Lecture 23: Moser’s approach to the mean value inequality
1
The mean value inequality: Moser’s Approach
In this lecture we will give an alternative proof of the mean value inequality. As before we
take L a uniformly elliptic second order operator in divergence form, and u an L harmonic
function on Rn . Let x0 ∈ Rn , take an open ball Bs (x0 ),k > 1 a positive constant, and η
be a cutoff function (ie η : Bs (x0 ) → R with η = 0 on the boundary). We have
�
Aij
Bs (x0 )
∂u ∂η 2 uk
=0
∂xi ∂xj
(1)
(this is from the weak definition of L harmonic). Therefore
�
�
∂u ∂η
∂u ∂u
+2
ηuk Aij
.
0=k
η 2 uk−1 Aij
∂xi ∂xj
∂xi ∂xj
Bs (x0 )
Bs (x0 )
Note that
∂uβ ∂uβ
∂u ∂u
= β 2 u2β−2 Aij
∂xi ∂xj
∂xi ∂xj
Aij
for all constants β. If we pick β =
�
k+1
2
k+1
2 we
u
k−1
(3)
get
k+1
�2
(2)
k+1
∂u ∂u
∂u 2 ∂u 2
Aij
= Aij
∂xi ∂xj
∂xi ∂xj
(4)
and applying this to 2 gives
4k
(k + 1)2
k+1
�
k+1
∂u 2 ∂u 2
η Aij
= −2
∂xi ∂xj
Bs (x0 )
2
�
ηuk Aij
Bs (x0 )
∂u ∂η
.
∂xi ∂xj
(5)
Let k = l1 + l2 and apply Cauchy Schwarz to the right hand side to get
4k
(k + 1)2
�
k+1
k+1
∂u 2 ∂u 2
η 2 Aij
≤2
∂xi ∂xj
Bs (x0 )
��
∂u ∂u
Aij η 2 u2l1
∂xi ∂xj
Bs (x0 )
1
�1/2 ��
∂η ∂η
Aij u2l2
∂xi ∂xj
Bs (x0 )
(6)
�1/2
.
Apply 3 with β = l1 + 1 to get
4k
(k + 1)2
k+1
∂u 2 ∂u 2
2
η 2 Aij
≤
l1 + 1
∂xi ∂xj
Bs (x0 )
Choosing l1 =
2l2 − 1
l2 2
k+1
�
k−1
2 , l2
=
k+1
2
��
∂ul1 +1 ∂ul1 +1
Aij η 2
∂xi
∂xj
Bs (x0 )
�1/2 ��
∂η ∂η
Aij u2l2
∂xi ∂xj
Bs (x0 )
(7)
we have
�1/2 ��
�1/2
∂η
∂η
Aij u2l2
,
∂xi ∂xj
Bs (x0 )
(8)
��
2
∂ul2 ∂ul2
≤
η 2 Aij
l2
∂xi ∂xj
Bs (x0 )
∂ul2 ∂ul2
Aij η 2
∂xi ∂xj
Bs (x0 )
�
so dividing and squaring gives
�
2l2 − 1
l2
�2 �
∂ul2 ∂ul2
η Aij
≤4
∂xi ∂xj
Bs (x0 )
�
2
Aij u2l2
Bs (x0 )
∂η ∂η
.
∂xi ∂xj
(9)
Use uniform ellipticity to simplify this to
�
2l2 − 1
l2
�2 �
4Λ
η |�u | ≤
λ
Bs (x0 )
2
Since k ≥ 1, we have l2 ≥ 1 and so
�
�
l2 2
2l2 −1
l2
�2
We need to estimate
�
�
l2 2
Bs (x0 ) |�(ηu
l2 )|2 .
u2l2 |�η|2 .
(10)
Bs (x0 )
≥ 1. Thus
4Λ
η |�u | ≤
λ
Bs (x0 )
2
�
u2l2 |�η|2 .
(11)
Bs (x0 )
Note that
|�(ηul2 )|2 ≤ 2u2l2 |�η|2 + 2η 2 |�ul2 |2 .
(12)
Apply 11 to get
�
�
l2
|�(ηu )| ≤
Bs (x0 )
8Λ
2+
λ
��
u2l2 |�η|2 .
(13)
Bs (x0 )
From this we can use Sobolev to estimate
��
l2
(ηu )
2n
n−2
� n−2
n
�
≤ c
Bs (x0 )
|�(ηul2 )|2
(14)
Bs (x0 )
�
8Λ
≤ c 2+
λ
2
��
Bs (x0 )
u2l2 |�η|2
(15)
�1/2
.
for some dimensional constant c. Pick r < s and let η be the usual linear cutoff function,
i.e.
φ=
⎧
⎨ 1
s−|x|
s−r
⎩
0
on Br (x0 )
on Bs (x0 ) \ Br (x0 ), and
outside Bs (x0 )
.
(16)
This gives
��
(ul2 )
2n
n−2
� n−2
��
n
(ηul2 )
≤
2n
n−2
� n−2
n
(17)
Bs (x0 )
Br (x0 )
�
u2l2|�η|2
≤ c̃
(18)
Bs (x0 )\Br (x0 )
�
c̃
u2l2
(s − r)2 Bs (x0 )\Br (x0 )
�
c̃
u2l2
(s − r)2 Bs (x0 )
≤
≤
for an appropriate constant c̃ that depends on L and n. Define χ =
We’ve shown that
��
�1/pχ
upχ
Br (x0 )
�
≤
c̃
(s − r)2
(20)
> 1 and p = 2l2 .
�1/p
�1/p ��
up
.
(21)
Bs (x0 )
We can define the Lq norm of a function f by
��
||f ||Lq (BR (x0 )) =
n
n−2
(19)
�1/q
f
q
.
(22)
BR (x0 )
This is useful notion of length for integrable functions, and in this notation we have
�
�1/p
c̃
||u||Lp (Bs (x0 ))
(23)
||u||Lpχ (Br (x0 )) ≤
(s − r)2
for all r < s. Let rm = 1 + 2−m and pm = 2χm . Then
�
�1/pm
||u||Lpm+1 (Brm+1 (x0 )) ≤ c̃22m+2
||u||Lpm (Brm (x0 )) ,
(24)
so, by induction,
||u||Lpm+1 (Brm+1 (x0 ))
�
�
1
m
2i+2 2χi
||u||L2 (B2 (x0 ))
≤
Πi=0 (2
c̃)
Pm
≤ 2
i+1
i=0 2χm
3
Pm
c
c
i=0 2χm
||u||L2 (B2 (x0 )) .
(25)
(26)
Both of these sums converge as m → ∞ by the ration test. Therefore if the L2 norm
of u is finite on B2 (x0 ) then the L∞ norm onB1 (x0 ) is finite, and indeed, is bounded by a
dimensional constant.
4