Five inequalities for Harmonic functions
In this lecture we will prove five inequalities for harmonic functions.
1
Bounding integrals of Harmonic functions
Proposition 1.1 Let r and s be real numbers with 0 < r ≤ s, and x ∈ Rn . There are
constants ci such that
�
� r �n �
f 2 ≤ c1
f 2,
(1)
s
Br (x)
Bs (x)
�
� r �n+2 �
(f − Ax,r )2 ≤ c2
(f − Ax,s )2 ,
(2)
s
Br (x)
Bs (x)
�
� r �n �
|�f |2 ,
(3)
|�f |2 ≤ c3
s
Bs (x)
Br (x)
and
�
2
|�f − (�f )x,r | ≤ c4
Br (x)
� r �n+2 �
s
|�f − (�f )x,s |2 .
(4)
Bs (x)
for all functions f that are harmonic on Bs (x) with Ax,t , (�f )x,t the averages of f and �f
over Bt (x) respectively.
Before proving these we will prove another inequality, the mean value inequality.
Proposition 1.2 If f is harmonic on B2r (x) then
2n
sup f ≤
vol B2r (x)
Br (x)
2
�
f 2.
Proof Pick y ∈ Br (x). By the mean value property (from lecture 1)
�
1
f (y) =
f,
vol Br (y) Br (y)
so
1
(5)
Br (x)
(6)
�
2
f (y) =
�
=
�2
1
vol Br (y)
�
1
vol Br (y)
�2 ��
f
(7)
Br (y)
�2
f
(8)
Br (y)
� ��
�
�2 ��
1
2
2
f
1
vol Br (y)
Br (y)
Br (y)
�
1
f2
vol Br (y) Br (y)
�
≤
≤
(9)
(10)
by Cauchy Schwarz. Note that Br (y) ⊂ B2r (x), so we can expand the area of integration
to get
�
1
f2
vol Br (y) B2r (x)
�
2n
f 2.
vol B2r (x) B2r (x)
f 2 (y) ≤
≤
(11)
(12)
Therefore
sup f 2 ≤
Br (x)
2n
vol B2r (x)
�
f2
(13)
B2r (x)
as required.
Now we’ll use this to get our first inequality. If r ≤ s ≤ 2r then
�
2
�
f2
f ≤
Br (x)
(14)
Bs (x)
� �
2r n
≤
f2
s
Bs (x)
� �n �
n r
≤ 2
f 2.
s
Bs (x)
�
(15)
(16)
If instead 2r ≤ s then
sup f 2 ≤
Br (x)
≤
sup f 2
Bs/2 (x)
2n
vol Bs (x)
2
(17)
�
Bs (x)
f2
(18)
by the mean value inequality Therefore
1
vol Br (x)
�
f2 ≤
Br (x)
≤
and the ration of the volumes is
�
� r �n
s
1
vol Br (x)
�
2n
vol Bs (x)
�
∂
∂xi �f
2n
vol Bs (x)
Br (x)
�
�
f2
(19)
Bs (x)
f 2,
(20)
Bs (x)
, so
f 2 ≤ 2n
� r �n �
s
Br (x)
for large s as well.
∂f
=
Note that � ∂x
i
�
f2
(21)
Bs (x)
= 0. Therefore 3 follows immediately from 1. Now we’ll
∂f
∂xi
prove 2. First consider the case 4r ≤ s. Since
inequality to get
is harmonic we can apply the mean value
1
sup |�f | ≤
vol B2r (x)
Br (x)
2
�
|�f |2 .
(22)
B2r (x)
Now apply this. By the intermediate value theorem there is y ∈ Br (x) with f (y) = Ax,r .
Pick z ∈ Br (x). Clearly |f (z) − f (y)| ≤ |z − y | supBr (x) |�f | ≤ 2r supBr (x) |�f |. Therefore
�
1
vol Br (x)
1
vol Br (x)
(f − Ax,r )2 ≤
Br (x)
�2
�
�
2r sup |�f |
Br (x)
(23)
Br (x)
≤ 4r2 sup |�f |2
(24)
Br (x)
≤ 4r2 sup |�f |2
(25)
Bs/4 (x)
1
≤ 4r
vol Bs/2 (x)
2
�
|�f |2 .
�
�
Apply Cacciopolli to get B (x) |�f |2 ≤ s12 Bs (x) (f − Ax,s )2 , so
s/2
�
�
1
4r2
2
(f − Ax,r ) ≤ 2
(f − Ax,s )2 ,
s vol Bs/2 (x) Bs (x)
vol Br (x) Br (x)
and
�
(f − Ax,r )2 ≤ 2n+2
� r �n+2 �
Br (x)
s
as required. For r ≤ s ≤ 4r we simply note that
3
Bs (x)
(26)
Bs/2 (x)
(f − Ax,s )2
(27)
(28)
�
2
n+2
(f − Ax,r ) ≤ 4
Br (x)
� r �n+2 �
s
2
(f − Ax,r ) ≤ 4
Br (x)
n+2
� r �n+2 �
s
(f − Ax,r )2 . (29)
Bs (x)
This completes the proof of 2. The final inequality, 4, follows from 2 in exactly the same
way that 3 follows from 1.
We can also prove 1, 2, 3, and 4 for L harmonic operators when L is a uniformly elliptic
2u
. In this case the constants ci depend on the operator.
operator taking Lu = Aij ∂x∂i ∂x
j
Proofs are omitted.
4