πiξzf(z)𝑑z
=-e-2πiξ(x+ib)f(x+ib)𝑑x
=e2πξb-e-2πiξxf(x+ib)𝑑x.

Corollary 2.

If a>0 and fFa, then for any 0b<a there is some B such that

|f^(ξ)|Be-2πb|ξ|,ξ.
Proof.

Because f𝔉a there is some A>0 such f satisfies (1). Put

B=A-11+x2𝑑x=πA.

By Theorem 1,

|f^(ξ)| e-2π|ξ|b-|e-2πiξxf(x-isgnξb)|𝑑x
=e-2π|ξ|b-|f(x-isgnξb)|𝑑x
e-2π|ξ|b-A1+x2𝑑x
=e-2π|ξ|bB.

Define

𝔉=a>0𝔉a.

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

Theorem 3.

If fF, then

f(x)=-e2πixξf^(ξ)𝑑ξ,x.
Proof.

Say f𝔉a, write

-e2πixξ𝑑ξ=0e2πixξ𝑑ξ+-0e2πixξ𝑑ξ=I1+I2,

and take 0<b<a. First we handle I1. By Theorem 1, for ξ>0,

f^(ξ)=e-2πξb-e-2πiξuf(u-ib)𝑑u=-e-2πiξ(u-ib)f(u-ib)𝑑u,

with which, because ξb>0,

0e2πixξf^(ξ)𝑑ξ =0e2πixξ(-e-2πiξ(u-ib)f(u-ib)𝑑u)𝑑ξ
=-f(u-ib)0e-2πiξ(u-ib-x)𝑑ξ𝑑u
=-f(u-ib)12πi(u-ib-x)𝑑u
=12πiL1f(ζ)ζ-x𝑑ζ.

where L1={u-ib:u} traversed left to right. Now we handle I2. By Theorem 1, for ξ<0,

f^(ξ)=e2πξb-e-2πiξuf(u+ib)𝑑x=-e-2πiξ(u+ib)f(u+ib)𝑑x,

with which, because ξb<0,

-0e2πixξf^(ξ)𝑑ξ =-0e2πixξ(-e-2πiξ(u+ib)f(u+ib)𝑑u)𝑑ξ
=-f(u+ib)-0e-2πiξ(u+ib-x)𝑑ξ𝑑u
=-f(u+ib)-12πi(u+ib-x)𝑑u
=-12πiL2f(ζ)ζ-x𝑑ξ,

where L2={u+ib:u} traversed left to right. Thus

-e2πixξf^(ξ)𝑑ξ=12πiL1f(ζ)ζ-x𝑑ζ-12πiL2f(ζ)ζ-x𝑑ξ. (2)

Let γ-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 Sa, on which f is holomorphic, by the residue theorem we have, for R>|x|,

γRf(ζ)ζ-x𝑑ζ=2πiResζ=xf(ζ)ζ-x=2πif(x).

We estimate the integrand on the vertical sides of γR. For the left side, taking A such that f satisfies (1),

|-R+ib-R-ibf(ζ)ζ-x𝑑ζ|-bb|f(-R+iy)-R+iy-x|𝑑y-bbA1+R21R-|x|𝑑y=O(R-3).

For the right side,

|R-ibR+ibf(ζ)ζ-x𝑑ζ|-bb|f(R+iy)R+iy-x|𝑑y-bbA1+R21R-|x|𝑑y=O(R-3).

Thus, taking R we get

L1f(ζ)ζ-x𝑑ζ-L2f(ζ)ζ-x𝑑ζ=2πif(x),

which by (2) is

-e2πixξf^(ξ)𝑑ξ=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 fF, then

nf(n)=nf^(n).
Proof.

Say f𝔉a, take 0<b<a, and for N a positive integer let γN be the rectangle starting at -N-12-ib, going to N+12-ib, going to N+12+ib, going to -N-12+ib, going to -N-12-ib. Because f𝔉a, f(z)e2πiz-1 is meromorphic on a region containing γN and its interior, and has poles at z=-N,,N, with residues

Resz=nf(z)e2πiz-1=f(n)2πie2πin=f(n)2πi.

Thus the residue theorem gives us

γNf(z)e2πiz-1𝑑z=2πi|n|Nf(n)2πi=|n|Nf(n). (3)

For the left side of γN, with z=-N-12+iy, -byb,

|e2πiz-1|=|e-2πiN-πi-2πy-1|=|-e-2πy-1|1,

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

|-N-12+ib-N-12-ibf(z)e2πiz-1𝑑z| -bb|f(-N-12+iy)|𝑑y
-bbA1+(-N-12)2𝑑y
=O(N-2).

Likewise,

|N+12-ibN+12+ibf(z)e2πiz-1𝑑z|=O(N-2).

Therefore, taking N, (3) becomes

L1f(z)e2πiz-1𝑑z-L2f(z)e2πiz-1𝑑z=nf(n),

where L1={x-ib:x}, traversed left to right, and L2={x+ib:x}, traversed left to right. Then, as b>0,

nf(n) =L1f(z)e-2πiz1-e-2πiz𝑑z+L2f(z)11-e2πiz𝑑z
=L1f(z)e-2πizn=0(e-2πiz)ndz+L2f(z)n=0(e2πiz)n
=n=0L1e-2πi(n+1)zf(z)𝑑z+n=0L2e2πinzf(z)𝑑z
=n=1-e-2πin(x-ib)f(x-ib)𝑑x+n=0-e2πin(x+ib)f(x+ib)𝑑x
=n=1e-2πnb-e-2πinxf(x-ib)𝑑x
+n=0e-2πnb-e2πinxf(x+ib)𝑑x.

Using Theorem 1 this becomes

nf(n)=n=1f^(n)+n=0f^(-n)=nf^(n),

proving the claim. ∎

Take as granted that

-e-2πiξxe-πx2𝑑x=e-πξ2,ξ.

For t>0 and a, with y=t1/2(x+a),

-e-2πiξxe-πt(x+a)2𝑑x =-e-2πiξ(t-1/2y-a)e-πy2t-1/2𝑑y
=e2πiξat-1/2-e-2πiξt-1/2ye-πy2𝑑y
=e2πiξat-1/2e-πξ2t-1.

With f(x)=e-πt(x+a)2, this shows us that

f^(ξ)=e2πiξat-1/2e-πξ2t-1,

and applying the Poisson summaton gives

ne-πt(n+a)2=ne2πinat-1/2e-πn2t-1. (4)

Define

ϑ(t)=ne-πn2t,t>0.

Using (4) with a=0 gives

ϑ(t)=t-1/2ϑ(1t).