POSITIVE HARMONIC FUNCTIONS IN THE UNIT BALL
YIFEI PAN
1. Introduction
Let B = {x ∈ Rn : |x| < 1} be the unit ball in Rn . S n−1 = ∂B. We will prove
the following properties of positive harmonic functions in B.
Theorem 1.1. Let u be a positive harmonic function in B. If 0 ≤ r1 ≤ r2 < 1,
ω ∈ S n−1 , then
(1 − r1 )n−1
(1 − r2 )n−1
u(r2 ω) ≤
u(r1 ω),
1 + r2
1 + r1
(1 + r2 )n−1
(1 + r1 )n−1
u(r2 ω) ≥
u(r1 ω).
1 − r2
1 − r1
In other words, if we let, for each ω ∈ S n−1 , 0 ≤ r < 1
(1 − r)n−1
u(rω),
1+r
(1 + r)n−1
u(rω),
J(r, ω) =
1−r
then, I(r, ω) is decreasing in r and J(r, ω) is increasing in r. In particular, their
limits exist when r → 1. Denote by
I(r, ω) =
a(ω) = lim I(r, ω),
r→1
b(ω) = lim J(r, ω).
r→1
We notice by Harnack’s inequality that 0 ≤ a(ω) ≤ u(0) and u(0) ≤ b(ω) ≤ ∞.
The next theorem is concerned with how a(ω), b(ω) are related to positive measures
associated with positive harmonic functions.
Theorem 1.2. Let u be a positive harmonic function in B, and µ its positive
Borel measure, so that u = P [µ]. Then for each ω ∈ S n−1 ,
a(ω) = µ({ω}),
Z
b(ω) =
f (η, ω)dµ(η),
S n−1
where
f (η, ω) =
(
∞
2n
|η−ω|n
Date: June 10, 2006.
1
η=ω
η 6= ω.
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YIFEI PAN
Here P [µ] is the Poisson integral of µ. The next result deals with the maximum
value of a(ω), which is u(0). A natural question is that if a(ω) = u(0) for some
ω ∈ S n−1 , is it possible
u(x) = u(0)P (x, ω)?
1 − |x|2
is the Poisson kernel for B. We answer this question
|x − ω|n
below for n = 2 using complex analysis.
where P (x, ω) =
Theorem 1.3. Let u(z) be a positive harmonic function in the unit disk |z| < 1.
If a(ω) = u(0) for some ω with |ω| = 1, then
u(z) = u(0)
1 − |z|2
.
|z − ω|2
It seems to be interesting to prove the same result for n > 2. The proof of
Theorem 1.1 is based on the following sharp normal derivative estimate for postive
harmonic functions.
Theorem 1.4. Let u be a positive harmonic function in B. For ω ∈ S n−1 , x ∈ B,
|x| = r, then
−
∂
n + (n − 2)|x|
n − (n − 2)|x|
u(x) ≤
u(x) ≤
u(x).
1 − |x|2
∂r
1 − |x|2
The equalities can hold for the Poisson kernel of B.
2. Proof of Theorem 1.1
Here we are going to prove Theorem 1.1 and 1.4. We first begin with the
following lemma.
Lemma 2.1. For x ∈ Rn , |x| = r, y ∈ S n−1 ,
−
∂ 1 − |x|2
n + (n − 2)|x|
n − (n − 2)|x|
≤
(
)≤
n
n
|x − y|
∂r |x − y|
|x − y|n
.
Proof. Since r2 = x1 2 + ... + xn 2 , x = (x1 , ..., xn ), it follows that
dx1
dxn
+ ... + xn
dr
dr
n−1
Also, we claim that, for y ∈ S
, r > 0,
(2.1)
r = x1
dx1
dxn
1
+ ... + yn
= hx, yi.
dr
dr
r
n−1
Indeed write x = rω, where ω ∈ S
, therefore, xi = rωi , i = 1, ..., n, and
ω1 2 + ... + ωn 2 = 1. It follows
dxi
= ωi ,
dr
and
(2.2)
y1
POSITIVE HARMONIC FUNCTION IN UNIT BALL
3
dx1
dxn
+ ... + yn
= y1 ω1 + ... + yn ωn
dr
dr
1
= (y1 rω1 + ... + yn rωn )
r
1
= hx, yi.
r
We also note, by (2.1), and (2.2)
y1
n
∂
∂
|x − y|n =
((x1 − y1 )2 + ... + (xn − yn )2 ) 2
∂r
∂r
dx1
dxn
= n((x1 − y1 )
+ ... + (xn − yn )
)|x − y|n−2
dr
dr
dx1
dxn
dx1
dxn
= n(x1
+ ... + xn
− y1
− ... − yn
)|x − y|n−2
dr
dr
dr
dr
1
= n(|x| −
hx, yi)|x − y|n−2 .
|x|
Now, for |x| > 0.
∂
−2|x||x − y|n − (1 − |x|2 ) ∂r
|x − y|n
∂ 1 − |x|2
(
)
=
∂r |x − y|n
|x − y|2n
−2|x||x − y|n − n(1 − |x|2 )(|x| −
(2.3)
=
1
|x| hx, yi)|x
− y|n−2
|x − y|2n
1
|x| hx, yi)(1
y|n+2
−2|x||x − y|2 − n(|x| −
=
|x −
− |x|2 )
.
In order to show the right hand side of the lemma, it suffices to show, by (2.3),
that
(2.4)
−2|x||x − y|2 − n(|x| −
1
hx, yi)(1 − |x|2 ) ≤ (n + (n − 2)|x|)|x − y|2
|x|
Multiplying (2.4) by |x|, and simplifying, one gets
−2|x|2 |x − y|2 − n(|x|2 − hx, yi)(1 − |x|2 ) ≤ (n|x| + (n − 2)|x|2 )|x − y|2 ,
which is equivalent to
−(|x|2 − hx, yi)(1 − |x|2 ) ≤ (|x| + |x|2 )|x − y|2 ,
which is equivalent again, to, using |y| = 1,
−|x|2 (1 − |x|2 ) + hx, yi(1 − |x|2 ) ≤ (|x| + |x|2 )(|x|2 − 2hx, yi + 1),
which is equivalent to
hx, yi(1 − |x|2 ) ≤ |x|(1 + |x|)2 − 2hx, yi|x|(1 + |x|),
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YIFEI PAN
which is equivalent to, after dividing by 1 + |x|,
hx, yi(1 − |x|) ≤ |x|(1 + |x|) − 2hx, yi|x|),
which is equivalent to
hx, yi(1 + |x|) ≤ |x|(1 + |x|),
which is finally equivalent to
hx, yi ≤ |x|,
which is of course true since |hx, yi| ≤ |x||y| = |x|.
In order to prove the left hand side, it suffices to prove that by (2.3)
1
hx, yi)(1 − |x|2 ) ≥ −(n − (n − 2)|x|)|x − y|2 .
(2.5) −2|x||x − y|2 − n(|x| −
|x|
Using the similar steps as above, (2.5) can be seen to be equivalent to
−hx, yi ≤ |x|,
which is of course true since |hx, yi| ≤ |x|. As far as for equality cases in the lemma,
if x = |x|y, y ∈ S n−1 , we have hx, yi = h|x|y, yi = |x|hy, yi = |x|. By the above
argument, it is obvious that the equality holds. Similarly if x = −|x|y, then the
left hand equality holds since −hx, yi = |x|.
Now we can give the proof of Theorem 1.4. First assume u is harmonic in B.
By the Poisson integral representation of u in B, one has
Z
1 − |x|2
u(x) =
u(y)dσ(y),
n
S n−1 |x − y|
where σ(y) is the normalized surface area measure on S n−1 (so that σ(S n−1 ) = 1).
It follows
Z
∂ 1 − |x|2
∂u(x)
=
(
)u(y)dσ(y).
n
∂r
S n−1 ∂r |x − y|
By Lemma 2.1, and u > 0, it follows
n + (n − 2)|x|
u(y)dσ(y)
|x − y|n
Z
n + (n − 2)|x|
1 − |x|2
=
u(y)dσ(y)
n
1 − |x|2
S n−1 |x − y|
n + (n − 2)|x|
=
u(x).
1 − |x|2
∂u(x)
≤
∂r
Z
S n−1
1 − |x|2
, it is
|x − y|n
well-known, u(x, y) is harmonic in Rn \ {y} for y ∈ S n−1 . A simple calculation
shows that the equalities hold for x = |x|y, x = −|x|y respectively.
For a general function u, consider its dilate uR (x) = u(Rx). Then uR (x) is also
harmonic. Apply above, to uR (x) for R < 1. Then take the limit as R → 1 to get
the desired results.
Now we prove Theorem 1.1.
The other side of the inequality is proved similarly. If u(x, y) =
POSITIVE HARMONIC FUNCTION IN UNIT BALL
Consider ϕ(r) =
5
(1 − r)n−1
(1 + r)n−1
, ψ(r) =
for 0 ≤ r < 1. It follows
1+r
1−r
ϕ0
n + (n − 2)r
,
=−
ϕ
1 − r2
ψ0
n − (n − 2)r
=
.
ψ
1 − r2
In particular, ϕ0 < 0, and ψ 0 > 0 for 0 < r < 1, i.e., ϕ is decreasing, ψ is increasing.
Given ω ∈ S n−1 , consider
I(r, ω) = ϕ(r)u(rω),
J(r, ω) = ψ(r)u(rω).
To show Theorem 1.1, it suffices to show I(r, ω) is non-increasing, and J(r, ω) is
non-decreasing in r. In fact, by Theorem 1.5,
ϕ0 u 0
+
ϕ
u
n + (n − 2)r u0
=−
+
1 − r2
u
n + (n − 2)r n + (n − 2)r
≤−
+
1 − r2
1 − r2
≤ 0.
(log I(r, ω))0 =
Therefore log I(r, ω) is non-increasing, so is I(r, ω). Similarly,
ψ 0 u0
+
ψ
u
n − (n − 2)r u0
+
=
1 − r2
u
n − (n − 2)r n − (n − 2)r
≥
−
1 − r2
1 − r2
≥ 0.
(log J(r, ω))0 =
Hence, J(r, ω) is non-decreasing in r. This completes the proof of Theorem 1.1.
It is well-known that there exists a positive harmonic function u in B(n > 2)
such that
lim sup u(x) = ∞
x→ζ
n−1
for every ζ ∈ S n−1 . (see [1]). However, by Theorem 1.1, (1−r)
u(rω) has a limit
1+r
n−1
as r → 1 for every ω ∈ S
. We also remark that the monotone property of
positive harmonic functions is no longer true for other harmonic functions. For
example, let u(x) be a homogeneous harmonic polynomial of degree m. Then
I(r, ω) =
(1 − r)n−1
(1 − r)n−1 m
u(rω) =
r u(ω).
1+r
1+r
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YIFEI PAN
(1 − r)n−1 m
r . A simple calculation shows that f 0 (r) > 0 for
1+r
0 < r < r1 and f 0 (r) < 0 for r1 < r < 1, where
√
−n + 9n2 − 8n
r1 =
.
4n − 4
So if u(ω) 6= 0 for some ω, then I(r, ω) changes monotonicity at r1 .
Consider f (r) =
3. Applications of Theorem 1.1
Here we will prove Theorem 1.2 among other things. The Harnack inequality
for positive harmonic functions in B is the following,
1 + |x|
1 − |x|
u(0) ≤ u(x) ≤
u(0).
(1 + |x|)n−1
(1 − |x|)n−1
Theorem 1.1 gives a little more information about this. For ω ∈ S n−1 , lim I(r, ω)
r→1
exists. We denote
a(ω) = lim I(r, ω).
r→1
We remark, 0 ≤ a(ω) ≤ u(0). Similarly, we denote
b(ω) = lim J(r, ω).
r→1
We remark, by Harnack’s inequality,
b(ω) ≥ u(0).
By Theorem 1.1, we have
Proposition 3.1. For any positive harmonic function u in B, ω ∈ S n−1 , 0 ≤ r <
1, it follows
1+r
1+r
a(ω)
≤ u(rω) ≤ u(0)
,
n−1
(1 − r)
(1 − r)n−1
1−r
1−r
u(0)
≤ u(rω) ≤ b(ω)
.
n−1
(1 + r)
(1 + r)n−1
Proof. Since I(r, ω) is non-increasing, I(r, ω) ≥ a(ω), it follows,
1+r
u(rω) ≥ a(ω)
.
(1 − r)n−1
Remark: when a(ω) = 0 or b(ω) = ∞ for a ω ∈ S n−1 , there is nothing new in
1−|x|2
n−1 , the Poisoon kernel, we can see
above.. For u(x) = |x−y|
n, y ∈ S
(
0 ω 6= y
a(ω) =
1 ω = y,
and
(
b(ω) =
2n
|ω−y|n
ω 6= y
1
ω = −y.
POSITIVE HARMONIC FUNCTION IN UNIT BALL
7
When a(ω) = u(0) for a ω ∈ S n−1 , we have by Proposition 3.1
u(rω) = u(0)
1+r
.
(1 − r)n−1
2
1−|x|
It would be interesting to see if u(x) = |x−ω|
2 u(0). We are able to answer this
positively when n = 2 in Theorem 1.3, which will be given in the next section.
Now we point out that a(ω) can take any value between 0 and u(0). Here is a
simple example using the Poisson kernel.
u(x) = a
1 − |x|2
1 − |x|2
+
b
,
|x − ω|2
|x − η|2
where a, b > 0 and ω, η ∈ S n−1 . The next example shows that a(ω) can be positive
for a countable dense set of S n−1 . The following function is one
u(x) =
∞
X
k=0
where
∞
P
∞
P
ai
1 − |x|2
|x − ωi |2
n−1 . We note u(0) =
ai < ∞ and {ωi }∞
0 is a countable dense set of S
k=0
ai and by the Harnack principle, u is a positive harmonic function. It is easy
k=0
to verify that a(ωi ) = ai for i = 0, 1, 2... and otherwise a(ω) = 0. We also notice
for this function that b(−ωi ) = ai and otherwise b(ω) = ∞.
Because of the above example, we may ask the following question. If a(ω) > 0
for some ω ∈ S n−1 , can we find v a positive harmonic function so that
u(x) = a(ω)
1 − |x|2
+ v(x)
|x − ω|2
where of course a(ω) for the function v is 0. This question may be regarded as
the decomposition of u in terms of the Poisson kernels.
Proposition 3.2. Let u be a positive harmonic function in B. Then for 0 ≤ r1 ≤
r2 < 1, one has
(1 − r2 )n−1
(1 − r1 )n−1
max u(x) ≤
max u(x),
1 + r2 |x|=r2
1 + r1 |x|=r1
(1 + r2 )n−1
(1 − r1 )n−1
min u(x) ≥
min u(x).
1 − r2 |x|=r2
1 + r1 |x|=r1
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YIFEI PAN
Proof. By the maximum principle, there is ω0 ∈ S n−1 , max u(x) = max u(r2 ω0 ),
|x|=r2
therefore, by Theorem 1.1,
(1 − r2 )n−1
(1 − r2 )n−1
max u(x) =
u(r2 ω0 )
1 + r2 |x|=r2
1 + r2
(1 − r1 )n−1
u(r1 ω0 )
1 + r1
(1 − r1 )n−1
≤
max u(x)
1 + r1 |x|=r1
≤
The second inequality is proved using the minimum principle.
We note that
(1 − r)n−1
maxu(x),
r→1
1 + r |x|=r
a = lim
(1 + r)n−1
maxu(x)
r→1
1 − r |x|=r
b = lim
exist.
Proposition 3.3. There exist ω1 ∈ S n−1 , ω2 ∈ S n−1 , so that,
(1 − r)n−1
u(rω1 ),
r→1
1+r
a = lim
(1 + r)n−1
u(rω2 ).
r→1
1−r
b = lim
Proof. Choose rk → 1, rk < 1, then ωk ∈ S n−1 exist.
(1 − rk )n−1
max u(x)
rk →1
1 + rk |x|=rk
a = lim
(1 − rk )n−1
u(rk ωk )
rk →1
1 + rk
For a fixed r ∈ (0, 1) and k large enough,
= lim
(1 − r)n−1
(1 − rk )n−1
u(rωk ) ≥
u(rk ωk ).
1+r
1 + rk
Assume ωnk → ω1 ∈ S n−1 , therefore, letting k → ∞,
(1 − r)n−1
u(rω1 ) ≥ a.
1+r
But,
(1 − r)n−1
(1 − r)n−1
maxu(x) ≥
u(rω1 ) ≥ a,
1 + r |x|=r
1+r
and letting r → 1, we have
(1 − r)n−1
u(rω1 )
r→1
1+r
a = lim
POSITIVE HARMONIC FUNCTION IN UNIT BALL
9
The second inequality can be proved similarly.
We note that a ≥ a(ω), b ≤ b(ω) for every ω ∈ S n−1 . Here we define ω1 the
maximal growth direction of u, ω2 the minimal growth direction of u. It would
be interesting to know if these directions are unique for a given positive harmonic
function. If a > 0, we say u is of maximal growth. Similarly, if b < ∞, we say u
is of minimal growth.
Proposition 3.4. Let u be a positive harmonic function in B. For ω ∈ S n−1 ,
and r1 ≤ r2 < 1, one has
1 − r1 n−1 1 + r2
1 − r2 1 + r1 n−1 u(r2 ω)
≤
.
≤
1 − r1 1 + r2
u(r1 ω)
1 − r2
1 + r1
Proof. By Theorem 1.1,
(3.1)
(1 − r2 )n−1
(1 − r1 )n−1
u(r2 ω) ≤
u(r1 ω).
1 + r2
1 + r1
It follows
u(r2 ω)
≤
u(r1 ω)
1 − r1
1 − r2
n−1
1 + r2
.
1 + r1
Since
(3.2)
(1 + r1 )n−1
(1 + r2 )n−1
u(r2 ω) ≥
u(r1 ω),
1 − r2
1 − r1
it follows
u(r2 ω)
≥
u(r1 ω)
1 + r1
1 + r2
n−1
1 − r2
.
1 − r1
Lemma 3.5. Let ϕ(x, y) = (1 − x)n−1 (1 + y) − (1 − y)n−1 (1 + x). Then
n−1
k−2
X
X
k n
ϕ(x, y) = (y−x)(n+1− (−1)
(xk−1 +(1+x) xj y k−j−1 )) = (y−x)U (x, y).
k
j=0
k=2
Proposition 3.6. Let u be a positive harmonic function in B. For ω ∈ S n−1 ,
and r1 < r2 < 1,
U (−r1 , −r2 )(1 + r1 )
u(r2 ω) − u(r1 ω)
u(0) ≤
n−1
n
(1 + r2 )
(1 − r1 )
r2 − r 1
U (r1 , r2 )
≤
u(0),
(1 − r2 )n−1 (1 − r1 )n−1
where U (x, y) is defined as in Lemma 3.5.
Proof. By (3.1), it follows
u(r2 ω) − u(r1 ω) ≤
1 − r1
1 − r2
n−1
!
1 + r2
− 1 u(r1 ω).
1 + r1
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YIFEI PAN
Since ϕ(r) =
(1−r)n−1
1+r
is decreasing, therefore
1 − r1
1 − r2
n−1
1 + r2
− 1 ≥ 0,
1 + r1
and so that
u(r2 ω) − u(r1 ω) ≤
1 − r1
1 − r2
n−1
!
1 + r2
1 + r1
−1
u(0)
1 + r1
(1 − r1 )n−1
(1 − r1 )n−1 (1 + r2 ) − (1 + r2 )(1 − r2 )n−1
u(0)
(1 − r2 )n−1 (1 − r1 )n−1
U (r1 , r2 )
u(0).
=
(1 − r2 )n−1 (1 − r1 )n−1
=
Similarly, by (3.2),
u(r2 ω) − u(r1 ω) ≥
Since ψ(r) =
1−r
(1+r)n−1
1 + r1
1 + r2
n−1
!
1 − r2
− 1 u(r1 ω).
1 − r1
is increasing, we see
Therefore, using u(r1 ω) ≤
1 + r1
1 + r2
n−1
1 − r2
− 1 ≤ 0.
1 − r1
1+r1
u(0),
(1−r1 )n−1
u(r2 ω) − u(r1 ω) ≥
=
1 + r1
1 + r2
we obtain,
n−1
!
1 − r2
1 + r1
−1
1 − r1
(1 − r1 )n−1
U (−r1 , −r2 )(1 + r1 )
.
(1 + r2 )n−1 (1 − r1 )n
Now we give the proof of Theorem 1.2.
Proof. Let P (x, y) =
1−|x|2
|x−y|n .
u = P [µ]. Notice that
(1 − r)n−1
lim
P (x, y) =
r→1
1+r
(
1
0
if x = ry
x 6= ry.
POSITIVE HARMONIC FUNCTION IN UNIT BALL
11
n−1
Also (1−r)
P (x, y) ≤ 1 by the Harnack inequality. By the Lebesque Dominated
1+r
Convergence Theorem,
(1 − r)n−1
u(rω)
r→1
1+r
Z
(1 − r)n−1
= lim
P (rω, y)dµ(y)
r→1 S n−1
1+r
Z
(1 − r)n−1
lim
=
P (rω, y)dµ(y)
1+r
n−1 r→1
ZS
δω dµ(y)
=
a(ω) = lim
S n−1
= µ({ω}).
Note
(1 + r)n−1
P (rω, y) = f (y, ω),
r→1
1−r
lim
n−1
P (rω, y) is increasing in r by Theorem 1.1, so by the Monotone Conand (1+r)
1−r
vergence Theorem,
(1 + r)n−1
u(rω)
r→1
1−r
Z
(1 + r)n−1
= lim
P (rω, y)dµ(y)
r→1 S n−1
1−r
Z
(1 + r)n−1
=
lim
P (rω, y)dµ(y)
1−r
S n−1 r→1
Z
=
f (y, ω)dµ(y).
b(ω) = lim
S n−1
To end this section, we prove a result on positive harmonic functions on the
upper half space H = {(x, y) : x ∈ Rn−1 , y > 0}.
Theorem 3.7. Let u(x, y) be a positive harmonic function on the upper half space
u(x, y)
H. Then for each x ∈ Rn−1 ,
is a decreasing function in y > 0. In
y
particular the limits
u(x, y)
lim
y→0+
y
and
lim
y→+∞
exist including possibly ∞.
u(x, y)
y
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YIFEI PAN
Proof. By Theorem 7.26 in [1], there exists a positive Borel measure µ on Rn−1
and a nonnegative constant c such that
u(x, y) = PH [µ](x, y) + cy
for all (x, y) ∈ H. Here P [µ] is the Poisson integral of µ in H. A simple proof by
calculation shows that
∂u(x, y)
u(x.y)
≤
,
∂y
y
u(x, y)
which means for x ∈ Rn−1 ,
is a decreasing function in y > 0.
y
We point out if one uses the Kelvin transformation one can obtain some monotonicity for u on some circles in H.
4. Proof of Theorem 1.3
Let D be the unit disk in C, and T = ∂D. Let P be the class of analytic
functions p(z) in D with <p(z) > 0, and B be the class of analytic functions ϕ(z)
1+ϕ
with |ϕ(z)| < 1. We note if ϕ ∈ B, then P = 1−ϕ
∈ P, and if P ∈ P, then
1−P
ϕ = 1+P ∈ B.
Lemma 4.1. If P ∈ P, then for ζ ∈ T,
0
2
P (z)ζ
2
−
≤<
≤
2
1−r
P (z)
1 − r2
Proof. First we recall the well-known inequality
2
(4.1)
|P 0 (z)| ≤
<P (z).
1 − r2
Here we give a proof, based on the Schwarz Lemma. Consider ϕ =
ϕ ∈ B, and
|ϕ0 | ≤
(4.2)
But ϕ0 =
ζ ∈ T,
−2P 0
,
(1+P )2
1 − |ϕ|2 =
4<P
.
|1+P |2
1−P
1+P ,
then
1 − |ϕ|2
.
1 − |z|2
(4.1) follows from (4.2) immediately. Now for
P 0 (z)ζ P 0 (z) <(
≤
≤ 2 .
)
P (z) P (z) 1 − r2
Lemma 4.2. If P ∈ P, then for 0 ≤ r1 ≤ r2 < 1, ζ ∈ T,
1 − r2
1 − r1
|P (r2 ζ)| ≤
|P (r1 ζ)|,
1 + r2
1 + r1
1 + r2
1 + r1
|P (r2 ζ)| ≥
|P (r1 ζ)|.
1 − r2
1 − r1
POSITIVE HARMONIC FUNCTION IN UNIT BALL
13
Proof. Consider
0 1−r
1−r 0
log
|P (rζ)| = log
+ (log |P (rζ)|)0
1+r
1+r
0
2
P (z)ζ
=−
+<
1 − r2
P (z)
2
2
≤−
+
= 0.
1 − r2 1 − r2
1−r
So, log 1+r
|P (rζ)| is decreasing in r. It follows
other inequality can be proved equally.
1−r
1+r |P (rζ)|
is also decreasing. The
Lemma 4.3. If P ∈ P, then for ζ ∈ T,
<(P 0 (z)ζ) 2
<P (z) ≤ 1 − r2 .
Proof. By (4.1)
|<(P 0 (z)ζ)| ≤ |P 0 (z)ζ| = |P 0 (z)| ≤
Corollary 4.4. If P ∈ P, then for ζ ∈ T,
1+r
1−r <P (rζ) is increasing in r.
2
<P (z).
1 − r2
1−r
1+r <P (rζ)
is decreasing in r,
Proof.
0
1−r
2
<P 0 (rζ)ζ
<P (rζ)
=−
+
≤ 0.
2
1+r
1−r
<P (rζ)
This gives a simple proof of Theorem 1.1 when n=2 since for any positive harmonic
function u in D, there is a harmonic conjugate v, so that u + iv ∈ P.
log
We consider for P ∈ P, and ζ ∈ T,
1−r
<(rζ),
r→1 1 + r
1+r
|P (rζ)|.
β(ζ) = lim
r→1 1 − r
The following two lemmas are well-known and can be found in [2].
α(ζ) = lim
Lemma 4.5. If P ∈ P, and P (0) = 1, then |P (z)| ≤
Lemma 4.6. If P ∈ P, and P (z) = 1 +
∞
P
1+|z|
1−|z| ,
z ∈ D.
cn z n , then |cn | ≤ 2.
n=1
Lemma 4.7. If P ∈ P, then for z ∈ D, |z| = r
!
P (z) − i=P (0) 2
1 − r2
.
<P (z) ≥
<P (0) 1 + 2(1 + r2 )
<P (0)
14
YIFEI PAN
Proof. First, we prove if P (0) = 1, then
<P (z) ≥
(4.3)
Indeed, we consider P =
1−ϕ
1+ϕ ,
<P =
1 − r2
(1 + |P (z)|2 ).
2(1 + r2 )
for some ϕ ∈ B, such that ϕ(0) = 0. Then
2(1 + |ϕ|2 )
1 − |ϕ|2
2
,
1
+
|P
|
=
.
|1 + ϕ|2
|1 + ϕ|2
To prove (4.3), it suffices to prove
1 − |ϕ|2
1 − r2 2(1 + |ϕ|2 )
≥
,
|1 + ϕ|2
2(1 + r2 ) |1 + ϕ|2
or
1 − |ϕ|2 ≥
or
1 − r2
(1 + |ϕ|2 ),
1 + r2
(1 + r2 )(1 − |ϕ|2 ) ≥ (1 − r2 )(1 + |ϕ|2 ),
which is equivalent to |ϕ(z)| ≤ |z|, which in turn is true since ϕ(0) = 0 by the
Schwarz lemma. Now when P (0) 6= 1, consider
P (z) − i=P (0)
Pe(z) =
,
<P (0)
and we see that Pe ∈ P, Pe(0) = 1. Apply (4.3) to Pe to get the desired result.
Now we are ready to prove the relationship between α(ζ) and β(ζ).
Theorem 4.8. For P ∈ P,
1
2
<P (0) β (ζ)
≤ α(ζ) ≤ β(ζ) ≤ <P (0).
Proof. First, we observe if P ∈ P, then
1 + |z|
+ |=P (0)|.
(4.4)
|P (z)| ≤ <P (0)
1 − |z|
Indeed, consider Pe(z) =
4.5
P (z)−i=P (0)
.
<P (0)
Then Pe(z) ∈ P, and Pe(0) = 1, by Lemma
|Pe(z)| ≤
(4.5)
1 + |z|
.
1 − |z|
But P (z) = <P (0)Pe(z) + i=P (0), (4.4) follows from (4.5). It follows from (4.4),
we see that
α(ζ) ≤ β(ζ) ≤ <P (0).
Now by Lemma 4.7,
(4.6)
!
P (z) − i=P (0) 2
1 − r2
.
<P (z) ≥
<P (0) 1 + 2(1 + r2 )
<P (0)
Also
|P (z) − i=P (0)|2 = |P (z)|2 + 2=P (0)=P (z) + (=P (0))2
POSITIVE HARMONIC FUNCTION IN UNIT BALL
Multiplying
1−r
1+r
15
on both sides of (4.6), and letting r → 1, we see
α(ζ) ≥
β 2 (ζ)
.
<P (0)
1−r
Proposition 4.9. If P ∈ P, then lim 1+r
=P (rζ) exists and lim 1−r
=P (rζ) =
r→1
r→1 1+r
p
± β 2 (ζ) − α2 (ζ).
Lemma 4.10. Let ϕ ∈ B, Then for ζ ∈ T,
|<(ϕz ϕζ)| ≤
1 − |ϕ(z)|2
.
1 − |z|2
Proof.
|<(ϕz ϕζ)| ≤ |ϕz ϕζ|
≤ |ϕz | ≤
1 − |ϕ(z)|2
.
1 − |z|2
Theorem 4.11. Let ϕ ∈ B, Then ζ ∈ T. The function
decreasing in r and
1+r
(1 − |ϕ(rζ)|2 ) is increasing in r.
1−r
1−r
(1 − |ϕ(rζ)|2 ) is
1+r
Proof.
0
1−r
2
(1 − |ϕ(rζ)| )
log
1+r
1−r 0
= log
+ [log(1 − |ϕ(rζ)|2 )]0
1+r
−2<(ϕz ϕζ)
2
=−
+
1 − r2 1 − |ϕ(rζ)|2
≤ 0.
The second case follows as well.
Lemma 4.12. Let ϕ ∈ B, Then for z, ζ ∈ D,
ϕ(z) − ϕ(ζ) 2 1 − |ϕ(z)|2 1 − |ϕ(ζ)|2
≤
.
z−ζ
1 − |z|2
1 − |ζ|2
Proof. By the Schwarz lemma
ϕ(z) − ϕ(ζ) z − ζ .
≤
1 − ϕ(ζ)ϕ(z) 1 − ζz 16
YIFEI PAN
Squaring both sides, we see
ϕ(z) − ϕ(ζ) 2
z − ζ 2
≤1−
1 − 1 − ϕ(ζ)ϕ(z) 1 − ζz
(1 − |ζ|2 )(1 − |z|2 )
(1 − |ϕ(z)|2 )(1 − |ϕ(ζ)|2 )
≤
.
|1 − ζz|2
|1 − ϕ(ζ)ϕ(z)|2
It follows
|1 − ϕ(ζ)ϕ(z) 2 1 − |ϕ(z)|2 1 − |ϕ(ζ)|2
.
≤
1 − |z|2
1 − |ζ|2
1 − ζz
But
2
ϕ(z) − ϕ(ζ) 2 1 − ϕ(ζ)ϕ(z) ≤
.
z−ζ
1 − ζz
Theorem 4.13.
If ϕ ∈ B, then for ζ ∈ T,
1 − r 1−ϕ(rζ) a)
is decreasing in r.
1 + r 1+ϕ(rζ) 1 + r 1−ϕ(rζ) b)
is increasing in r.
1 − r 1+ϕ(rζ)
1 − r 1−|ϕ(rζ)|2
c)
2 is decreasing in r.
1 + r |1−ϕ(rζ)|
1 + r 1−|ϕ(rζ)|2
d)
2 is increasing in r.
1 − r |1−ϕ(rζ)|
1+r
|1 − ϕ(rζ)| is increasing in r.
e)
1−r
2
1−|ϕ|
Proof. If ϕ ∈ B, Then P = 1−ϕ
1+ϕ ∈ P, and <P = |1−ϕ|2 . Then a), b), c), and d)
follow from above. We only need to prove e). For 0 ≤ r1 ≤ r2 < 1, by a)
1 − r2 1 − ϕ(r2 ζ) 1 − r1 1 − ϕ(r1 ζ) ≤
1 + r2 1 + ϕ(r2 ζ) 1 + r1 1 + ϕ(r1 ζ) Since
1−r2
1+r2
≤
1−r1
1+r1 ,
then
# #
"
2 "
1 − ϕ(r2 ζ) 2
1 − ϕ(r1 ζ) 2
1 − r2 2
1
−
r
1
≤
.
1 + 1 + 1 + r2
1 + ϕ(r2 ζ) 1 + r1
1 + ϕ(r1 ζ) (4.7)
1 − r2
1 + r2
2
1 + |ϕ(r2 ζ)|2
≤
|1 − ϕ(r2 ζ)|2
1 − r1
l
1 + r1
2
1 + |ϕ(r1 ζ)|2
|1 − ϕ(r1 ζ)|2
By c),
1 − r2 1 − |ϕ(r2 ζ)|2
1 − r1 1 − |ϕ(r1 ζ)|2
≤
.
2
1 + r2 |1 − ϕ(r2 ζ)|
1 + r1 |1 − ϕ(r1 ζ)|2
POSITIVE HARMONIC FUNCTION IN UNIT BALL
Since
1−r2
1+r2
≤
17
1−r1
1+r1 ,
(4.8)
multiplying the above
1 − r2 2 1 − |ϕ(r2 ζ)|2
1 − r1 2 1 − |ϕ(r1 ζ)|2
≤
1 + r2
|1 − ϕ(r2 ζ)|2
1 + r1
|1 − ϕ(r1 ζ)|2
Adding (4.7) and (4.8), we have
1 − r2 2
1
1 − r1 2
1
≤
.
2
1 + r2
|1 − ϕ(r2 ζ)|
1 + r1
|1 − ϕ(r1 ζ)|2
This is equivalent to
1 + r1 2
1 + r2 2
|1 − ϕ(r1 ζ)|2 ≤
|1 − ϕ(r2 ζ)|2 .
1 − r1
1 − r2
This proves e).
Corollary 4.14. If ϕ ∈ B, ζ ∈ T, then for 0 ≤ r1 ≤ r2 < 1,
1 + r1 1 − r2
2(r2 − r1 )
|ϕ(r2 ζ)|2 ≤
|ϕ(r1 ζ)|2 +
.
1 − r1 1 + r2
(1 − r1 )(1 + r2 )
In particular,
1 + r1 1 − r2
2(r2 − r1 )
max |ϕ(z)|2 ≤
max |ϕ(z)|2 +
.
1
−
r
1
+
r
(1
−
r1 )(1 + r2 )
|z|=r2
1
2 |z|=r1
Proof. By Theorem 4.10, for ζ ∈ T, 0 ≤ r1 ≤ r2 < 1,
1 + r2
1 + r1
(1 − |ϕ(r2 ζ)|2 ) ≥
(1 − |ϕ(r1 ζ)|2 ),
1 − r2
1 − r1
which is equivalent to
1 + r2
1 + r1
1 + r2 1 + r1
|ϕ(r2 ζ)|2 ≤
|ϕ(r1 ζ)|2 +
−
.
1 − r2
1 − r1
1 − r2 1 − r1
Dividing
1+r2
1−r2 ,
the proof is complete.
Now we are ready to give a proof of the well-known Julia lemma, which is key
to the proof of Theorem 1.3.
Theorem 4.15. Let ϕ ∈ B, for η, ζ ∈ T. The following holds
1 − |ϕ(rζ)|
|η − ϕ(rζ)|
β = lim
= lim
.
r→1
1−r
1−r
r→1−
If β < ∞, then for z ∈ D,
|η − ϕ(z)|2
|ζ − z|2
≤
β
.
1 − |ϕ(z)|2
1 − |z|2
Proof. First we show lim 1−|ϕ(rζ)|
exists. Since
1−r
Theorem 4.10,
r→1
1+r
lim 1−r
(1 − |ϕ(rζ)|2 )
r→1−
lim
r→1−
1+r
1−r (1
− |ϕ(rζ)|2 ) is increasing by
exists and is denoted by a. If a = +∞, then,
1 − |ϕ(rζ)|
= ∞.
1−r
18
YIFEI PAN
If a < +∞, then |ϕ(rζ)| → 1 as r → 1. So
1 − |ϕ(rζ)|
1
1+r
= lim (1 − |ϕ(rζ)|2 )
r→1
r→1
1−r
1 − r (1 + |ϕ(rζ)|)(1 + r)
a
exists and equals 4 .
1+r
1+r
lim 1−r
Now, 1−r
|η − ϕ(rζ)| = 1+r
|η − ϕ(rζ)|
1−r |1 − ηϕ(rζ)| is increasing, therefore r→1
exists.
Denoting β = lim |η−ϕ(rζ)|
, we want to show β = a4 .
1−r
lim
r→1
First, we see β ≥ a4 . If β = ∞, then β = a4 . If β < +∞, then, ϕ(rζ) → η as
r → 1. By Lemma 4.12,
ϕ(rζ) − ϕ(z) 2 1 − |ϕ(rζ)|2 1 − |ϕ(z)|2
≤
.
rζ − z
1 − r2
1 − r2
Letting r → 1
η − ϕ(z) 2 a 1 − |ϕ(z)|2
(4.9)
ζ − z ≤ 4 1 − |z|2
letting z = rζ, and r → 1, we see from (4.9)
aa
.
β2 ≤
44
Therefore β ≤ a4 , and in turn β = a4 . The inequality in the lemma follows from
(4.9).
Theorem 4.16. Let P (z) =
some ζ ∈ T . Then for n ≥ 1
∞
P
n=0
cn z n ∈ P, and let β(ζ) = lim 1−r
1+r |P (rζ)| for
r→1
n
|cn − 2β(ζ)ζ | ≤ 2(<P (0) − β(ζ)).
Proof. If β(ζ) = 0, then it suffices to prove |cn | ≤ 2 This follows from Pe =
∞
P
P (z)−i=P (0)
cn z n
∈ P, Pe(0) = 1, and Pe(z) = 1 +
, by Lemma 4.6, | cn | ≤ 2.
<P (0)
n=0
<P (0)
<P (0)
Now we assume β(ζ) > 0.
Consider ϕ(z) = PP (z)−1
(z)+1 , we have |ϕ(z)| <, z ∈ D. Since β(ζ) > 0, then
|P (rζ)| → ∞ as r → 1, and
ϕ(rζ) =
P (rζ) − 1
,
P (rζ) + 1
|1 − ϕ(rζ)|
1
2
1
=
→
< +∞.
1−r
1 − r |P (rζ) + 1|
β(ζ)
By the Julia lemma (Theorem 4.15),
|1 − ϕ(z)|2
1 |ζ − z|2
≤
.
1 − |ϕ(z)|2
β(ζ) 1 − |z|2
β(ζ)
1 − |z|2
1 − |ϕ(z)|2
≤
2
|ζ − z|
|1 − ϕ|2
z∈D
POSITIVE HARMONIC FUNCTION IN UNIT BALL
or
19
which is
ζ +z
1+ϕ
β(ζ)<
≤<
z ∈ D.
ζ −z
1−ϕ
ζ +z
< P (z) − β(ζ)
≥ 0.
ζ −z
But,
∞
P (z) − β(ζ)
X
ζ +z
= c0 − β(ζ) +
(cn − 2β(ζ)η n )z n .
ζ −z
n=1
By above ,
|cn − 2β(ζ)η n | ≤ 2<(c0 − β(ζ))
= 2(<P (0) − β(ζ)).
We can give a uniqueness result when β(ζ) = <P (0).
1+r
|P (rζ)| = <P (0), then P (z) = <P (0)
Corollary 4.17. If β(ζ) = lim 1−r
r→1
1 + ζz
+
1 − ζz
i=P (0).
Proof. By Theorem 4.16, cn = 2β(ζ)η n , n = 1, 2, ..., and then
∞
X
P (z) = c0 +
2β(ζ)η n )z n
n=1
= <P (0)(1 +
∞
X
2(ηz)n ) + i=P (0)
n=1
1 + ηz
= <P (0)
+ i=P (0).
1 − ηz
The following is Theorem 1.3.
Corollary 4.18. If u is a positive harmonic function in D, and α(ζ) = lim 1−r
1+r u(rζ) =
r→1
u(0) for some ζ ∈ T, then
u(z) = u(0)<(
1 + ζz
).
1 − ζz
Proof. Let v be a harmonic conjugate of u, so P (z) = u(z) + iv(z) ∈ P. By
Theorem 4.8
β 2 (ζ)
≤ α(ζ) ≤ β(ζ) ≤ u(0).
u(0)
If α(ζ) = u(0), then β(ζ) = u(0) = <P (0) . By Corollary 4.17,
P (z) = <P (0)
1 + ζz
+ i=P (o).
1 − ζz
20
YIFEI PAN
Therefore u = <P (z) = u(0)<( 1+ζz
).
1−ζz
Corollary 4.19. If u is a positive harmonic function in D, then
!
u(z)
u(z) 2
1 − |z|2
1+
≥
.
u(0)
2(1 + |z|2 )
u(0)
Also, this inequality is equivalent to Harnack inequality.
Proof. Let v be a positive harmonic conjugate of u, then P (z) = u + iv ∈ P. By
Lemma 4.7,
!
P (z) − i=P (0) 2
1 − r2
<P (z) ≥
<P (0) 1 + 2(1 + r2 )
<P (0)
u2 + (v − i=P (0))2
1 − r2
<P (0) 1 +
=
2(1 + r2 )
u2 (0)
!
1 − r2
u(z) 2
≥
u(0)
1
+
.
2(1 + r2 )
u(0)
To show that the inequality is equivalent to the Harnack inequality, consider
ϕ(y) =
1 − r2
(1 + y 2 ) − y.
2(1 + r2 )
1−r
A simple calculation shows that the two roots of φ(y) are y1 = 1+r
, y2 =
observe, ϕ(y) ≤ 0 if and only if y1 ≤ y ≤ y2 . This proves the corollary.
1+r
1−r .
we
References
[1]
[2]
S. Axler and etc., Harmonic Function Theory, 2nd ed. Springer, 2001
C. Pommerenke, Univalent Functions, Gottingen, 1975
Department of Mathematical Sciences, Indiana University - Purdue University
Fort Wayne, Fort Wayne, IN 46805-1499
School of Mathematics and Informatics, Jiangxi Normal University, Nanchang,
China
E-mail address: [email protected]