Mean Value Properties

1. Harmonic Functions

1.1 Guide

In this chapter we will use various methods to study harmonic functions. These

include mean value properties, fundamental solutions, maximum principles and energy

methods. Four sections in this chapter are relatively independent of each other.

The materials in this chapter are rather elementary, but they contain several im-

portant ideas on the whole subject, and thus should be covered thoroughly. While doing

Sections 1.2-1.3, the classic book of Protter and Weinberger [13] may be a very good

reference. Also, when one reads Section 1.4, some statements concern- ing the Hopf

maximal principle in Section 2.2 can be selected as exercises. The interior gradient esti-

mates of Section 2.4 follow from the same arguments as those in the proof of Proposition

1.31 in Section 1.4.

1.2 Mean Value Properties

We begin this section with the denition of mean value properties. We assume that

Ω is a connected domain in R

Denition 1.1. For u ∈ C(Ω) we dene

(i) u satises the rst mean value property if

u(x) =

n−1

∂B

(x)

u(y)dS

for any B

(x) ⊂ Ω ;

(ii) u satisfes the second mean value property if

u(x) =

(x)

u(y)dy for any B

(x) ⊂ Ω

where ω

denotes the surface area of the unit sphere in R

Remark 1.2. These two denitions are equivalent. In fact, if we write (i) as

Chapter 1: Harmonic Functions

u(x)r

n−1

∂B

(x)

u(y)dS

we may integrate to get (ii). If we write (ii) as

u(x)r

(x)

u(y)dy ,

Remark 1.3. We may write the mean value properties in the following equivalent ways:

(i) u satisfes the fırst mean value property if

u(x) =

|w|=1

u(x + rw)dS

for anyB

(x) ⊂ Ω ;

(ii) u satisfes the second mean value property if

u(x) =

|z|≤1

u(x + rz)dz for any B

(x) ⊂ Ω .

Now we prove the maximum principle for the functions satisfying mean value prop-

erties.

Proposition 1.4. If u ∈ C(

Ω) satisfres the mean value property in Ω, then u assumes

its maximum and minimum only on ∂Ω unless u is constant.

Proof. We only prove for the maximum. Set

Σ =



x ∈ Ω; u(x) = M ≡ max

Ω



⊂ Ω .

It is obvious that Σ is relatively closed. Next we show that Σ is open. For any x

∈ Σ,

take

) ⊂ Ω for some r > 0. By the mean value property we have

M = u(x

) =

)

u(y)dy ≤ M

)

dy = M .

This implies u = M in B

). Hence Σ is both closed and open in Ω. Therefore

either Σ = ∅ or Σ = Ω.

Denition 1.5. A function u ∈ C

(Ω) is harmonic if ∆u = 0 in Ω.

Theorem 1.6. Let u ∈ C

(Ω) be harmonic in Ω. Then u satises the mean value

property in Ω.

Proof. Take any ball B

(x) ⊂ Ω. For ρ ∈ (0, r)�we apply the divergence theorem in

(x) and get

Section 1.2: Mean Value Properties

(x)

∆u(y)dy =

∂B

∂u

∂ν

dS = ρ

n−1

|w|=1

∂u

∂ρ

(x + ρw)dS

= ρ

n−1

∂

∂ρ

|w|=1

u(x + ρw)dS

(∗)

Hence for harmonic function u we have for any ρ ∈ (0, r)

∂

∂ρ

|w|=1

u(x + ρw)dS

= 0 .

Integrating from 0 to r we obtain

|w|=1

u(x + rw)dS

|w|=1

u(x)dS

= u(x)ω

u(x) =

|w|=1

u(x + rw)dS

n−1

∂B

(x)

u(y)dS

Remark 1.7. For a function u satisfying the mean value property, u is not required to

be smooth. However a harmonic function is required to be C

. We prove these two are

equivalent.

Theorem 1.8. If u ∈ C(Ω) has mean value property in Ω, then u is smooth.

Proof. Choose φ ∈ C

∞

(0)) with

(0)φ = 1 and φ(x) = ψ(|x|)�that is,

n−1

ψ(r)dr = 1 .

We dene φ

(z) =

φ(

) for ε > 0. Now for any x ∈ Ω consider ε <dist(x, ∂Ω). Then

we have

Ω

u(y)φ

(y − x)dy =

u(x + y)φ

(y)dy =

|y |<ε

u(x + y)φ

|y |< 1

u(x + εy)φ(y)dy

n−1

∂B

(0)

u(x + εrw)φ(rw)dS

ψ(r)r

n−1

|w|=1

u(x + εrw)dS

= u(x)ω

ψ(r)r

n−1

dr = u(x)

Chapter 1: Harmonic Functions

where in the last equality we used the mean value property. Hence we Get

u(x) = (φ

∗ u)( x) for any x ∈ Ω

= {y ∈ Ω; d(y, ∂Ω) > ε}.

Therefore u is smooth. Moreover, by formula (∗) in the proof of Theorem 1.2 and the

mean value property we have

(x)

∆u = r

n−1

∂

∂r

|w|=1

u(x + rw)dS

= r

n−1

∂

∂r



u(x)



= 0 for any B

(x) ⊂ Ω .

This implies ∆u = 0 in Ω.

Remark 1.9. By combining Theorem 1.6 and Theorem 1.8,we conclude that harmonic

functions are smooth and satisfy the mean value property. Hence harmonic functions

satisfy the maximum principle, a consequence of which is the uniqueness of solution to

the following Dirichlet problem in a bounded domain

∆u = f in Ω

u = φ on ∂Ω

for f ∈ C(Ω) and φ ∈ C( ∂Ω). In general uniqueness does not hold for an unbounded

domain. Consider the following Dirichlet problem in the unbounded domain Ω

∆u = 0 in Ω

u = 0 on ∂Ω .

First consider the case Ω = {x ∈ R

; |x| > 1}. For n = 2, u(x) = log |x| is a solution.

Note u → ∞ as r → ∞. For n ≥ 3, u(x) = |x|

2−n

− 1 is a solution. Note u → −1 as

r → ∞. Hence u is bounded. Next, consider the upper half space Ω = {x ∈ R

; x

> 0}.

Then u(x) = x

is a nontrivial solution, which is unbounded.

In the following we discuss the gradient estimates.

Lemma 1.10. Suppose u ∈ C(

) is harmonic in B

= B

). Then there holds

|Du(x

)| ≤

max

|u| .

Proof. For simplicity we assume u ∈ C

(

). Since u is smooth, then ∆(D

u) = 0,

that is, D

u is also harmonic in B

. Hence D

u satises the mean value property. By

the divergence theorem we have

Section 1.2: Mean Value Properties

u(x

) =

)

u(y)dy =

∂B

)

u(y)v

which implies

u(x

)| ≤

max

∂B

|u| · ω

n−1

≤

max

|u| .

Lemma 1.11. Suppose u ∈ C(

) is a nonnegative harmonic function in B

).T hen there holds

|Du(x

)| ≤

u(x

) .

Proof. As before by the divergence theorem and the nonnegativeness of u we have

u(x

)| ≤

∂B

)

u(y) dS

u(x

)

where in the last equality we used the mean value property.

Corollary 1.12. A harmonic function in R

bounded from above or below is constant.

Proof. Suppose u is a harmonic function in R

. We will prove that u is a constant if

u ≥ 0. In fact, for any x ∈ R

we apply Lemma l.11 to u in B

(x) and then let R → ∞.

We conclude that Du(x) = 0 for any x ∈ R

Proposition 1.13. Suppose u ∈ C(

) is harmonic in B

= B

).T hen there holds

for any multi-index � with |α| = m

u(x

)| ≤

m−1

max

|u| .

Proof. We prove by induction. It is true for m = 1 by Lemma 1.10. Assume it holds for

m. Consider m + 1. For 0 < θ < 1�dene r = (1 − θ)R ∈ (0, R).We apply Lemma 1.10

to u in B

and get



D

m+1

u(x

)



≤

max

u| .

By the induction assumption we have

max

u| ≤

· e

m−1

· m!

(R − r)

max

|u| .

Hence we obtain



m+1

u(x

)



≤

m−1

(R − r)

max

|u| =

m+1

m−1

m+1

(1 − θ)

max

|u| .

Chapter 1: Harmonic Functions

Take θ =

m+1

. This implies

(1 − θ)



1 +



(m + 1) < e(m + 1) .

Hence the result is established for any single derivative. For any multi-index α =

(α

, . . . , α

) we have

! · · · α

! ≤ (|α|)! .

Theorem 1.14. Harmonic function is analytic.

Proof. Suppose u is a harmonic function in Ω. For xed x ∈ Ω�take B

(x) ⊂ Ω and

h ∈ R

with |h| ≤ R. We have by Taylor expansion

u(x + h) = u(x) +

m−1

i=1



∂

∂x

+ · · · + h

∂

∂x



(x) + R

(h)

where

(h) =



∂

∂x

+ · · · + h

∂

∂x





+ θh

, . . . , x

+ θh

)

for some θ ∈ (0, 1). Note x + h ∈ B

(x) for |h| < R. Hence by Proposition 1.13 we

obtain

(h)| ≤

|h|

· n

m−1

max

|u| ≤



|h|n



max

|u| .

Then for any h with |h|n

e < R/2 there holds R

(h) → 0 as m → ∞.

Next we prove the Harnack inequality.

Theorem 1.15. Suppose u is harmonic in Ω. Then for any compact subset K of Ω

there exists a positive constant C = (Ω, K) such that if u ≥ 0 in Ω�then

u(y) ≤ u(x) ≤ Cu(y) for any x, y ∈ K .

Proof. By mean value property, we can prove if B

) ⊂ Ω�then

u(y) ≤ u(x) ≤ cu(y) for anyx, y ∈ B

)

where c is a positive constant depending only on n. Now for the given compact subset

K�take x

, . . . , x

∈ K such that {B

)} covers K with 4

R <dist(

K, ∂Ω). Then we

can choose C = c

Section 1.2: Mean Value Properties

We nish this section by proving a result, originally due to Weyl. Suppose u is

harmonic in Ω. Then integrating by parts we have

Ω

u∆φ = 0 for any φ ∈ C

(Ω) .

The converse is also true.

Theorem 1.16. Suppose u ∈ C(Ω) satisfres

Ω

u∆φ = 0 for any φ ∈ C

(Ω) . (1.1)

Then u is harmonic in Ω.

Proof. We claim for any B

(x) ⊂ Ω there holds

∂B

(x)

u(y)dS

= n

(x)

u(y)dy . (1.2)

Then we have

n−1

∂B

(x)

u(y)dS







(x)

u(y)dy







(

−

n+1

(x)

u(y)dy +

∂B

(x)

u(y)dS

)

= 0 .

This implies

n−1

∂B

(x)

u(y)dS

= const .

This constant is u(x) if we let r → 0. Hence we have

u(x) =

n−1

∂B

(x)

u(y)dS

for any B

(x) ⊂ Ω .

Next we prove (2) for n ≥ 3. For simplicity we assume that x = 0. Set

φ(y, r) =







(|y|

− r

)

|y| ≤ r

0 |y| > r

and then φ

(y, r) = (|y|

− r

)

n−k

(2(n − k + 1)|y|

+ n(|y|

− r

)) for |y| ≤ r and

Chapter 1: Harmonic Functions

k = 2 , 3, . . . , n. Direct calculation shows φ(·, r) ∈ C

(Ω) and

∆

φ(y, r) =







2nφ

(y, r) |y| ≤ r

0 |y| > r .

By assumption (1) we have

(0)

u(y)φ

(y, r)dy = 0 .

Now we prove if for some k = 2, . . . , n − 1�

(0)

u(y)φ

(y, r)dy = 0 (1.3)

then

(0)

u(y)φ

k+1

(y, r)dy = 0 . (1.4)

In fact, we dierentiate (3) with respect to r and get

∂B

(0)

u(y)φ

(y, r)dy +

(0)

u(y)

∂φ

∂r

(y, r)dy = 0 .

For 2 ≤ k < n, φ

(y, r) = 0 for |y| = r. Then we have

(0)

u(y)

∂φ

∂r

(y, r)dy = 0 .

Direct calculation yields

∂φ

∂r

(y, r) = (−2r)(n − k + 1)φ

k+1

(y, r). Hence we have (4).

Therefore by taking k = n − 1 in (4) we conclude

(0)

u(y)



(n + 2)|y|

− nr



dy = 0 .

Dierentiating with respect to r again we get (2).