Maximum Principles

Section 1.4: Maximum Principles

1.4 Maximum Principles

In this section we will use the maximum principle to derive the interior gradient

estimate and the Harnack inequality.

Theorem 1.29. Suppose u ∈ C

) ∩ C(B

) is a subharmonic function in B

; that

is, ∆u ≥ 0. Then there holds

sup

u ≤ sup

∂B

Proof. For ε > 0 we consider u

(x) = u(x) + ε|x|

in B

. Then simple calculation yields

∆u

= ∆u + 2nε ≥ 2nε > 0.

It is easy to see, by a contradiction argument, that u

cannot have an interior maximum,

in particular,

sup

≤ sup

∂B

Therefore we have

sup

u ≤ sup

≤ sup

∂B

u + ε.

We nish the proof by letting ε → 0.

Remark 1.30. The result still holds if B

is replaced by any bounded domain.

The next result is the interior gradient estimate for harmonic functions. The method

is due to Bernstein back in 1910.

Proposition 1.31. Suppose u is harmonic in B

. Then there holds

sup

1/2

|Du| ≤ c sup

∂B

|u|

where c = c(n) is a positive constant. In particular, for any α ∈ [0, 1] there holds

|u(x) − u(y)| ≤ c|x − y|

sup

∂B

|u| for any x, y ∈ B

1/2

where c = c(n, α) is a positive constant.

Proof. Direct calculation shows that

∆(|Du|

) = 2

i,j=1

+ 2

i=1

(∆u) = 2

i,j=1

where we used ∆u = 0 in B

. Hence |Du|

is a subharmonic function. To get interior

Chapter 1: Harmonic Functions

estimates we need a cuto function. For any φ ∈ C

) we have

∆(φ|Du|

) = (∆φ)|Du|

+ 4

i,j

φD

u + 2φ

i,j

By taking φ = η

for some η ∈ C

) with η = 1 in B

1/2

, we obtain by the Hölder

inequality

∆(η

|Du|

) = 2η∆η|Du|

+ 2|Dη|

|Du|

+ 8η

i,j=1

ηD

u + 2η

i,j=1

≥ (2η∆η − 6|Dη|

)|Du|

≥ −C| Du|

where C is a positive constant depending only on η. Note that ∆(u

) = 2|Du|

+2u∆u =

2|Du|

since u is harmonic. By taking α large enough we get

∆(η

|Du|

+ αu

) ≥ 0.

We may apply Theorem 1.29 (the maximum principle) to get the result.

Next we derive the Harnack inequality.

Lemma 1.32. Suppose u is a nonnegative harmonic function in B

. Then there holds

sup

1/2

|D log u| ≤ C

where C = C(n) is a positive constant.

Proof. We may assume u > 0 in B

. Set v = log u. Then direct calculation shows

∆v = −|Dv|

We need the interior gradient estimate on v. Set w = |Dv|

. Then we get

∆w + 2

i=1

w = 2

i,j=1

As before we need a cuto function. First note

i,j=1

≥

i=1

≥

(∆v)

|Dv|

. (1.6)

Section 1.4: Maximum Principles

Take a nonnegative function φ ∈ C

). We obtain by the Hölder inequality

∆(φw) + 2

(φw) = 2φ

i,j

+ 4

i,j

φD

v + 2w

φD

v + (∆φ)w

≥ φ

i,j=1

− 2|Dφ||Dv|

−



|∆φ| + C

|Dφ|



|Dv|

if φ is chosen such that |Dφ|

/φ is bounded in B

. Choose φ = η

for some η ∈ C

Hence for such xed η we obtain by (1.1)

∆(

) + 2

i=1

(

)

≥

|Dv|

− Cη

|Dη||Dv|

− 4η

(η∆η + C|Dη|

)|Dv|

≥

|Dv|

− Cη

|Dv|

− Cη

|Dv|

where C is a positive constant depending only on n and η. Hence we get by the Hölder

inequality

∆(η

w) + 2

i=1

(η

w) ≥

− C

where C is a positive constant depending only on n and η. Suppose η

w attains its

maximum at x

∈ B

. Then D(η

w) = 0 and ∆(η

w) ≤ 0 at x

. Hence there holds

) ≤ C(n, η).

If w(x

) ≥ 1, then η

w(x

) ≤ C(n). Otherwise η

w(x

) ≤ w(x

) ≤ η

). In both

cases we conclude

w ≤ C(n, η) in B

Corollary 1.33. Suppose u is a nonnegative harmonic function in B

. Then there holds

u(x

) ≤ Cu(x

) for any x

, x

∈ B

1/2

where C is a positive constant depending only on n.

Proof. We may assume u > 0 in B

. For any x

, x

∈ B

1/2

by simple integration we

obtain with Lemma 1.32

log

u(x

)

u(x

)

≤ |x

− x

|D log u(tx

+ (1 − t)x

)|dt ≤ C|x

− x

Chapter 1: Harmonic Functions

Next we prove a quantitative Hopf lemma.

Proposition 1.34. Suppose u ∈ C(B

) is a harmonic function in B

= B

(0). If

u(x) < u(x

) for any x ∈ B

and some x

∈ ∂B

, then there holds

∂u

∂n

) ≥ C(u(x

) − u(0))

where C is a positive constant depending only on n.

Proof. Consider a positive function v in B

dened by

v(x) = e

−α|x|

− e

−α

It is easy to see

∆v(x) = e

−α|x|

(−2αn + 4α

|x|

) > 0 for any |x| ≥

if α ≥ 2n + 1 . Hence for such xed α the function v is subharmonic in the region

A = B

\ B

1/2

. Now dene for ε > 0

(x) = u(x) − u(x

) + εv(x).

This is also a subharmonic function, that is, ∆h

≥ 0 in A. Obviously h

≤ 0 on

∂B

and h

) = 0. Since u(x) < u(x

) for |x| =

we may take ε > 0 small such

that h

(x) < 0 for |x| =

. Therefore by Theorem 1.29 h

assumes at the point x

its

maximum in A. This implies

∂h

∂n

) ≥ 0 or

∂u

∂n

) ≥ −ε

∂v

∂n

) = 2αεe

−α

> 0.

Note that so far we have only used the subharmonicity of u. We estimate ε as follows.

Set w(x) = u(x

) − u(x) > 0 in B

. Obviously w is a harmonic function in B

. By

Corollary 1.33 (the Harnack inequality) there holds

inf

1/2

w ≥ c(n)w(0) or max

1/2

u ≤ u(x

) − c(n)(u(x

) − u(0)).

Hence we may take

ε = δc(n)(u(x

) − u(0))

for δ small, depending on n. This nishes the proof.

To nish this section we prove a global Hölder continuity result.

Lemma 1.35. Suppose u ∈ C(B

) is a harmonic function in B

with u = φ on ∂B

. If

Section 1.4: Maximum Principles

φ ∈ C

(∂B

) for some α ∈ (0, 1), then u ∈ C

α/2

). Moreover, there holds

kuk

α/2

)

≤ Ckφk

(∂B

)

where C is a positive constant depending only on n and α.

Proof. First the maximum principle implies that inf

∂B

φ ≤ u ≤ sup

∂B

φ in B

. Next

we claim that for any x

∈ ∂B

there holds

sup

x∈B

|u(x) − u(x

|x − x

α/2

≤ 2

α/2

sup

x∈∂B

|φ(x) − φ(x

|x − x

. (1.7)

Lemma 1.35 follows easily from (1.7). For any x, y ∈ B

, set d

= dist(x, ∂B

) and

= dist(y, ∂B

). Suppose d

≤ d

. Take x

, y

∈ ∂B

such that |x − x

| = d

and

|y − y

| = d

. Assume rst that |x −y| ≤ d

/2.

Then y ∈

(x) ⊂ B

. We apply Proposition 1.31 (scaled version) to

u − u(x

) in B

(x) and get by (1.7)

α/2

|u(x) − u(y)|

|x − y|

α/2

≤ Cku − u(x

∞

(x))

≤ Cd

α/2

kφk

(∂B

)

Hence we obtain

|u(x) − u(y)| ≤ C|x − y|

α/2

kφk

(∂B

)

Assume now that d

≤ d

≤ 2|x − y|. Then by (1.7) again we have

|u(x) − u(y)| ≤ |u(x) − u(x

)| + |u(x

) − u(y

)| + |u(y

) − u(y)|

≤ C(d

α/2

+ |x

− y

α/2

+ d

α/2

)kφk

(∂B

)

≤ C|x − y|

α/2

kφk

(∂B

)

since |x

− y

| ≤ d

+ |x − y| + d

≤ 5|x − y|.

In order to prove (1.7) we assume B

= B

((1, 0, . . . , 0)), x

= 0, and φ(0) = 0.

Dene K = sup

x∈∂B

|φ(x)|/|x|

. Note |x|

= 2x

for x ∈ ∂B

. Therefore for x ∈ ∂B

there holds

φ(x) ≤ K|x|

≤ 2

α/2

Dene v(x) = 2

α/2

in B

. Then we have

∆v(x) = 2

α/2

K ·



− 1



α/2−2

< 0 in B

Theorem 3.1 implies

u(x) ≤ v(x) = 2

α/2

≤ 2

α/2

K|x|

α/2

for any x ∈ B

Chapter 1: Harmonic Functions

Considering −u similarly, we get

|u(x)| ≤ 2

α/2

K|x|

α/2

for any x ∈ B

This proves (1.7).