JOHN-NIRENBERG ESTIMATE
JAN MALÝ
1. Lemma. Let v ∈ W 1,1 (B(0, 1)) be a function of zero median. Suppose that
Z
|∇v(y)| dy ≤ rn−1
B(x,r)
for each x ∈ B(0, 1) and r > 0. Then
Z
eλ|v(x)| dx ≤ A,
B(0,1)
where A, λ depend only on n.
Proof. We may assume that v ≥ 0, otherwise we split into positive and negative parts. Write B = B(0, 1),
(
|∇v(x)|, x ∈ B,
f (x) =
0,
x∈
/ B.
We know that
Z
f (y)
dy.
|y − x|n−1
v(x) ≤ C
B
For k = 2, 3, . . . we use the Hölder inequality to obtain
Z
k−1
k1 Z
f (y)
f (y)
k
v(x) ≤ C
1 dy
k−1 dy
n−1−
n−1+
k
k
B |y − x|
B |y − x|
We estimate
Z
f (y)
1
B
|y − x|n−1− k
dy = n − 1 −
= n−1−
≤ n−1−
≤ n−1−
1
k
1
Z
B
∞
Z
k
1
t k −n
Z
1
t
1
k −n
Z
0
1
Z
k
f (y) dy dt
B∩B(x,t)
Z
k
1
t k −n dt dy
|y−x|
0
1
=k n−1−
∞
Z
f (y)
1
Z
f (y) dy dt +
B(x,t)
Z ∞
1
t k −1 dt +
0
1
k
∞
1
t k −n
1
1
t k −n dt
1
+ 1 ≤ kn.
Thus
Z
k
k k−1
v (x) dx ≤ C k
Z Z
B
B
B
Z Z
f (y)
n−1+ k−1
k
|y − x|
f (y)
dy dx
= C k k k−1
dx
dy
1
n− k
B
B |y − x|
Z
= C k kk
f (y) dy ≤ C k k k .
B
For k = 1 we use the Poincaré inequality to estimate
Z
Z
v(x) dx ≤ C
f ≤ C.
B
B
1
Z
B
f (y) dy dt
Together
Z
eλv(x) dx =
B
≤
≤
Z X
∞
λk v k (x) B k=0
∞
X kZ
k=0
∞
X
λ
k!
dx
v k (x) dx
B
(Cλe)k
k=0
Since
k!
kk
.
ek k!
kk
≤ 1,
ek k!
the sum converges if we choose λ so small that
Cλe < 1.
2. Theorem. Let u ∈ W
1,1
(B(0, R)) be a function of zero median, u ≥ 0, and K ≥ 0. Suppose that
Z
|∇u(y)| dy ≤ Krn−1
B(x,r)
for each x ∈ B(0, 1) and r > 0. Then
Z
eλu(x)/K dx ≤ ARn ,
B(0,R)
where A, λ are as in Lemma 1.
Proof. Set
v(y) =
1
u(Ry),
K
Then
∇v(y) =
y ∈ B(0, 1).
R
∇u(Ry).
K
We have
Z
1
|∇v(y)| dy =
K
B(z,ρ)
Z
1 1−n
|R∇u(Ry)| dy =
R
K
B(z,ρ)
Z
|∇u(x)| dx ≤ ρn−1 .
B(Rz,Rρ)
Hence for λ, A as in Lemma 1,
R
−n
Z
e
λu(x)/K
Z
dx =
B(0,R)
eλv(y) dy ≤ A.
B(0,1)
2