Invariant Mean Value Property andHarmonic Functions

Jinman Kim and M. W. Wong∗

Department of Mathematics and StatisticsYork University

4700 Keele StreetToronto, Ontario M3J 1P3

Canada

Abstract We give conditions on the functions σ and u on Rn suchthat if u is given by the convolution of σ and u, then u is harmonicon Rn.

Keywords: Harmonic functions; Mean value property, Convolutiontransform; Heat kernel; Fourier transforms; Paley-Wiener theorem;Fourier hyperfunctions, Gelfand–Shilov spaces; Weyl’s lemma

2000 Mathematics Subject Classifications: Primary 46F12, 46F15;Secondary 30D15

1 Introduction

A function u on Rn is said to be harmonic if it is in C2(Rn) and such that

(∆u)(x) = 0, x ∈ Rn,∗Corresponding author. Email: mwwong@mathstat.yorku.ca

where ∆ =∑n

j=1∂2

∂x2j. One of the best characterizations of harmonic func-

tions is perhaps the mean value property to the effect that

1

µ(Sn−1)

∫Sn−1

u(x+ rω) dµ(ω) = u(x), x ∈ Rn, (1.1)

where µ is the Lebesgue measure on the unit sphere Sn−1 with center at theorigin in Rn. Now, let σ ∈ L1(Rn) be a radial function given by

σ(x) = Σ(|x|), x ∈ Rn.

Let u be a harmonic function for which the convolution transform Kσu ofu with kernel σ given by

Kσu = σ ∗ u

is defined. Then

(Kσu)(x) = µ(Sn−1)

(∫ ∞0

rn−1Σ(r) dr

)u(x), x ∈ Rn. (1.2)

If µ(Sn−1)∫∞

0rn−1Σ(r) dr = 1, then Kσu = u. In other words, u is invariant

under the convolution transform with kernel σ. This prompts us to look atthe following problem in this paper.

Problem Find conditions on the kernel σ and, if necessary, on the functionu such that Kσu = u implies that u is harmonic.

Some positive answers can be found in [1] by Ahern, Flores and Rudin, [4]by Englis and [2] by Ben Natan and Weit.

One of the main results in this paper is the following theorem.

Theorem 1.1 Let σ be a nonnegative and radial function in C20(Rn) such

that ∫Rn

σ(x) dx = 1.

Then for all tempered functions u on Rn with

Kσu = u,

u is harmonic on Rn.

2

For tempered functions and tempered distributions used in this paper,see, for instance, Chapter 4 of the book [7] by Wong.

The following example, due to M. Englis, points out the need for condi-tions on u for some kernels σ.

Example 1.2 Let σ(x) = 12πe−|x|

2/2, x ∈ R2. Then∫R2 σ(x) dx = 1. Let u

be the function on R2 defined by

u(x1, x2) = eax1+bx2 , x1, x2 ∈ R,

where a and b are complex numbers such that a2+b2 = 4πi. Then ∆u = 4πiuand an easy contour integration gives σ ∗ u = u.

It is important to realize that the kernel σ in Example 1.2 is the heatkernel evaluated at time t = 1

2. We give in the following theorem a condition

on the function u to ensure that the harmonicity of u is a consequence ofthe invariance of u under the convolution transform with the heat kernel attime t = 1

2.

Theorem 1.3 Let u ∈ L1loc(R

n) be such that∥∥u e−ε|·|∥∥L∞(Rn)

<∞

for every positive number ε and

Kσu = u,

whereσ(x) = (2π)−n/2e−|x|

2/2, x ∈ Rn.Then u is harmonic on Rn.

We conclude this introduction with a genesis of this paper. Theorem 1.1is proved in Section 2. This is achieved using the Paley–Wiener theorem forthe Fourier transform. The Fourier transform f of a function f in L1(Rn)is taken to be the one defined by

f(ξ) = (2π)−n/2∫Rn

e−ix·ξf(x) dx, ξ ∈ Rn.

Fourier hyperfunctions, which are used in the proof of Theorem 1.3, willbe recalled with proofs in Section 3. The proof of Theorem 1.3 is given inSection 4.

3

2 Proof of Theorem 1.1

Lemma 2.1 Let σ be as in Theorem 1.1. Then

(I −Kσ)S = ∆S,

where S is the Schwartz space on Rn.

Proof Without loss of generality, we assume that n = 1. Since ∧ : S → Sis a homeomorphism, it is sufficient to prove that the multiplication operatorq : S → S is surjective, where

q(ξ) =1− (2π)1/2σ(ξ)

−ξ2, ξ ∈ R.

In view of the Paley-Wiener theorem, σ is an even entire function on C and

σ(0) = (2π)−1/2

∫ ∞−∞

σ(x) dx = (2π)−1/2.

Using the Taylor series expansion of σ, we get

σ(z) =∞∑n=0

a2nz2n, z ∈ C,

where

a2n =1

(2n)!(2π)−1/2

∫ ∞−∞

(−ix)2nσ(x) dx, n = 1, 2, . . . .

Since a0 = (2π)−1/2, it follows that 0 is a removable singularity of thefunction q defined on the punctured plane C \ {0} by

q(z) = −(2π)1/2

∞∑n=1

a2nz2n−2, z 6= 0.

To prove that qϕ ∈ S for all ϕ ∈ S, it is enough to prove that for everynonnegative integer m, there exist positive constants Cm and Nm such that

|q(m)(x)| ≤ Cm(1 + |x|)Nm , x ∈ R.

4

But, by Cauchy’s integral formula,

q(m)(x) =m!

2πi

∫|z−x|=1

q(z)

(z − x)m+1dz, x ∈ R, (2.1)

for m = 0, 1, 2, . . . . If we assume that supp(σ) ⊆ [−R,R], then, by thePaley-Wiener theorem, there exists a positive constant C such that

|q(z)| ≤ CeR|Im z|, z ∈ C. (2.2)

Thus, by (2.1) and (2.2), there exists a positive constant C ′ such that

|q(m)(x)| ≤ C ′m!, m = 0, 1, 2, . . . .

Next,ϕ

q∈ S, ϕ ∈ S.

Indeed, using the fact that σ is even, we get for all nonzero x in R,

1− (2π)1/2σ(x) =

∫ ∞−∞

σ(ξ) dξ −∫ ∞−∞

e−ixξσ(ξ) dξ

=

∫ ∞−∞

(1− e−ixξ)σ(ξ) dξ

= 2

∫ ∞0

(1− cosxξ)σ(ξ) dξ > 0.

By Leibniz’ formula (see, for instance, the formula (0.4) in the book [7] byWong), we get for m = 0, 1, 2, . . . ,(

1

q

)(m)

=∑

m1+m2+···+mk=m

q(m1)q(m2) · · · q(mk)

qk+1,

wherem1,m2, . . . ,mk form a partition ofm. Therefore for allm = 0, 1, 2, . . . ,there exists a positive constant Cm such that∣∣∣∣∣

(1

q

)(m)∣∣∣∣∣ ≤ Cm

5

and we are done. 2

Proof of Theorem 1.1 Let ϕ ∈ S. Then using Kσu = u and Lemma 2.1,there exists a function ψ in S such that

(∆u)(ϕ) = u(∆ϕ) = u((I −Kσ)ψ) = (u−Kσ)(ψ) = 0.

Therefore ∆u = 0, and by Weyl’s lemma, u is harmonic on Rn.

3 Fourier Hyperfunctions

Let F be the set of all functions ϕ in C∞(Rn) such that there exist positiveconstants C, h and k for which

|(∂αϕ)(x)|ek|x| ≤ Ch|α|α!

for all x in Rn and all multi-indices α.Let {ϕj}∞j=1 be a sequence of functions in F . Then we say that ϕj → 0

in F as j →∞ if there exist positive constants h and k such that

supx,α

|(∂αϕj)(x)|ek|x|

h|α|α!→ 0

as j →∞.Let u be a linear functional on F such that u(ϕj) → 0 as j → ∞ for

every sequence {ϕj}∞j=1 in F with ϕj → 0 in F as j → ∞. Then we call ua Fourier hyperfunction on Rn. The set of all Fourier hyperfunctions on Rn

is denoted by F ′.It is obvious that F ⊆ S. Thus, tempered distributions on Rn are

contained in F ′.Using the fact proved by Chung, Chung and Kim in [3], F is the same

as the Gelfand–Shilov space S11 in Chapter 4 of the book [5] by Gelfand and

Shilov.

Lemma 3.1 Let ϕ ∈ C∞(R) be such that we can find a positive constantH and a positive constant Cε for every positive number ε for which

supm∈N0

∣∣∣∣ϕ(m)(x)

Hmm!

∣∣∣∣ ≤ Cεeε|x|, x ∈ R,

where N0 is the set of all nonnegative integers. Then ϕψ ∈ F for all ψ ∈ F .

6

Proof Let ψ ∈ F . Then there exist positive constants A and k such that

supm∈N0,x∈R

|ψ(m)(x)ek|x||Amm!

<∞.

In other words, there exists a positive constant C such that

|ψ(m)(x)| ≤ CAmm!e−k|x|, x ∈ R, m ∈ N0.

So, using Leibniz’ formula and the hypothesis on ϕ, we get a positive con-stant H and a positive number Cε for every positive number ε such that

|(ϕψ)(m)(x)|

=

∣∣∣∣∣m∑j=0

(m

j

)ϕ(j)(x)ψ(m−j)(x)

∣∣∣∣∣≤ C

m∑j=0

Cε

(m

j

)Hjj!Am−j(m− j)!eε|x|e−k|x|

for all x in R and all m in N0. Let ε = k2. Then

|(ϕψ)(m)(x)| ≤ Cm∑j=0

C k2

(m

j

)Hjj!Am−j(m− j)!e−k|x|/2

for all x in R and all m in N0. So, there exists a positive constant C ′ suchthat

|(ϕψ)(m)| ≤ C ′m∑j=0

m!{max(H,A)}me−k|x|/2, x ∈ R, m ∈ N0.

This shows that ∑m∈N0,x∈R

|(ϕψ)(m)(x)ek|x|/2|Kmm!

<∞,

where K = max(H,A). This completes the proof. 2

Lemma 3.2 For t > 0, let kt be the heat kernel given by

kt(x) = (4πt)−n/2e−|x|2

/4t, x ∈ Rn.

Then for all ϕ ∈ F , ϕt → ϕ in F as t→ 0+, where ϕt = kt ∗ ϕ.

7

Proof Since ϕ ∈ F , we can find positive constants C, h and k such that

|(∂αϕ)(x)| ≤ Ch|α|α!e−k|x|

for all x in Rn and all multi-indices α. Let δ be any positive number. Thenfor all x in Rn and all multi-indices α,

(∂αϕt)(x)− (∂αϕ)(x) = I1 + I2 − I3,

where

I1 =

∫|y|≤δ

kt(y){(∂αϕ)(x− y)− (∂αϕ)(x)}dy,

I2 =

∫|y|≥δ

kt(y)(∂αϕ)(x− y) dy

and

I3 =

∫|y|≥δ

kt(y)(∂αϕ)(x) dy.

Now, for |y| ≤ δ,, we can use the mean value theorem with integral remain-der (see Theorem 6.3 in the book [7] by Wong) to get for all x in Rn andall multi-indices α,

(∂αϕ)(x− y)− (∂αϕ)(x) =∑|γ|=1

(−y)γ∫ 1

0

(∂γ∂αϕ)(x− θy) dθ,

and hence there exist positive constants C ′ and H such that

|(∂αϕ)(x− y)− (∂αϕ)(x)| ≤ Cekδ|y|h|α|+1(|α|+ 1)!e−k|x|

≤ C ′|y|(Hh)|α||α|!e−k|x|.

Let ε be a positive number. Then we can find a positive number δ such thatC ′δ < ε

3. Hence

|I1|ek|x|

(Hh)|α||α|!<ε

3.

Next, for all x in Rn and all multi-indices α,

|I2|ek|x|

h|α|α!≤

∫|y|≥δ

ht(y)|(∂αϕ)(x− y)ek|x−y|ek|y|

hαα!dy

≤ C

∫|y|≥δ

ht(y)ek|y|dy <ε

3

8

provided that t is small enough. Finally, for all x in Rn and all multi-indicesα,

|I3|ek|x|

h|α|α!≤

∫|y|≥δ

ht(y)|(∂αϕ)(x)|ek|x|

h|α|α!dy

≤ C

∫|y|≥δ

ht(y) dy <ε

3

provided that t is small enough. 2

We need the following version of Weyl’s lemma for Fourier hyperfunc-tions.

Theorem 3.3 Let u ∈ F ′ be such that ∆u = 0. Then u is harmonic on Rn.

Proof For t > 0, the convolution u ∗ kt is a C∞ function given by

(u ∗ kt)(x) = u(kt(x− ·)), x ∈ Rn, t > 0.

Now, for t > 0,

0 = (∆u) ∗ kt = ∆(u ∗ kt) = u ∗ (∆kt) = u ∗ (∂tkt) = ∂t(u ∗ kt) = 0.

So, for every x in Rn,

(u ∗ kt)(x) = v(x), t ∈ (0,∞),

where v is a C∞ function on Rn. Since kt → δ in F ′ as t→ 0, it follows that

v = u ∗ kt → u ∗ δ = u

as t→ 0. Therefore u ∈ C∞(Rn) and the proof is complete. 2

4 Proof of Theorem 1.3

We begin with a careful look at M. Englis’ counterexample with respect tothe growth rate at infinity. Indeed,

|u(x)| = |eax1+bx2| ≤ e(|a|+|b|)|x|, x ∈ R2.

9

Thus, u is of an exponential growth at infinity. Now, we show that thisgrowth condition is in fact an optimal criterion for the implication

Kσu = u⇒ ∆u = 0.

To do this, we make use of Fourier analysis on Gelfand–Shilov spaces. Fornonnegative real numbers α and β, we define the subspace Sβα to be the setof all functions ϕ in S(R) for which there exist positive constants A, B andC such that for all nonnegative integers m and n,

supx∈R|xmϕ(n)(x)| ≤ C AmBn(m!)α(n!)β.

As is noted in Section 3, S11 = F by a result in Chung, Chung and Kim [3].

So, we can denote S11 by F and let F ′ be its dual space. Let u ∈ F ′. Then

we define the Fourier transform of u by

u(ϕ) = u(ϕ), ϕ ∈ F .

Lemma 4.1 The Fourier transforms ∧ : F → F and ∧ : F ′ → F ′ arehomeomorphisms.

According to a result in Kaneko [6], a function u in L1loc(R

n) is a Fourierhyperfunction if and only if for every positive number ε,∥∥u e−ε|·|∥∥

L∞(Rn)<∞.

We are now ready for a proof of Theorem 1.3.

Proof of Theorem 1.3 Again, we assume that n = 1. First we prove that

(I −Kσ)F = ∆F .

By Lemma 3.1, this is the same as proving that the multiplication operatorq : F → F is surjective, where q is given by

q(x) =1− (2π)1/2σ(x)

−x2=

1− e−x2/2

−x2, x ∈ R.

As in the complex-analytic proof of Lemma 1.2, we get positive constantsA and C such that

|q(m)(x)| =∣∣∣∣m!

2πi

∫|z−x|=A−1

q(z)

(z − x)m+1dz

∣∣∣∣ ≤ C m!Am, m = 0, 1, 2, . . . .

10

Thus, qϕ ∈ F for all ϕ in F . This follows from a fact in [6], which statesthat if a function η in C∞(R) is such that for every positive number ε, thereexist positive constants A and Cε satisfying

|η(m)(x)|Amm!

≤ Cεeε|x|, x ∈ R,

for m = 0, 1, 2, . . . . Then ηϕ ∈ F for all ϕ ∈ F . To estimate(

1q

)(m)

, we

first note that1− e−z2/2 = 0⇔ e−z

2/2 = 1,

where z = x+ iy. To solve the equation for z, we write the equation as

e(y2−x2)/2e−ixy = 1.

Therefore ∣∣∣e(y2−x2)/2∣∣∣ = 1,

and we get x = ±y. So,

−ix(±x) = 2kπi, k ∈ Z.

Hencex = ±

√2kπ, k = 0, 1, 2, . . . .

Thus, on the strip R given by

R = {x+ iy : x ∈ R, |y| ≤√

2π/4},

e−z2/2 = 1⇔ z = 0.

Since the function1

q(z)=

−z2

1− e−z2/2

has a removable singularity at z = 0, Cauchy’s integral formula gives(1

q

)(m)

(x) =m!

2πi

∫|z−x|=

√2π/4

(1/q)(z)

(z − x)m+1dz, x ∈ R.

11

Now, to estimate(

1q

)(x), we note that for |x| sufficiently large, say, |x| ≥ η,

|1− e−z2/2| ≥ 1

2,

and so, ∣∣∣∣(1

q

)(z)

∣∣∣∣ =

∣∣∣∣ −z2

1− e−z2/2

∣∣∣∣ ≤ 2|z|2 ≤ 4x2.

Hence there exists a positive constant C such that∣∣∣∣(1

q

)(z)

∣∣∣∣ ≤ C(x2 + 1), z ∈ R.

So, using Leibniz’ formula as in the proof of Theorem 1.1, we get positiveconstants B and C ′ such that∣∣∣∣(1

q(x)

)∣∣∣∣ ≤ C ′m!Bm(x2 + 1), x ∈ R.

Since for every positive number ε, we can choose a positive constant Cε suchthat

x2 + 1 ≤ Cεeε|x|, x ∈ R,

it follows from Lemma 3.1 that 1qϕ ∈ F for all ϕ in F . As in the proof

of Theorem 2.1, we get ∆u = 0 in the sense of Fourier hyperfunctions.Using the analytic hypoellipticity of Fourier hyperfunctions, u is a classicalharmonic function, and this completes the proof.

Acknowledgment This research has been partially supported by the Nat-ural Sciences and Engineering Research Council of Canada.

References

[1] P. Ahern, M. Flores and W. Rudin, An invariant volume-mean-value-property, J. Funct. Anal. 111 (1993), 380–397.

[2] Y. Ben Natan and Y. Weit, Integrable harmonic functions on Rn, J.Funct. Anal. 150 (1997), 471–477.

12

[3] J. Chung, S. Y. Chung and D. Kim, Characterizations of the Gelfand-Shilov spaces via Fourier transforms, Proc. Amer. Math. Soc. 124(1996), 2101–2108.

[4] M. Englis, Functions invariant under the Berezin transform, J. Funct.Anal. 121 (1994), 233–254.

[5] I. M. Gelfand and G. E. Shilov, Generalized Functions II, AcademicPress, 1968.

[6] A. Kaneko, Introduction to Hyperfunctions, Iwanami, 1978.

[7] M. W. Wong, An Introduction to Pseudo-Differential Operators, Sec-ond Edition, World Scientific, 1999.

13