Homogeneous Distributions

2.4 More About Distributions and the Fourier Transform

2.4.3 Homogeneous Distributions

The fundamental solutions of the Laplacian are locally integrable functions onRⁿ and also homogeneous of degree 2−nwhenn≥3. Since homogeneous distribu- tions often arise in applications, it is desirable to pursue their study. Here we do not undertake such a study in depth, but we discuss a few important examples.

Definition 2.4.5.Forz∈Cwe define a distributionu_zas follows:

uz,f

= Z

Rⁿ

π

z+n 2

Γ ^z+n₂ |x|^zf(x)dx. (2.4.6) Clearly theu_z’s coincide with the locally integrable functions

π

z+n

2 Γ ^z+n₂ −1

|x|^z

when Rez>−nand the definition makes sense only for that range ofz’s. It follows from its definition thatu_zis a homogeneous distribution of degreez.

We would like to extend the definition ofu_zforz∈C. Let Rez>−nfirst. FixN to be a positive integer. Given f∈S(Rⁿ), write the integral in (2.4.6) as follows:

Z

|x|<1

π^z+n2 Γ(^z+n₂ )

(

f(x)−

∑

|α|≤N

(∂^αf)(0) α! x^α

)

|x|^zdx

+ Z

|x|>1

π

z+n 2

Γ(^z+n₂ )f(x)|x|^zdx+ Z

|x|<1

π

z+n 2

Γ(^z+n₂ )

∑

|α|≤N

(∂^αf)(0)

α! x^α|x|^zdx. The preceding expression is equal to

Z

|x|<1

π

z+n 2

Γ(^z+n₂ ) (

f(x)−

∑

|α|≤N

(∂^αf)(0) α! x^α

)

|x|^zdx

+ Z

|x|>1

π

z+n 2

Γ(^z+n₂ )f(x)|x|^zdx

+

∑

|α|≤N

(∂^αf)(0) α!

π^z+n2 Γ(^z+n₂ )

Z 1 r=0

Z

Sⁿ⁻¹

(rθ)^αr^z+n−1dr dθ, where we switched to polar coordinates in the penultimate integral. Now set

b(n,α,z) = π

z+n 2

Γ(^z+n₂ ) 1 α!

Z Sⁿ⁻¹

θ^αdθ Z 1

r=0

r^|α|+n+z−1dr

= π

z+n 2

Γ(^z+n₂ )

1 α!

Z

Sⁿ⁻¹

θ^αdθ

|α|+z+n ,

whereα = (α₁, . . . ,αn)is a multi-index. These coefficients are zero when at least oneαjis odd. Consider now the case that all theαj’s are even; then|α|is also even.

The functionΓ(^z+n₂ )has simple poles at the points

z=−n, z=−(n+2), z=−(n+4), and so on;

see Appendix A.5. These poles cancel exactly the poles of the function z7→(|α|+z+n)⁻¹

atz=−n− |α|when|α|is an even integer in[0,N]. We therefore have u_z,f

= Z

|x|≥1

π^z+n2

Γ(^z+n₂ )f(x)|x|^zdx+

∑

|α|≤N

b(n,α,z)

∂^αδ₀,f

+ Z

|x|<1

π

z+n 2

Γ(^z+n₂ ) (

f(x)−

∑

|α|≤N

(∂^αf)(0) α! x^α

)

|x|^zdx.

(2.4.7)

Both integrals converge absolutely when Rez>−N−n−1, since the expression inside the curly brackets above is bounded by a constant multiple of|x|^N+1, and the resulting function ofzin (2.4.7) is a well defined analytic function in the range Rez>−N−n−1.

SinceNwas arbitrary, u_z,f

has an analytic extension to all ofC. Therefore,u_z is a distribution-valued entire function ofz.

Next we would like to calculate the Fourier transform ofu_z. We know by Exercise 2.3.9 thatub_zis a homogeneous distribution of degree−n−z. The choice of constant in the definition ofu_zwas made to justify the following result:

Theorem 2.4.6.For all z∈Cwe haveub_z=u−n−z.

Proof. The idea of the proof is straightforward. First we show that for a certain range ofz’s we have

Z

Rⁿ

|ξ|^zϕ(ξb )dξ =C(n,z) Z

Rⁿ

|x|^−n−zϕ(x)dx, (2.4.8) for some fixed constantC(n,z)and allϕ∈S(Rⁿ). Next we pick a specificϕ to evaluate the constantC(n,z). Then we use analytic continuation to extend the va- lidity of (2.4.8) for allz’s. Use polar coordinates by settingξ =ρ ϕ andx=rθ in (2.4.8). We have

Z

Rⁿ

|ξ|^zϕb(ξ)dξ

= Z ∞

0

ρ^z+n−1 Z ∞

0 Z

Sⁿ⁻¹

ϕ(rθ) Z

Sⁿ⁻¹

e−2πirρ(θ·ϕ)dϕ

dθrⁿ⁻¹dr dρ

= Z ∞

0

_{Z ∞}

0 σ(rρ)ρ^z+n−1dρ Z

Sⁿ⁻¹ϕ(rθ)dθ

rⁿ⁻¹dr

=C(n,z) Z ∞

0

r^−z−n _Z

Sⁿ⁻¹

ϕ(rθ)dθ

rⁿ⁻¹dr

=C(n,z) Z

Rⁿ

|x|^−n−zϕ(x)dx, where we set

σ(t) = Z

Sⁿ⁻¹

e^{−2πit(θ·ϕ)}dϕ= Z

Sⁿ⁻¹

e^{−2πit(ϕ1)}dϕ, (2.4.9) C(n,z) =

Z ∞

0

σ(t)t^z+n−1dt, (2.4.10)

and the second equality in (2.4.9) is a consequence of rotational invariance. It re- mains to prove that the integral in (2.4.10) converges for some range ofz’s.

Ifn=1, then σ(t) =

Z

S⁰

e^−2πitϕdϕ=e^−2πit+e^2πit=2 cos(2πt) and the integral in (2.4.10) converges conditionally for−1<Rez<0.

Let us therefore assume thatn≥2. Since|σ(t)| ≤ωn−1, the integral converges near zero when−n<Rez. Let us study the behavior ofσ(t)fortlarge. Using the formula in Appendix D.2 and the definition of Bessel functions in Appendix B.1, we write

σ(t) = Z ₁

−1e^2πitsωn−2

p1−s²n−2 ds

√

1−s² =c_nJn−2 2

(2πt),

for some constantc_n. Using the asymptotics for Bessel functions (Appendix B.7), we obtain that|σ(t)| ≤ct^−1/2 when n−2>−1/2 and t≥1. In either case the integral in (2.4.10) converges absolutely near infinity when Rez+n−1−1/2<−1, i.e., when Rez<−n+1/2.

We have now proved that when−n<Rez<−n+1/2 we have ub_z=C(n,z)u_−n−z

for some constantC(n,z)that we wish to compute. Insert the functionϕ(x) =e^−π|x|2 in (2.4.8). Example 2.2.9 gives that this function is equal to its Fourier transform.

Use polar coordinates to write ωn−1

Z ∞

0

r^z+n−1e^−πr2dr=C(n,z)ω_n−1 Z ∞

0

r^{−z−n+n−1}e^−πr2dr.

Change variabless=πr² and use the definition of the gamma function to obtain that

C(n,z) =Γ(^z+n₂ ) Γ(−₂^z)

π^−z+n2 π

z 2

. It follows thatub_z=u−n−zfor the range ofz’s considered.

At this point observe that for everyf ∈S(Rⁿ), the functionz7→

ubz−u−z−n,f is entire and vanishes for−n<Rez<−n+1/2. Therefore, it must vanish every-

where and the theorem is proved.

Homogeneous distributions were introduced in Exercise 2.3.9. We already saw that the Dirac mass onRⁿis a homogeneous distribution of degree−n. There is another important example of a homogeneous distributions of degree−n, which we now discuss.

LetΩ be an integrable function on the sphereSⁿ⁻¹with integral zero. Define a tempered distributionW_Ω onRⁿby setting

W_Ω,f

=lim

ε→0 Z

|x|≥ε

Ω(x/|x|)

|x|ⁿ f(x)dx. (2.4.11) We check thatW_Ω is a well defined tempered distribution on Rⁿ. Indeed, since Ω(x/|x|)/|x|ⁿhas integral zero over all annuli centered at the origin, we obtain

W_Ω,ϕ =

ε→0lim Z

ε≤|x|≤1

Ω(x/|x|)

|x|ⁿ (ϕ(x)−ϕ(0))dx+ Z

|x|≥1

Ω(x/|x|)

|x|ⁿ ϕ(x)dx

≤

∇ϕ L^∞

Z

|x|≤1

|Ω(x/|x|)|

|x|ⁿ⁻¹ dx+

sup

x∈Rⁿ

|x| |ϕ(x)|

Z

|x|≥1

|Ω(x/|x|)|

|x|ⁿ⁺¹ dx

≤C₁

∇ϕ L^∞

Ω

L¹(Sⁿ⁻¹)+C₂

∑

|α|≤1

ϕ(x)x^α L^∞

Ω L¹(Sⁿ⁻¹),

for suitable constantsC₁andC₂in view of (2.2.2).

One can verify thatW_Ω ∈S⁰(Rⁿ)is a homogeneous distribution of degree−n just like the Dirac mass at the origin. It is an interesting fact that all homogeneous distributions onRⁿof degree−nthat coincide with a smooth function away from the origin arise in this way. We have the following result.

Proposition 2.4.7.Suppose that m is a C^∞ function onRⁿ\ {0} that is homoge- neous of degree zero. Then there exist a scalar b and aC^∞functionΩ onSⁿ⁻¹with integral zero such that

m^∨=bδ0+W_Ω, (2.4.12)

where W_Ω denotes the distribution defined in (2.4.11).

To prove this result we need the following proposition, whose proof we postpone until the end of this section.

Proposition 2.4.8.Suppose that u is aC^∞function onRⁿ\ {0}that is homogeneous of degree z∈C. Thenu is ab C^∞function onRⁿ\ {0}.

We now prove Proposition 2.4.7 using Proposition 2.4.8.

Proof. Letabe the integral of the smooth functionmoverSⁿ⁻¹. The functionm−a is homogeneous of degree zero and thus locally integrable onRⁿ; hence it can be thought of as a distribution that we call bu (the Fourier transform of a tempered distributionu). Since bu is a C^∞ function on Rⁿ\ {0}, Proposition 2.4.8 implies that u is also aC^∞ function onRⁿ\ {0}. LetΩ be the restriction ofu onSⁿ⁻¹. ThenΩ is a well definedC^∞function onSⁿ⁻¹. Sinceuis a homogeneous function of degree−n that coincides with the smooth functionΩ onSⁿ⁻¹, it follows that u(x) =Ω(x/|x|)/|x|ⁿforxinRⁿ\ {0}.

We show that Ω has mean value zero over Sⁿ⁻¹. Pick a nonnegative, radial, smooth, and nonzero functionψonRⁿsupported in the annulus 1<|x|<2. Switch- ing to polar coordinates, we write

u,ψ

= Z

Rⁿ

Ω(x/|x|)

|x|ⁿ ψ(x)dx=c_ψ Z

Sⁿ⁻¹

Ω(θ)dθ, u,ψ

= bu,ψb

= Z

Rⁿ

(m(ξ)−a)ψb(ξ)dξ=c⁰_ψ Z

Sⁿ⁻¹

m(θ)−a

dθ=0, and thusΩ has mean value zero overSⁿ⁻¹(sincecψ6=0).

We can now legitimately define the distributionW_Ω, which coincides with the functionΩ(x/|x|)/|x|ⁿonRⁿ\ {0}. But the distributionualso coincides with this function onRⁿ\ {0}. It follows thatu−W_Ω is supported at the origin. Proposition 2.4.1 now gives that u−W_Ω is a sum of derivatives of Dirac masses. Since both distributions are homogeneous of degree−n, it follows that

u−W_Ω =cδ₀.

Butu= (m−a)^∨ =m^∨−aδ₀, and thusm^∨= (c+a)δ₀+W_Ω.This proves the propo-

sition.

We now turn to the proof of Proposition 2.4.8.

Proof. Letu∈S⁰be homogeneous of degreezandC^∞ onRⁿ\ {0}. We need to show that ubis C^∞ away from the origin. We prove thatubis C^M for all M. Fix M∈Z⁺and letαbe any multi-index such that

|α|>n+M+Rez. (2.4.13)

Pick aC^∞functionϕ onRⁿthat is equal to 1 when|x| ≥2 and equal to zero for

|x| ≤1. Writeu₀= (1−ϕ)uandu_∞=ϕu. Then

∂^αu=∂^αu₀+∂^αu_∞ and thus ∂d^αu=∂[^αu₀+∂[^αu_∞,

where the operations are performed in the sense of distributions. Sinceu₀is com- pactly supported, Theorem 2.3.21 implies that∂[^αu₀isC^∞. Now Leibniz’s rule gives that

∂^αu_∞=v+ϕ ∂^αu,

wherevis a smooth function supported in the annulus 1≤ |x| ≤2. ThenbvisC^∞ and we need to show only thatϕ ∂\^αu isC^M. The functionϕ ∂^αuis actually C^∞, and by the homogeneity of ∂^αu (Exercise 2.3.9(c)) we obtain that (∂^αu)(x) =

|x|^−|α|+z(∂^αu)(x/|x|). Sinceϕis supported away from zero, it follows that

|ϕ(x)(∂^αu)(x)| ≤ C_α

(1+|x|)^|α|−Rez (2.4.14) for someC_α>0. It is now straightforward to see that if a function satisfies (2.4.14), then its Fourier transform isC^Mwhenever (2.4.13) is satisfied. See Exercise 2.4.1.

We conclude that ∂[^αu_∞ is a C^M function whenever (2.4.13) is satisfied; thus so is∂d^αu. Since∂d^αu(ξ) = (2πiξ)^αu(ξb ), we deduce smoothness forbuaway from the origin. Letξ 6=0. Pick a neighborhoodV ofξ that does not meet the jth co- ordinate axis for some 1≤ j≤n. Thenη_j6=0 whenη∈V. Letα be the multi- index (0, . . . ,M, . . . ,0) with M in the jth coordinate and zeros elsewhere. Then (2πiη_j)^Mbu(η)is aC^M function onV, and thus so isu(ηb ), since we can divide by η^M_j . We conclude thatu(ξb )isC^MonRⁿ\ {0}. SinceMis arbitrary, the conclusion

follows.

We end this section with an example that illustrates the usefulness of some of the ideas discussed in this section.

Example 2.4.9.Let η be a smooth function on Rⁿ that is equal to 1 on the set

|x| ≥1/2 and vanishes on the set|x| ≤1/4. Let 0<Re(α)<n. Let g(ξ) = η(x)|x|^−α

b(ξ).

The functiongdecays faster than the reciprocal of any polynomial at infinity and g(ξ)−π^α−n2Γ(^n−α₂ )

Γ(^α₂) |ξ|^α−n

is aC^∞function onRⁿ. Therefore,g is integrable onRⁿ. This example indicates the interplay between the smoothness of a function and the decay of its Fourier transform. The smoothness of the functionη(x)|x|^−α near zero is reflected by the decay ofg near infinity. Moreover, the function η(x)|x|^−α is not affected by the bumpηnear infinity, and this results in a behavior ofg(ξ)near zero similar to that of(|x|^−α)b(ξ).

To see these assertions, first observe that∂^γ(η(x)|x|^−α)is integrable and thus (−2πiξ)^γg(ξ)is bounded ifγis large enough. This gives the decay ofgnear infin- ity. We now use Theorem 2.4.6 to obtain

g(ξ) =π^α−n2Γ(^n−α₂ )

Γ(^α₂) |ξ|^α−n+ϕ(ξb ),

whereϕ(ξb ) = (η(x)−1)|x|^−α

b(ξ), which isC^∞ as the Fourier transform of a compactly supported distribution.

Exercises

2.4.1.Suppose that a functionf satisfies the estimate

|f(x)| ≤ C_α (1+|x|)^N, for someN>n. Then bf isC^M when 1≤M≤[N−n].

2.4.2.Use Corollary 2.4.3 to prove Liouville’s theorem that every bounded har- monic function onRⁿmust be a constant. Derive as a consequence thefundamental theorem of algebra, stating that every polynomial onCmust have a complex root.

2.4.3.Prove thate^xis not inS⁰(R)but thate^xe^iex is inS⁰(R).

2.4.4.Show that the Schwartz functionx7→sech(πx),x∈R, coincides with its Fourier transform.

Hint:Integrate the functione^iazover the rectangular contour with corners(−R,0), (R,0),(R,iπ), and(−R,iπ).

2.4.5.(Ismagilov [137]) Construct an uncountable family of linearly independent Schwartz functions f_a such that|f_a|=|f_b|and|bf_a|=|bf_b|for all f_aand f_bin the family.

Hint:Letwbe a smooth nonzero function whose Fourier transform is supported in the interval[−1/2,1/2]and letϕ be a real-valued smooth nonconstant periodic function with period 1. Then take f_a(x) =w(x)e^iϕ(x−a)fora∈R.

2.4.6.LetPybe the Poisson kernel defined in (2.1.13). Prove that for f ∈L^p(Rⁿ), 1≤p<∞, the function

(x,y)7→(P_y∗f)(x)

is a harmonic function onRⁿ⁺¹₊ . Use the Fourier transform and Exercise 2.2.11 to prove that(P_y1∗P_y2)(x) =P_y1+y₂(x)for allx∈Rⁿ.

2.4.7.(a) For a fixedx₀∈Rⁿ, show that the function v(x;x₀) = 1− |x|²

|x−x₀|ⁿ is harmonic onRⁿ\ {x0}.

(b) For fixedx0∈Sⁿ⁻¹, prove that the family of functionsθ7→v(θ;rx0), 0<r<1, defined on the sphere satisfies

limr↑1 Z

θ∈Sⁿ⁻¹

|θ−x₀|>δ

v(θ;rx₀)dθ=0

uniformly inx₀. The functionv(θ;rx₀)is called thePoisson kernel for the sphere.

(c) Let f be a continuous function onSⁿ⁻¹. Prove that the function u(x) = 1

ωn−1

(1− |x|²) Z

Sⁿ⁻¹

f(θ)

|x−θ|ⁿdθ

solves the Dirichlet problem∆(u) =0 on|x|<1 andu=f on|x|=1.

2.4.8.Fix a real numberλ, 0<λ<n.

(a) Prove that

Z

Sⁿ

|ξ−η|^−λdξ=2^n−λπ^π2Γ(

n−λ 2 )

Γ(n−^λ₂). (b) Prove that

Z

Rⁿ

|x−y|^−λ(1+|x|²)^λ2−ndx=2^n−λπ^π2Γ(

n−λ 2 )

Γ(n−^λ₂)(1+|y|²)^−λ2 . Hint:Use the stereographic projection in Appendix D.6.

2.4.9.Prove the followingbeta integral identity:

Z

Rⁿ

dt

|x−t|^α1|y−t|^α2 =π

n

2 Γ ^n−α₂¹

Γ ^n−α₂²

Γ ^α1+α₂²⁻ⁿ Γ ^α₂¹

Γ ^α₂²

Γ n−^α1+α₂ ² |x−y|^{n−α1−α2}, where 0<α1,α₂<n,α1+α2>n.

2.4.10.(a) Prove that if a function f onRⁿ(n≥3) is constant on the spheresrSⁿ⁻¹ for allr>0, then so is its Fourier transform.

(b) If a function onRⁿ(n≥2) is constant on all(n−2)–dimensional spheres orthog- onal toe₁= (1,0, . . . ,0), then its Fourier transform possesses the same property.

2.4.11.(Grafakos and Morpurgo [108]) Suppose that 0<d1,d2,d3<n satisfy d1+d₂+d3=2n.Prove that for any distinctx,y,z∈Rⁿwe have the identity

Z

Rⁿ

|x−t|^−d2|y−t|^−d3|z−t|^−d1dt

=π

n 2

3

∏

j=1

Γ n−^d₂^j

Γ ^d₂^j |x−y|^d1−n|y−z|^d2−n|z−x|^d3−n. Hint: Reduce matters to the case that z=0 and y=e₁. Then take the Fourier transform inxand use Exercise 2.4.10.

2.4.12.(a) Integrate the functione^iz2over the contour consisting of the three pieces P1={(x,0): 0≤x≤R},P2={(Rcosθ,Rsinθ): 0≤θ ≤^π₄}, andP3={(t,t): tbetween ^R

√ 2

2 and 0}to obtain theFresnel integral identity:

R→∞lim Z +R

−R e^ix2dx=

√ 2π 2 (1+i).

(b) Use the result in part (a) to show that the Fourier transform of the functione^iπ|x|2 inRⁿis equal toe^iπn4e^−iπ|ξ|2.

Hint:Part (a): OnP₂we havee^{−R2sin(2θ)}≤e^−4πR2θ, and the integral overP₂tends to 0. Part (b): Try firstn=1.