Fourier Analysis, Distributions, and Constant-Coe–cient ...mtaylor.web.unc.edu/files/2018/04/fourier.pdfwith Fourier analysis on ﬂnite cyclic groups. We study this both as an approximation

Fourier Analysis, Distributions, andConstant-Coefficient Linear PDE

Michael E. Taylor

Contents0. Introduction1. Fourier series2. Harmonic functions and holomorphic functions in the plane3. The Fourier transform4. Distributions and tempered distributions5. The classical evolution equations6. Radial distributions, polar coordinates, and Bessel functions7. The method of images and Poisson’s summation formula8. Homogeneous distributions and principal value distributions9. Elliptic operators10. Local solvability of constant-coefficient PDE11. The discrete Fourier transform12. The fast Fourier transformA. The mighty Gaussian and the sublime gamma function

Introduction

Fourier analysis is perhaps the most important single tool in the study oflinear partial differential equations. It serves in several ways, the mostbasic—and historically the first—being to give specific formulas for solu-tions to various linear PDE with constant coefficients, particularly the threeclassics, the Laplace, wave, and heat equations:

(0.1) ∆u = f,∂2u

∂t2−∆u = f,

∂u

∂t−∆u = f,

with ∆ = ∂2/∂x21 + · · · + ∂2/∂x2

n. The Fourier transform accomplishesthis by transforming the operation of ∂/∂xj to the algebraic operation ofmultiplication by iξj . Thus the equations (0.1) are transformed to algebraicequations and to ODE with parameters.

Before introducing the Fourier transform of functions on Euclidean spaceRn, we discuss the Fourier series associated to functions on the torus Tn

in §1. Methods developed to establish the Fourier inversion formula for

2

Fourier series, in the special case of the circle S1 = T1, provide for free adevelopment of the basic results on harmonic functions in the plane, andwe give such results in §2, noting that these results specialize further toyield standard basic results in the theory of holomorphic functions of onecomplex variable, such as power-series expansions and Cauchy’s integralformula.

In §3 we define the Fourier transform of functions on Rn and prove theFourier inversion formula. The proof shares with the argument for Fourierseries in §1 the property of simultaneously yielding explicit solutions to aPDE, this time the heat equation.

It turns out that representations of solutions to such PDE as listed in(0.1) are most naturally done in terms of objects more general than func-tions, called distributions. We develop the theory of distributions in §4;Fourier analysis works very naturally with the class of distributions knownas tempered distributions.

Section 5, in some sense the heart of this chapter, derives explicit so-lutions to the classical linear PDE (0.1) via Fourier analysis. The use ofFourier analysis and distribution theory to represent solutions to these PDEgives rise to numerous interesting identities, involving both elementaryfunctions and “special functions,” such as the gamma function and Besselfunctions, and we present some of these identities here, only a smatteringfrom a rich area of classical analysis. Further development of harmonicanalysis in Chapter 8 will bring in additional studies of special functions.

Fourier analysis and distribution theory are also useful tools for generalinvestigations of linear PDE, in cases where explicit formulas might not beobtainable. We illustrate a couple of cases of this in the present chapter,discussing the existence and behavior of “parametrices” for elliptic PDEwith constant coefficients, and applications to smoothness of solutions tosuch PDE, in §9, and proving local solvability of general linear PDE withconstant coefficients in §10. Fourier analysis and distribution theory willacquire further power in the next chapter as tools for investigations ofexistence and qualitative properties of solutions to various classes of PDE,with the development of Sobolev spaces.

Sections 11 and 12 deal with the discrete Fourier transform, particularlywith Fourier analysis on finite cyclic groups. We study this both as anapproximation to Fourier analysis on the torus and Euclidean space, some-times useful for numerical work, and as a subject with its intrinsic interest,and with implications for number theory. In §12 we give a brief descriptionof “fast” algorithms for computing discrete Fourier transforms.

1. Fourier series

Let f be an integrable function on the torus Tn, naturally isomorphic toRn/Zn and to the Cartesian product of n copies S1×· · ·×S1 of the circle.

1. Fourier series 3

Its Fourier series is by definition a function on Zn given by

(1.1) f(k) =1

(2π)n

∫

Tn

f(θ)e−ik·θ dθ,

where k = (k1, . . . , kn), k · θ = k1θ1 + · · ·+ knθn. We use the notation

(1.2) Ff(k) = f(k).

Clearly, we have a continuous linear map

(1.3) F : L1(Tn) −→ `∞(Zn),

where `∞(Zn) denotes the space of bounded functions on Zn, with the supnorm. If f ∈ C∞(Tn), then we can integrate by parts to get

(1.4) kαf(k) =1

(2π)n

∫

Tn

(Dαf)(θ) e−ik·θ dθ,

where kα = kα11 · · · kαn

n , and

(1.5) Dα = Dα11 · · ·Dαn

n , Dj =1i

∂

∂θj.

It follows easily that

(1.6) F : C∞(Tn) −→ s(Zn),

where s(Zn) consists of functions u on Zn which are rapidly decreasing, inthe sense that, for each N ,

(1.7) pN (u) = supk∈Zn

〈k〉N |u(k)| < ∞.

Here, we use the notation

〈k〉 =(1 + |k|2)1/2

,

where |k|2 = k21 + · · ·+ k2

n. If we use the inner product

(1.8) (f, g) = (f, g)L2 =1

(2π)n

∫

Tn

f(θ)g(θ) dθ,

for f, g ∈ C∞(Tn), or more generally for f, g ∈ L2(Tn), and if on s(Zn), ormore generally on `2(Zn), the space of square summable functions on Zn,we use the inner product

(1.9) (u, v) = (u, v)`2 =∑

k∈Zn

u(k)v(k),

we have the formula

(1.10) (Ff, u)`2 = (f,F∗u)L2 ,

4

valid for f ∈ C∞(Tn), u ∈ s(Zn), where

(1.11) F∗ : s(Zn) −→ C∞(Tn)

is given by

(1.12)(F∗u)

(θ) =∑

k∈Zn

u(k) eik·θ.

Another identity that we will find useful is

(1.13)1

(2π)n

∫

Tn

eik·θ e−i`·θ dθ = δk`,

where δk` = 1 if k = ` and δk` = 0 otherwise.Our main goal here is to establish the Fourier inversion formula

(1.14) f(θ) =∑

k∈Zn

f(k) eik·θ,

the sum on the right in (1.14) converging in the appropriate function space,depending on the nature of f . Let us single out another space of functionson Tn, the trigonometric polynomials:

(1.15) T P = ∑

k∈Zn

a(k)eik·θ : a(k) = 0 except for finitely many k.

Clearly,

(1.16) F : T P −→ c00(Zn),

where c00(Zn) consists of functions on Zn which vanish except at a finitenumber of points; this follows from (1.13). The formula (1.12) gives

(1.17) F∗ : c00(Zn) −→ T P,

and the formula (1.13) easily yields

(1.18) FF∗ = I on c00(Zn),

and even

(1.19) FF∗ = I on s(Zn).

By comparison, the inversion formula (1.14) states

(1.20) F∗F = I,

on C∞(Tn), or some other space of functions on Tn, as specified below.Before getting to this, let us note one other implication of (1.13), namely,if

(1.21) fj(θ) =∑

k

ϕj(k)eik·θ

1. Fourier series 5

are elements of T P, or more generally, if ϕj ∈ s(Zn), then we have theParseval identity

(1.22) (f1, f2)L2 =∑

k∈Zn

ϕ1(k)ϕ2(k);

in particular, the Plancherel identity

(1.23) ‖fj‖2L2 =∑

k∈Zn

|ϕj(k)|2,

for fj ∈ T P, or more generally for any fj of the form (1.21) with ϕj ∈s(Zn). In particular, the map F∗ given by (1.12), and satisfying (1.11) and(1.17), has a unique continuous extension to `2(Zn), and

(1.24) F∗ : `2(Zn) −→ L2(Tn)

is an isometry of `2(Zn) onto its range. Part of the inversion formula willbe that the map (1.24) is also surjective.

Let us note that if fj ∈ T P, satisfying (1.21), then (1.13) implies fj(k) =ϕj(k), so we have directly in this case:

(1.25) F∗F = I on T P.

One approach to more general inversion formulas would be to establish thatT P is dense in various function spaces, on which F∗F can be shown to actcontinuously. For more details on this approach, see the exercises at theend of §1 and §2 in the Functional Analysis appendix. Here, we will takea superficially different approach. We will make use of such basic resultsfrom real analysis as the denseness of C(Tn) in Lp(Tn), for 1 ≤ p < ∞.

Our approach to (1.14) will be to establish the following Abel summabilityresult. Consider

(1.26) Jrf(θ) =∑

k∈Zn

f(k) r|k| eik·θ,

where |k| = |k1|+ · · ·+ |kn|, r ∈ [0, 1). We will show that

(1.27) Jrf → f, as r 1,

in the appropriate spaces. The operator Jr in (1.26) is defined for anyf ∈ L1(Tn), if r < 1, and we have the formula

(1.28) Jrf(θ) = (2π)−n

∫

Tn

f(θ′)∑

k∈Zn

r|k| eik·(θ−θ′) dθ′.

The sum over Zn inside the integral can be written as

(1.29)

∑

k∈Zn

r|k| eik·(θ−θ′) = Pn(r, θ − θ′)

= p(r, θ1 − θ′1) · · · p(r, θn − θ′n),

6

where

(1.30)

p(r, θ) =∞∑

k=−∞r|k| eikθ

= 1 +∞∑

k=1

(rkeikθ + rke−ikθ

)

=1− r2

1− 2r cos θ + r2.

Then we have the explicit integral formula

(1.31)

Jrf(θ) = (2π)−n

∫

Tn

f(θ′)Pn(r, θ − θ′) dθ′

= (2π)−n

∫

Tn

f(θ − θ′)Pn(r, θ′) dθ′.

Let us examine p(r, θ). It is clear that the numerator and denominator onthe right side of (1.30) are positive, so p(r, θ) > 0 for each r ∈ [0, 1), θ ∈ S1.Of course, as r 1, the numerator tends to 0; as r 1, the denominatortends to a nonzero limit, except at θ = 0. Since it is clear that

(1.32) (2π)−1

∫

S1

p(r, θ) dθ = (2π)−1

∫ π

−π

∑r|k|eikθ dθ = 1,

we see that, for r close to 1, p(r, θ) as a function of θ is highly peaked nearθ = 0 and small elsewhere, as in Fig. 1.1.

Figure 1.1

We are now prepared to prove the following result giving Abel summa-bility (1.27).

1. Fourier series 7

Proposition 1.1. If f ∈ C(Tn), then

(1.33) Jrf → f uniformly on Tn as r 1.

Furthermore, for any p ∈ [1,∞), if f ∈ Lp(Tn), then

(1.34) Jrf → f in Lp(Tn) as r 1.

The proof of (1.33) is an immediate consequence of (1.31) and the peakednature of p(r, θ) near θ = 0 discussed above, together with the observationthat, if f is continuous at θ, then it does not vary very much near θ. Theconvergence in (1.34) is in the Lp-norm, defined by

(1.35) ‖g‖Lp =[(2π)−n

∫

Tn

|g(θ)|p dθ]1/p

.

We have the well-known triangle inequality in such a norm:

(1.36) ‖g1 + g2‖Lp ≤ ‖g1‖Lp + ‖g2‖Lp ,

and this implies, via (1.31) and (1.32),

(1.37)

‖Jrf‖Lp = (2π)−n∥∥∥∫

Tn

Pn(r, θ′)τθ′f dθ′∥∥∥

Lp

≤ (2π)−n

∫Pn(r, θ′)‖τθ′f‖Lp dθ′

= ‖f‖Lp ,

where

(1.38) τθ′f(θ) = f(θ − θ′),

which implies ‖τθ′f‖Lp = ‖f‖Lp . In other words,

(1.39) ‖Jr‖L(Lp) ≤ 1, 1 ≤ p < ∞,

where we are using the operator norm on Lp:

(1.40) ‖T‖L(Lp) = sup ‖Tf‖Lp : ‖f‖Lp ≤ 1.Using this, we can deduce (1.34) from (1.33), and the denseness of C(Tn)in each space Lp(Tn), for 1 ≤ p < ∞. Indeed, given f ∈ Lp(Tn), andgiven ε > 0, find g ∈ C(Tn) such that ‖f − g‖Lp < ε. Note that, generally,‖g‖Lp ≤ ‖g‖sup. Now we have

(1.41)‖Jrf − f‖Lp ≤ ‖Jr(f − g)‖Lp + ‖Jrg − g‖Lp + ‖g − f‖Lp

< ε + ‖Jrg − g‖L∞ + ε,

making use of (1.39). By (1.33), the middle term is < ε if r is close enoughto 1, so this proves (1.34).

8

Corollary 1.2. If f ∈ C∞(Tn), then the Fourier inversion formula (1.14)holds.

Proof. In such a case, as noted, we have f ∈ s(Zn), so certainly the rightside of (1.14) is absolutely convergent to some f# ∈ C(Tn). In such a case,one a fortiori has

(1.42) limr1

∑

k∈Zn

f(k)r|k|eik·θ = f#(θ).

But now Proposition 1.1 implies (1.42) is equal to f(θ) (i.e., f# = f), sothe inversion formula is proved for f ∈ C∞(Tn).

As a result, we see that

(1.43) F∗ : s(Zn) −→ C∞(Tn)

is surjective, as well as injective, with two-sided inverse F : C∞(Tn) →s(Zn). This of course implies that the map (1.24) has dense range inL2(Tn); hence

(1.44) F∗ : `2(Zn) −→ L2(Tn) is unitary.

Another way of stating this is

(1.45) eik·θ : k ∈ Zn is an orthonormal basis of L2(Tn),

with inner product given by (1.8). Also, the inversion formula

(1.46) F∗F = I on C∞(Tn)

implies

(1.47) ‖Ff‖`2 = ‖f‖L2 ,

so therefore F extends by continuity from C∞(Tn) to a map

(1.48) F : L2(Tn) −→ `2(Zn), unitary,

inverting (1.44). The denseness C∞(Tn) ⊂ L2(Tn) ⊂ L1(Tn) implies thatthis F coincides with the restriction to L2(Tn) of the map (1.3). Notethat the fact that (1.44) and (1.48) are inverses of each other extends theinversion result of Corollary 1.2.

We devote a little space to conditions implying that the Fourier se-ries (1.14) is absolutely convergent, weaker than the hypothesis that f ∈C∞(Tn). Note that since |eik·θ| = 1, the absolute convergence of (1.14)implies uniform convergence. By (1.4), we see that

(1.49) f ∈ C`(Tn) =⇒ |f(k)| ≤ C〈k〉−`,

which in turn clearly gives absolute convergence provided

(1.50) ` ≥ n + 1.

Exercises 9

Using Plancherel’s identity and Cauchy’s inequality, we can do somewhatbetter:

Proposition 1.3. If f ∈ C`(Tn), then the Fourier series for f is absolutelyconvergent provided

(1.51) ` >n

2.

Proof. We have

(1.52)

∑

k

|f(k)| =∑

k

〈k〉−`〈k〉`|f(k)|

≤[∑

k

〈k〉−2`]1/2

·[∑

k

〈k〉2`|f(k)|2]1/2

≤ C[∑

k

〈k〉2`|f(k)|2]1/2

,

as long as (1.51) holds. The square of the right side is dominated by

(1.53)C ′

∑

k

∑

|γ|≤`

∣∣kγ f(k)∣∣2 = C ′

∑

|γ|≤`

‖Dγf‖2L2

≤ C ′′‖f‖2C` ,

so the proposition is proved.

Sharper results on absolute convergence of Fourier series will be given inChapter 4. See also some of the exercises below for more on convergencewhen n = 1.

Exercises

1. Given f, g ∈ L1(Tn), show that

f(k)g(k) = u(k),

with

u(θ) = (2π)−n

∫

Tn

f(ϕ)g(θ − ϕ) dϕ.

2. Given f, g ∈ C(Tn), show that

(fg)(k) =∑m

f(k −m)g(m).

3. Using the proof of Proposition 1.3, show that every f ∈ Lip(S1) has anabsolutely convergent Fourier series.

10

4. Show that for any f ∈ L1(Tn), f(k) → 0 as |k| → ∞.(Hint: Given ε > 0, pick fε ∈ C∞(Tn), ‖f − fε‖L1 < ε. Compare fε(k) andf(k).)This result is known as the Riemann-Lebesgue lemma.

5. For f ∈ L1(S1), set

(1.54) SNf(θ) =

N∑

k=−N

f(k)eikθ.

Show that SNf(θ) = (1/2π)∫ π

−πf(θ − ϕ)DN (ϕ) dϕ, where

(1.55) DN (θ) =

N∑

k=−N

eikθ =sin(N + 1

2)θ

sin 12θ

.

(Hint: To evaluate the sum, recall how to sum a finite geometrical series.)DN (θ) is called the Dirichlet kernel. See Fig. 1.2.

Figure 1.2

6. Let f ∈ L1(S1) have the following property of “vanishing” at θ = 0:

f(θ)

sin 12θ

= g(θ) ∈ L1(−π, π).

Show that SNf(0) → 0 as N →∞.(Hint. Adapt the Riemann-Lebesgue lemma to show that

g ∈ L1(−π, π) ⇒∫ π

−π

g(θ) sin(N + 12)θ dθ → 0 as N →∞.)

7. Deduce that if f ∈ L1(S1) is Lipschitz continuous at θ0, then SNf(θ0) →f(θ0) as N →∞. Furthermore, if f is Lipschitz on an open interval J ⊂ S1,then SNf → f uniformly on compact subsets of J .

Exercises 11

8. Let f ∈ L∞(S1) be piecewise Lipschitz, with a finite number of simple jumps.Show that SNf(θ) → f(θ) at points of continuity. If f has a jump at θj , withlimiting values f±(θj), show that

(1.56) SNf(θj) → 1

2

[f+(θj) + f−(θj)

],

as N →∞.(Hint: By Exercise 7, it remains only to establish (1.56). Show that this canbe reduced to the case θj = π, f(θ) = θ, for −π ≤ θ < π. Verify that thisfunction has Fourier series

2

∞∑

k=1

(−1)k

ksin kθ.)

Alternative: Reduce to the case θj = 0 and note that SNf(0) depends onlyon the even part of f, (1/2)[f(θ) + f(−θ)].

9. Work out the Fourier series of the function f ∈ Lip(S1) given by

f(θ) = |θ|, −π ≤ θ ≤ π.

Examining this at θ = 0, establish that

(1.57)

∞∑

k=1

1

k2=

π2

6.

10. One can obtain Fourier coefficients of functions θk and |θ|k on [−π, π] interms of the Fourier coefficients of

qk(θ) = θk on [0, π],

0 on [−π, 0].

Show that, for n 6= 0,

∞∑

k=0

1

k!qk(n) (is)k = − 1

2πin

[(−1)n · eπis − 1

](1− s

n

)−1

,

and use this to work out the Fourier series for these functions. Apply this toExercise 9, and to the calculation at the end of Exercise 8.

11. Assume that g ∈ L1(S1) has uniformly convergent Fourier series (SNg → g)on compact subsets of an open interval J ⊂ S1. Show that whenever f ∈L1(S1) and f = g on J , then f also has uniformly convergent Fourier serieson compact subsets of J .(Hint: Apply Exercise 7 to f − g.)This result is called the localization principle for Fourier series.

12. Suppose f is Holder continuous on S1, that is, f ∈ Cr(S1), for some r ∈ (0, 1),which means

|f(ϕ + θ)− f(ϕ)| ≤ C|θ|r.Show that f has uniformly convergent Fourier series on S1.(Hint: We have

∣∣∣∣f(θ + ϕ)− f(ϕ)

sin 12θ

∣∣∣∣ ≤ C∣∣∣sin 1

2θ∣∣∣−(1−r)

.

12

Apply Exercise 6.)13. If ω : [0,∞) → [0,∞) is continuous and increasing and ω(0) = 0, we say a

function f on S1 is continuous with modulus of continuity ω provided

|f(ϕ + θ)− f(ϕ)| ≤ Cω(|θ|).

Formulate the most general condition you can to establish uniform conver-gence of the Fourier series of a function with such a modulus of continuity.Note that Exercise 12 deals with the case ω(s) = sr, r ∈ (0, 1).

14. Consider the Cesaro sum of the Fourier series of f :

CNf(θ) =

N∑

k=−N

(1− |k|

N

)f(k)eikθ =

1

N

N−1∑

`=0

S`f(θ).

Show that CNf(θ) = (1/2π)∫ π

−πf(θ − ϕ)FN (ϕ)dϕ, where

(1.58) FN (θ) =1

N

N−1∑

`=0

D`(θ) =1

N sin 12θ

N−1∑

`=0

sin

(` +

1

2

)θ =

1

N

(sin N

2θ

sin 12θ

)2

.

The function FN (θ) is called the Fejer kernel (see Fig. 1.3). Modify the proofof Proposition 1.1 to show that

CNf → F in B, for f ∈ B,

where B is one of the Banach spaces C(S1) or Lp(S1), 1 ≤ p < ∞.(Hint: To evaluate the second sum in (1.58), use sin(` + 1

2)θ = Im eiθ/2ei`θ

and sum a finite geometrical series. Also use the identity 2 sin2 z = 1−cos 2z.)

Figure 1.3

2. Harmonic functions and holomorphic functions in the plane 13

2. Harmonic functions and holomorphic functions in theplane

The method of proof of the Abel summability (1.26)–(1.27) of Fourier se-ries, specialized to T1 = S1, has important implications for the theory ofharmonic functions on a domain Ω ⊂ R2, which we will discuss here. Inthe case of S1, let us rewrite (1.26),

(2.1) Jrf(θ) =∞∑

k=−∞f(k) r|k| eikθ,

as

(2.2) (PI f)(r, θ) =∞∑

k=−∞f(k) r|k| eikθ.

The function u(r, θ) = PI f(r, θ) is called the Poisson integral of f . If weuse polar coordinates in the complex plane C = R2,

(2.3) z = r eiθ,

then (2.2) becomes

(2.4)(PI f)(z) =

∞∑

k=0

f(k)zk +∞∑

k=1

f(−k)zk

= (PI+f)(z) + (PI−f)(z),

defined on the unit disk |z| < 1. Note also, from (1.30), that

(2.5) PI f(z) =1− |z|2

2π

∫

S1

f(w)|w − z|2 ds(w),

the integral being with respect to arclength on S1. Recall that if f ∈L1(S1), the function f(k) is bounded, so both power series in (2.4) haveradius of convergence at least 1. Clearly, on the unit disk, v(z) = (PI+f)(z)is holomorphic and w(z) = (PI−f)(z) is antiholomorphic. In other words,v and w belong to C∞(D), where D = z ∈ C : |z| < 1, and

(2.6)∂v

∂z= 0,

∂w

∂z= 0 on D,

where

(2.7)∂

∂z=

12

( ∂

∂x+ i

∂

∂y

),

∂

∂z=

12

( ∂

∂x− i

∂

∂y

).

Note that∂

∂z

∂

∂z=

∂

∂z

∂

∂z=

14∆,

14

where ∆ is the Laplace operator on R2, a special case of the Laplace oper-ator introduced in Chapter 2. Since v, w ∈ C∞(D), we have ∆v = 0 and∆w = 0, and hence

(2.8) ∆(PI f) = 0.

In light of the results of §1, we have the following.

Proposition 2.1. If f ∈ C(S1), then

(2.9) u = PI f(z) ∈ C∞(D) ∩ C(D)

is harmonic, with boundary value f , that is, u solves the Dirichlet problem

(2.10) ∆u = 0 in D, u|∂D = f.

One should expect that if f has extra smoothness on S1, so does PI fon D. The following result is crude compared to results established inChapters 4 and 5, but it will be of some interest.

Proposition 2.2. For ` = 1, 2, 3, . . . , we have

(2.11) PI : C`+1(S1) −→ C`(D).

Proof. We begin with the case ` = 1. Since we know from (2.4) thatPI f ∈ C∞(D), we need merely check smoothness near ∂D = S1. Clearly,

(2.12)∂

∂θPI f = PI

∂f

∂θ,

so if f ∈ C1(S1), then ∂f/∂θ ∈ C(S1) and we have (∂/∂θ)PI f continuouson D. Also, by (2.2),

(2.13) r∂

∂rPI f = PI (Nf),

where Nf is characterized by the Fourier series representation

(2.14) Nf(θ) =∞∑

k=−∞f(k)|k|eikθ.

Thus

(2.15) Nf = −i∂

∂θHf = −iH

∂f

∂θ,

where H has the Fourier series representation

(2.16) Hg(θ) =∞∑

k=−∞(sgn k)g(k)eikθ.


We claim that, for ` ≥ 0,

(2.17) H : C`+1(S1) −→ C`(S1),

and hence N : C`+2(S1) → C`(S1). Given this, for f ∈ C2(S1), thequantity (2.13) is seen to belong to C(D), and this finishes the ` = 1 caseof (2.11).

In turn, since H commutes with ∂/∂θ, it suffices to establish the ` = 0case of (2.17). Now, by Proposition 1.3, the Fourier series for g ∈ C1(S1)is absolutely convergent, giving (2.17).

To prove the general case of (2.11), a short calculation yields

(2.18)(r

∂

∂r

)j( ∂

∂θ

)k

PI f = PI(( ∂

∂θ

)k

N jf).

Note that N j = (−i)j(∂/∂θ)jHo(j), where o(j) is zero if j is even and oneif j is odd, so, for ` ≥ 0,

N j : C`+j+1(S1) −→ C`(S1),

the left side being improved to C`+j if j is even. Therefore, if f ∈ C`+1(S1)and j + k = `, the right side of (2.18) is PI fjk with fjk ∈ C(S1), whichproves (2.11) in general.

The implication f ∈ C`(S1) ⇒ PI f ∈ C`(D) does not quite work, as wewill see later, essentially because H does not map C`(S1) to itself. It is truethat f ∈ C`,α(S1) ⇒ PI f ∈ C`,α(D), for α ∈ (0, 1). This is a special caseof Holder estimates that will be established in §7 of Chapter 13. Similarlythere are “sharp” results on regularity of PI f in Sobolev spaces, discussedin Chapter 4, and in much greater generality, in Chapter 5.

It is important to know that PI f provides the unique solution to theDirichlet problem (2.10). We will establish several general uniqueness re-sults, starting with the following.

Proposition 2.3. Let Ω ⊂ Rn be a bounded region with smooth bound-ary, say, Ω = D. Suppose u, v ∈ C2(Ω), with u = v = f on ∂Ω, and∆u = ∆v = 0 in Ω. Then u = v on all of Ω.

Proof. Set w = u − v ∈ C2(Ω);w = 0 on ∂Ω. We can apply the Greenidentity (3.15) of Chapter 2, to write

(2.19) (dw, dw) = −(∆w, w) +∫

∂Ω

w∂w

∂νdS.

By hypothesis the right side of (2.19) is 0. Thus w is constant on eachcomponent of Ω, and the boundary condition forces w = 0.

In view of Proposition 2.2, we could apply this to u = PI f if f ∈ C3(S1),but this is not a satisfactory result, and we will do much better below.

16

A result related to our uniqueness question is the mean-value property,a special case of which is the following.

Proposition 2.4. If f ∈ C(S1), u = PI f , then

(2.20) u(0) =12π

∫ π

−π

f(θ) dθ.

Proof. It follows from the series (2.2) that u(0) = f(0), which gives (2.20).

A more general result is the following.

Proposition 2.5. If BR ⊂ Rn is the open ball of radius R, centered at theorigin, with ∂BR = SR, of area A(R), then for u ∈ C2(BR)∩C(BR), ∆u =0, we have

(2.21) u(0) =1

A(R)

∫

SR

u(x) dS.

Proof. We apply Green’s identity

(2.22)∫

Ω

[u∆v − v∆u

]dx =

∫

∂Ω

[u

∂v

∂ν− v

∂u

∂ν

]dS,

to Ω = Br, 0 < r < R, v(x) = |x|2, with ∆v = 2n, to get, when ∆u = 0,

(2.23) n

∫

Br

u(x) dx = r

∫

Sr

u(x) dS,

noting that substituting v = 1 in (2.22) gives∫

∂Ω(∂u/∂ν) dS = 0. If we let

ϕ(r) =∫

Bru(x) dx, this implies ϕ′(r) = (n/r)ϕ(r), and hence ϕ(r) = Krn,

i.e., V (r)−1∫

Bru(x) dx is constant. Passing to the limit r → 0 gives (2.21).

Second Proof. Define v ∈ C2(BR) ∩ C(BR) by

v(x) =∫

SO(n)

u(gx) dg,

where dg is Haar measure on the rotation group SO(n), defined in §5 ofAppendix B. The Laplace operator is invariant under rotations, so ∆v = 0on BR. The finction v is radial; v(x) = v(|x|). The formula

∆ =∂2

∂r2+

n− 1r

∂

∂r+

1r2

∆S ,


(cf. (4.17) of Chapter 2), for ∆ in polar coordinates, where ∆S is theLaplace operator on the unit sphere Sn−1, gives

( d2

dr2+

n− 1r

d

dr

)v(r) = 0.

This is an Euler equation, whose solutions are

A + Br2−n, n ≥ 3,

A + B log r, n = 2.

Since v does not blow up at 0, we have B = 0, so v is constant. Clearlyv(x) equals the right side of (2.21) for |x| = R, and v(0) = u(0), so weagain have (2.21).

Corollary 2.6. For any Ω ⊂ Rn open, any u ∈ C2(Ω) harmonic, any ballBp, centered at p and contained in Ω, we have

(2.24) u(p) = Avg∂Bpu(z).

We can now prove the following important maximum principle for har-monic functions. Much more general versions of this will be given in Chap-ter 5.

Proposition 2.7. Let Ω ⊂ Rn be connected and open, and let u ∈ C2(Ω)be harmonic and real-valued. Then u has no interior maximum, or mini-mum, unless u is constant. In particular, if Ω is bounded and u ∈ C(Ω),then

(2.25) supp∈Ω

u(p) = supq∈∂Ω

u(q).

Also, even for u complex-valued,

(2.26) supp∈Ω

|u(p)| = supq∈∂Ω

|u(q)|.

Proof. That a non-constant, real harmonic function has no interior ex-tremum is an obvious consequence of (2.24), and the other consequences,(2.25) and (2.26), follow immediately.

Corollary 2.8. The uniqueness result of Proposition 2.3 holds for anybounded open Ω ⊂ Rn, with no smoothness on ∂Ω, and for any harmonic

(2.27) u, v ∈ C2(Ω) ∩ C(Ω).

Proof. Apply the maximum principle to u− v.

18

Here, we have used the mean-value property to prove the maximumprinciple. The more general maximum principle established in Chapter 5will not use the mean-value property; indeed, together with a symmetryargument, it can be made a basis for a proof of the mean-value property.We will leave these considerations until Chapter 5.

With our uniqueness result in hand, we can easily establish the followinginterior regularity result for harmonic functions.

Proposition 2.9. Let Ω ⊂ R2 be open, and let u ∈ C2(Ω) be harmonic.Then in fact, u ∈ C∞(Ω); u is even real analytic on Ω.

Proof. By translations and dilations, we can reduce to the case Ω =D, u ∈ C2(D). The uniqueness result of Corollary 2.8 implies

(2.28) u = PI f, where f = u|S1 .

Then the conclusion that u is real analytic on D follows directly from thepower-series expansion (2.4).

Parenthetically, we remark that, by Corollary 2.8, the identity (2.28)holds for any u ∈ C2(D) ∩ C(D) harmonic in D.

Using the results we have developed, via Fourier series, about harmonicfunctions, we can quickly draw some basic conclusions about holomorphicfunctions. If Ω ⊂ C is open, f : Ω → C is by definition holomorphic if andonly if f ∈ C1(Ω) and ∂f/∂z = 0, where ∂/∂z is given by (2.7). Clearly, iff ∈ C2(Ω) is holomorphic, then it is also harmonic, and so are its real andimaginary parts. Suppose u ∈ C2(D) ∩ C(D) is holomorphic in D. Thenthe series representation (2.4) is valid, since (2.28) holds. This series is asum of two terms:

(2.29)u(z) =

∞∑

k=0

f(k)zk +∞∑

k=1

f(−k)zk

= u1(z) + u2(z),

where ∂u1/∂z = 0 and ∂u2/∂z = 0. But if we are given that ∂u/∂z = 0,then also ∂u2/∂z = 0, so ∂u2/∂x = ∂u2/∂y = 0. Thus u2 is constant,and since u2(0) = 0, this forces u2 = 0. In other words, the holomorphicfunction u(z) has the power series

(2.30) u(z) =∞∑

k=0

akzk, z ∈ D,

where ak = f(k), k ≥ 0. Note that differentiation of (2.30) gives

(2.31) ak =1k!

( ∂

∂z

)k

u(0).


By the usual method of translating and dilating coordinates, we deducethe following.

Proposition 2.10. If Ω ⊂ C is open, u ∈ C2(Ω) holomorphic, p ∈ Ω, andDp a disk centered at p, Dp ⊂ Ω, then on Dp, u(z) is given by a convergentpower-series expansion

(2.32) u(z) =∞∑

k=0

bk(z − p)k.

We can relax the C2-hypothesis to u ∈ C1(Ω). As much stronger andmore general results are given in Chapter 5, we omit the details here.

For further use, we record the following result, whose proof is trivial.

Lemma 2.11. If u ∈ C∞(Ω) is holomorphic, and

P =∑

ajkDjxDk

y

is any constant-coefficient differential operator, then Pu is holomorphic inΩ.

Proof. (∂/∂z)Pu = P (∂u/∂z).

We can use the power-series representation (2.30)–(2.32) to prove thefundamental result on uniqueness of analytic continuation, which we givebelow. Here is the first result, of a very general nature.

Proposition 2.12. Let Ω ⊂ Rn be open and connected, and let u be areal-analytic function on Ω. If p ∈ Ω and all derivatives Dαu(p) = 0, thenu = 0 on all of Ω.

Proof. Let K = x ∈ Ω : Dαu(x) = 0 for all α ≥ 0. Since u ∈ C∞(Ω), Kis closed in Ω. However, for each p ∈ K, since u is given in a neighborhoodof p by a power series

u(q) =∑

α≥0

u(α)(p)α!

(q − p)α,

we also see that K is open in Ω. This proves the proposition.

Our basic corollary for holomorphic functions is the following.

Corollary 2.13. Let Ω ⊂ C be open and connected, u holomorphic on Ω.Let γ be a line segment contained in Ω. If u|γ = 0, then u = 0 on Ω.

20

Proof. Translating and rotating, we can assume γ is a segment in the realaxis, with 0 ∈ γ. Near 0, u(z) has a power-series expansion of the form(2.30), with ak given by (2.31). Using Lemma 2.11, we see that

(2.33)( ∂

∂z

)k

u(0) =( ∂

∂x

)k

u(0),

which vanishes for all k. Thus u = 0 on a nonempty open set in Ω, and therest follows by Proposition 2.12.

Actually, a much stronger result is true. If Ω ⊂ C is connected, pj ∈ Ωare distinct, pj → p ∈ Ω, u is holomorphic in Ω, and u(pj) = 0 for eachj, then u must vanish identically. In other words, u can have only isolatedzeros if it does not vanish identically. Indeed, say u(p) = 0. If u is notidentically zero, some coefficient in the series (2.32) is nonzero; let bm bethe first such coefficient:

(2.34)u(z) = (z − p)m

∞∑

k=0

bm+k(z − p)k

= (z − p)mv(z),

where v(z) is holomorphic on Dp and v(0) = bm 6= 0. Thus, by continuity,v(ζ) 6= 0 for |ζ − p| < ε if ε is sufficiently small, which implies u(ζ) 6= 0 ifζ 6= p but |ζ − p| < ε.

A typical use of Corollary 2.13 is in computations of integrals. We willsee an example of this in the next section.

We end this section by recalling the classical Cauchy integral theoremand integral formula. Throughout, Ω will be a bounded open domain in Cwith smooth boundary. Stokes’ formula, proved in Chapter 1, §13, states

(2.35)∫∫

Ω

dα =∫

∂Ω

α,

for a 1-form α with coefficients in C1(Ω). If α = p dx+ q dy, this gives theclassical Green’s formula

(2.36)∫

∂Ω

p dx + q dy =∫∫

Ω

( ∂q

∂x− ∂p

∂y

)dx dy.


If u(x, y) ∈ C1(Ω) is a complex-valued function, we consequently have

(2.37)

∫

∂Ω

u dz =∫

∂Ω

(u dx + iu dy)

=∫∫

Ω

(i∂u

∂x− ∂u

∂y

)dx dy

= 2i

∫∫

Ω

∂u

∂zdx dy.

In the special case when u is holomorphic in Ω, we have the Cauchy integraltheorem:

Theorem 2.14. If Ω ⊂ C is bounded with smooth boundary and u ∈C1(Ω) is holomorphic, then

(2.38)∫

∂Ω

u(z) dz = 0.

Using various limiting arguments, one can relax the hypotheses on smooth-ness of ∂Ω and of u near ∂Ω; we won’t go into this here. Next we proveCauchy’s integral formula.

Proposition 2.15. With Ω as above, u ∈ C2(Ω) holomorphic in Ω, wehave

(2.39) u(ζ) =1

2πi

∫

∂Ω

u(z)z − ζ

dz, for ζ ∈ Ω.

Proof. Write

(2.40)u(z)(ζ − z)−1 = (ζ − z)−1

[u(z)− u(ζ)

]+ u(ζ)(ζ − z)−1

= v(z) + u(ζ)(ζ − z)−1.

By the series expansion for u(z) about ζ, we see that v(z) is holomorphicnear ζ; clearly, it is holomorphic on the rest of Ω, and it belongs to C1(Ω),so

∫∂Ω

v(z) dz = 0. Thus, to prove (2.39), it suffices to show that

(2.41)∫

∂Ω

(z − ζ)−1 dz = 2πi, for ζ ∈ Ω.

Indeed, if ε is small enough that B(ζ, ε) = z ∈ C : |z−ζ| ≤ ε is containedin Ω, then Cauchy’s theorem implies that the left side of (2.41) is equal to

(2.42)∫

∂B(ζ,ε)

(z − ζ)−1 dz

22

since (ζ − z)−1 is holomorphic in z for z 6= ζ. Making a change of variable,we see that (2.42) is equal to

∫ 2π

0

(ε eiθ

)−1iε eiθ dθ = 2πi,

so the proof is complete.

As stated before, the C2-hypothesis can be relaxed to C1.A function u(z) of the form u(z) = v(z)/(z − a)k, k ∈ Z+, where v is

holomorphic on a neighborhood O of a, is said to have a pole of order k atz = a if v(a) 6= 0. In such a case, a variant of the preceding calculationsyields

12πi

∫

γ

u(z) dz =1

(k − 1)!v(k−1)(a),

the coefficient of (z − a)k−1 in the power series of v(z) about z = a, if γis a smooth, simple, closed curve about a such that v is holomorphic on aneighborhood of the closed region bounded by γ. This quantity is calledthe residue of u(z) at z = a.

One can often evaluate integrals by evaluating residues. We give a simpleillustration here; others are given in (2.48), (3.32), (A.14), and (A.15). Herewe evaluate

(2.43)∫ ∞

−∞

dx

1 + x2= lim

R→∞

∫

γR

dz

1 + z2,

where γR is the closed curve, going from −R to R along the real axis, thenfrom R to −R counterclockwise on the circle of radius R centered at 0, thatis, γR = ∂OR, where OR = z : Re z > 0, |z| < R. There is just one poleof (1+z2)−1 in OR, located at z = i. Since (1+z2)−1 = (z + i)−1(z− i)−1,we see that the residue of (1 + z2)−1 at z = i is 1/2i, so

∫ ∞

−∞

dx

1 + x2= π.

Exercises

1. Suppose u satisfies the following Neumann boundary problem in the disk D:

(2.44) ∆u = 0 in D,∂u

∂r= g on S1.

If u = PI(f), show that f and g must be related by g(k) = |k|f(k), for allk ∈ Z, that is,

(2.45) g = Nf,

Exercises 23

with N defined by (2.13).2. Define the function kN by

kN (θ) =∑

k 6=0

|k|−1eikθ.

Show that kN ∈ L2(S1) ⊂ L1(S1). Also show that, provided g ∈ L2(S1) and

(2.46)

∫

S1

g(θ) dθ = 0,

a solution to (2.45) is given by

(2.47) f(θ) = (2π)−1

∫

S1

kN (θ − ϕ)g(ϕ) dϕ = Tg(θ).

3. If T is defined by (2.47), show that T : Lp(S1) → Lp(S1) for p ∈ [1,∞) andT : C`(S1) → C`(S1) for ` = 0, 1, 2, . . . .

4. Given g ∈ C1(S1), show that (2.44) has a solution u ∈ C1(D) if and only if(2.46) holds. If g ∈ C`(S1), show that (2.44) has a solution u ∈ C`(D).Note: Regularity results of a more precise nature are given in Chapter 4, §4,in Exercise 1. See also Chapter 5, §7, for more general results.

5. Let Ω ⊂ R2 be a smooth, bounded, connected region. Show that if w ∈C2(Ω), ∆w = 0 on Ω, and ∂w/∂ν = 0 on ∂Ω, then w is constant. (Hint: Use(2.19).)Note: One can weaken the C2-hypothesis to w ∈ C1(Ω); compare Proposition2.2 of Chapter 5. For another type of relaxation, see Chapter 4, §4, Exercise3.

6. Show that a C1-function f : Ω → C is holomorphic if and only if, at eachz ∈ Ω, Df(z), a priori a real linear map on R2, is in fact complex linear onC.Note: This exercise has already been given in Chapter 1, §1.

7. Let f be a holomorphic function on Ω ⊂ C, with f : Ω → O, and let u beharmonic on O. Show that v = u f is harmonic on Ω. (Hint: For a shortproof, write u locally as a sum of a holomorphic and an anti-holomorphicfunction.)

8. Let g(z) =∑∞

1 akzk, and form the harmonic function u = 2 Re g = g + g.Show that, under appropriate hypotheses on (ak), g|S1 = P+(u|S1), whereP+ is given by

P+f(θ) =

∞∑

k=0

f(k)eikθ.

9. Find a holomorphic function on D that is unbounded but whose real part hasa continuous extension to D.Reconsider this problem after reading §6 of Chapter 5.

10. Hence show that P+ does not map C(S1) to itself, nor does it map anyC`(S1) to itself, for any integer ` ≥ 0.

11. Hence find f ∈ C1(S1) such that PI f /∈ C1(D).12. Use the method of residues to calculate

(2.48)

∫ ∞

−∞

eiξx

1 + x2dx, ξ ≥ 0.

24

(Hint: Write this as limR→∞∫

γReiξz/(1+z2) dz, with γR as in (2.43), given

ξ ≥ 0. Then find the residue of eiξz/(1 + z2) at z = i.)13. Use the Poisson integral formula (2.5) to prove the following. Let u ∈ C2(D)∩

C(D) be harmonic in the disk D = z ∈ C : |z| < 1. Assume u ≥ 0 on Dand u(0) = 1. Then

(2.49) |z| = a ∈ [0, 1) =⇒ u(z) ≥ 1− a

1 + a.

This result is known as a Harnack inequality. Hint. Use the inequality

|w| = 1, |z| = a ∈ [0, 1) =⇒ 1− |z|2|w − z|2 ≥

1− a

1 + a.

Note. By translating and scaling, if u is harmonic and ≥ 0 on DR(p) = z ∈C : |z − p| < R, then

|z − p| = a ∈ [0, R) =⇒ u(z) ≥ R− a

R + au(p).

14. Using Exercise 13, show that if u is harmonic in the entire plane C and u ≥ 0on C, then u is constant. More generally, if there exists a constant K suchthat u ≥ K on C, then u is constant. This is a version of Liouville’s theorem.See Proposition 4.6 for another version.

15. Using Exercise 13, show that there exists A ∈ (0,∞) with the followingproperty. Let u be harmonic on DR(0). Assume

u(0) = 0, u(z) ≤ M on DR(0).

Then

u(z) ≥ −AM on DR/2(0).

(Hint. Set v(z) = M − u(z), so v(z) ≥ 0 on DR(0), v(0) = M . Say p ∈DR/2(0), u(p) = infDR/2(0) u. Deduce that

v(z) ≥ 1

3(M − u(p)) on DR/4(p),

and from there that

AvgDR(0) v ≥ 1

16· 1

3(M − u(p)),

while this average is equal to v(0) = M .)16. Assume u is harmonic on C and

u(z) ≤ C0 + C1|z|k, ∀ z ∈ C,

with C0, C1 ∈ (0,∞). Take A from Exercise 15. Show that there exists C2

such that

u(z) ≥ −C2 −AC1|z|k, ∀ z ∈ C.

Note. In conjunction with Proposition 4.6, one gets that u(z) must be apolynomial of degree ≤ k in x and y.

17. Let Ω be the strip Ω = z ∈ C : 0 < Re z < 1, with closure Ω. Let f becontinuous on Ω and holomorphic on Ω. Show that if f is bounded,

|f(z)| ≤ A on ∂Ω =⇒ |f(z)| ≤ A on Ω.

3. The Fourier transform 25

(Hint. First prove the implication under the additional hypothesis that|f(z)| → 0 as |z| → ∞. Then consider fε(z) = eεz2

f(z).)18. In the setting of Exercise 17, show that if θ ∈ (0, 1),

|f(iy)| ≤ A, |f(1 + iy)| ≤ B, ∀ y ∈ R =⇒ |f(θ + iy)| ≤ A1−θBθ, ∀ y ∈ R.

This result is known as the Hadamard three-lines lemma. (Hint. Considerg(z) = Az−1B−zf(z).)

3. The Fourier transform

The Fourier transform is defined by

(3.1) Ff(ξ) = f(ξ) = (2π)−n/2

∫f(x)e−ix·ξ dx

when f ∈ L1(Rn). It is clear that

(3.2) F : L1(Rn) −→ L∞(Rn).

This is analogous to (1.3). The analogue for C∞(Tn), and simultaneouslyfor s(Zn), of §1, in this case is the Schwartz space of rapidly decreasingfunctions:

(3.3) S(Rn) =u ∈ C∞(Rn) : xβDαu ∈ L∞(Rn) for all α, β ≥ 0

,

where xβ = xβ11 · · ·xβn

n , Dα = Dα11 · · ·Dαn

n , with Dj = −i∂/∂xj . It is easyto verify that

(3.4) F : S(Rn) −→ S(Rn)

and

(3.5) ξαDβξFf(ξ) = (−1)|β|F(Dαxβf)(ξ).

We define F∗ by

(3.6) F∗f(ξ) = f(ξ) = (2π)−n/2

∫f(x)eix·ξ dx,

which differs from (3.1) only in the sign of the exponent. It is clear thatF∗ satisfies the mapping properties (3.2), (3.4), and

(3.7) (Fu, v) = (u,F∗v),

for u, v ∈ S(Rn), where (u, v) denotes the usual L2-inner product, (u, v) =∫u(x) v(x) dx.As in the theory of Fourier series, the first major result is the Fourier

inversion formula. The following is our first version.

Proposition 3.1. We have the inversion formula

(3.8) F∗F = FF∗ = I on S(Rn).

26

As in the proof of the inversion formula for Fourier series, via Proposition1.1, in the present proof we will sneak up on the inversion formula bythrowing in a convergence factor that will allow interchange of orders ofintegration (in the proof of Proposition 1.1, the orders of an integral andan infinite series were interchanged). Also, as we will see in §5, this methodwill have serendipitous applications to PDE. So, let us write, for f ∈ S(Rn),

(3.9)F∗Ff(x) = (2π)−n

∫ [∫f(y)e−iy·ξdy

]eix·ξ dξ

= (2π)−n limε0

∫∫f(y) e−ε|ξ|2 ei(x−y)·ξ dy dξ.

We can interchange the order of integration on the right for any ε > 0, toobtain

(3.10) F∗Ff(x) = limε0

∫f(y)p(ε, x− y) dy,

where

(3.11) p(ε, x) = (2π)−n

∫e−ε|ξ|2+ix·ξ dξ.

Note that

(3.12) p(ε, x) = ε−n/2 q(ε−1/2x),

where q(x) = p(1, x). In a moment we will show that

(3.13) p(ε, x) = (4πε)−n/2 e−|x|2/4ε.

The derivation of this identity will also show that

(3.14)∫

Rn

q(x) dx = 1.

From this, it follows as in the proof of Proposition 1.1 that

(3.15) limε0

∫f(y)p(ε, x− y) dy = f(x),

for any f ∈ S(Rn), even for f bounded and continuous, so we have provedF∗F = I on S(Rn); the proof that FF∗ = I on S(Rn) is identical.

It remains to verify (3.13). We observe that p(ε, x), defined by (3.11), isan entire holomorphic function of x ∈ Cn, for any ε > 0. It is convenientto verify that

(3.16) p(ε, ix) = (4πε)−n/2 e|x|2/4ε, x ∈ Rn,


from which (3.13) follows by analytic continuation. Now

(3.17)

p(ε, ix) = (2π)−n

∫e−x·ξ−ε|ξ|2dξ

= (2π)−ne|x|2/4ε

∫e−|x/2

√ε+√

εξ|2dξ

= (2π)−nε−n/2e|x|2/4ε

∫

Rn

e−|ξ|2dξ.

To prove (3.16), it remains to show that

(3.18)∫

Rn

e−|ξ|2dξ = πn/2.

Indeed, if

(3.19) A =∫ ∞

−∞e−ξ2

dξ,

then the left side of (3.18) is equal to An. But for n = 2 we can use polarcoordinates:

(3.20) A2 =∫

R2

e−|ξ|2dξ =

∫ 2π

0

∫ ∞

0

e−r2r dr dθ = π.

This completes the proof of the identity (3.16) and hence of (3.13).In light of (3.7) and the Fourier inversion formula (3.8), we see that, for

u, v ∈ S(Rn),

(3.21) (Fu,Fv) = (u, v) = (F∗u,F∗v).

Thus F and F∗ extend uniquely from S(Rn) to isometries on L2(Rn) andare inverses to each other. Thus we have the Plancherel theorem:

Proposition 3.2. The Fourier transform

(3.22) F : L2(Rn) −→ L2(Rn)

is unitary, with inverse F∗.

The inversion formulas of Propositions 3.1 and 3.2 do not provide for theinversion of F in (3.2). We will obtain this as a byproduct of the Fourierinversion formula for tempered distributions, in the next section.

We make a remark about the computation of the Fourier integral (3.11),done above via analytic continuation. The following derivation does notmake any direct use of complex analysis. It suffices to handle the caseε = 1/2, that is, to show

(3.23) G(ξ) = e−|ξ|2/2 if G(x) = e−|x|

2/2, on Rn.

28

We have interchanged the roles of x and ξ compared to those in (3.11)and (3.13). It suffices to get (3.23) in the case n = 1, by the obviousmultiplicativity. Now the Gaussian function G(x) = e−x2/2 satisfies thedifferential equation

(3.24)( d

dx+ x

)G(x) = 0.

By the intertwining property (3.5), it follows that (d/dξ + ξ)G(ξ) = 0, anduniqueness of solutions to this ODE yields G(ξ) = Ce−ξ2/2. The constantC is evaluated via the identity (3.20); C = 1; and we are done.

As for the necessity of computing the Fourier integral (3.11) to prove theFourier inversion formula, let us note the following. For any g ∈ S(Rn)with g(0) = 1, (g(ξ) = e−|ξ|

2being an example), we have (replacing ε by

δ2), just as in (3.9),

(3.25)F∗Ff(x) = (2π)−n lim

δ0

∫∫f(y)g(δξ)ei(x−y)·ξ dy dξ

= limδ0

∫f(y)hδ(x− y) dy,

where

(3.26)hδ(x) = (2π)−n

∫g(δξ)eix·ξ dξ

= (2π)−n/2δ−ng(δ−1x).

By the peaked nature of hδ as δ → 0, we see that the limit in (3.25) isequal to

(3.27) C f(x),

where

(3.28) C =∫

h1(x) dx = (2π)−n/2

∫g(x) dx.

The argument (3.25)–(3.27) shows that C is independent of the choice ofg ∈ S(Rn), and we need only find a single example g such that g(x) canbe evaluated explicitly and then the integral on the right in (3.28) can beevaluated explicitly. In most natural examples one picks g to be even, sog = g.

We remark that one does not need to have g ∈ S(Rn) in the argumentabove; it suffices to have g ∈ L1(Rn), bounded and continuous, and suchthat g ∈ L1(Rn). An example, in the case n = 1, is

(3.29) g(ξ) = e−|ξ|.

In this case, elementary calculations give

(3.30) g(x) =( 2

π

)1/2 1x2 + 1

;


compare (5.21). In this case, (3.28) can be evaluated in terms of the arct-angent. Another example, in the case n = 1, is

(3.31)g(ξ) = 1− |ξ| if |ξ| ≤ 1,

0 if |ξ| ≥ 1.

In this case,

(3.32) g(x) = (2π)−1/2( sin 1

2x12x

)2

,

and (3.28) can be evaluated by the method of residues. The calculation of(3.32) can be achieved by evaluating

∫ 1

0

(1− ξ) cos xξ dξ

via an integration by parts, though there is a more painless way, mentionedbelow.

We now make some comments on the relation between the Fourier trans-form and convolutions. The convolution u ∗ v of two functions on Rn isdefined by

(3.33)u ∗ v(x) =

∫u(y)v(x− y) dy

=∫

u(x− y)v(y) dy.

Note that u ∗ v = v ∗ u. If u, v ∈ S(Rn), so is u ∗ v. Also

‖u ∗ v‖Lp(Rn) ≤ ‖u‖L1‖v‖Lp ,

so the convolution has a unique, continuous extension to a bilinear map

(3.34) L1(Rn)× Lp(Rn) −→ Lp(Rn),

for 1 ≤ p < ∞; one can directly perceive that this also works for p =∞. Note that the right side of (3.10), for any ε > 0, is an example of aconvolution. Computing the Fourier transform of (3.33) leads immediatelyto the formula

(3.35) F(u ∗ v)(ξ) = (2π)n/2u(ξ)v(ξ).

We also note that if

(3.36) P =∑

|α|≤k

aαDα

is a constant-coefficient differential operator, we have

(3.37) P (u ∗ v) = (Pu) ∗ v = u ∗ (Pv)

30

if u, v ∈ S(Rn). This also generalizes; if u ∈ S(Rn), v ∈ Lp(Rn), the firstidentity continues to hold; as we will see in the next section, so does thesecond identity, once we are able to interpret what it means.

We mention the following simple application of (3.35), to a short calcu-lation of (3.32). With g given by (3.31), we have g = g1 ∗ g1, where

(3.38) g1(ξ) = 1 for ξ ∈[−1

2,12

], 0 otherwise.

Thus

(3.39) g1(x) = (2π)−1/2

∫ 1/2

−1/2

e−ixξ dξ = (2π)−1/2 sin 12x

12x

,

and then (3.32) follows immediately from (3.35).

Exercises

1. Show that F : L1(Rn) → C0(Rn), where C0(Rn) denotes the space of func-tions v, continuous on Rn, such that v(ξ) → 0 as |ξ| → ∞.(Hint: Use the denseness of S(Rn) in L1(Rn).)This result is the Riemann-Lebesgue lemma for the Fourier transform.

2. Show that the Fourier transforms (3.1) and (3.22) coincide on L1(Rn) ∩L2(Rn).

3. For f ∈ L1(R), set SRf(x) = (2π)−1/2∫ R

−Rf(ξ)eixξdξ. Show that

SRf(x) = DR ∗ f(x) =

∫ ∞

−∞DR(x− y)f(y) dy,

where

DR(x) = (2π)−1

∫ R

−R

eixξ dξ =sin Rx

πx.

Compare Exercise 5 of §1.4. Show that f ∈ L2(R) ⇒ SRf → f in L2-norm as R →∞.5. Show that there exist f ∈ L1(R) such that SRf /∈ L1(R) for any R ∈ (0,∞).

(Hint: Note that DR /∈ L1(R).)6. For f ∈ L1(R), set

CRf(x) = (2π)−1/2

∫ R

−R

(1− |ξ|

R

)f(ξ)eixξ dξ.

Show that CRf(x) = ER ∗ f(x), where

ER(x) = (2π)−1

∫ R

−R

(1− |ξ|

R

)eixξ dξ =

2

πR

[sin 1

2Rx

x

]2

.

Note that ER ∈ L1(R). Show that, for 1 ≤ p < ∞,

f ∈ Lp(R) =⇒ CRf → f in Lp-norm, as R →∞.

We say the Fourier transform of f is Cesaro-summable if CRf → f as R →∞.

Exercises 31

In Exercises 7–13, suppose f ∈ S(R) has the following properties: f ≥ 0,∫∞−∞ f(x) dx = 1, and

∫∞−∞ xf(x) dx = 0. Set F (ξ) = (2π)1/2f(ξ). The point

of the exercises is to obtain a version of the central limit theorem.7. Show that F (0) = 1, F ′(0) = 0, F ′′(0) = −2a < 0. Also, ξ 6= 0 ⇒ |F (ξ)| < 1.8. Set Fn(ξ) = F (ξ/

√n)n. Relate (2π)−1/2Fn(x) to the convolution of n copies

of f .9. Show that there exist A > 0 and G ∈ C∞([−A, A]) such that f(ξ) =

e−aξ2G(ξ) for |ξ| ≤ A, and G(0) = 1, G′(0) = G′′(0) = 0. Hence

Fn(ξ) = e−aξ2G(n−1/2ξ)n, for |ξ| ≤ A

√n.

10. Show that∣∣∣G(ξ/

√n)n − 1

∣∣∣ ≤ Cn−α if |ξ| ≤ n(1/2−α)/3, for n large.

Fix α ∈ (0, 12), and set γ = (1/2− α)/3 ∈ (0, 1

6).

11. Show that, for |ξ| ≥ nγ ,∣∣∣F (ξ/

√n)

∣∣∣ ≤ 1 − 12an−(1−2γ), for n large, so

|Fn(ξ)| ≤ e−an2γ/4 = δn. Deduce that∫

|ξ|≥nγ

|Fn(ξ)| dξ ≤ C δ(n−1)/nn

√n → 0, as n →∞.

12. From Exercises 9–11, deduce that Fn → e−aξ2in L1(R) as n →∞.

13. Deduce now that (2π)−1/2Fn → (4πa)−1/2e−x2/4a in both C0(R) and L1(R),as n → ∞. Relate this to the central limit theorem of probability theory.Weaken the hypotheses on f as much as you can.(Hint: In passing from the C0-result to the L1-result, positivity of Fn will beuseful.)

14. With pε(x) = (4πε)−1/2e−x2/4ε, as in (3.13) for n = 1, show that, for anyu(x), continuous and compactly supported on R, pε ∗ u → u uniformly asε → 0. Show that for each ε > 0, pε ∗u(x) is the restriction to R of an entireholomorphic function of x ∈ C.

15. Using Exercise 14, prove the Weierstrass approximation theorem:Any f ∈ C([a, b]) is a uniform limit of polynomials.

(Hint: Extend f to u as above, approximate u by pε ∗ u, and expand this ina power series.)

16. Suppose f ∈ S(Rn) is supported in BR = x ∈ Rn : |x| ≤ R. Show thatf(ξ) is holomorphic in ξ ∈ Cn and satisfies

(3.40) |f(ξ + iη)| ≤ CN 〈ξ〉−N eR|η|, ξ, η ∈ Rn.

17. Conversely, suppose g(ξ) = f(ξ) ∈ S(Rn) has a holomorphic extension to Cn

satisfying (3.40). Show that f is supported in |x| ≤ R.(Hint: With ω = x/|x|, r ≥ 0, write

(3.41) f(x) = (2π)−n/2

∫

Rn

f(ξ + irω)eix·ξ−rx·ω dξ,

with

|f(ξ + irω) eix·ξ−rx·ω| ≤ CN 〈ξ〉−N er(R−|x|),

and let r → +∞.)This is a basic case of the Paley-Wiener theorem.

32

18. Given f ∈ L1(Rn), show that f is supported in BR if and only if f(ξ) isholomorphic in Cn, satisfying

|f(ξ + iη)| ≤ C eR|η|, ξ, η ∈ Rn.

Reconsider this problem after reading §4.19. Show that

f(x) =1

cosh x=⇒ f(ξ) =

√π/2

cosh π2ξ.

(Hint: See (A.13)–(A.15).)

4. Distributions and tempered distributions

L. Schwartz’s theory of distributions has proved to be not only a wonderfultool in partial differential equations, but also a device that lends clarity tomany aspects of Fourier analysis. We sketch the basic concepts of distribu-tion theory here, making use of such basic concepts as Frechet spaces andweak topologies, which are treated in Appendix A, Functional Analysis.

We begin with the concept of a tempered distribution. This is a contin-uous linear functional

(4.1) w : S(Rn) −→ C,

where S(Rn) is the Schwartz space defined in §3. The space S(Rn) has atopology, determined by the seminorms

(4.2) pk(u) =∑

|α|≤k

supx∈Rn

〈x〉k|Dαu(x)|.

The distance function

(4.3) d(u, v) =∞∑

k=0

2−k pk(u− v)1 + pk(u− v)

makes S(Rn) a complete metric space; with such a topology it is a Frechetspace. For a linear map w as in (4.1) to be continuous, it is necessary andsufficient that, for some k,C,

(4.4) |w(u)| ≤ C pk(u), for all u ∈ S(Rn).

The action of w is often written as follows:

(4.5) w(u) = 〈u,w〉.The set of all continuous linear functionals on S(Rn) is denoted

(4.6) S ′(Rn)

and is called the space of tempered distributions.The space S ′(Rn) has a topology, called the weak∗ topology, or sometimes

simply the weak topology, in terms of which a directed family wγ converges

4. Distributions and tempered distributions 33

to w weakly in S ′(Rn) if and only if, for each u ∈ S(Rn), 〈u,wγ〉 → 〈u, w〉.One can also consider the strong topology on S ′(Rn), the topology of uni-form convergence on bounded subsets of S(Rn), but we will not considerthis explicitly. For more on the topology of S and S ′, see [H], [Sch], and[Yo]. We now consider examples of tempered distributions.

There is a natural injection

(4.7) Lp(Rn) → S ′(Rn),

for any p ∈ [1,∞], given by

(4.8) 〈u, f〉 =∫

u(x)f(x) dx, u ∈ S(Rn), f ∈ Lp(Rn).

Similarly any finite measure on Rn gives an element of S ′(Rn). The basicexample is the Dirac “delta function” δ, defined by

(4.9) 〈u, δ〉 = u(0).

Also, each differential operator Dj = −i∂/∂xj acts on S ′(Rn), by thedefinition

(4.10) 〈u,Djw〉 = −〈Dju,w〉, u ∈ S, w ∈ S ′.Iterating, we see that each Dα = Dα1

1 · · ·Dαnn acts on S ′:

(4.11) Dα : S ′(Rn) −→ S ′(Rn),

and we have

(4.12) 〈u,Dαw〉 = (−1)|α|〈Dαu,w〉for u ∈ S, w ∈ S ′. Similarly,

〈u, fw〉 = 〈fu,w〉defines fw for w ∈ S ′, provided that f and each of its derivatives is poly-nomially bounded.

To illustrate, consider on R the Heaviside function

(4.13)H(x) = 1 if x ≥ 0,

0 if x < 0.

Then H ∈ L∞(R) ⊂ S ′(R), and the definition (4.10) gives

(4.14)d

dxH = δ

as a consequence of the fundamental theorem of calculus. The derivativeof δ is characterized by

(4.15) 〈u, δ′〉 = −u′(0)

in this case.

34

The Fourier transform F : S(Rn) → S(Rn), studied in §3, extends to S ′by the formula

(4.16) 〈u,Fw〉 = 〈Fu,w〉;we can also set

(4.17) 〈u,F∗w〉 = 〈F∗u,w〉to get

(4.18) F , F∗ : S ′(Rn) −→ S ′(Rn).

The maps (4.18) are continuous when S ′(Rn) is given the weak∗ topology,as follows easily from the definitions.

The Fourier inversion formula of Proposition 3.1 yields:

Proposition 4.1. We have

(4.19) F∗F = FF∗ = I on S ′(Rn).

Proof. Using (4.16) and (4.17), if u ∈ S, w ∈ S ′,〈u,F∗Fw〉 = 〈F∗u,Fw〉 = 〈FF∗u,w〉 = 〈u,w〉,

and a similar analysis works for FF∗w.

As an example of a Fourier transform of a tempered distribution, thedefinition gives directly

(4.20) Fδ = (2π)−n/2;

the Fourier transform of the delta function is a constant function. One hasthe same result for F∗δ. By the Fourier inversion formula,

(4.21) F1 = (2π)n/2δ.

Next, let us consider on the line R, for any ε > 0,

(4.22)Hε(x) = e−εx for x ≥ 0,

0 for x < 0.

We have, by elementary calculation,

(4.23) Hε(ξ) = (2π)−1/2

∫ ∞

0

e−εx−iξx dx = (2π)−1/2(ε + iξ)−1,

for each ε > 0. Now it is clear that

(4.24) Hε → H as ε 0, in S ′(R),

in the weak∗ topology. It follows that

(4.25) H(ξ) = (2π)−1/2 limε0

(ε + iξ)−1 in S ′(R).


In particular, the limit on the right exists in S ′(R). Changing the sign ofx in (4.22)–(4.24) and noting that H(−x) = 1−H(x), we also have

(4.26) (2π)1/2δ − H = (2π)−1/2 limε0

(ε− iξ)−1 in S ′(R).

Let us set

(4.27) (ξ ± i0)−1 = limε0

(ξ ± iε)−1.

Then (4.25) and (4.26) give

(4.28) H = −i(2π)−1/2(ξ − i0)−1

and

(4.29) (ξ + i0)−1 − (ξ − i0)−1 = −2πiδ.

The last identity is often called the Plemelj jump relation. Also, subtracting(4.25) from (4.26) gives

(4.30) (ξ + i0)−1 + (ξ − i0)−1 = (2π)1/2i sgn(ξ),

where

(4.31)sgn(x) = 1 if x ≥ 0,

−1 if x < 0.

It is also an easy exercise to show that

(4.32) (ξ + i0)−1 + (ξ − i0)−1 = 2 PV(1

ξ

),

where the “principal value” distribution

(4.33) PV( 1

x

)∈ S ′(R)

is defined by

(4.34)

⟨u, PV

( 1x

)⟩= lim

h0

∫

R\(−h,h)

u(x)x

dx

= limh0

∫ ∞

h

[u(x)x

− u(−x)x

]dx

=∫ ∞

0

u(x)− u(−x)x

dx.

Note that if we replace the left side of (4.29) by

1ξ + iε

− 1ξ − iε

= − 2iεξ2 + ε2

= −2i1ε· 1(ξ/ε)2 + 1

,

the conclusion (4.29) is a special case of the following obvious result.

36

Proposition 4.2. If f ∈ L1(Rn),∫

f(x)dx = C0, then, as ε → 0,

(4.35) ε−nf(ε−1x) → C0δ in S ′(Rn).

That δ is the limit of a sequence of elements of S(Rn) is a special case ofthe fact that S(Rn) is dense in S ′(Rn), which will be established shortly.

Given w ∈ S ′(Rn), if Ω ⊂ Rn is open, one says w vanishes on Ω provided〈u, w〉 = 0 for all u ∈ C∞0 (Ω). By a partition of unity argument, it followsthat if w vanishes on Ωj , then it vanishes on their union; w is said to besupported on a closed set K ⊂ Rn if w vanishes on Rn \K. The smallestclosed set K on which w is supported exists; it is denoted supp w. Notethat if w vanishes on all of Rn, then w = 0, since C∞0 (Rn) is dense in S(Rn).(See the first part of the proof of Proposition 4.4 below.) If w ∈ S ′(Rn) issupported on a compact set K ⊂ Rn, we say w has compact support. Thespace of compactly supported distributions on Rn is denoted by

(4.36) E ′(Rn).

If w ∈ S ′(Rn) is supported on a compact set K ⊂ Rn, then w can beextended to a continuous linear functional

(4.37) w : C∞(Rn) −→ C

by setting

(4.38) 〈u,w〉 = 〈χu,w〉, u ∈ C∞(Rn),

for any χ ∈ C∞0 (Rn) such that χ = 1 on a neighborhood of K. The spaceC∞(Rn) is also a Frechet space, with topology defined by the seminorms

(4.39) pR,k(u) = sup|x|≤R

∑

|α|≤k

|Dαu(x)|.

To say a linear map (4.37) is continuous is to say there exist R, k, and Csuch that

(4.40) |w(u)| ≤ C pR,k(u), for all u ∈ C∞(Rn).

Such a linear functional restricts to S(Rn) ⊂ C∞(Rn), so it defines anelement of S ′(Rn), and from (4.40) it follows that such an element must besupported in the compact set x ∈ Rn : |x| ≤ R. Thus the space (4.36) isprecisely the dual space of C∞(Rn).

Fourier transforms of compactly supported distributions have some spe-cial properties.

Proposition 4.3. If w ∈ E ′(Rn), then w ∈ C∞(Rn) and, with eξ(x) =e−ix·ξ,

(4.41) w(ξ) = (2π)−n/2〈eξ, w〉,


for all ξ ∈ Rn. Furthermore, w extends to an entire holomorphic functionof ξ ∈ Cn.

Proof. For any u ∈ S(Rn), 〈u, w〉 = 〈u, w〉. Now we can write

(4.42) u(x) = (2π)−n/2

∫u(ξ) eξ(x) dξ,

the integral converging in the Frechet space topology of C∞(Rn), and thecontinuity of w acting on C∞(Rn) implies

〈u, w〉 = (2π)−n/2

∫u(ξ)〈eξ, w〉 dξ,

which gives (4.41). The right side of (4.41) is clearly holomorphic in ξ ∈ Cn.

We next obtain the promised denseness of S(Rn) in S ′(Rn).

Proposition 4.4. C∞0 (Rn) is dense in S ′(Rn), with its weak∗ topology.

Proof. Pick ϕ ∈ C∞0 (Rn), ϕ(0) = 1. It is easy to see that, given w ∈S ′(Rn), ϕw is well defined, by 〈u, ϕw〉 = 〈ϕu, w〉. Also, if ϕj(x) = ϕ(x/j),then for u ∈ S(Rn),

(4.43) ϕju → u in S(Rn),

which gives sequential denseness of C∞0 (Rn) in S(Rn). Hence ϕjw → win S ′(Rn) as j → ∞. Since F and F∗ are continuous on S ′(Rn), we haveF∗(ϕjw) → w as j → ∞. Now for each j, wjk = ϕk(F∗ϕjw) → F∗(ϕjw)as k →∞. But by Proposition 4.3, F∗(ϕjw) is smooth, so wjk ∈ C∞0 (Rn),and the result follows.

One useful result is the following classification of distributions supportedat a single point.

Proposition 4.5. If w ∈ S ′(Rn) is supported by 0, then there exist kand complex numbers aα such that

(4.44) w =∑

|α|≤k

aαDαδ.

Proof. We can suppose w satisfies the estimate (4.4). Thus w extendsto Bk, the closure of S(Rn) in the space of Ck-functions on Rn for whichthe norm pk is finite. By hypothesis, w annihilates the linear space E0 ofelements of C∞0 (Rn) vanishing on a neighborhood of 0; thus w annihilatesthe closure of E0 in Bk; call this closure Ek. It is not hard to prove that

(4.45) Ek = u ∈ Bk : Dαu(0) = 0 for |α| ≤ k.

38

See Exercise 7 below for some hints. Now, for general u ∈ Bk, write

(4.46) u(x) = χ[ ∑

|α|≤k

u(α)(0)α!

xα]

+ ub(x),

where χ ∈ C∞0 (Rn), χ(x) = 1 for |x| < 1, and ub ∈ Ek. Applyingw to both sides, we have an expression of the form (4.44), with aα =(−1)|α|〈χxα, w〉/α!.

As an application of Proposition 4.5, we establish the following result,which is an extension of the classical Liouville theorem for harmonic func-tions.

Proposition 4.6. Suppose u ∈ S ′(Rn) satisfies

(4.47) ∆u = 0 in Rn.

Then u is a polynomial in (x1, . . . , xn).

Proof. As in §3, the identity

(4.48) ∆u = f ∈ S ′(Rn)

is equivalent to

(4.49) −|ξ|2u = f in S ′(Rn).

In particular, (4.47) for u ∈ S ′(Rn) implies

(4.50) |ξ|2u = 0 in S ′(Rn).

This of course implies

(4.51) supp u ⊂ 0.By Proposition 4.5, u must have the form (4.44), and the result follows.

It is clear that any nonconstant polynomial blows up, so we have:

Corollary 4.7. If u is harmonic on Rn and bounded, then u is constant.

This is the classical version of the Liouville theorem. We remind thereader of one of its uses. If p(z) is a polynomial on C, and if it has nozeros, then q(z) = 1/p(z) is holomorphic (hence harmonic) on all of C;clearly, |q(z)| → 0 as |z| → ∞ if deg p ≥ 1. Corollary 4.7 yields theobvious contradiction that q(z) would have to be constant. This proves thefundamental theorem of algebra: Any nonconstant polynomial p(z) musthave a complex root.

Consider the function

(4.52) Ψ(z) =1z.


This is holomorphic on C\0. It is integrable near 0 and bounded outsidea neighborhood of 0, and hence it defines an element of S ′(R2). (∂/∂z)Ψ ∈S ′(R2) is supported at 0. In fact, we claim:


(4.53)∂

∂z

1z

= πδ.

Proof. Let u ∈ C∞0 (R2). We have

(4.54)

⟨u,

∂

∂z

1z

⟩= −

∫∫

R2

∂u

∂z

1z

dx dy

= − limε→0

∫∫

R2\Bε

∂(z−1u)∂z

dx dy,

where Bε = (x, y) ∈ R2 : x2 + y2 < ε2. By Green’s formula, in the form(2.37), the right side is equal to

(4.55)12i

∫

∂Bε

u

zdz,

which is clearly equal in the limit ε → 0 to πu(0). This proves the propo-sition.

We say (πz)−1 is a fundamental solution of (∂/∂z). We will say moreabout the use of fundamental solutions later. Let us look at the task ofproducing a fundamental solution for the Laplace operator ∆ on Rn. Inview of the rotational invariance of ∆, we are led to look for a function ofr = |x|, for x 6= 0. The form

(4.56) ∆ =∂2

∂r2+

n− 1r

∂

∂r+

1r2

∆S ,

for ∆ in polar coordinates, where ∆S is the Laplace operator on the unitsphere Sn−1, shows that, for n ≥ 3,

(4.57) |x|2−n

is harmonic for x 6= 0. As it is locally integrable near 0, it defines anelement of S ′(Rn). We have the following result.

Proposition 4.9. If n ≥ 3,

(4.58) ∆(|x|2−n

)= Cnδ on Rn,

with Cn = −(n− 2)·Area(Sn−1). Also,

(4.59) ∆(log |x|) = C2δ on R2,

40

with C2 = 2π.

Proof. This will use Green’s formula, in the form

(4.60)∫

Ω

(∆u · v − u ·∆v

)dx =

∫

∂Ω

[v∂u

∂ν− u

∂v

∂ν

]dS.

Let u ∈ C∞0 (Rn) be arbitrary. Let v = |x|2−n, and let Ω = Ωε = Rn \ Bε,where Bε = x ∈ Rn : |x| < ε. We have

(4.61)

〈∆u, |x|2−n〉 = limε→0

∫

Ωε

∆u · |x|2−n dx

= limε→0

∫

Ωε

[∆u · |x|2−n − u ·∆|x|2−n

]dx

since ∆|x|2−n = 0 for x 6= 0. Applying (4.60), we have this equal to

(4.62) − limε→0

∫

∂Bε

[ε2−n ∂u

∂r− (2− n)ε1−nu

]dS.

Since the area of ∂Bε is εn−1·Area Sn−1, this limit is seen to be

(4.63) −(n− 2)u(0) ·Area Sn−1,

which proves (4.58). The proof of (4.59) is similar.

Calculations yielding expressions for the area of Sn−1 will be given inthe appendix to this chapter.

Note that the equation

(4.64) ∆Φ = δ on Rn,

with Φ ∈ S ′(Rn), is equivalent to

(4.65) −|ξ|2Φ = (2π)−n/2.

If n ≥ 3, |ξ|−2 ∈ L1loc(Rn) and one solution to (4.65) is

(4.66) Φ(ξ) = −(2π)−n/2|ξ|−2 ∈ S ′(Rn),

in such a case. We can relate this directly to (4.58) as follows. The orthog-onal group O(n) acts on S(Rn), by

(4.67) π(g)u(x) = u(g−1x), x ∈ Rn, g ∈ O(n),

and this extends to an action on S ′(Rn), via 〈u, π(g)v〉 = 〈π(g−1)u, v〉. Thisaction commutes with F . Thus the Fourier transform of an element like|x|2−n which is invariant under the O(n)-action will also be O(n)-invariant.There is also the dilation action on S(Rn),

(4.68) D(s)u(x) = u(sx), s > 0, x ∈ Rn,


which extends to S ′(Rn), via 〈u,D(t)v〉 = t−n〈D(1/t)u, v〉. We have

(4.69) D(s)F = s−nFD(s−1).

The element |x|2−n ∈ S ′(Rn) is homogeneous of degree 2− n, that is,

(4.70) D(s)(|x|2−n) = s2−n|x|2−n,

so F(|x|2−n) will be homogeneous of degree −2. This establishes thatΦ(x) = Cn|x|2−n satisfies (4.66), up to a constant factor. Note thatδ ∈ S ′(Rn) is homogeneous of degree −n. Since the Laplace operator∆ decreases the order of homogeneity of a distribution by two units, theseconsiderations of orthogonal invariance and homogeneity directly suggesta constant times |x|2−n as a suitable candidate for a fundamental solutionfor the Laplace operator on Rn.

We mention some extensions of the convolution

(4.71) u ∗ v(x) =∫

u(y)v(x− y) dy,

which gives a bilinear map

(4.72) S(Rn)× S(Rn) → S(Rn).

Note that if u, v, w ∈ S(Rn),

(4.73) 〈u ∗ v, w〉 = 〈u, v# ∗ w〉,where v#(x) = v(−x), so the convolution extends in a straightforward wayto

(4.74) S(Rn)× S ′(Rn) → S ′(Rn),

with S(Rn)× E ′(Rn) → S(Rn), and hence

(4.75) E ′(Rn)× S ′(Rn) → S ′(Rn).

In either case, the identity

(4.76) F(u ∗ v) = (2π)n/2u(ξ)v(ξ)

continues to hold. For more on this, see Exercises 11–13 below. If P is anyconstant-coefficient differential operator, then

(4.77) P (u ∗ v) = (Pu) ∗ v = u ∗ Pv

in cases (4.74) and (4.75). For example, if Φ ∈ S ′(Rn) and

(4.78) PΦ = δ

(we say Φ is a fundamental solution of P ), then a solution to

(4.79) Pu = f,

for any given f ∈ E ′(Rn), is provided by

(4.80) u = f ∗ Φ.

42

An object more general than a tempered distribution is a distribution.In general, a distribution on Rn is a continuous linear map

(4.81) w : C∞0 (Rn) −→ C.

Here, continuity can be characterized as follows. For each ϕ ∈ C∞0 (Rn),the identity 〈u, ϕw〉 = 〈ϕu,w〉 makes ϕw a linear functional on C∞(Rn).We require that each such linear functional be continuous, in the sensespecified in (4.40). For further discussion, including a direct discussion ofthe natural “inductive limit” topology on C∞0 (Rn), see [RS] and [Sch]. Thespace of all distributions on Rn is denoted by

(4.82) D′(Rn).

More generally, if M is any smooth, paracompact manifold, the spaceof continuous linear functionals on C∞(M) is denoted by E ′(M) and thespace of continuous linear functionals on C∞0 (M) is denoted by D′(M). Ofcourse, if M is compact, E ′(M) = D′(M).

The case M = Tn is of interest, with respect to Fourier series. Givenw ∈ D′(Tn), we can define

(4.83) Fw(k) = w(k) = (2π)−n〈ek, w〉,where

(4.84) ek(θ) = e−ik·θ ∈ C∞(Tn).

Since w must satisfy an estimate of the form

(4.85) |〈u,w〉| ≤ C‖u‖C`(Tn),

it is clear that

(4.86) F : D′(Tn) −→ s′(Zn),

where

(4.87) s′(Zn) = a : Zn → C : |a(k)| ≤ C〈k〉N for some C, Nconsists of polynomially bounded functions on Zn. Note that s′(Zn) is thedual space to s(Zn), defined in §1, and the map F in (4.86) is the adjointof the map F∗ : s(Zn) −→ C∞(Tn) given by (1.11) and (1.12). Herewe use the Hermitian inner product (u,w) = 〈u,w〉 = 〈u,w〉. The mapF : C∞(Tn) → s(Zn) given by (1.1) and (1.6) also has an adjoint

(4.88) F∗ : s′(Zn) → D′(Tn),

extending the map (1.11)–(1.12), which, we recall, is

(4.89) (F∗a)(θ) =∑

k∈Zn

a(k)eik·θ.

The Fourier inversion formulas

(4.90) F∗F = I on C∞(Tn), FF∗ = I on s(Zn),

Exercises 43

extend by duality (or by continuity, and denseness of C∞(Tn) in D′(Tn)and of s(Zn) in s′(Zn)) to

(4.91) F∗F = I on D′(Tn), FF∗ = I on s′(Zn);

consequently, the map (4.86) is an isomorphism.

Exercises

1. Define Mfu by 〈v, Mfu〉 = 〈fv, u〉, for v ∈ S(Rn), u ∈ S ′(Rn). Mfuis also denoted by fu. Show that Mf : S ′(Rn) → S ′(Rn) continuously,provided f ∈ C∞(Rn) and each derivative is polynomially bounded, that is,|Dαf(x)| ≤ Cα〈x〉N(α).

2. Show that the identity ξαDβξFf(ξ) = (−1)|β|F(Dαxβf)(ξ) from §3 continues

to hold for f ∈ S ′(Rn).3. Calculate F of xα and of Dαδ.4. Verify the identity (4.32) involving PV (1/ξ).5. Give a proof of Proposition 4.4, that S(Rn) is dense in S ′(Rn), using convo-

lutions in place of the Fourier transform.6. Show the denseness of S(Rn) in S ′(Rn) follows from the Hahn-Banach the-

orem. See Appendix A for a discussion of the Hahn-Banach theorem. Onthe other hand, sharpen the proof of Proposition 4.4, to obtain sequentialdenseness.

7. Prove the identity (4.45), used in the proof of Proposition 4.5.(Hint: Fix η ∈ C∞(Rn) so that η(x) = 0 for |x| ≤ 1/2, 1 for |x| ≥ 1. Showthat η(Rx)u → u in Bk, as R → ∞, for any u ∈ Ek. This, plus a couple offurther approximations, yields (4.45).)

8. Let f(x, y) = 1/z ∈ S ′(R2). Using (4.53), compute Ff ∈ S ′(R2). UsingProposition 4.9, compute the Fourier transform of log |x| on R2, and of|x|2−n on Rn, n ≥ 3. Reconsider this problem after reading §8.

9. Let u ∈ E ′(Rn), and suppose 〈p, u〉 = 0 for every polynomial p on Rn. Showthat u = 0. (Hint: Show that Dαu(0) = 0 for all α; but u is analytic.)Show that this result implies the Weierstrass approximation theorem, dis-cussed in Exercise 15 of §3.

10. Let f ∈ C∞(Rn) be real-valued, Σ = x : f(x) = 0. Define δ(f(x)) ∈D′(Rn) to be

(4.92) δ(f(x)) = limε0

δε(f(x)),

where δε(x) = 1/ε for |t| < ε/2, 0 otherwise, provided this limit exists, withrespect to the weak∗ topology on D′(Rn). Show that if ∇f 6= 0 on Σ, thelimit does exist and that, for u ∈ C∞0 (Rn),

⟨u, δ(f(x))

⟩=

∫

Σ

u(y)|∇f(y)|−1 dS(y),

where dS is the (n−1)-dimensional measure on Σ. Consider cases where thelimit in (4.92) exists though ∇f vanishes on a variety in Σ.

44

11. Using an argument like that in the proof of Proposition 4.5, show that ifu ∈ S ′(Rn) has support in a closed ball B, then, for some C, k,

|〈f, u〉| ≤ C supx∈B,|α|≤k

|Dαf(x)|.

(Hint: Establish the following analogue of (4.45). If EB is the linear space ofelements of C∞0 (Rn) vanishing on a neighborhood of B, and Ek is the closureof EB in Bk, then

Ek = u ∈ Bk : u = 0 on B.Then show that if u : Bk → C is continuous and supp u ⊂ B,

〈f, u〉 = 〈Eρ(f), u〉,where ρ(f) = f

∣∣∣B

and E : Ck(B) → Bk is an extension operator. For help

in constructing E, look ahead to §4 of Chapter 4.)12. If u ∈ E ′(Rn), show that u ∈ C∞(Rn) satisfies an estimate

(4.93) |u(ξ)| ≤ C〈ξ〉m, ξ ∈ Rn,

for some m ∈ R. More generally, show that if a distribution u has supportin BR = x ∈ Rn : |x| ≤ R, then

(4.94) |u(ξ + iη)| ≤ C〈ξ + iη〉m eR|η|, ξ, η ∈ Rn.

(Hint. Use (4.39)–(4.41). For (4.94), use the result of Exercise 11.)13. Given the formula (4.76) for F(u∗v) when u ∈ S(Rn), v ∈ S ′(Rn), show that

if u ∈ S(Rn) and v ∈ E ′(Rn), then F(u ∗ v) ∈ S(Rn), hence u ∗ v ∈ S(Rn),as asserted above (4.75). (Hint: Use (4.93).)

14. Show that the convolution product extends to

E ′(Rn)×D′(Rn) −→ D′(Rn), E ′(Rn)× E ′(Rn) → E ′(Rn).

15. Given u ∈ E ′(Rn), show that there exist k ∈ Z+, f ∈ L2(Rn) such thatu = (1−∆)kf .(Hint: Obtain 〈ξ〉−2k+m ∈ L2(Rn).)Show that there exist compactly supported fα ∈ L2(Rn) such that u =∑|α|≤k Dαfα.

16. Assume that u ∈ S ′(Rn) and u is holomorphic in Cn and satisfies (4.94).Show that u is supported in the ball BR. This is the distributional versionof the Paley-Wiener theorem.(Hint: Pick ϕ ∈ C∞0 (Rn), supported in B1,

∫ϕ dx = 1, let ϕε(x) =

ε−nϕ(x/ε), and consider uε = ϕε ∗ u ∈ S(Rn). Apply Exercises 17 and18 of §3.)

5. The classical evolution equations

In this section we analyze solutions to the classical heat equation on R+ ×Rn, to the Laplace equation on Rn+1

+ , and to the wave equation on R×Rn.We begin with the heat equation for u = u(t, x),

(5.1)∂u

∂t−∆u = 0,

5. The classical evolution equations 45

where ∆ is the Laplace operator on Rn,

(5.2) ∆u =∂2u

∂x21

+ · · ·+ ∂2u

∂x2n

.

We pose an initial condition

(5.3) u(0, x) = f(x).

We suppose that f ∈ S ′(Rn), and we look for a solution u ∈ C∞(R+,S ′(Rn)),

via Fourier analysis. Taking the Fourier transform of u with respect to x,we obtain the ODE with parameters

(5.4)∂u

∂t= −|ξ|2u(t, ξ),

with initial condition

(5.5) u(0, ξ) = f(ξ).

The unique solution to (5.4)–(5.5) is

(5.6) u(t, ξ) = e−t|ξ|2 f(ξ).

Set

(5.7) G(t, ξ) = e−t|ξ|2 .

Note that, for each t > 0, G(t, ·) ∈ S(Rn). By (4.76), we have

(5.8) u(t, x) = (2π)−n/2G(t, ·) ∗ f(x).

The computation of the Fourier transform of such a Gaussian function wasmade in §3. From (3.18), we deduce that

(5.9) u(t, x) = p(t, ·) ∗ f(x),

where

(5.10) p(t, x) = (4πt)−n/2e−|x|2/4t,

for t > 0. The function p(t, x) is called the fundamental solution to theheat equation. It satisfies

(5.11)(∂/∂t−∆)p = 0, for t > 0,

limt0

p(t, x) = δ(x) in S ′(Rn).

We record what Fourier analysis has yielded for the heat equation.

Proposition 5.1. The heat equation (5.1)–(5.3), with f ∈ S ′(Rn), has a

unique solution u ∈ C∞(R+,S ′(Rn)), given by (5.9)–(5.10). The solution

is C∞ on (0,∞)× Rn. If f ∈ S(Rn), then u ∈ C∞(R+,S(Rn)).

Note carefully that uniqueness of the solution is asserted only within theclass C∞(R+

,S ′(Rn)); this entails bounds on the solution considered, near

46

infinity. If one removes such growth restrictions, uniqueness fails. Thereexist nontrivial solutions to

(5.12)( ∂

∂t−∆

)v = 0, for t > 0, v(0, x) = 0,

outlined in Exercise 2 at the end of this section. For fixed t > 0, suchsolutions blow up too fast to belong to S ′(Rn).

In view of the boundedness and continuity properties of (5.7), we havethe following:

Corollary 5.2. Suppose f ∈ L2(Rn). Then the solution u to (5.1)–(5.3)

of Proposition 5.1 belongs to C(R+, L2(Rn)).

We cannot say that u ∈ C∞(R+, L2(Rn)), or even that ∂u/∂t belongs

to C(R+, L2(Rn)), in such a case, without further restrictions on f . The

appropriate behavior of ∂ju/∂tj in such a case is best described in termsof Sobolev spaces, which will be discussed in Chapter 4.

Next we look at the following boundary problem for functions harmonicin an upper half space:

( ∂2

∂y2+ ∆

)u(y, x) = 0, for y > 0, x ∈ Rn,(5.13)

u(0, x) = f(x).(5.14)

Here, ∆ is given by (5.2). In view of such simple examples as

(5.15) u(y, x) = y,

which satisfy (5.13) and (5.14) with f = 0, we will need to make appropriaterestrictions on u in order to obtain uniqueness. As before, we suppose thatf ∈ S ′(Rn) and look for u ∈ C∞(R+

,S ′(Rn)). Fourier transforming withrespect to x gives the second-order ODE, with parameters,

( d2

dy2− |ξ|2

)u(y, ξ) = 0,(5.16)

u(0, ξ) = f(ξ).(5.17)

The general solution to (5.16)–(5.17), for any fixed ξ 6= 0, is

(5.18) u(y, ξ) = c0(ξ)ey|ξ| + c1(ξ)e−y|ξ|,

with c0(ξ) + c1(ξ) = f(ξ). Let us restrict attention to f such that f(ξ) iscontinuous, and look for u(y, ξ) continuous in (y, ξ); as (5.15) illustrates,things can be more complicated if u(y, ·) is a singular distribution nearξ = 0. In view of the blow-up of ey|ξ| as y ∞, it is natural to requirethat c0(ξ) = 0, so

(5.19) u(y, ξ) = e−y|ξ|f(ξ).


In partial analogy with Proposition 5.1, we have obtained the followingresult.

Proposition 5.3. Let f ∈ S ′(Rn), and suppose f is continuous. Thenthere is a unique solution u(y, x) of (5.13) and (5.14), belonging to the space

C∞(R+,S ′(Rn)) and satisfying the condition that u(y, ξ) is continuous on

R+ × Rn, and furthermore that u(y, ·) is a bounded function of y takingvalues in S ′(Rn). It is given by (5.19).

Note that f(ξ) is continuous provided f is a finite measure. It is alsocontinuous if f ∈ E ′(Rn).

We want to find the “fundamental solution” P (y, x) for (5.13)–(5.14),corresponding to f = δ. In other words,

(5.20) P (y, ξ) = (2π)−n/2e−y|ξ|.

This computation is elementary in the case n = 1. We have

(5.21)

P (y, x) = (2π)−1

∫ ∞

−∞e−y|ξ|+ixξ dξ

= (2π)−1[∫ ∞

0

e−yξ+ixξ dξ +∫ ∞

0

e−yξ−ixξ dξ]

= (2π)−1[(y − ix)−1 + (y + ix)−1

]

=1π

y

y2 + x2.

For n > 1, a direct calculation of such a Fourier transform is not so el-ementary. One way to perform the computation is to use the followingsubordination identity:

(5.22) e−yA =y

2π1/2

∫ ∞

0

e−y2/4t e−tA2t−3/2 dt, A > 0, y > 0.

We will give a proof of (5.22) shortly. First we will show how it leads toa computation of the Fourier transform of (5.20). We let A = |ξ|, and weuse our prior computation of the Fourier transform of e−|ξ|

2. Thus, for any

n ≥ 1,

(5.23)P (y, x) = (2π)−n

∫e−y|ξ|+ix·ξ dξ

= (2π)−n y

2π1/2

∫ ∞

0

e−y2/4t[∫

e−t|ξ|2+ix·ξdξ]t−3/2 dt,

48

and substituting in the calculation (5.7)–(5.10), we have

(5.24)P (y, x) = (4π)−(n+1)/2y

∫ ∞

0

e−y2/4te−|x|2/4tt−(n+3)/2 dt

= cny

(y2 + |x|2)(n+1)/2,

where the last integral is evaluated using the substitution s = 1/t. Theconstant cn is given by

(5.25) cn = π−(n+1)/2Γ(n + 1

2

),

where Γ(z) =∫∞0

e−ssz−1ds is Euler’s gamma function, which is discussedin the appendix to this chapter. Note that c1 = 1/π, so the calculation(5.24) agrees with (5.21), in the case n = 1.

The observation that the calculations (5.21) and (5.24) coincide for n = 1can be used to provide a simple proof of the subordination identity (5.22).Indeed, with |ξ| substituted for A in (5.22), we know that the operationof Fourier multiplication by the left side coincides with the operation ofFourier multiplication by the right side, so the two functions of |ξ| (ξ ∈ R)must coincide.

There are other proofs of (5.22). In Appendix A we note the equivalenceof such an identity and a classical identity involving Euler’s gamma func-tion. While the proof of (5.22) given above is complete, it leaves one withan unsatisfied feeling, since the right side of the formula (5.22) seems tohave been pulled out of a hat. We want to introduce a setting where such aformula arises naturally, a setting involving the use of operator notations,as follows. For a decent function f defined on [0,∞), define f(

√−∆) onS(Rn), or on L2(Rn), or even on S ′(Rn), when it makes sense, by

(5.26)(f(√−∆)u

)(ξ) = f(|ξ|)u(ξ).

Thus, the content of (5.10) is

(5.27) et∆δ(x) = (4πt)−n/2e−|x|2/4t,

for t > 0. The formula (5.24) is a formula for e−y√−∆δ(x), x ∈ Rn, and

the formula (5.21) is the special case of this for n = 1.We will approach the subordination identity via the formula

(5.28) (λ2 −∆)−1 =∫ ∞

0

e−(λ2−∆)t dt,

for the resolvent (λ2−∆)−1 of the Laplace operator ∆. This identity followsvia the Fourier transform, as in (5.26), from

(5.29) (λ2 + |ξ|2)−1 =∫ ∞

0

e−(λ2+|ξ|2)t dt.

In order to derive the subordination identity, we will apply both sides of(5.28) to δ; we will do this in the special case of ∆ = d2/dx2 acting on


S ′(R1). For the special case ξ ∈ R1, we have the Fourier integral formula

(5.30)∫ ∞

−∞(λ2 + ξ2)−1eixξ dξ =

π

λe−λ|x| (λ > 0),

a fact that can be established either by residue calculus or by applying theFourier inversion formula to the computation (5.21). Thus applying bothsides of (5.28) to δ ∈ S ′(R) and using et∆δ = (4πt)−1/2e−x2/4t in this case,we have

(5.31) λ−1e−λ|x| =1√π

∫ ∞

0

e−x2/4te−λ2tt−1/2 dt.

Making the change of variables y = |x|, A = λ, and taking the y-derivativeof the resulting identity give (5.22). Also note that taking the A-derivativeof (5.22) gives (5.31). The identity (5.31) is also called the subordinationidentity. One can see that it arises very naturally in this context, from(5.28). We will return to the calculation of (λ2 −∆)−1δ(x) in case n > 1,later in this section.

We next consider the wave equation on R× Rn:

(5.32)∂2u

∂t2−∆u = 0,

with initial data

(5.33) u(0, x) = f(x), ut(0, x) = g(x).

As before, we suppose that f and g belong to S ′(Rn) and look for a solutionu in C∞(R,S ′(Rn)). Taking the Fourier transform with respect to x againyields an ODE with parameters:

(5.34)d2u

dt2+ |ξ|2u = 0,

(5.35) u(0, ξ) = f(ξ), ut(0, ξ) = g(ξ).

The general solution to (5.28) (for ξ 6= 0) is of the form

u(t, ξ) = c2(ξ) sin t|ξ|+ c1(ξ) cos t|ξ|,which it is convenient to write as

u(t, ξ) = c0(ξ)|ξ|−1 sin t|ξ|+ c1(ξ) cos t|ξ|,since the right side is well defined for any c0, c1 ∈ S ′(Rn). The initialconditions (5.35) imply c1 = f , c0 = g, so the solution to (5.34)–(5.35) is

(5.36) u(t, ξ) = g(ξ)|ξ|−1 sin t|ξ|+ f(ξ) cos t|ξ|.This is clearly the unique solution in C∞(R,S(Rn)), if f, g ∈ S(Rn). Theuniqueness in C∞(R,S ′(Rn)) for general f, g ∈ S ′(Rn) will be provedshortly.

50

If f = 0, g = δ, the solution given by (5.36) is called the fundamentalsolution, or the “Riemann function.” Of course, it is actually a distribution.It is characterized by

(5.37) R(t, ξ) = (2π)−n/2|ξ|−1 sin t|ξ|.We want a direct formula for R(t, x). We will be able to deduce thisformula from the formula (5.24) for the Fourier transform of e−y|ξ|, viaanalytic continuation. To bring in the factor |ξ|−1, integrate (5.24) withrespect to y. Thus, if

(5.38) F (y, ξ) = (2π)−n/2|ξ|−1e−y|ξ|,

which belongs to S ′(Rn) for each y ≥ 0 if n ≥ 2, we have

(5.39) F (y, x) = c′n(y2 + |x|2)−(n−1)/2

with

(5.40) c′n =cn

n− 1=

12π−(n+1)/2Γ

(n− 12

).

This has been verified for real y > 0. But (5.38) is holomorphic in y, withvalues in S ′(Rn), for all y such that Re y > 0. Also, it is continuous inthe right half-plane y ∈ C : Re y ≥ 0. In view of the continuity of theFourier transform on S ′(Rn), we deduce that if

(5.41) Φ(t, ξ) = (2π)−n/2|ξ|−1eit|ξ|,

t ∈ R, then

(5.42) Φ(t, x) = limε0

c′n(|x|2 − (t− iε)2

)−(n−1)/2,

the limit existing in S ′(Rn) for each t ∈ R, since Φ(t, ·) = limε0 Φ(t−iε, ·)in S ′(Rn). Consequently, for the Riemann function, we have

(5.43) R(t, x) = limε0

c′n Im(|x|2 − (t− iε)2

)−(n−1)/2.

Note that if |x| > |t|, limε0

(|x|2− (t− iε)2)−(n−1)/2 = (|x|2− t2)−(n−1)/2

is real, so

(5.44) R(t, x) = 0, for |x| > |t|.This is a reflection of the finite propagation speed, which was discussed inChapter 2. Note also that if n is odd, then (n−1)/2 is an integer, so (5.43)vanishes also for |x| < |t|. In other words,

(5.45) n ≥ 3 odd ⇒ supp R(t, ·) ⊂ x ∈ Rn : |x| = |t|.This is the strict Huygens principle. Of course, it does not hold when n iseven. When n = 2, the computation of the limit in (5.43) is elementary.


We have

(5.46)R(t, x) = c′2(t

2 − |x|2)−1/2 · sgn(t), for |x| < |t|,0, for |x| > |t|,

for n = 2. For n = 3, the Plemelj jump relation (4.29) yields

(5.47) R(t, x) = (4πt)−1δ(|x| − |t|).The discussion above has to be modified for n = 1, since (5.41) is notlocally integrable near ξ = 0 in that case. This simple case (n = 1) wastreated in §1 of Chapter 2; see (1.24)–(1.28) in that chapter.

The solution to (5.32)–(5.33) given by (5.36) can be expressed as

(5.48) u(t, ·) = R(t, ·) ∗ g +∂

∂tR(t, ·) ∗ f.

We record our result on solutions to the wave equation.

Proposition 5.4. Given f, g ∈ S ′(Rn), there is a unique solution u ∈C∞(R,S ′(Rn)) to the initial-value problem (5.32)–(5.33). It is given by(5.48).

Proof. The only point remaining to be established is the uniqueness. Sup-pose u ∈ C∞(R,S ′(Rn)) solves (5.32)–(5.33) with f = g = 0. Then

v(t, ·) = u(t, ·) ∗ ϕ

solves the same equation, for any ϕ ∈ C∞0 (Rn); ϕ = ϕ(x). We havev ∈ C∞(R × Rn). Thus the energy estimates of Chapter 2 are applicableto v, and we have v = 0 everywhere. Taking a sequence ϕj ∈ C∞0 (Rn)approaching δ, we have vj → u; since each vj = 0, it follows that u = 0,and the proof is complete.

We note that the argument above yields uniqueness for u in the classC∞(R,D′(Rn)). We also remark that any u ∈ D′(R × Rn) solving (5.32)actually belongs to C∞(R,D′(Rn)), and that (5.48) gives the unique solu-tion to (5.32)–(5.33) for any f, g ∈ D′(Rn). Justification of these statementsis left as an exercise.

Returning to the operator notation (5.26), we have

(5.49) R(t, x) = (−∆)−1/2 sin t√−∆ δ(x)

and

(5.50)∂

∂tR(t, x) = cos t

√−∆ δ(x).

We also denote (5.49) by R(t) and (5.50) by R′(t).Having introduced in (5.28) the notion of synthesizing some operators

from other operators, we want to mention the particular desirability of syn-thesizing functions of the Laplace operator from the fundamental solution

52

of the wave equation. If ϕ(s) is an even function, the Fourier inversionformula implies

(5.51) ϕ(√−∆) = (2π)−1/2

∫ ∞

−∞ϕ(t) cos t

√−∆ dt.

Note that

(5.52) cos t√−∆ u = R′(t) ∗ u,

where R(t) is the Riemann function constructed above. We have the fol-lowing rather general calculation of ϕ(

√−∆)δ, using the formula (5.37) forthe Riemann function on Rn.

Proposition 5.5. Let ϕ ∈ S(R) be even. Then, on Rn, we have

(5.53) ϕ(√−∆)δ(x) =

1√2π

[− 1

2πr

∂

∂r

]k

ϕ(r)

if n = 2k + 1 is odd, and

(5.54) ϕ(√−∆)δ(x) =

√2π

∫ ∞

r

[− 1

2πs

∂

∂s

]k

ϕ(s) s√

s2 − r2ds

if n = 2k is even. Here, r = |x|.

Proof. When n = 3, we have

(5.55)ϕ(√−∆)δ = − 2√

2π

∫ ∞

0

ϕ′(t)(4πt)−1δ(r − t) dt

=1√2π

[− 1

2πr

∂

∂r

]ϕ(r),

giving (5.53) in this case. When n = 2, we have, from (5.46),

(5.56) ϕ(√−∆)δ = − 2c′2√

2π

∫ ∞

r

ϕ′(t)(t2 − r2)−1/2 dt,

giving (5.54) in this case, once one checks that c′2 = (1/2)π−3/2Γ(1/2) =1/2π. To pass to the general case, we note that if Rn(t, r) denotes theformula (5.43) for the Riemann function, in view of the evaluation of c′n in(5.40), we have the formal relation

(5.57) Rn+2(t, r) =[− 1

2πr

∂

∂r

]Rn(t, r),

so (5.53) and (5.54) follow by induction.

It is clear that Proposition 5.5 holds for a more general class of evenfunctions ϕ than those in S(R), by simple limiting arguments. For example,the function ϕ(s) = (λ2 + s2)−1, λ > 0, giving the resolvent of ∆, can be


treated. We leave the formulation of general results on classes of ϕ whichcan be treated as an exercise.

In the case of using Proposition 5.5 to treat the resolvent of ∆, we havethe following formula. With ϕ(s) = (λ2 + s2)−1, from (5.30) we have

(5.58) ϕ(t) =(π

2

)1/2

λ−1e−λ|t|,

to plug into (5.53)–(5.54). For example, for n = 3, we have

(5.59) (λ2 −∆)−1δ = (4π|x|)−1e−λ|x| on R3.

Note that computing (λ2 − ∆)−1δ by evaluating the right side of (5.28)gives in this case

(5.60) (λ2 −∆)−1δ =∫ ∞

0

e−|x|2/4te−λ2t(4πt)−3/2 dt;

comparing (5.59) and (5.60) again reveals the subordination identity, in theoriginal form (5.22).

The fact that the answer comes out “in closed form” for n odd is aconsequence of the strict Huygens principle. For n even, one tends not toget elementary functions. Note that, for n = 2, the formula (5.54) gives

(5.61) (λ2 −∆)−1δ = c

∫ ∞

|x|e−λs(s2 − r2)−1/2 ds on R2;

the formula (5.28) gives

(5.62) (λ2 −∆)−1δ =14π

∫ ∞

0

e−|x|2/4t−t t−1 dt on R2.

Both of these integrals can be expressed in terms of the modified Besselfunction K0; we say a little more about this in the next section; see (6.46)–(6.54).

In general, the use of results on the wave equation together with (5.51)provides a tool of tremendous power and flexibility in the analysis of numer-ous functions of the Laplace operator. We will see more of this in Chapter8.

To end this section, we re-derive formulas for the solution to the waveequation (5.32)–(5.33), and then re-derive the formulas (5.53)–(5.54). Re-call that the solution to the wave equation on R× Rn is given by

(5.63) u(t, x) = cos t√−∆ f(x) +

sin t√−∆√−∆

g(x).

We first derive formulas for these solution operators, in case

(5.64) n = 2k + 1,

by comparing two formulas for et∆f(x). This approach follows material in[PT].

54

The first formula for et∆ is

(5.65)

et∆f(x) = (4πt)−n/2

∫

Rn

e−|y|2/4tf(x− y) dy

= (4πt)−n/2An−1

∫ ∞

0

fx(r)rn−1e−r2/4t dr,

where An−1 is the area of the unit sphere Sn−1 in Rn, and

(5.66) fx(r) =1

An−1

∫

Sn−1

f(x + rω) dS(ω).

Note that fx(r) is well defined for all r ∈ R, and fx(−r) = fx(r). Thefirst identity in (4) follows, via Fourier analysis, from the evaluation of theGaussian integral

(5.67)∫

Rn

e−t|ξ|2+ix·ξ dξ =(π

t

)n/2

e−|x|2/4t.

The second identity in (5.65) follows by switching to spherical polar coor-dinates, y = rω, and using dy = rn−1 dr dS(ω).

The second formula for et∆ is

(5.68) et∆f(x) =1√2π

∫ ∞

−∞ht(s) cos s

√−∆ f(x) ds,

with ht(σ) = e−tσ2, hence, by (5), ht(s) = (2t)−1/2e−s2/4t. This is a special

case of (5.51).Setting 4t = 1/λ and comparing the formulas (5.65) and (5.68), we have

(with v(s, x) = cos s√−∆f(x))

(5.69)∫ ∞

0

v(s, x)e−λs2ds =

An−1

2

(λ

π

)(n−1)/2∫ ∞

0

fx(r)rn−1e−λr2dr,

for all λ > 0. The key to getting a formula for v(s, x) from this is to makethe factor λ(n−1)/2 on the right side of (5.69) disappear.

Bringing in the hypothesis (5.64), we use the identity

(5.70) − 12r

d

dre−λr2

= λe−λr2

to write the right side of (5.69) as

(5.71) Cn

∫ ∞

0

r2kfx(r)(− 1

2r

d

dr

)k

e−λr2dr.

Repeated integration by parts shows that this is equal to

(5.72) Cn

∫ ∞

0

r( 1

2r

d

dr

)k[r2k−1fx(r)

]e−λr2

dr.


Now it follows from uniqueness of Laplace transforms (see the exercises)that

(5.73) cos t√−∆f(x) = Cnt

( 12t

d

dt

)k[t2k−1fx(t)

],

for well behaved functions f on Rn, when n = 2k + 1. By (5.69), we have

(5.74) Cn =12π−(n−1)/2An−1.

We can also compute Cn directly in (5.73), by considering f = 1. Thenfx = 1 and u = 1, so

(5.75) 1 = Cnt( 1

2t

d

dt

)k

t2k−1 = Cn

(k − 1

2

)(k − 3

2

)· · · 1

2,

i.e.,

(5.76) Cn =1

(k − 12 )(k − 3

2 ) · · · 12

, n = 2k + 1.

This simply means

(5.77) A2k =2πk

(k − 12 )(k − 3

2 ) · · · 12

,

a formula that is frequently derived by looking at Gaussian integrals. Seeformulas (A.4)–(A.9) at the end of this chapter.

To compute (sin t√−∆)/

√−∆, we use

(5.78)sin t

√−∆√−∆g(x) =

∫ t

0

cos s√−∆ g(x) ds.

From (5.73), if k ≥ 1,

(5.79) cos t√−∆ g(x) =

Cn

2d

dt

( 12t

d

dt

)k−1[t2k−1gx(t)

],

so (5.78) becomes

(5.80)sin t

√−∆√−∆g(x) =

Cn

2

( 12t

d

dt

)k−1[t2k−1gx(t)

].

The formulas (5.73) and (5.80) are for t > 0. For arbitrary t ∈ R, use

(5.81) cos t√−∆ = cos(−t)

√−∆, sin t

√−∆ = − sin(−t)

√−∆.

The case k = 0 is exceptional. Then (5.79) does not work. Instead, wehave

(5.82) cos t√−∆ g(x) =

12[g(x + t) + g(x− t)],

56

and (5.78) gives

(5.83)

sin t√−∆√−∆

g(x) =12

∫ t

0

[g(x + s) + g(x− s)] ds

=12

∫ t

−t

g(x + s) ds,

for n = 1.Let us take another look at the case k = 1, when n = 3. From (5.76),

C2 = 2, and then (5.80) gives

(5.84)

sin t√−∆√−∆

g(x) = tgx(t)

=t

4π

∫

S2

g(x + tω) dS(ω)

=1

4πt

∫

|y|=|t|

g(x + y) dS(y),

which is equivalent to (5.47).To solve the wave equation (5.32)–(5.33) for u = u(t, x), t ∈ R, x ∈

Rn, n = 2k, we can use the following device, known as the method ofdescent. Set

(5.85) F (x, xn+1) = f(x), G(x, xn+1) = g(x),

and solve for U = U(t, x, xn+1) the wave equation

(5.86) ∂2t U −∆n+1U = 0, U(0) = F, ∂tU(0) = G.

Then U is independent of xn+1 and

(5.87) u(t, x) = U(t, x, 0).

In particular,

(5.88) cos t√−∆ f(x) = cos t

√−∆n+1F (x, 0),

and

(5.89)sin t

√−∆√−∆g(x) =

sin t√−∆n+1√−∆n+1

G(x, 0).

Using the formula (5.73) for waves on Rn+1 = R2k+1, we have, forv(t, x) = cos t

√−∆f(x),

(5.90) v(t, x) = Cn+1t( 1

2t

d

dt

)k[t2k−1f#

x (t)],


where

(5.91)

f#x (r) = F (x,0)(r)

=1

An

∫

Sn

F ((x, 0) + rω) dS(ω)

=2

An

∫

Sn+

f(x + rωb) dS(ω),

with

(5.92) Sn+ = ω = (ωb, ωn+1) ∈ Sn : ωn+1 ≥ 0.

Here An is the n-dimensional area of Sn and, we recall,

(5.93) Cn+1 =12π−n/2An.

Note that f#x (r) = f#

x (−r). To proceed, map B = y ∈ Rn : |y| ≤ 1 toSn

+ by

(5.94) y 7→ (y, ψ(y)), ψ(y) =√

1− |y|2.Then

(5.95) dS(ω) =√

1 + |∇ψ(y)|2 dy =dy√

1− |y|2 ,

and we get

(5.96) f#x (r) =

2An

∫

|y|≤1

f(x + ry)dy√

1− |y|2 .

Using the identity

(5.97)∫

|y|≤1

h(y) dy =∫ 1

0

∫

Sn−1

H(ρω)ρn−1 dS(ω) dρ,

we get (for r > 0)

(5.98)

f#x (r) =

2An

∫ 1

0

∫

Sn−1

f(x + rρω)ρn−1

√1− ρ2

dS(ω) dρ

= 2An−1

An

∫ 1

0

fx(ρr)ρn−1

√1− ρ2

dρ

= 2An−1

An

1rn−1

∫ r

0

fx(s)sn−1

√r2 − s2

ds.

58

Plugging this in (5.90), we get, for a function f on Rn = R2k, t > 0,

(5.99) cos t√−∆f(x) = 2

An−1

AnCn+1t

( 12t

d

dt

)k∫ t

0

fx(s)sn−1

√t2 − s2

ds.

Note that

(5.100) 2An−1

AnCn+1 = π−n/2An−1 =

2Γ(n/2)

=2

Γ(k).

Similarly, we have

(5.101)sin t

√−∆√−∆g(x) =

Cn+1

2

( 12t

d

dt

)k−1[t2k−1g#

x (t)],

where g#x (|t|) is as in (5.86), with G in place of F , hence as in (5.98), with

gx(s) in place of fx(s). Consequently, for t > 0,

(5.102)sin t

√−∆√−∆g(x) =

An−1

AnCn+1

( 12t

d

dt

)k−1∫ t

0

gx(s)sn−1

√t2 − s2

ds,

and (An−1/An)Cn+1 = 1/Γ(k).If we specialize to n = 2 (k = 1), we get

(5.103)

sin t√−∆√−∆

g(x) =∫ t

0

gx(s)s√

t2 − s2ds

=12π

∫ t

0

∫

S1

g(x− sω)s√

t2 − s2dS(ω) ds

=12π

∫

|y|≤t

g(x− y)√t2 − |y|2 dy,

which is equivalent to (5.46).We turn to a re-derivation of the formulas in Proposition 5.5. As before,

we use (5.51), plus formulas for cos t√−∆. This time, we combine (5.51)

with the formula (5.79), for a function f on Rn = R2k+1, where Cn is givenby (5.74). We get

(5.104)ϕ(√−∆)f(x) =

Cn√2π

∫ ∞

−∞ϕ(t)

( d

dt

12t

)k[tn−1fx(t)

]dt

=Cn√2π

∫ ∞

−∞

(− 1

2t

d

dt

)k

ϕ(t) · tn−1fx(t) dt.


Now

(5.105)

∫ ∞

0

Φ(r)rn−1fx(r) dr

=1

An−1

∫ ∞

0

∫

Sn−1

f(x− rω)Φ(r)rn−1 dS(ω) dr

=1

An−1

∫

Rn

f(x− y)Φ(|y|) dy.

Hence, using (5.74), we obtain from (5.104) that

(5.106) ϕ(√−∆)f(x) =

1√2π

∫

Rn

Φn(|y|)f(x− y) dy,

for n = 2k + 1, where

(5.107) Φ2k+1(r) =(− 1

2πr

d

dr

)k

ϕ(r).

Another way to write (5.106) is

(5.108) ϕ(√−∆)δ(x) =

1√2π

Φn(|x|), x ∈ Rn,

which is equivalent to (5.53).

Remark. The case k = 0 of (5.107) can be seen directly by the Fourierinversion theorem, without use of the calculations (5.104)–(5.105).

We seek an analogous formula for ϕ(√−∆)f(x) when f is a function on

Rn with n = 2k. We get this by the following extension of the method ofdescent, which gives

(5.109) cos t√−∆f(x) = cos t

√−∆n+1F (x, 0),

with

(5.110) F (x, xn+1) = f(x).

From this and (5.51), we get(5.111)

ϕ(√−∆)f(x) = ϕ(

√−∆n+1)F (x, 0)

=1√2π

∫

R2k+1

Φ2k+1

(|(y, yn+1)|)F (x− y, yn+1) dy dyn+1

=1√2π

∫

R2k+1

Φ2k+1

((|y|2 + s2)1/2

)f(x− y) dy ds,

60

or

(5.112) ϕ(√−∆)f(x) =

1√2π

∫

Rn

Φ2k(|y|)f(x− y) dy,

where

(5.113) Φ2k(r) =∫ ∞

−∞Φ2k+1(

√r2 + s2) ds,

and Φ2k+1 is given by (5.107). The change of variable t =√

r2 + s2 gives

(5.114) Φ2k(r) = 2∫ ∞

r

Φ2k+1(t)t√

t2 − r2dt.

Another way to write (5.112) is

(5.115) ϕ(√−∆)δ(x) =

√2π

Φ2k(|x|),

which is equivalent to (5.54).

Exercises

1. A function g ∈ C∞(Rn) is said to be in the Gevrey class Gσ(Rn) provided,for each compact K ⊂ Rn, there exist C and R, such that

(5.116) |Dαg(x)| ≤ CRkkσk, |α| = k, x ∈ K.

The class G1(Rn) is equal to the space of real-analytic functions on Rn. Ifσ > 1, show that there exist compactly supported elements of Gσ(Rn), notidentically zero.Remark: This is part of the Denjoy-Carleman theorem; see [Ru].

2. Consider the following “sideways heat equation” for u = u(t, x) on R×R:

ut = uxx, u(t, 0) = g(t), ux(t, 0) = 0.

Show that if g ∈ Gσ(R) for some σ ∈ (1, 2), then a solution on all of R×R isgiven by the convergent series

(5.117) u(t, x) =

∞∑

k=0

x2k

(2k)!g(k)(t),

which is the power series for the “formal” object(cos x

√−∂/∂t

)g. Using

Exercise 1, find nontrivial solutions to ut = uxx which are supported in a stripa ≤ t ≤ b. This construction is due to J. Rauch.(Hint: To prove convergence, use Stirling’s formula:

(5.118) n! ∼ (2πn)1/2 · e−n · nn as n →∞.)

3. Given the formula (5.37) for R(t, ξ), that is,

R(t, ξ) = (2π)−n/2 |ξ|−1 sin t|ξ|, ξ ∈ Rn,

6. Radial distributions, polar coordinates, and Bessel functions 61

show that the fact that R(t, ·) ∈ S ′(Rn) is supported in B|t| = x ∈ Rn : |x| ≤|t| follows from the Paley-Wiener theorem for distributions, given in Exercise16 of §4.

Exercises 4–7 provide justification for passing from (5.69)–(5.71) to (5.73).4. Take v ∈ L1(R+) and assume

∫ ∞

0

v(s)e−λs2ds = 0, ∀λ > 0.

Deduce that v ≡ 0. (Hint. Use the Stone-Weierstrass theorem to show that if

eλ(s) = e−λs2, then the linear span of eλ : λ > 0 is dense in C∗(R

+), the

space of continuous functions on R+= [0,∞) that vanish at ∞. Hence the

hypothesis implies∫∞0

v(s)f(s) = 0 for all f ∈ C∗(R+).)

5. Show that if instead we assume v ∈ L∞(R+), the result of Exercise 4 stillholds. (Hint. Consider vε(s) = e−εs2

v(s).)6. Show that if f, g ∈ S(Rn), then the solution u to (5.32)–(5.33) is bounded and

continuous on R×Rn. Hence deduce the validity of (5.73) for f ∈ S(Rn).7. Extend the range of validity of (5.73) from f ∈ S(Rn) to other function spaces,

including f ∈ C∞(Rn).

6. Radial distributions, polar coordinates, and Besselfunctions

The rotational invariance of the Laplace operator on Rn directly suggeststhe use of polar coordinates; one has

(6.1) ∆ =∂2

∂r2+

n− 1r

∂

∂r+

1r2

∆S ,

where ∆S is the Laplace operator on the unit sphere Sn−1. This formulahas been used in (4.56) and follows from the formula given in Chapter 2for the Laplace operator in a general coordinate system; see (4.4) in thatchapter.

Related is the fact that in treating the equations of §5 via Fourier anal-ysis, one computes the Fourier transforms of various radial functions, suchFourier transforms also being radial functions (or rotationally invarianttempered distributions). Bessel functions arise naturally in either ap-proach, and we will develop a little of the theory of Bessel functions here.More results on Bessel functions will appear in Chapter 8, which discussesspectral theory, and in Chapter 9, which treats scattering theory. One canfind a great deal more material on this subject in the treatise [Wat].

62

We begin by considering the Fourier transform of a radial function,F (x) = f(r), r = |x|. We have

(6.2) F (ξ) = (2π)−n/2

∫ ∞

0

f(r)ψn(r|ξ|)rn−1 dr,

where

(6.3) ψn(|ξ|) = Ψn(ξ)

with

(6.4) Ψn(ξ) =∫

Sn−1

eiξ·ω dS(ω).

In other words, with An−2 the volume of Sn−2,

(6.5) ψn(r) = An−2

∫ 1

−1

eirs(1− s2)(n−3)/2 ds.

From (A.4) we have An−2 = 2π(n−1)/2/

Γ((n − 1)/2

). It is common to

write

(6.6) ψn(r) = (2π)n/2r1−n/2Jn/2−1(r),

where, for general ν satisfying Re ν > −1/2, the Bessel function Jν(z) isdefined to be

(6.7) Jν(z) =[Γ(1

2

)Γ(ν +

12

)]−1(z

2

)ν∫ 1

−1

(1− t2)ν−1/2eizt dt.

For example, ψ2(r) = 2πJ0(r). Since (1 − t2)ν−1/2 is even in t, one canreplace eizt by cos zt in this formula. Now, (6.2) becomes

(6.8) F (ξ) = |ξ|1−n/2

∫ ∞

0

f(r)Jn/2−1(r|ξ|)rn/2 dr.

We want to consider the ODE, known as Bessel’s equation, solved byJν(r). First we consider the case ν = n/2 − 1. Since Ψn is the Fouriertransform of a measure supported on the unit sphere, we have that

(6.9) (∆ + 1)Ψn = 0.

Using the polar coordinate expression (6.1) for ∆, we have

(6.10)( d2

dr2+

n− 1r

d

dr+ 1

)ψn(r) = 0.

Substituting (6.6) yields Bessel’s equation

(6.11)[ d2

dr2+

1r

d

dr+

(1− ν2

r2

)]Jν(r) = 0

in case ν = n/2 − 1. We want to verify this for all ν, from the integralformula (6.7). This is an exercise, but we will present the details to one


approach, which yields further interesting identities for Bessel functions.For notational simplicity, let us set

c(ν) =[Γ(1

2

)Γ(ν +

12

)]−1

· 2−ν .

Differentiating (6.7) with respect to z yields

(6.12)

d

dzJν(z) =

(νc(ν)z

)zν

∫ 1

−1

eizt(1− t2)ν−1/2 dt

+ ic(ν)zν

∫ 1

−1

eiztt(1− t2)ν−1/2 dt.

The first term on the right is equal to (ν/z)Jν(z). The second is equal to

c(ν)z

zν

∫ 1

−1

( d

dteizt

)t(1− t2)ν−1/2 dt

= −c(ν)z

zν

∫ 1

−1

eizt[(1− t2)ν−1/2 − (2ν − 1)t2(1− t2)ν−1−1/2

]dt

= −c(ν)z

zν

∫ 1

−1

eizt[2ν(1− t2)ν−1/2 − (2ν − 1)(1− t2)ν−1−1/2

]dt

= −2ν

zJν(z) +

(2ν − 1)c(ν)c(ν − 1)

Jν−1(z).

Since c(ν)/c(ν − 1) = 1/(2ν − 1), we have the formula

(6.13)d

dzJν(z) = −ν

zJν(z) + Jν−1(z),

or

(6.14)( d

dz+

ν

z

)Jν(z) = Jν−1(z).

As we have stated, the formula (6.7) for Jν(z) is convergent for Re ν >−1/2. The formula (6.14) provides an analytic continuation for all complexν. In fact, one can see directly that the integral in (6.7) is meromorphicin ν, with simple poles at ν + 1/2 = −1,−2, . . . . The factor Γ(ν + 1/2)−1

cancels these poles. This serves to explain the desirability of throwing inthis factor in the definition (6.7) of Jν(z). Of course, the factor Γ(1/2)−1

is more arbitrary.Next, we note that

Jν+1(z) = c(ν + 1)zν+1

∫ 1

−1

eizt(1− t2)ν+1−1/2 dt

= −ic(ν + 1)zν

∫ 1

−1

( d

dteizt

)(1− t2)ν+1−1/2 dt

= −ic(ν + 1)(2ν + 1)zν

∫ 1

−1

eiztt(1− t2)ν−1/2 dt,

64

and since c(ν + 1) = c(ν)/(2ν + 1), this is equal to the negative of thesecond term on the right in (6.12). Hence we have

(6.15)( d

dz− ν

z

)Jν(z) = −Jν+1(z),

complementing (6.14). Putting together (6.14) and (6.15), we have

(6.16)( d

dz− ν − 1

z

)( d

dz+

ν

z

)Jν(z) = −Jν(z),

which is equivalent to Bessel’s equation, (6.11). Note that adding andsubtracting (6.14) and (6.15) produce the identities

(6.17)2J ′ν(z) = Jν−1(z)− Jν+1(z),

2ν

zJν(z) = Jν−1(z) + Jν+1(z).

Note that, by analytic continuation, J−ν(z) is also a solution to Bessel’sequation. This equation, for each ν, has a two-dimensional solution space.We will examine when Jν(z) and J−ν(z) are linearly independent. First,we will obtain a power-series expansion for Jν(z). This is done by replacingeitz by its power-series expansion in (6.7). To simplify the expression forthe coefficients, one uses identities for the beta function and the gammafunction established in Appendix A. From (6.7), we have(6.18)

Jν(z) =[Γ(1

2

)Γ(ν +

12

)]−1(z

2

)ν ∞∑

k=0

1(2k)!

∫ 1

−1

(izt)2k(1− t2)ν−1/2 dt.

The identity (A.24) implies∫ 1

−1

t2k(1− t2)ν−1/2 dt =Γ(k + 1

2 )Γ(ν + 12 )

Γ(k + ν + 1),

so

Jν(z) =(z/2)ν

Γ( 12 )Γ(ν + 1

2 )

∑ 1(2k)!

(iz)2k Γ(k + 12 )Γ(ν + 1

2 )Γ(k + ν + 1)

.

Setting (2k)! = Γ(2k +1), and using the duplication formula (A.22), whichimplies

Γ(k + 12 )

Γ( 12 )Γ(2k + 1)

=2−2k

Γ(k + 1),

we obtain the formula

(6.19) Jν(z) =(z

2

)ν ∞∑

k=0

(−1)k

k!Γ(k + ν + 1)

(z

2

)2k

.


This follows from (6.18) if Re ν > −1/2, and then for general ν by analyticcontinuation. In particular, we note the leading behavior as z → 0,

(6.20) Jν(z) =zν

2νΓ(ν + 1)+ O(zν+1)

and

(6.21) J ′ν(z) =zν−1

2νΓ(ν)+ O(zν).

The leading coefficients are nonzero as long as ν is not a negative integer(or 0, for (6.21)).

From the expression (6.19) it is clear that Jν(z) and J−ν(z) are linearlyindependent provided ν is not an integer. On the other hand, comparisonof power series shows

(6.22) J−n(z) = (−1)nJn(z), n = 0, 1, 2, . . . .

We want to construct a basis of solutions to Bessel’s equation, uniformlygood for all ν. This construction can be motivated by a calculation of theWronskian.

Generally, for a pair of solutions u1 and u2 to a second-order ODE

a(z)u′′ + b(z)u′ + c(z)u = 0,

u1 and u2 are linearly independent if and only if their Wronskian

(6.23) W (z) = W (u1, u2)(z) = u1u′2 − u2u

′1

is nonvanishing. Note that the Wronskian satisfies the first-order ODE

(6.24) W ′(z) = − b

aW (z).

In the case of Bessel’s equation, (6.11), this becomes

(6.25) W ′(z) = −W (z)z

,

so

(6.26) W (z) =K

z,

for some K (independent of z, but perhaps depending on ν). If u1 =Jν , u2 = J−ν , we can compute K by considering the limiting behavior ofW (z) as z → 0. From (6.20) and (6.21), we get(6.27)

W (Jν , J−ν)(z) = −[ 1Γ(ν)Γ(1− ν)

− 1Γ(ν + 1)Γ(−ν)

]1z

= −2sin πν

πz,

making use of the identity (A.10). This recaptures the observation that Jν

and J−ν are linearly independent, and consequently a basis of solutions to(6.11), if and only if ν is not an integer.

66

To construct a basis of solutions uniformly good for all ν, it is naturalto set

(6.28) Yν(z) =Jν(z) cos πν − J−ν(z)

sin πν

when ν is not an integer, and define

(6.29) Yn(z) = limν→n

Yν(z) =1π

[∂Jν(z)∂ν

− (−1)n ∂J−ν(z)∂ν

]∣∣∣ν=n

.

We have

(6.30) W (Jν , Yν)(z) =2πz

,

for all ν. Another important pair of solutions to Bessel’s equation is thepair of Hankel functions

(6.31) H(1)ν (z) = Jν(z) + iYν(z), H(2)

ν (z) = Jν(z)− iYν(z).

For H(1)ν , there is the integral formula

(6.32) H(1)ν (z) =

2e−πiν

i√

πΓ(ν + 12 )

(z

2

)ν∫ ∞

1

eizt(t2 − 1)ν−1/2 dt,

for Re ν > −1/2, Im z > 0. Another formula, valid for Re ν > 12 and Re

z > 0, is(6.33)

H(1)ν (z) =

( 2πz

)1/2 ei(z−πν/2−π/4)

Γ(ν + 12 )

∫ ∞

0

e−ssν−1/2(1− s

2iz

)ν−1/2

ds.

To prove these identities, one can show as above that each of the rightsides of (6.32) and (6.33) satisfies the same recursion formulas as Jν(z)and hence solves the Bessel equation; thus it is a linear combination ofJν(z) and Yν(z). The coefficients can be found by examining the limitingbehavior as z → 0, to establish the asserted identity. Hankel functions areimportant in scattering theory; see Chapter 9.

It is worth pointing out that the Bessel functions Jk+1/2(z), etc., for k aninteger, are elementary functions, particularly since they arise in analysison odd-dimensional Euclidean space. For ν = k + 1/2, the integrand in(6.7) involves (1−t2)k, so the integral can be evaluated explicitly. We have,in particular,

(6.34) J1/2(z) =( 2

πz

)1/2

sin z.

Then (6.14) gives

(6.35) J−1/2(z) =( 2

πz

)1/2

cos z,


which by (6.28) is equal to −Y1/2(z). Applying (6.16) and (6.14) repeatedlygives

(6.36) Jk+1/2(z) = (−1)k k∏

j=1

( d

dz− j − 1

2

z

) sin z√2πz

and the same sort of formula for J−k−1/2(z), with the (−1)k removed, andsin z replaced by cos z. Similarly,

(6.37) H(1)1/2(z) = −i

( 2πz

)1/2

eiz,

with a formula for H(1)k+1/2(z) similar to (6.36).

We now make contact between the formulas (6.2)–(6.6) and some of theformulas of §5, particularly from Proposition 5.5. Note that if F (x) =f(|x|), then

(6.38) F (x) = (2π)n/2f(√−∆)δ.

Thus, as in (5.51), we have

(6.39) F (x) = (2π)n/2−1

∫∫f(r)eitrR′n(t, x) dt dr,

where R′n(t, x) = (∂/∂t)Rn(t, x), and Rn(t, x) is the Riemann functiongiven by (5.43). Comparison with (6.2) gives

ψn(r|x|) = (2π)n−1r1−n

∫ ∞

−∞eitrR′n(t, x) dt

or, equivalently,

(6.40) Jn/2−1(r|x|) = 2(2π)n/2−1( |x|

r

)n/2−1∫ ∞

−∞(sin tr)Rn(t, x) dt.

Note that using R3(t, x) = (4πt)−1δ(|t|−|x|) gives again the formula (6.34)for J1/2(r). Note also that the recursive formula

Rn+2(t, s) = − 12πs

∂

∂sRn(t, s)

used in the proof of Proposition 5.5, when substituted into (6.40), givesrise to the formula (6.15), in the case ν = n/2− 1.

Instead of synthesizing functions of ∆ via the formula (5.51), we coulduse

(6.41) g(−∆) = (2π)−1/2

∫g(t)e−it∆ dt,

where the operator e−it∆ is obtained from the solution operator et∆ to theheat equation by analytic continuation:

(6.42) e−it∆δ(x) = (−4πit)−n/2e|x|2/4it, t 6= 0.

68

If f(r) = g(r2), with g real-valued and even, we get

(6.43) g(t) =4√2π

∫ ∞

0

f(r)(cos r2t

)r dr

and hence

(6.44) g(−∆)δ =2π

∫ ∞

−∞

∫ ∞

0

(−4πit)−n/2(cos r2t

)e|x|

2/4itf(r)r dr dt.

Comparison of this with (6.2) gives

(6.45) ψn(r|x|) = cnr2−n

∫ ∞

−∞

(cos r2t

)e|x|

2/4itt−n/2 dt,

where, for n odd, we take t−n/2 = limε0(t− iε)−n/2. Note that (6.45) isan improper integral near t = 0.

We will not look in detail at implications of (6.41)–(6.44), which aregenerally not as incisive as those of Proposition 5.5, but we will brieflymake a connection with the idea, used in §5, of synthesizing operatorsfrom the heat semigroup. Recall particularly the formula for the resolvent:

(6.46) (λ2 −∆)−1 =∫ ∞

0

e−(λ2−∆)t dt.

Generalizing (5.28)–(5.31), we have, for λ > 0,

(6.47) (λ2 −∆)−1δ =∫ ∞

0

e−|x|2/4t−λ2t(4πt)−n/2 dt.

A superficial resemblance with (6.45) suggests that this function is relatedto Bessel functions. This is consistent with the fact that the resolventkernel Rλ(x) = (λ2 −∆)−1δ = Rb

n,λ(|x|) satisfies the ODE

(6.48)[ d2

dr2+

n− 1r

− λ2]Rb

n,λ(r) = 0, r > 0,

as a consequence of the formula (6.1) for the Laplace operator in polarcoordinates; this is similar to (6.10), with 1 replaced by −λ2. In fact, thereis the following result. From (6.47),

(6.49) (λ2 −∆)−1δ = (2π)−n/2( |x|

λ

)1−n/2

Kn/2−1(λ|x|), x ∈ Rn \ 0,

where Kν(r) is defined by

(6.50) Kν(r) =12

(r

2

)ν∫ ∞

0

e−r2/4t−t t−1−ν dt.

Exercises 69

Simple manipulations of (6.50) produce the following analogues of (6.14)–(6.16):

(6.51)

( d

dr− ν

r

)Kν(r) = −Kν+1(r),

Kν+1(r)−Kν−1(r) =2ν

rKν(r),

so we have the ODE

(6.52)[ d2

dr2+

1r

d

dr−

(1 +

ν2

r2

)]Kν(r) = 0, r > 0,

which differs from Bessel’s equation (6.11), only in one sign. The ODE(6.52) is solved by Jν(ir) and by Yν(ir), so Kν(r) must be a linear combi-nation of these functions. In fact,

(6.53) Kν(r) =12πieπiν/2 H(1)

ν (ir), r > 0.

A proof of (6.53) can be found in [Leb], Chapter 5; see also Exercise 4below. When ν = 1/2, (6.53) follows from (6.33) and (6.34) together withthe identity

(6.54) K1/2(r) =( π

2r

)1/2

e−r,

which in turn, given (6.50), follows from the subordination identity. Thenthe recursion relation (6.51) and analogues for the Bessel functions imply(6.53) for ν of the form ν = k + 1/2, when k is a positive integer.

We mention that it is customary to take a second, linearly independent,solution to (6.52) to be

(6.55) Iν(r) = e−πiν/2Jν(ir), r > 0.

The functions Kν(r) and Iν(r) are called modified Bessel functions; alsoKν(r) is sometimes called MacDonald’s function.

Exercises

1. Using the integral formula (6.7), show that, for fixed ν > −1/2, as z → +∞,

(6.56) Jν(z) =

(2

πz

)1/2

cos

(z − νπ

2− π

4

)+ O(z−3/2).

(Hint: The endpoint contributions from the integral give exponentials timesFourier transforms of functions with simple singularities at the origin.)Reconsider this problem after reading §§7 and 8.

Similarly, using (6.33), show that, for fixed ν > −1/2, as z → +∞,

(6.57) H(1)ν (z) =

(2

πz

)1/2

ei(z−νπ/2−π/4) + O(z−3/2).

70

2. Using the integral formula (6.50), show that, for fixed ν, as r → +∞,

Kν(r) =

(π

2r

)1/2

e−r[1 + O(r−1)

].

(Hint: Use the Laplace asymptotic method, such as applied to the gammafunction in the appendix to this chapter; compare (A.34)–(A.39). To imple-ment this, rewrite (6.50) as

Kν(r) = 2−1−νr−1

∫ ∞

0

e−r(s+1/4s)s−1−ν ds.

Note that ϕ(s) = s + 1/4s has its minimum at s = 1/2.)3. Using the definition (6.55) for Iν(r), and plugging z = ir into the integral

formula (6.7) for Jν(z), show that, for fixed ν > −1/2, as r → +∞,

Iν(r) =

(1

2πr

)1/2

er[1 + O(r−1)

].

4. Show that, for r > 0,

Kν(r) =1

Γ( 12)Γ(ν + 1

2)

(r

2

)ν ∫ ∞

1

e−rt(t2 − 1)ν−1/2 dt,

by showing that the function on the right solves the modified Bessel equation(6.52) and has the same asymptotic behavior as r → +∞ as Kν(r) does,according to Exercise 2. Hence establish the identity (6.53).

5. For y, a > 0, ξ ∈ Rn, consider

F (ξ) = e−y(|ξ|2+a2)1/2.

Applying the subordination identity (5.22) to A2 = |ξ|2 + a2, and takingFourier transforms of both sides of the resulting identity, show that

F (x) = cny

∫ ∞

0

e−(y2+|x|2)/4te−ta2t−(n+3)/2 dt

= c′n y aν r−ν Kν(ar),

with ν = (n + 1)/2, r2 = |x|2 + y2.6. Using analytic continuation involving both y and a, find an expression for the

fundamental solution to

utt + 2aut −∆u = 0,

for u = u(t, x), t ∈ R, x ∈ Rn, where a is a real number. Be explicit in thecase n = 2, using the elementary character of K3/2(z).

7. Show that, under the change of variable u(r) = rαf(cr), Bessel’s equation(6.11), u′′(r) + (1/r)u′(r) + (1− ν2/r2)u(r) = 0, is transformed to

(6.58) f ′′(r) +A

rf ′(r) +

(µ2 − λ2

r2

)f(r) = 0,

with

A = 2α + 1, µ = c−1, λ2 = ν2 − α2.

8. Suppose in particular that v(x) = f(r)w(θ), θ ∈ O ⊂ Sn−1, and ∆Sw =−λ2w. Show that the equation ∆v = −µ2v is equivalent to (6.58), with

7. The method of images and Poisson’s summation formula 71

A = n − 1, so α = n/2 − 1. Thus f is a linear combination of r1−n/2Jν(µr)

and r1−n/2H(1)ν (µr), with ν =

[λ2 + (n− 2)2/4

]1/2

.

9. Show that, complementary to (6.20), we have, for ν > 0,

H(1)ν (z) ∼ −i

Γ(ν)

π

(2

z

)ν

, z 0.

7. The method of images and Poisson’s summation formula

We discuss here techniques for solving such problems as the Dirichlet prob-lem for the heat equation on a rectangular solid in Rn, defined by

(7.1) Ω = x ∈ Rn : 0 ≤ xj ≤ aj , 1 ≤ j ≤ n.That is, we want to solve

(7.2)∂u

∂t−∆u = 0,

for u = u(t, x), t ≥ 0, x ∈ Ω, subject to the boundary condition

(7.2) u∣∣R+×∂Ω

= 0

and the initial condition

(7.4) u(0, x) = f(x), x ∈ Ω.

There are two ways of doing this. One involves using Fourier series onthe torus Rn/2Γ, where Γ is the lattice in Rn generated by ajej (ej beingthe standard basis of Rn). The other is to use the solution on R+ × Rn

constructed in §5 together with the method of images, described below.Comparing these methods provides interesting analytical identities.

The method of images works as follows. Let u# solve the heat equation

(7.5)∂u#

∂t−∆u# = 0 on R+ × Rn,

with initial data

(7.6) u#(0, x) = f#(x),

where f# = f on Ω and f# is odd with respect to reflections across the wallsof all the translates of Ω by elements of Γ. The set of such translates is a setof rectangles tiling Rn, and f# is uniquely determined by this prescription.Since reflections are isometries, it follows that, for each t > 0, u#(t, ·) isodd with respect to such reflections; since u# is smooth for t > 0, it musttherefore vanish on all these walls. The restriction of u#(t, x) to R+×Ω ishence the desired solution to (7.2)–(7.4).

72

The same sort of technique works for the wave equation on R× Ω,

(7.7)∂2u

∂t2−∆u = 0 on R× Ω,

with Dirichlet boundary condition

(7.8) u(t, x) = 0, for x ∈ ∂Ω,

and initial condition

(7.9) u(0, x) = f(x), ut(0, x) = g(x) on Ω.

One takes odd extensions f#, g#, as above.One can apply the method of images to regions other than rectangular

solids. It applies when Ω is a half-space, for example; in that case, onlyone reflection, across the hyperplane ∂Ω, is involved. Similarly, one cantreat slabs, bounded by parallel hyperplanes, quadrants, and so on. Onecan also treat different boundary conditions. If one extends f above to beeven with respect to these reflections, one obtains solutions with Neumannboundary condition satisfied on ∂Ω.

Another type of boundary condition to impose is a periodic boundarycondition:

(7.10) u(t, x) = u(t, x + γ) if x, x + γ ∈ ∂Ω, γ ∈ Γ.

The solution to (7.1), (7.4), (7.10) is obtained as follows. Let f0(x) = f(x)for x ∈ Ω, let f0(x) = 0 for x /∈ Ω, set

(7.11) f b(x) =∑

γ∈Γ

f0(x + γ),

and let ub(t, x) be the solution to

(7.12)∂ub

∂t−∆ub = 0 on R+ × Rn, ub(0, x) = f b(x).

Note that if u0(t, x) is defined by

(7.13)∂u0

∂t−∆u0 = 0 on R+ × Rn, u0(0, x) = f0(x),

then

(7.14) ub(t, x) =∑

γ∈Γ

u0(t, x + γ).

In this case, u(t, x) is the restriction of ub(t, x) to R+ × Ω.Let us specialize to the case Γ = (2πZ)n, f = δ. We have the funda-

mental solution, satisfying periodic boundary conditions, given by

(7.15) H(t, x) = (4πt)−n/2∑

k∈Zn

e−|x+2πk|2/4t.

7. The method of images and Poisson’s summation formula 73

On the other hand, identifying Rn/(2πZ)n with Tn, we obtain via Fourierseries

(7.16) H(t, x) = (2π)−n∑

`∈Zn

e−t|`|2+i`·x.

Comparing these formulas gives the following important case of Poisson’ssummation formula:

(7.17)∑

k∈Zn

e−|x+2πk|2/4t =( t

π

)n/2 ∑

`∈Zn

ei`·x−|`|2t.

We now show how the special case of this for n = 1 implies the famousfunctional equation for the Riemann zeta function. With n = 1, x = 0,and t = π/τ , (7.17) yields the identity

(7.18)∞∑

n=−∞e−n2πτ =

(1τ

)1/2 ∞∑n=−∞

e−πn2/τ .

In other words, with g1(τ) denoting the left side of (7.18), we have g1(τ) =τ−1/2g1(1/τ). This is a transformation formula of Jacobi. It follows thatif

(7.19) g(t) =∞∑

n=1

e−n2πt,

then

(7.20) g(t) = −12

+12t−1/2 + t−1/2g(t−1).

Now (7.19) is related to the Riemann zeta function

(7.21) ζ(s) =∞∑

n=1

n−s (Re s > 1)

via the Mellin transform, discussed briefly in Appendix A, at the end ofthis chapter. Indeed, we have

(7.22)

∫ ∞

0

g(t)ts−1 dt =∫ ∞

0

∞∑n=1

e−n2πt ts−1 dt

=∞∑

n=1

n−2sπ−s

∫ ∞

0

e−tts−1 dt

= ζ(2s) π−s Γ(s).

Consequently, for Re s > 1,

(7.23)Γ(s

2

)π−s/2ζ(s) =

∫ ∞

0

g(t)ts/2−1 dt

=∫ 1

0

g(t)ts/2−1 dt +∫ ∞

1

g(t)ts/2−1 dt.

74

Now, into the integral over [0, 1], substitute the right side of (7.20) for g(t),to obtain

(7.24)Γ(s

2

)π−s/2ζ(s) =

∫ 1

0

(−1

2+

12t−1/2

)ts/2−1 dt

+∫ 1

0

g(t−1)ts/2−3/2 dt +∫ ∞

1

g(t)ts/2−1 dt.

We evaluate the first integral on the right, and replace t by 1/t in thesecond integral, to obtain, for Re s > 1,

(7.25) Γ(s

2

)π−s/2ζ(s) =

1s− 1

− 1s

+∫ ∞

1

[ts/2 + t(1−s)/2

]g(t)t−1 dt.

Note that g(t) ≤ Ce−πt for t ∈ [1,∞), so the integral on the right is anentire analytic function of s. Since 1/Γ(s/2) is entire, with simple zeros ats = 0,−2,−4, . . . , as shown in Appendix A at the end of this chapter, thisimplies that ζ(s) is continued as a meromorphic function on C, with onesimple pole, at s = 1. The punch line is this: The right side of (7.25) isinvariant under replacing s by 1 − s. Thus we have Riemann’s functionalequation

(7.26) Γ(s

2

)π−s/2ζ(s) = Γ

(1− s

2

)π−(1−s)/2ζ(1− s).

The functional equation is often written in an alternative form, obtainedby multiplying both sides by Γ

((1 + s)/2

), and using the identities

(7.27)Γ(1− s

2

)Γ(1 + s

2

)=

π

sin 12π(1− s)

,

Γ(s

2

)Γ(1 + s

2

)= 2−s+1π1/2Γ(s),

which follow from (A.10) and (A.22). We obtain

(7.28) ζ(1− s) = 21−sπ−s(cos

πs

2

)Γ(s)ζ(s).

Exercises

1. Apply the method of images to find the solution to the heat equation on ahalf-line:

∂u

∂t= uxx, t ≥ 0, x ≥ 0,

u(0, x) = f(x), u(t, 0) = 0.

2. Similarly, treat the wave equation on a half-space:

utt −∆u = 0, t ∈ R, x1 > 0,

u(0, x) = f(x), ut(0, x) = g(x), u(t, 0, x′) = 0,

8. Homogeneous distributions and principal value distributions 75

where x = (x1, x′) ∈ Rn.

3. Given u ∈ S(Rn), define f ∈ C∞(Tn) by f(x) =∑

ν∈Zn u(x + 2πν). Showthat, for ` ∈ Zn, we have f(`) = (2π)−n/2u(`) and hence

(7.29)∑

k

u(x + 2πk) = (2π)−n/2∑

`

u(`)ei`·x.

Show that this generalizes the identity (7.17).4. Let (a`) be polynomially bounded, and consider v =

∑`∈Zn a`e

i`·x, picturedas a 2πZn-periodic (tempered) distribution on Rn rather than as a distributionon Tn; v ∈ S ′(Rn). Show that

(7.30) v = (2π)n/2∑

`∈Zn

a` δ` ∈ S ′(Rn).

Relate this to the result in Exercise 3.5. Show that ζ(s) satisfies the identity

ζ(s) =∏p

(1− p−s

)−1

, Re s > 1,

the product taken over all the primes. This is known as the Euler productformula.

8. Homogeneous distributions and principal valuedistributions

Recall from §4 that the fundamental solution of the Laplace operator ∆on Rn is cn|x|2−n (if n ≥ 3), which is homogeneous. It is useful to con-sider homogeneous distributions in general. The notion of homogeneity isdetermined by the action of the group of dilations,

(8.1) D(t)f(x) = f(tx), t > 0.

Note that D(t) : S(Rn) → S(Rn). Also, if f, g ∈ S(Rn), a change ofvariable gives

(8.2)∫

g(x) D(t)f(x) dx = t−n

∫f(x) D(t−1)g(x) dx.

Thus we can define

(8.3) D(t) : S ′(Rn) −→ S ′(Rn)

by

(8.4) 〈f, D(t)u〉 = t−n〈D(t−1)f, u〉,for f ∈ S(Rn), u ∈ S ′(Rn). We say that u ∈ S ′(Rn) is homogeneous ofdegree m if

(8.5) D(t)u = tm u, for all t > 0.

76

Here, m can be any complex number. Let us denote the space of elementsof S ′(Rn) which are homogeneous of degree m by Hm(Rn). It is easy tosee that if F is the Fourier transform, then

(8.6) FD(t) = t−nD(t−1)F ,

so

(8.7) F : Hm(Rn) −→ H−m−n(Rn).

Before we delve any further into Hm(Rn), we should aver that one’s realinterest is in elements of Hm(Rn) which are smooth outside the origin, sowe consider

(8.8) H#m(Rn) = u ∈ Hm(Rn) : u ∈ C∞(Rn \ 0).

It is easy to see that

(8.9) u ∈ H#m(Rn) ⇒ Dαu ∈ H#

m−|α|(Rn) and xαu ∈ H#

m+|α|(Rn).

We claim (8.7) can be strengthened as follows.


(8.10) F : H#m −→ H#

−m−n(Rn).

The only point left to prove is that if u ∈ H#m(Rn), then u is smooth

on Rn \ 0. Taking ϕ ∈ C∞0 (Rn), ϕ(x) = 1 for |x| ≤ 1, we can writeu = ϕu + (1 − ϕ)u = u1 + u2 with u1 ∈ E ′(Rn) and u2 ∈ C∞(Rn),homogeneous for |x| large. We know that u1 ∈ C∞(Rn), so it suffices toshow that u2 ∈ C∞(Rn\0). This is a special case of the following importantresult.

For m ∈ R, we define the class Sm1 (Rn) of C∞-functions by

(8.11) p ∈ Sm1 (Rn) ⇐⇒ |Dα

x p(x)| ≤ Cα〈x〉m−|α|, for all α ≥ 0.

Clearly, Sm1 (Rn) ⊂ S ′(Rn). It is also clear that u2 ∈ SRe m

1 (Rn), so theproof of Proposition 8.1 is finished once we establish the following.

Proposition 8.2. If p ∈ Sm1 (Rn), then p ∈ C∞(Rn \ 0). Also, if ϕ ∈

C∞0 (Rn), and ϕ(x) = 1 for |x| < a (a > 0), then (1− ϕ)p ∈ S(Rn).

Proof. We will show that if β is large, then xβ p is bounded and continuous,and so are lots of derivatives, which will suffice. Clearly,

F : Sµ1 (Rn) −→ L∞(Rn) ∩ C(Rn), for µ < −n.

Now, given p ∈ Sm1 (Rn), then Dβp ∈ S

m−|β|1 (Rn), so

xβ p = F(Dβp) ∈ L∞ ∩ C, for |β| > m + n,


and more generally xαDβp ∈ Sm−|β|+|α|1 (Rn), so

(8.12) Dα(xβ p) = F(xαDβp) ∈ L∞ ∩ C, for |β| > m + n + |α|.This proves Proposition 8.2.

Generally, there is going to be a singularity at the origin for an elementof H#

m(Rn). In fact, there is the following result, whose proof we leave asan exercise.

Proposition 8.3. If there is a nonzero u ∈ H#m(Rn)∩C∞(Rn), then m is

a nonnegative integer and u is a homogeneous polynomial.

Let us consider other examples of homogeneous distributions. It is easyto see from the definition (8.4) of the action of D(t) that

(8.13) δ ∈ H#−n(Rn).

Of course, δ is zero on Rn \ 0! Since Fδ = (2π)−n/2 ∈ H#0 (Rn), this result

is consistent with Proposition 8.1. For more examples, choose any

(8.14) w ∈ C∞(Sn−1),

and consider, for any m ∈ C,

(8.15) um(x) = |x|m w(|x|−1x), x ∈ Rn \ 0.

If Re m > −n, then um ∈ L1loc(Rn), so it defines in a natural manner an

element of S ′(Rn), which belongs to H#m(Rn). Thus

(8.16) Dαum ∈ H#m−|α|(R

n) (Re m > −n).

If Re m ≤ −n, then um /∈ L1loc(Rn). In the borderline case m = −n, it is

significant that there is a natural identification of um with an element ofS ′(Rn), under the further condition that

(8.17)∫

Sn−1

w(x) dS = 0.

The element of S ′(Rn) is called a principal value distribution and is denotedPV um. We establish this as follows. Pick any radial ϕ ∈ S(Rn) such thatϕ(0) = 1, such as ϕ(x) = e−|x|

2. Then, for any v ∈ S(Rn), with u−n as in

(8.15), u−n(x)[v(x)− v(0)ϕ(x)

]belongs to L1(Rn), so we can define

(8.18) 〈v, PV u−n〉 =∫

Rn

u−n(x)[v(x)− v(0)ϕ(x)

]dx.

Note that (8.17) is precisely what is required to guarantee that the rightside of (8.18) is independent of the choice of ϕ (satisfying the conditions

78

given above). Thus we can write, for any t > 0,

(8.19)

〈D(t)v, PV u−n〉 =∫

Rn

u−n(x)[v(tx)− v(0)ϕ(tx)

]dx

= t−n

∫

Rn

u−n(x/t)[v(x)− v(0)ϕ(x)

]dx

= 〈v, PV u−n〉.In light of (8.4), this implies

(8.20) PV u−n ∈ H#−n(Rn),

provided (8.17) holds. By Proposition 8.1, we have

(8.21) F(PV u−n) ∈ H#0 (Rn).

In particular, this Fourier transform is bounded. Consequently, the convo-lution operator

(8.22) Tv = (PV u−n) ∗ v,

a priori taking S(Rn) to S ′(Rn), has the property that

(8.23) T : L2(Rn) −→ L2(Rn).

Continuity properties of such a convolution operator on Lp(Rn), for 1 <p < ∞, will be demonstrated in Chapter 13.

The special one-dimensional case of a principal value distribution hasbeen discussed in §4. In analogy with (4.34), we have the following.

Proposition 8.4. Under the hypothesis (8.17), we have, for v ∈ S(Rn),

(8.24) 〈v, PV u−n〉 = limε→0

∫

Rn\Bε

v(x) u−n(x) dx,

where Bε = x ∈ Rn : |x| < ε.

Proof. Since u−n(x)[v(x)− v(0)ϕ(x)

] ∈ L1(Rn), via (8.18), we have

〈v, PV u−n〉 = limε→0

∫

Rn\Bε

u−n(x)[v(x)− v(0)ϕ(x)

]dx,

so (8.24) follows from the observation that if (8.17) holds, then∫

Rn\Bε

u−n(x)ϕ(x) dx = 0, for all ε > 0,

for any radial ϕ ∈ S(Rn).


In general, if u(x) has the form (8.15) with m = −n, then u is a sum ofa term to which (8.17) applies and a constant times r−n. Now one can stilldefine a distribution in S ′(Rn), equal to r−n on Rn \ 0, by the prescription

(8.25) 〈v,Eϕr−n〉 =∫

Rn

r−n[v(x)− v(0)ϕ(x)

]dx,

for any given radial ϕ ∈ S(Rn) satisfying ϕ(0) = 1. This time, Eϕr−n ∈S ′(Rn) depends on the choice of ϕ. One has

(8.26) Eϕr−n − Eψr−n =(∫ [

ψ(x)− ϕ(x)]r−ndx

)δ.

Also, Eϕr−n is not homogeneous. Instead, one has

(8.27) D(t)(Eϕr−n) = t−nED(t)ϕr−n,

and by (8.26) this yields, after a brief calculation,

(8.28) D(t)(Eϕr−n) = t−nEϕr−n + An−1t−n(log t)δ,

where An−1 = vol(Sn−1). This implies for the Fourier transform of Eϕr−n

that

(8.29) D(t)F(Eϕr−n) = F(Eϕr−n) + (2π)−n/2An−1 log t,

which, in view of rotational invariance, implies that

(8.30) F(Eϕr−n)(ξ) = (2π)−n/2An−1 log |ξ|+ B,

where B is a constant, depending in an affine manner on ϕ. A “canonical”choice of Eϕr−n would be one for which B = 0; such a distribution Eϕr−n ∈S ′(Rn) is denoted PF r−n (for “finite part”); we have

(8.31) F(PF r−n)(ξ) = (2π)−n/2An−1 log |ξ|.Note that this is consistent with (4.59) when n = 2.

It turns out that r−m, which is holomorphic in m ∈ C : Re m >−n, with values in S ′(Rn), has a meromorphic continuation. This canbe perceived as follows. First note that if −n < Re m < 0, then bothrm and r−n−m belong to L1

loc(Rn), so from Proposition 8.1 and rotationalinvariance we deduce that

(8.32) F(rm) = c(m) r−m−n,

for a certain factor c(m), which we want to work out. We claim that

(8.33) F(rm) = 2m+n/2 Γ( 12 (m + n))Γ(− 1

2m)r−m−n,

for −n < Re m < 0. This can be deduced from (8.32) and Parseval’sidentity, which gives

(8.34) 〈u, rm〉 = c(m)〈u, r−m−n〉.

80

If we plug in u(x) = e−|x|2/2 = u(x), both sides of (8.34) can be evaluated

by integrating in polar coordinates. The left side is(8.35)

An−1

∫ ∞

0

rm+n−1e−r2/2 dr = 2(m+n−1)/2An−1

∫ ∞

0

s(m+n)/2−1 e−s ds

= 2(m+n−1)/2Γ(

12 (m + n)

)An−1,

and the right side of (8.34) is similarly evaluated, giving (8.33).Now the left side of (8.33) extends to be holomorphic in Re m > −n,

with values in S ′(Rn), while the right side extends to be meromorphic inRe m < 0, with poles at m = −n,−n − 2,−n − 4, . . . , due to the factorΓ((m + n)/2). Thus we have the desired meromorphic continuation. Withrm so defined,

(8.36) rm ∈ H#m(Rn), m 6= −n,−n− 2,−n− 4, . . . ;

indeed, D(t)rm − tm rm is a meromorphic function of m which vanisheson a nonempty open set. As we have seen, PF r−n can be defined by a“renormalization,” though it does not belong to H#

−n(Rn).Let us now consider the possibility of extending um, of the form (8.15),

to an element of S ′(Rn), in case

(8.37) m = −n− j, j = 1, 2, 3, . . . .

In analogy with (8.18) and (8.25), we can define Ej,ϕum in this case by

(8.38) 〈v, Ej,ϕum〉 =∫

um(x)[v(x)−

∑

|α|≤j

v(α)(0)α!

xαϕ(x)]

dx,

provided ϕ ∈ S(Rn) is a radial function such that ϕ(0) = 1 and 1 − ϕvanishes to order at least j at 0; for example, we could require ϕ(x) = 1for |x| ≤ c. The dependence on ϕ is given by

(8.39) Ej,ϕum − Ej,ψum =∑

|α|≤j

βα(ϕ− ψ) δ(α),

where

(8.40) βα(ϕ− ψ) = − 1α!

∫xα

[ϕ(x)− ψ(x)

]um(x) dx.

In analogy with (8.27), we have

(8.41) D(t)Ej,ϕum = tmEj,D(t)ϕum,

and hence, given (8.37), by a calculation similar to that establishing (8.28),

(8.42)

D(t)(Ej,ϕum) = tmEj,ϕum + tm∑

|α|<j

γα

(t|α|−j − 1

)δ(α)

+ tm log t∑

|α|=j

γα δ(α),

Exercises 81

for certain constants γα, which depend in an affine fashion on ϕ. Conse-quently, if we set

(8.43) Eum = Ej,ϕum −∑

|α|<j

γαδ(α),

we have another element of S ′(Rn) which agrees with um on Rn \ 0, and

(8.44) D(t)(Eum) = tmEum + tm(log t)∑

|α|=j

γα δ(α).

It follows for the Fourier transform F(Eum) that

D(t)F(Eum) = tjF(Eum) + tj(log t)∑

|α|=j

γ′α ξα.

Consequently, if F(Eum) = ω(ξ) for |ξ| = 1, we have

F(Eum)(tξ) = tjω(ξ) + (log t)∑

|α|=j

γ′α(tξ)α, for |ξ| = 1,

and hence

(8.45) F(Eum)(ξ) = wj(ξ) + pj(ξ) log |ξ|,where

(8.46) wj ∈ H#j (Rn) and pj is a homogeneous polynomial, of degree j.

We leave it as an exercise to the reader to construct a similar extensionof um to an element of S ′(Rn), when Re m ≤ −n and m is not an integer.In such a case one can produce an element of H#

m(Rn); log terms do notarise.

Exercises

1. More generally than Sm1 (Rn), for 0 < ρ ≤ 1, define Sm

ρ (Rn) by

p ∈ Smρ (Rn) ⇐⇒ |Dα

x p(x)| ≤ Cα〈x〉m−ρ|α|, for all α ≥ 0.

Show that p ∈ C∞(Rn \ 0) in this case, as in Proposition 8.2.2. Define p ∈ C∞(Rn \ 0) by p(ξ) = (iξ1 + |ξ′|2)−1, ξ = (ξ1, ξ2, . . . , ξn) = (ξ1, ξ

′).Show that p(ξ) agrees outside any neighborhood of the origin with a memberof S−1

1/2(Rn).

3. Prove Proposition 8.3.4. If −n < Re m < 0 and um is of the form (8.15), then um and um belong to

L1loc(Rn), with um ∈ H#

m, um ∈ H#−n−m. Hence

u(x) = |x|−n−mWm(|x|−1x), Wm ∈ C∞(Sn−1).

Study the transformation w 7→ Wm. Use this to produce a meromorphiccontinuation of um.

82

5. Study the residue of the meromorphic distribution-valued function rz at z =−n, and relate this to the failure of PF r−n to be homogeneous.

6. In case n = 1 and m = −s, the formula (8.33) says

F(r−s) = 21/2−s Γ( 1−s2

)

Γ( s2)

rs−1, for 0 < Re s < 1,

while Riemann’s functional equation (7.26) can be written

ζ(s)

ζ(1− s)= πs−1/2 Γ( 1−s

2)

Γ( s2)

.

Is this a coincidence? (See [Pat], Chapter 2.) Note that these formulas yield

(2π)s/2

ζ(s)F(r−s) =

(2π)(1−s)/2

ζ(1− s)rs−1, 0 < Re s < 1.

9. Elliptic operators

A partial differential operator P (D) of order m,

(9.1) P (D) =∑

|α|≤m

aαDα,

is said to be elliptic provided

(9.2) |P (ξ)| ≥ C|ξ|m, for |ξ| large.

Here P (ξ) =∑

aαξα. The paradigm example is the Laplace operator∆ = P (D), with P (ξ) = −|ξ|2, which is elliptic of order 2. In this sectionwe consider some important properties of solutions to

(9.3) P (D)u = f

when P (D) is elliptic.The hypothesis (9.2) implies the following. If (9.2) holds for |ξ| ≥ C1,

and if ϕ ∈ C∞0 (Rn) is equal to 1 for |ξ| ≤ C1, then

(9.4) q(ξ) =(1− ϕ(ξ)

)P (ξ)−1 ∈ S−m

1 (Rn),

where Sm1 (Rn) is the space defined by (8.11); we call it a space of “symbols.”

Now consider

(9.5) E = (2π)−n/2q ∈ S ′(Rn).

By Proposition 8.2, we know that E is smooth on Rn \ 0 and rapidlydecreasing as |x| → ∞. If we set

(9.6) v = P (D)E,

then

(9.7) v(ξ) = (2π)−n/2(1− ϕ(ξ)

).

9. Elliptic operators 83

In other words,

(9.8) P (D)E = δ + w,

with

(9.9) w = −(2π)−n/2ϕ ∈ S(Rn).

We say E is a parametrix for P (D). It is almost as useful as a fundamentalsolution, for some qualitative purposes. For example, it enables us to saya great deal about the singular support of a solution u to (9.3), givenf ∈ D′(Rn). The singular support of a general distribution u ∈ D′(Rn) isdefined as follows. Let Ω ⊂ Rn be open. We say u is smooth on Ω if thereexists v ∈ C∞(Ω) such that u = v on Ω. The smallest set K for which u issmooth on Rn \K is the singular support of u, denoted

sing supp u.

For example, sing supp δ = 0; also sing supp |x|2−n = 0, if n 6= 2.Now, suppose that u ∈ E ′(Rn) and (9.3) holds. Then

(9.10) E ∗ f = E ∗ P (D)u = (P (D)E) ∗ u = u + w ∗ u

and, of course,

w ∗ u ∈ C∞(Rn).

On the other hand, it is easy to see that, for any f ∈ E ′(Rn),

(9.11) sing supp E ∗ f ⊂ sing supp f,

provided sing supp E ⊂ 0. More generally, for any f1, f2 ∈ E ′(Rn), ifsing supp fj ⊂ Kj , then

(9.12) sing supp f1 ∗ f2 ⊂ K1 + K2,

a result we leave as an exercise.Noting that we can multiply distributions by cut-offs χ ∈ C∞0 (Rn), equal

to 1 on an arbitrarily large set, we deduce the following result, known aselliptic regularity.

Proposition 9.1. For any u ∈ D′(Rn), if (9.3) holds with P (D) elliptic,then

(9.13) sing supp u = sing supp f.

Finally, we want to make a detailed analysis of the behavior of the singu-larity at the origin of the parametrix E for an elliptic operator P (D). SinceE is given by (9.4) and (9.5), with P (ξ) =

∑|α|≤m aαξα a polynomial, it

follows that, for |ξ| large,

(9.14) q(ξ) ∼∑

j≥0

qj(ξ),

84

where each qj ∈ C∞(Rn), and, for |ξ| ≥ C, qj(ξ) is homogeneous in ξ ofdegree −m− j. The meaning of (9.14) is that, for any N ,

(9.15) q(ξ)−N−1∑

j=0

qj(ξ) = rN (ξ) ∈ S−m−N1 (Rn).

Consequently,

(9.16) E ∼ (2π)−n/2∑

j≥0

qj

in the sense that, for any K, one can take N large enough that

(9.17) E − (2π)−n/2N−1∑

j=0

qj = (2π)−n/2rN ∈ CK(Rn).

Now, we can replace each qj by qbj ∈ C∞(Rn \ 0), equal to qj for |ξ| large,

and homogeneous of degree −m − j on Rn \ 0, and replace each qbj by

q#j ∈ S ′(Rn), equal to qb

j on Rn \ 0, such that q#j ∈ H#

−m−j if m + j < n,or in any event satisfying the counterpart of (8.44)–(8.46). Note that, foreach j, qj−q#

j ∈ E ′(Rn), so the Fourier transform of the difference belongsto C∞(Rn). We have established the following.

Proposition 9.2. A parametrix E for an elliptic operator P (D) of orderm satisfies the condition that E ∈ C∞(Rn \ 0), and the singularity is givenby

(9.18) E ∼∑

`≥0

(E` + p`(x) log |x|),

where

(9.19) E` ∈ H#m−n+`(R

n)

and p`(x) is a polynomial homogeneous of degree m − n + `; these logcoefficients appear only for ` ≥ n−m.

More generally, this result holds for E = (2π)−n/2q whenever q ∈ Sm1

has an expansion of the form (9.14), for any m ∈ R, and log terms do notarise if m is not an integer.

Exercises

1. Using Exercises 1 and 2 of §8, establish an analogue of the regularity result inProposition 9.1 when P (D) is the (nonelliptic) “heat operator”:

P (D) =∂

∂x1−

(∂2

∂x22

+ · · ·+ ∂2

∂x2n

).

10. Local solvability of constant-coefficient PDE 85

2. Give a detailed proof of (9.12), in order to deduce (9.11).(Hint: Use

f ∈ E ′(Rn), g ∈ C∞(Rn) =⇒ f ∗ g ∈ C∞(Rn).

Break up f1 and f2 into pieces. For nonsmooth pieces, establish and use

supp ϕj ⊂ Kj =⇒ supp ϕ1 ∗ ϕ2 ⊂ K1 + K2.)

10. Local solvability of constant-coefficient PDE

In the previous sections we have mainly used Fourier analysis as a tool toprovide explicit solutions to the classical linear PDEs. Here we use Fourierseries to prove an existence theorem for solutions to a general constant-coefficient linear PDE

(10.1) P (D)u = f.

We show that, given any f ∈ D′(Rn), and any R < ∞, there existsu ∈ D′(Rn) solving (10.1) on the ball |x| < R. This result was originallyestablished by Malgrange and Ehrenpreis. If f ∈ C∞(Rn), we produceu ∈ C∞(Rn). We do not produce a global solution, and other references,particularly [H] and [Tre], contain much more information on solutions to(10.1) than is presented here. Our method, due to Dadok and Taylor [DT],does have the advantage of being fairly straightforward and short.

For any α ∈ Rn, solving (10.1) on BR = x ∈ Rn : |x| < R is equivalentto solving

(10.2) P (D + α)v = g,

where v = e−iα·xu and g = e−iα·xf . To solve (10.2) on BR, we can cut off gto be supported on B3R/2 and work on Rn/2RZn. Without loss of general-ity (altering P (D)), we can rescale and suppose R = π, so Rn/2RZn = Tn.The following result then implies solvability on BR.

Proposition 10.1. For almost every α ∈ A = (α1, . . . , αn) : 0 ≤ αν ≤1,(10.3) P (D + α) : D′(Tn) −→ D′(Tn)

is an isomorphism, as is P (D + α) : C∞(Tn) → C∞(Tn).

In view of the characterizations of Fourier series of elements of D′(Tn)and of C∞(Tn), it suffices to establish the following.

Proposition 10.2. Let P (ξ) be a polynomial of order m on Rn. Foralmost all α ∈ A, there are constants C, N such that

(10.4) |P (k + α)−1| ≤ C〈k〉N , for all k ∈ Zn.

86

We will prove this using the following elementary fact about the behaviorof a polynomial near its zero set.

Lemma 10.3. Let P (ξ) be a polynomial of order m on Rn, not identicallyzero. Then there exists δ > 0 such that

(10.5) |P (ξ)|−δ ∈ L1loc(Rn).

Before proving Lemma 10.3, we show how it yields (10.4). First, we claimthat, for any polynomial of order m on Rn, not identically zero, there existδ > 0 and M such that

(10.6)∫|P (ξ)|−δ〈ξ〉−M dξ < ∞.

Indeed, Lemma 10.3 guarantees∫

|ξ|≤1

|P (ξ)|−δ dξ < ∞,

while, for M sufficiently large,∫

|ξ|≥1

|P (ξ)|−δ|ξ|−M dξ ≤ C

∫

|ξ|≤1

|P (|ξ|−2ξ)|−δ dξ,

and Lemma 10.3 also implies that, for δ > 0 small enough,

|P (|ξ|−2ξ)|−δ ∈ L1loc.

Now, using (10.6), note that

(10.7)∫

A

∑

k∈Zn

|P (k + α)|−δ〈k〉−M dα ≤ C

∫

Rn

|P (ξ)|−δ〈ξ〉−M dξ < ∞.

Thus, for almost all α ∈ A,

(10.8)∑

k∈Zn

|P (k + α)|−δ〈k〉−M < ∞,

which immediately gives (10.4).We now prove the lemma for any δ < 1/m. We must prove that |P (ξ)|−δ

is integrable on any bounded subset of Rn. Rotating coordinates, we cansuppose that P (ξ1, 0) is a polynomial of order exactly m:

(10.9) P (ξ1, 0) = amξm1 + · · ·+ a0, am 6= 0.

It follows that, with ξ′ = (ξ2, . . . , ξn),

(10.10) P (ξ) = amξm1 +

m−1∑

`=0

a`(ξ′)ξ`1,

11. The discrete Fourier transform 87

where a`(ξ′) is a polynomial on Rn−1 of order ≤ m− `. Consequently, wehave

(10.11) P (ξ) = am

m∏

j=1

(ξ1 − λj(ξ′)

).

Hence it is clear that, for any C1 < ∞, there is a C2 < ∞ such that ifδ < 1/m,

(10.12)∫ C1

−C1

|P (ξ)|−δ dξ1 ≤ C2, for |ξ′| ≤ C1.

This completes the proof.

Exercises

1. Consider the following boundary problem on [0, A]× Tn:

utt −∆u = 0,

u(0, x) = f1(x), u(A, x) = f2(x),

where fj ∈ C∞(Tn). Show that, for almost all A ∈ R+, this has a uniquesolution u ∈ C∞([0, A]×Tn), for all fj ∈ C∞(Tn). Show that, for a dense setof A, this solvability fails.

11. The discrete Fourier transform

When doing numerical work involving Fourier series, it is convenient todiscretize, and replace S1, pictured as the group of complex numbers ofmodulus 1, by the group Γn generated by ω = e2πi/n. One can also approx-imate Td by (Γn)d, a product of d copies of Γn. We will restrict attentionto the case d = 1 here; results for general d are obtained similarly.

The cyclic group Γn is isomorphic to the group Zn = Z/(n), but we willobserve a distinction between these two groups; an element of Γn is a certaincomplex number of modulus 1, and an element of Zn is an equivalence classof integers. For n large, we think of Γn as an approximation to S1 and Zn

as an approximation to Z. We note the natural dual pairing Γn ×Zn → Cgiven by (ωj , `) 7→ ωj`, which is well defined since ωjn = 1.

Now, given a function f : Γn → C, its discrete Fourier transform f# =Φnf , mapping Zn to C, is defined by

(11.1) f#(`) =1n

∑

ωj∈Γn

f(ωj)ω−j`.

88

Similarly, given a function g : Zn → C, its “inverse Fourier transform”gb : Γn → C is defined by

(11.2) gb(ωj) =∑

`∈Zn

g(`)ωj`.

The following is the Fourier inversion formula in this context.

Proposition 11.1. The map

(11.3) Φn : L2(Γn) −→ L2(Zn)

is a unitary isomorphism, with inverse defined by (11.2), so

(11.4) f(ωj) =∑

`∈Zn

f#(`)ωj`.

Here the space L2(Zn) is defined by counting measure and L2(Γn) by1/n times counting measure, that is,

(11.5) (u, v)L2(Γn) =1n

∑

ωj∈Γn

u(ωj)v(ωj).

Note that if we define functions ej on Γn by

(11.6) ej(ωk) = ωjk,

then Proposition 11.1 is equivalent to:

Proposition 11.2. The functions ej , 1 ≤ j ≤ n, form an orthonormalbasis of L2(Γn).

Proof. Since L2(Γn) has dimension n, we need only check that the ejs aremutually orthogonal. Note that

(ek, e`) =1n

∑

ωj∈Γn

ωmj , m = k − `.

Denote the sum by Sm. If we multiply by ωm, we have a sum of the sameset of powers of ω, so Sm = ωmSm. Thus Sm = 0 whenever ωm 6= 1, whichcompletes the proof. Alternatively, the series is easily summed as a finitegeometrical series.

Note that the functions ej in (11.6) are the restrictions to Γn of eijθ (i.e.,values at θ = 2πk/n). These restrictions depend only on the residue classof j mod n, which leads to the following simple but fundamental connectionbetween Fourier series on S1 and on Γn.


Proposition 11.3. If f ∈ C(S1) has absolutely convergent Fourier series,then

(11.7) f#(`) =∞∑

j=−∞f(` + jn).

We will use (11.7) as a tool to see how well a function on S1 is approxi-mated by discretization, involving restriction to Γn. Precisely, we considerthe operators

(11.8) Rn : C(S1) −→ L2(Γn), En : L2(Γn) −→ C∞(S1)

given by

(11.9) (Rnf)(ωj) = f(2πj

n

),

for f = f(θ), 0 ≤ θ ≤ 2π, and

(11.10) En

( ∑

`∈Zn

g(`)ωj`)

=ν−1∑

`=−ν

g(`)ei`θ, n = 2ν.

We assume n = 2ν is even; one can also treat n = 2ν − 1, changing theupper limit in the last sum from ν − 1 to ν. Clearly, RnEn is the identityoperator on L2(Γn). The question of interest to us is: How close is EnRnfto f , a function on S1? The answer depends on smoothness properties off and is expressed in terms involving (typically) negative powers of n.

We compare EnRn and the partial summing operator

(11.11) Pnf =ν−1∑

`=−ν

f(`)ei`θ

for Fourier series. Note that

(11.12) EnRnf(θ) =ν−1∑

`=−ν

f#(`)ei`θ.

Consequently,

(11.13) EnRnf = Pnf + Qnf,

with

(11.14) Qnf(θ) =ν−1∑

`=−ν

[f#(`)− f(`)

]ei`θ.

By (11.7), we have, for −ν ≤ ` ≤ ν − 1,

(11.15) f#(`)− f(`) =∑

j∈Z\0f(` + jn).

90

Consequently, the sup norm of Qnf is bounded by

(11.16)ν−1∑

`=−ν

∣∣f#(`)− f(`)∣∣ ≤

∑

|k|≥ν

|f(k)|.

The right side also dominates the sup norm of f − Pnf , proving:

Proposition 11.4. If f ∈ C(S1) has absolutely convergent Fourier series,then

(11.17) ‖f − EnRnf‖L∞ ≤ 2∑

|k|≥ν

|f(k)|.

The estimates of various norms of f − Pnf is an exercise in Fourieranalysis on S1. There are many estimates involving Sobolev spaces; seeChapter 4. Here we note a simple estimate, for m ≥ 1:

(11.18)‖f − Pnf‖C`(S1) ≤

∑

|k|≥ν

|k|`|f(k)|

≤ Cm`‖f‖C`+m+1(S1) · n−m,

the last inequality following from (1.49). As the reader can verify, use of theproof of Proposition 1.3 can lead to a sharper estimate. As for an estimateof the contribution of Qn to the discretization error, from (11.14)–(11.16)we easily obtain

(11.19)‖Qnf‖C`(S1) ≤

(n

2

)` ∑

|k|≥ν

|f(k)|

≤ C`m‖f‖C`+m+1(S1) · n−m.

We reiterate that sharper estimates are possible.Recall that solutions to a number of evolution equations are given by

Fourier multipliers on L2(S1), of the form

(11.20) F (D)u(θ) =∞∑

`=−∞F (`)u(`)ei`θ.

We want to compare such an operator with its discretized version onL2(Γn):

(11.21) F (Dn)[ ∑

`∈Zn

g(`)ωj`]

=ν−1∑

`=−ν

F (`)g(`)ωj`.

In fact, a simple calculation yields

(11.22) EnF (Dn)Rnu(θ) =ν−1∑

`=−ν

F (`)u#(`)ei`θ


and hence

(11.23) EnF (Dn)Rnu = PnF (D)u + Ψnu,

where

(11.24) Ψnu(θ) =ν−1∑

`=−ν

F (`)[ ∑

j∈Z\0u(` + jn)

]ei`θ.

This implies the estimate

(11.25) ‖Ψnu‖L∞ ≤[sup|`|≤ν

|F (`)|] ∑

|k|≥ν

|u(k)|.

Also, as in (11.18), we have, for m ≥ 1,

(11.26) ‖Ψnu‖C`(S1) ≤ C`m

[sup|`|≤ν

|F (`)|]‖u‖C`+m+1(S1) · n−m.

The significance of these statements is that, for u smooth and n large, thediscretized F (Dn) provides a very accurate approximation to F (D). Thisis of practical importance for a number of numerical problems.

Note the distinction between Dn and the centered difference operator∆n, defined by

(∆nf)(ωj) =n

4πi

[f(ωj+1)− f(ωj−1)

].

We have, in place of (11.21),

(11.27) F (∆n)[ ∑

`∈Zn

g(`)ωj`]

=ν−1∑

`=−ν

F( n

2πsin

(2π`

n

))g(`)ωj`,

so, for gb ∈ L2(Γn) given by (11.2),(11.28)

F (∆n)gb(ωj)− F (Dn)gb(ωj) =ν−1∑

`=−ν

[F

( n

2πsin

(2π`

n

))− F (`)]g(`)ωj`.

This identity leads to a variety of estimates, of which the following is asimple example. If |F ′(λ)| ≤ K for −ν ≤ λ ≤ ν, then

(11.29) ‖F (∆n)u− F (Dn)u‖L∞ ≤ 23· π2K

[ ν−1∑

`=−ν

|`|3 |u#(`)|]· n−2,

since, for −π ≤ x ≤ π, | sin x− x| ≤ (1/6)|x|3. The basic content of this isthat F (∆n) furnishes a second-order-accurate approximation to F (D) (asn → ∞). This is an improvement over the first-order accuracy one wouldget by using a one-sided difference operator, such as

(∆+n f)(ωj) =

n

2πi

[f(ωj+1)− f(ωj)

],

92

but not as good as the “infinite-order accuracy” one gets for F (Dn) as aconsequence of (11.23)–(11.26).

Similar to the case of functions on S1, we have, for u ∈ L2(Γn),

(11.30) F (Dn)u(ωj) = (kF ∗ u)(ωj) =1n

∑

`∈Zn

kF (ωj−`)u(ω`),

where

(11.31) kF (ωj) =ν−1∑

`=−ν

F (`)ωj`.

For example, with F (λ) = e−y|λ|, we get the discrete version of the Poissonkernel:

(11.32) kF (ωj) = py(ωj) =ν−1∑

`=−ν

e−y|`|ωj`,

which we can write as a sum of two finite geometrical series to get

(11.33) py(ωj) =1− r2 − 2rν+1(−1)j cos(2πj/n)

1 + r2 − 2r cos(2πj/n)+ rνω−jν ,

with r = e−y and, as usual, ω = e2πi/n, n = 2ν. Compare with (1.30).The reader can produce a similar formula for n odd.

As in the case of S1, the sum (11.31) for the (discretized) heat kernel,with F (`) = e−t`2 , cannot generally be simplified to an expression whosesize is independent of n. However, when t is an imaginary integer, such anevaluation can be performed. Such expressions are called Gauss sums, andtheir evaluation is regarded as one of the pearls of early nineteenth-centurymathematics. We present one such result here.

Proposition 11.5. For any n ≥ 1, even or odd,

(11.34)n−1∑

k=0

e2πik2/ne2πi`k/n =12(1 + i)e−πi`2/2n

[1 + (−1)`i−n

]n1/2.

Proof. The sum on the left is n · f#(−`), where f ∈ C(S1) is given by

f(θ) = einθ2/2π, 0 ≤ θ ≤ 2π.

Note that f is Lipschitz on S1, with a simple jump in its derivative, sof(k) = O(|k|−2). Hence Proposition 11.3 applies, and (11.7) yields

(11.35) f#(−`) =∞∑

j=−∞

∫ 1

0

e2πin[y2+(j+`/n)y] dy.

Exercises 93

To evaluate this, we use the “Gaussian integral” (convergent though notabsolutely convergent):

(11.36)∫ ∞

−∞e2πiny2

dy = n−1/2γ, γ =12(1 + i),

obtained from (3.20) by a change of variable and analytic continuation, asin (6.42). We will break up the real line as a countable union of intervals⋃

k [k +a, k +a+1], in two different ways, and then evaluate (11.35). Notethat

(11.37)∫ k+a+1

k+a

e2πiny2dy =

∫ 1

0

e2πin[y2+2(k+a)y] dy · e2πin(k+a)2 .

If we pick a = `/2n, then 2(k +a) = 2k + `/n, and as k runs over Z, we getthose integrands in (11.35) for which j is even. If we pick a = −1/2+ `/2n,then 2(k + a) = 2k − 1 + `/n. Furthermore, we have

(11.38) e2πin(k+a)2 = eπi`2/2n and eπi(`−n)2/2n,

respectively, for these two choices of a. Thus the sum in (11.35) is equalto n−1/2γ times e−πi`2/2n + e−πi(`−n)2/2n, which gives the desired formula,(11.34).

The basic case of this sum is the ` = 0 case:

(11.39)n−1∑

k=0

e2πik2/n =12(1 + i)(1 + i−n)n1/2 = σn · n1/2,

where σn is periodic of period 4 in n, with

(11.40) σ0 = 1 + i, σ1 = 1, σ2 = 0, σ3 = i.

This result, particularly when n = p is a prime, is used as a tool to obtainfascinating number-theoretical results. For more on this, see the exercisesand the references [Hua], [Land], and [Rad].

Exercises

1. Generalize the Gauss sum identity (11.34) to(11.41)

n−1∑

k=0

e2πik2m/ne2πi`k/n =1 + i

2

(n

m

)1/2

e−πi`2/2mn2m−1∑ν=0

e−πinν2/2me−πiν`/m.

(Hint: The left side is n · f#(−`), with

f(θ) = einmθ2/2π, 0 ≤ θ ≤ 2π.

94

For this, one has a formula like (11.35):

f#(−`) =

∞∑j=−∞

∫ 1

0

e2πinm[y2+(1/m)(j+`/n)y] dy.

Write j = 2mµ + ν, so

∞∑j=−∞

aj =

2m−1∑ν=0

∑

j=ν mod 2m

aj .

For fixed ν, the sum becomes a multiple of the Gaussian integral (11.36), withn replaced by nm, and the formula (11.41) arises.)Note the ` = 0 case of this:

n−1∑

k=0

e2πik2m/n =1 + i

2

(n

m

)1/2 2m−1∑ν=0

e−πinν2/2m.

2. Let ∆ be d2/dx2 on S1 = R/(2πZ). Using Fourier series, show that, fort = 2πm/n, where m and n are positive integers, e−it∆δ(x) = H(t, x) has theform

(11.42) H

(2π

m

n, x

)=

1

n

n−1∑

`=0

G(m, n, `) δ2π`/n(x),

where G(m, n, `) is given by the left side of (11.41). On the other hand,applying e−it∆, acting on S ′(R), to

∑ν δ(x − 2πν), show that (11.42) holds,

with G(m, n, `) given by the right side of (11.41). Hence deduce another proofof this Gauss sum identity.

3. Let #(`, n) denote the number of solutions k ∈ Zn to

` = k2 (mod n).

Show that, with ω = e2πi/n,

n−1∑

k=0

ωjk2=

n−1∑

`=0

#(`, n)ωj`.

4. Show that, more generally,

(n−1∑

k=0

ωjk2)ν

=

n−1∑

`=0

#(`, n; ν) ωj`,

where #(`, n; ν) denotes the number of solutions (k1, . . . , kν) ∈ (Zn)ν to

` = k21 + · · ·+ k2

ν (mod n).

5. Let p be a prime. The Legendre symbol (`|p) is defined to be +1 if ` = k2

mod p for some k and ` 6= 0, 0 if ` = 0, and −1 otherwise. If p is an oddprime, #(`, p) = (`|p) + 1. The Legendre symbol has the useful multiplicativeproperty: (`1`2|p) = (`1|p)(`2|p). Check this. Show that, with ω = e2πi/p, if pis an odd prime,

p−1∑

k=0

ωk2=

p−1∑

`=0

(`|p)ω`,

12. The fast Fourier transform 95

and, more generally,

p−1∑

k=0

ωjk2=

p−1∑

`=0

(`|p)ωj` + pδj0,

where δj0 = 1 if j = 0 (mod p), 0 otherwise. (Hint: Use Exercise 3.)6. Denoting

∑p−1k=0 ωk2

by Gp, p an odd prime, show that

p−1∑

k=0

ωjk2= (j|p) ·Gp + p · δj0.

(Hint: If 1 ≤ j ≤ p − 1, use∑p−1

`=0 (`|p)ω` =∑p−1

`=0 (j`|p)ωj` and (j`|p) =(j|p)(`|p).)

Denote by S(m, n) the Gauss sum

S(m, n) =

n−1∑

k=0

e2πik2m/n.

Then the content of Exercise 6 is that S(j, p) = (j|p)S(1, p), for 1 ≤ j ≤ p−1,when p is an odd prime.

7. Assume p and q are distinct odd primes. Show that

S(1, pq) = S(q, p)S(p, q).

(Hint: To resum∑pq−1

k=0 e2πik2/pq, use the fact that, as µ runs over 0, 1, . . . , p−1 and ν runs over 0, 1, . . . , q − 1, then k = µq + νp runs once over Z modpq.)

8. From Exercises 6 and 7, it follows that when p and q are distinct odd primes,

(p|q)(q|p) =S(1, pq)

S(1, p)S(1, q).

Use the evaluation (11.39) of S(1, n) to deduce the quadratic reciprocity law:

(p|q)(q|p) = (−1)(p−1)(q−1)/4.

This law, together with the complementary results

(−1|p) = (−1)(p−1)/2, (2|p) = (−1)(p2−1)/8,

allows for an effective computation of (`|p), as one application, but the sig-nificance of quadratic reciprocity goes beyond this. It and other implicationsof Gauss sums are absolutely fundamental in number theory. For material onthis, see [Hua], [Land], and [Rad].

12. The fast Fourier transform

In the last section we discussed some properties of the discrete Fouriertransform

(12.1) f#(`) =1n

∑

ωj∈Γn

f(ωj)ω−j`,

96

where ` ∈ Zn = Z/(n) and Γn is the multiplicative group of unit complexnumbers generated by ω = e2πi/n. We now turn to a discussion of the effi-cient numerical computation of the discrete Fourier transform. Note that,for any fixed `, computing the right side of (12.1) involves n− 1 additionsand n multiplications of complex numbers, plus n integer products j` = mand looking up ωm and f(ωj). If the computations for varying ` are doneindependently, the total effort to compute f# involves n2 multiplicationsand n(n− 1) additions of complex numbers, plus some further chores. Thefast Fourier transform (denoted FFT) is a method for computing f# inCn(log n) steps, in case n is a power of 2.

The possibility of doing this arises from observing redundancies in thecalculation of the Fourier coefficients f#(`). Let us illustrate this in thecase of Γ4. We can write

(12.2)4f#(0) =

[f(1) + f(i2)

]+

[f(i) + f(i3)

],

4f#(2) =[f(1) + f(i2)

]− [f(i) + f(i3)

],

and

(12.3)4f#(1) =

[f(1)− f(i2)

]− i[f(i)− f(i3)

],

4f#(3) =[f(1)− f(i2)

]+ i

[f(i)− f(i3)

].

Note that each term in square brackets appears twice. Note also that (12.2)gives the Fourier coefficients of a function on Γ2; namely, if

(12.4) 0f(1) = f(1) + f(−1), 0f(−1) = f(i) + f(i3),

then

(12.5) 2f#(2`) = 0f#(`), for ` = 0 or 1.

Similarly, if we set

(12.6) 1f(1) = f(1)− f(−1), 1f(−1) = −i[f(i)− f(i3)

],

then

(12.7) 2f#(2` + 1) = 1f#(`), for ` = 0 or 1.

This phenomenon is a special case of a more general result that leads to afast inductive procedure for evaluating the Fourier transform f#.

Suppose n = 2k; let us use the notation Gk = Γn. Note that Gk−1 is asubgroup of Gk. Furthermore, there is a homomorphism of Gk onto Gk−1,given by ωj 7→ ω2j . Given f : Gk → C, define the following functions 0fand 1f on Gk−1, with ω1 = ω2, generating Gk−1:

0f(ωj1) = f(ωj) + f(ωj+n/2),(12.8)

1f(ωj1) = ωj

[f(ωj)− f(ωj+n/2)

].(12.9)

Note that the factor ωj in (12.9) makes 1f(ωj1) well defined for j ∈ Zn/2,

that is, the right side of (12.9) is unchanged if j is replaced by j + n/2.

12. The fast Fourier transform 97

Then 0f# and 1f#, the discrete Fourier transforms of the functions 0f and1f , respectively, are functions on Zn/2 = Z/(2k−1).

Proposition 12.1. We have the following identities relating the Fouriertransforms of 0f, 1f , and f :

(12.10) 2f#(2`) = 0f#(`)

and

(12.11) 2f#(2` + 1) = 1f#(`),

for ` ∈ 0, 1, . . . , n/2− 1.

Proof. Recall that we set ω1 = ω2. Since ωn = 1 and ωn/21 = 1, we have

(12.12)

nf#(2`) =∑

ωj∈Gk

f(ωj)ω2j`

=∑

ωj1=ω2j∈Gk−1

[f(ωj) + f(ωj+n/2)

]ωj`

1 ,

proving (12.10), and, since ωn/2 = −1,

(12.13)

nf#(2` + 1) =∑

ωj∈Gk

f(ωj)ωj ω2j`

=∑

ωj1=ω2j∈Gk−1

ωj[f(ωj)− f(ωj+n/2)

]ωj`

1 ,

proving (12.11).

Thus the problem of computing f#, given f ∈ L2(Gk), is transformed,after n/2 multiplications and n additions of complex numbers in (12.8)and (12.9), to the problem of computing the Fourier transforms of twofunctions on Gk−1. After n/4 new multiplications and n/2 new additionsfor each of these functions 0f and 1f , that is, after an additional total ofn/2 new multiplications and n additions, this is reduced to the problem ofcomputing four Fourier transforms of functions on Gk−2. After k iterations,we obtain 2k functions on G0 = 1, which precisely give the Fouriercoefficients of f . Doing this hence takes kn = (log2 n)n additions andkn/2 = (log2 n)n/2 multplications of complex numbers, plus a comparablenumber of integer operations and fetching from memory values of given orpreviously computed functions.

To describe an explicit implementation of Proposition 12.1 for a compu-tation of f#, let us identify an element ` ∈ Zn (n = 2k) with a k-tuple L =(Lk−1, . . . , L1, L0) of elements of 0, 1 giving the binary expansion of theinteger in 0, . . . , n−1 representing ` (i.e., L0+L1 ·2+· · ·+Lk−1 ·2k−1 = `

98

mod n). To be a little fussy, we use the notation

(12.14) f#(`) = f##(L).

Then the formulas (12.10) and (12.11) state that

(12.15) 2f##(Lk−1, . . . , L1, 0) = 0f##(Lk−1, . . . , L1)

and

(12.16) 2f##(Lk−1, . . . , L1, 1) = 1f##(Lk−1, . . . , L1).

The inductive procedure described above gives, from 0f and 1f defined onGk−1, the functions

(12.17) 00f = 0(0f), 10f = 1(0f), 01f = 0(1f), 11f = 1(1f)

defined on Gk−2, and so forth, and we see from (12.15) and (12.16) that

(12.18) f#(`) =1n

Lf,

where Lf = Lf(1) is defined on G0 = 1. From (12.8) and (12.9) we havethe following inductive formula for Lm+1Lm···L1f on Gk−m−1:

(12.19)0Lm···L1f(ωj

m+1) = Lm···L1f(ωjm) + Lm···L1f(ωj+2k−m−1

m ),1Lm···L1f(ωj

m+1) = ωjm

[Lm···L1f(ωj

m)− Lm···L1f(ωj+2k−m−1

m )],

where ωm is the generator of Gk−m, defined by ω0 = ω = e2πi/n (n =2k), ωm+1 = ω2

m, that is, ωm = ω2m

.When doing computations, particularly in a higher-level language, it

may be easier to work with integers ` than with m-tuples (L1, . . . , L1).Therefore, let us set

(12.20) Lm···L1f(ωjm) = Fm(2m · j + `),

where

` = L1 + L2 · 2 + · · ·+ Lm · 2m−1 ∈ 0, 1, . . . , 2m − 1and

j ∈ 0, 1, . . . , 2k−m − 1.Note that this precisely defines Fm on 0, 1, . . . , 2k − 1. For m = 0, wehave

(12.21) F0(j) = f(ωj), 0 ≤ j ≤ 2k − 1.

The iterative formulas (12.19) give(12.22)

Fm+1(2m+1j + `) = Fm(2mj + `) + Fm(2mj + 2k−1 + `),

Fm+1(2m + 2m+1j + `) = ωjm

[Fm(2mj + `)− Fm(2mj + 2k−1 + `)

],

Exercises 99

for 0 ≤ j ≤ 2k−m−1−1, 0 ≤ ` ≤ 2m−1. It is easy to write a computer pro-gram to implement such an iteration. The formula (12.18) for the Fouriertransform of f becomes

(12.23) f#(`) = n−1 Fk(`), 0 ≤ ` ≤ 2k − 1.

While (12.21)–(12.23) provide an easily implementable FFT algorithm, itis not necessarily the best. One drawback is the following. In passing fromFm to Fm+1 via (12.22), you need two different arrays of n complex num-bers. A variant of (12.19), where Lm+1Lm···L1f is replaced by fL1···LmLm+1 ,leads to an iterative procedure where a transformation of the type (12.19)is performed “in place,” and only one such array needs to be used. If mem-ory is expensive and one needs to make the best use of it, this savings canbe important. At the end of such an iteration, one needs to perform a “bitreversal” to produce f#. Details, including sample programs, can be foundin [PFTV].

On any given computer, a number of factors would influence the choiceof the best FFT algorithm. These include such things as relative speedof memory access and floating-point performance, efficiency of computingtrigonometric functions (e.g., whether this is implemented in hardware),degree of accuracy required, and other factors. Also, special features, suchas computing the Fourier transform of a real-valued function or of a functionwhose Fourier transform is known to be real-valued, would affect specificcomputer programs designed for maximum efficiency. Working out howbest to implement FFTs on various computers presents many interestingproblems.

Exercises

1. Write a computer program to implement the FFT via (12.21)–(12.23). Try tomake it run as fast as possible.

2. Using the FFT, write a computer program to solve numerically the initial-value problem for the heat equation ∂u/∂t− uxx = 0 on R+ × S1.

3. Consider multidimensional generalizations of the discrete Fourier transform,and in particular the FFT. What size three-dimensional FFT could be handledby a computer with 4 megabytes of RAM? With 256 MB?

4. Generalize the FFT algorithm to a cyclic group Γn with n = 3k. Also, gener-alize to the case n = p1 · · · pk where pj are “small” primes.

100

A. The mighty Gaussian and the sublime gamma function

The Gaussian function e−|x|2

on Rn is an object whose study yields manywonderful identities. We will use the identity

(A.1)∫

Rn

e−|x|2dx = πn/2,

which was established in (3.18), to compute the area An−1 of the unit sphereSn−1 in Rn. This computation will bring in Euler’s gamma function, andother results will flow from this. Switching to polar coordinates for theright side of (A.1), we have

(A.2)

πn/2 = An−1

∫ ∞

0

e−r2rn−1 dr

=12An−1

∫ ∞

0

e−t tn/2−1 dt

=12An−1Γ

(n

2

),

where the gamma function is defined by

(A.3) Γ(z) =∫ ∞

0

e−t tz−1 dt,

for Re z > 0. Thus we have the formula

(A.4) An−1 =2πn/2

Γ( 12n)

.

To be satisfied with this, we need an explicit evaluation of Γ(n/2). Thiscan be obtained from Γ(1/2) and Γ(1) via the following identity:

(A.5)

Γ(z + 1) =∫ ∞

0

e−t tz dt

= −∫ ∞

0

d

dt(e−t) tz dt

= zΓ(z),

for Re z > 0, where we used integration by parts. The definition (A.3)clearly gives

(A.6) Γ(1) = 1.

Thus, for any integer k ≥ 1,

(A.7) Γ(k) = (k − 1)Γ(k − 1) = · · · = (k − 1)! .

Note that, for n = 2, we have A1 = 2π/Γ(1), so (A.6) agrees with the factthat the circumference of the unit circle is 2π (which, of course, figured

A. The mighty Gaussian and the sublime gamma function 101

into the proof of (3.18), via (3.20)). In case n = 1, we have A0 = 2, whichby (A.4) is equal to 2π1/2/Γ(1/2), so

(A.8) Γ(1

2

)= π1/2.

Again using (A.5), we see that, when k ≥ 1 is an integer,

(A.9)

Γ(k +

12

)=

(k − 1

2

)Γ(k − 1

2

)= · · ·

=(k − 1

2

)(k − 3

2

)· · ·

(12

)Γ(1

2

)

= π1/2(k − 1

2

)(k − 3

2

)· · ·

(12

).

In particular, Γ(3/2) = (1/2)Γ(1/2) = π1/2/2, so A2 = 2π3/2/(π1/2/2) =4π, which agrees with the well known formula for the area of the unit spherein R3.

Note that while Γ(z) defined by (A.3) is a priori holomorphic for Re zpositive, the equation (A.5) shows that Γ(z) has a meromorphic extensionto the entire complex plane, with simple poles at z = 0,−1,−2, . . . . Itturns out that Γ(z) has no zeros, so 1/Γ(z) is an entire analytic function.This is a consequence of the identity

(A.10) Γ(z)Γ(1− z) =π

sin πz,

which we now establish. From (A.4) we have (for 0 < Re z < 1)

(A.11)

Γ(z)Γ(1− z) =∫ ∞

0

∫ ∞

0

e−(s+t)s−ztz−1 ds dt

=∫ ∞

0

∫ ∞

0

e−uvz−1(1 + v)−1 du dv

=∫ ∞

0

(1 + v)−1vz−1 dv,

where we have used the change of variables u = s + t, v = t/s. Withv = ex, the last integral is

(A.12)∫ ∞

−∞(1 + ex)−1 exz dx,

which is holomorphic for 0 < Re z < 1, and we want to show that it isequal to the right side of (A.10) on this strip. It suffices to prove identityon the line z = 1/2+ iξ, ξ ∈ R; then (A.12) is equal to the Fourier integral

(A.13)∫ ∞

−∞

(2 cosh

x

2

)−1

eixξ dx.

To evaluate this, shift the contour of integration from the real line to theline Im x = −2π. There is a pole of the integrand at x = −πi, and we have

102

(A.13) equal to

(A.14) −∫ ∞

−∞

(2 cosh

x

2

)−1

e2πξeixξ dx− Residue · (2πi).

Consequently, (A.13) is equal to

(A.15) −2πiResidue1 + e2πξ

=π

cosh πξ,

and since π/

sin π(1/2 + iξ) = π/ cosh πξ, the demonstration of (A.10) iscomplete.

The integral (A.3) and also the last integral in (A.11) are special casesof the Mellin transform:

(A.16) Mf(z) =∫ ∞

0

f(t)tz−1 dt.

If we evaluate this on the imaginary axis:

(A.17) M#f(s) =∫ ∞

0

f(t)tis−1 dt,

given appropriate growth restrictions on f , this is related to the Fouriertransform by a change of variable:

(A.18) M#f(s) =∫ ∞

−∞f(ex)eisx dx.

The Fourier inversion formula and Plancherel formula imply

(A.19) f(r) = (2π)−1

∫ ∞

−∞(M#f)(s)r−is ds

and

(A.20)∫ ∞

−∞|M#f(s)|2 ds = (2π)

∫ ∞

0

|f(r)|2r−1 dr.

In some cases, as seen above, one evaluates Mf(z) on a vertical line otherthan the imaginary axis, which introduces only a slight wrinkle.

An important identity for the gamma function follows from taking theMellin transform with respect to y of both sides of the subordination iden-tity

(A.21) e−yA =12yπ−1/2

∫ ∞

0

e−y2/4te−tA2t−3/2 dt

(if y > 0, A > 0), established in §5; see (5.22). The Mellin transform ofthe left side is clearly Γ(z)A−z. The Mellin transform of the right side isa double integral, which is readily converted to a product of two integrals,each defining gamma functions. After a few changes of variables, there


results the identity

(A.22) π1/2Γ(2z) = 22z−1Γ(z)Γ(z +

12

),

known as the duplication formula for the gamma function. In view ofthe uniqueness of Mellin transforms, following from (A.18) and (A.19), theidentity (A.22) conversely implies (A.21). In fact, (A.22) was obtained first(by Legendre) and this argument produces one of the standard proofs ofthe subordination identity (A.21).

There is one further identity, which, together with (A.5), (A.10), and(A.22), completes the list of the basic elementary identities for the gammafunction. Namely, if the beta function is defined by

(A.23) B(x, y) =∫ 1

0

sx−1(1− s)y−1 ds =∫ 1

0

(1 + u)−x−yux−1 du

(with u = s/(1− s)), then

(A.24) B(x, y) =Γ(x)Γ(y)Γ(x + y)

.

To prove this, note that since

(A.25) Γ(z)p−z =∫ ∞

0

e−pttz−1 dt,

we have

(1 + u)−x−y =1

Γ(x + y)

∫ ∞

0

e−(1+u)ttx+y−1 dt,

so

B(x, y) =1

Γ(x + y)

∫ ∞

0

e−ttx+y−1

∫ ∞

0

e−utux−1 du dt

=Γ(x)

Γ(x + y)

∫ ∞

0

e−tty−1 dt

=Γ(x)Γ(y)Γ(x + y)

,

as asserted.The four basic identities proved above are the workhorses for most appli-

cations involving gamma functions, but fundamental insight is provided bythe identities (A.27) and (A.31) below. First, since 0 ≤ e−t− (1− t/n)n ≤

104

e−t · t2/n for 0 ≤ t ≤ n, we have, for Re z > 0,

(A.26)

Γ(z) =∫ ∞

0

e−ttz−1 dt

= limn→∞

∫ n

0

(1− t

n

)n

tz−1 dt

= limn→∞

nz

∫ 1

0

(1− s)nsz−1 ds.

Repeatedly integrating by parts gives

Γ(z) = limn→∞

nz n(n− 1) · · · 1z(z + 1) · · · (z + n− 1)

∫ 1

0

sz+n−1 ds,

which yields the following result of Euler:

(A.27) Γ(z) = limn→∞

nz 1 · 2 · · ·nz(z + 1) · · · (z + n)

.

Using the identity (A.5), analytically continuing Γ(z), we have (A.27) forall z, other than 0,−1,−2, . . . . We can rewrite (A.27) as

(A.28) Γ(z) = limn→∞

nz z−1(1 + z)−1(1 +

z

2

)−1

· · ·(1 +

z

n

)−1

.

If we denote by γ Euler’s constant:

(A.29) γ = limn→∞

(1 +

12

+ · · ·+ 1n− log n

),

then (A.28) is equivalent to(A.30)

Γ(z) = limn→∞

e−γzez(1+1/2+···+1/n)z−1(1 + z)−1(1 +

z

2

)−1

· · ·(1 +

z

n

)−1

,

that is, to the Euler product expansion

(A.31)1

Γ(z)= z eγz

∞∏n=1

(1 +

z

n

)e−z/n.

It follows that the entire analytic function 1/Γ(z)Γ(−z) has the productexpansion

(A.32)1

Γ(z)Γ(−z)= −z2

∞∏n=1

(1− z2

n2

).

Since Γ(1−z) = −zΓ(−z), by virtue of (A.10) this last identity is equivalentto the Euler product expansion

(A.33) sin πz = πz

∞∏n=1

(1− z2

n2

).


It is quite easy to deduce the formula (A.5) from the Euler product expan-sion (A.31). Also, to deduce the duplication formula (A.22) from the Eulerproduct formula is a fairly straightforward exercise.

Finally, we derive Stirling’s formula, for the asymptotic behavior of Γ(z)as z → +∞. The approach uses the Laplace asymptotic method, whichhas many other applications. We begin by setting t = sz and then s = ey

in the integral formula (A.3), obtaining

(A.34)Γ(z) = zz

∫ ∞

0

e−z(s−log s)s−1ds

= zz

∫ ∞

−∞e−z(ey−y) dy.

The last integral is of the form

(A.35)∫ ∞

−∞e−zϕ(y) dy,

where ϕ(y) = ey − y has a nondegenerate minimum at y = 0; ϕ(0) =1, ϕ′(0) = 0, ϕ′′(0) = 1. If we write 1 = A(y) + B(y), A ∈ C∞0 ((−2, 2)),A(y) = 1 for |y| ≤ 1, then the integral (A.35) is readily seen to be

(A.36)∫ ∞

−∞A(y)e−zϕ(y)dy + O(e−(1+1/e)z).

We can make a smooth change of variable x = ξ(y) such that ξ(y) =y + O(y2), ϕ(y) = 1 + x2/2, and the integral in (A.36) becomes

(A.37) e−z

∫ ∞

−∞A1(x)e−zx2/2 dx,

where A1 ∈ C∞0 (R), A1(0) = 1, and it is easy to see that, as z → +∞,

(A.38)∫ ∞

−∞A1(x)e−zx2/2 dx ∼

(2π

z

)1/2[1 +

a1

z+ · · ·

].

In fact, if z = 1/2t, then (A.38) is equal to (4πt)1/2u(t, 0), where u(t, x)solves the heat equation, ut − uxx = 0, u(0, x) = A1(x). Returning to(A.34), we have Stirling’s formula:

(A.39) Γ(z) =(2π

z

)1/2

zz e−z[1 + O(z−1)

].

Since n! = Γ(n + 1), we have in particular that

(A.40) n! = (2πn)1/2 nn e−n[1 + O(n−1)

]

as n →∞.Regarding this approach to the Laplace asymptotic method, compare

the derivation of the stationary phase method in Appendix B of Chapter6.

106

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Documents

Fourier Analysis, Distributions, and Constant-Coe–cient ...mtaylor.web.unc.edu/files/2018/04/fourier.pdfwith Fourier analysis on ﬂnite cyclic groups. We study this both as an approximation