on the lower bound for a class of harmonic functions in the half space

Acta Mathematica Scientia 2012,32B(4):1487–1494
http://actams.wipm.ac.cn
ON THE LOWER BOUND FOR A CLASS OF
HARMONIC FUNCTIONS IN THE HALF SPACE∗
)
Zhang Yanhui (
Department of Mathematics, Beijing Technology and Business University, Beijing 100048, China
E-mail: [email protected]
Deng Guantie (
)†
School of Mathematical Sciences, Beijing Normal University, Beijing 100875, China
E-mail: [email protected]
)
Kou Kit Ian (
Department of Mathematics, Faculty of Science and Technology, University of Macau, China
E-mail: [email protected]
Abstract The main objective is to derive a lower bound from an upper one for harmonic functions in the half space, which extends a result of B. Y. Levin from dimension
2 to dimension n ≥ 2. To this end, we first generalize the Carleman’s formula for harmonic functions in the half plane to higher dimensional half space, and then establish a
Nevanlinna’s representation for harmonic functions in the half sphere by using Hörmander’s
theorem.
Key words harmonic function; Carleman’s formula; Nevanlinna’s representation for half
sphere; lower bound
2010 MR Subject Classification
1
31B05; 35J05
Introduction and Main Results
A version of the well-know Liouville theorem states that if a real harmonic function
u(x) (x ∈ Rn ) has a finite upper bound, then u(x) is a constant in Rn . In this paper, we
derive an estimate of lower bound of harmonic functions from its upper bound in the upper half
space. This estimate is useful and important for studying harmonic functions and the growth
properties because its associated assumption is weaker than that of the Maximum Principle.
The method to derive an estimate of a lower bound for a harmonic function from its upper
bound in a half plane is using Carleman’s formula and Nevanlinna’s representation in the halfdisk. This motivates that to obtain a similar estimate in a higher dimension, one might need to
∗ Received September 10, 2010; revised October 26, 2011. Project supported by the Academic Human
Resources Development in Institutions of Higher Learning under the Jurisdiction of Beijing Municipality (IHLB
201008257) and Scientific Research Common Program of Beijing Municipal Commission of Education (KM
200810011005) and PHR (IHLB 201102) and research grant of University of Macau MYRG142(Y1-L2)-FST111KKI.
† Corresponding author: Deng Guantie.
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first generalize Carleman’s formula from the half plane to the half space in Rn and Nevanlinna’s
representation from the half disk to the half sphere in Rn for a harmonic function. We know
that Carleman’s formula for harmonic functions in half plane is derived by taking the real part of
contour integral for analytic functions. Taking the real part of the contour integral is impossible
in half space of Rn , but it is possible for us to use the contour integral to construct Green’s
function of half sphere in Rn by Hörmander’s method. In this paper, Carleman’s formula
for a harmonic function in the half space is established by Green’s function, Navanlinna’s
representation in the half sphere is obtained by using Hörmander’s theorem, and the estimate
of lower bound from an upper one in the half space is obtained by using Carleman’s formula
and Navanlinna’s representation in the half sphere. For the proof, we let Rn denote the ndimensional Euclidean space and |x| denote the Euclidean norm. Set x = (x1 , · · · , xn−1 , xn ) =
(x′ , xn ) with x′ = (x1 , · · · , xn−1 ). Then |x′ |2 = x21 +· · ·+x2n−1 , |x|2 = |x′ |2 +x2n . For simplicity,
a point x′ in Rn−1 is often identified with (x′ , 0) in Rn . The boundary and closure of a subset
Ω of Rn are denoted by ∂Ω and Ω respectively. Let Ω+ = {x = (x′ , xn ) : xn > 0, x ∈ Ω}. We
also write B(r) and S(r) for the open ball and the sphere of radius r(> 0) and centered at the
origin, and B(r, R) = B(R)\B(r) (R > r > 0).
In the sense of Lebesgue measure dx′ = dx1 · · · dxn−1 , dx = dx′ dxn . The mirror image of
x = (x′ , xn ) in the hyperplane Rn−1 × {0} = Rn−1 is denoted by x∗ = (x′ , −xn ).
A twice continuously differentiable function u(x) defined on Ω+ is harmonic if △u ≡ 0,
where △ = ∂12 + ∂22 + · · · + ∂n2 with ∂j2 denote the second partial derivative with respect to the
jth coordinate variable (j = 1, · · · , n).
The following theorem is well-known [1–4]:
Theorem A Let u(z) be a harmonic function in the upper half-plane C+ = {z = x+iy =
reiϕ , y > 0} with continuous boundary values on the real axis. Suppose that
u(z) ≤ Krρ ,
z ∈ C+ ,
r = |z| ≥ 1,
ρ > 1,
and
|u(z)| ≤ K,
|z| ≤ 1,
Imz ≥ 0.
Then
(1 + rρ )
,
reiϕ ∈ C+ ,
sin ϕ
where c does not depend on K, r, ϕ and the function u(z).
Our objective is to establish the following theorem.
Theorem 1 Let u(x) be a harmonic function in the half space Rn+ = {x ∈ Rn , xn > 0}
with continuous boundary values on the boundary ∂Rn+ . Suppose that
u(reiϕ ) ≥ −cK
u(x) ≤ Krρ ,
x ∈ Rn+ ,
r = |x| ≥ 1,
ρ > n − 1,
(1)
and
|u(x)| ≤ K,
|x| ≤ 1,
xn ≥ 0.
(2)
Then
(1 + rρ )
,
sinn−1 ϕ
where c is a constant independent of K, r, ϕ and the function u(x).
Remark When n = 2, the above theorem reduces to Theorem A.
u(x) ≥ −cK
(3)
No.4
Y.H. Zhang et al: ON THE LOWER BOUND FOR A CLASS OF HARMONIC FUNCTIONS
2
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Two Preliminary Lemmas
The following Lemma 1 is a generalization of the Carleman’s formula for harmonic functions
in the half plane to the higher dimensional Euclidean half space.
Lemma 1 (Carleman’s formula of harmonic functions in the half space) Let u(x) be a
n
harmonic function in Rn+ and continuous on R+ , and dσ(x) be the surface element of sphere
in Rn+ . Then, for R > r > 0, we have
Z
Z
nxn
1
1
′
u(x )
− n dx′ = Au (r, R),
u(x) n+1 dσ(x) +
R
|x′ |n
R
{x: r<|x′ |<R, xn =0}
{x: |x|=R, xn >0}
where Au (r, R) = R−n c1 (r) + c2 (r) is a function depending on r and R. Here, functions
c1 (r), c2 (r) are defined respectively by
Z
xn
∂u(x)
− xn
dσ(x),
c1 (r) =
u(x)
∂r
{x: |x|=R, xn >0}
Z
(n − 1)xn
∂u(x)
c2 (r) = −
− xn
dσ(x)
rn+1
∂r
{x: |x|=r, xn >0}
with
xn
|x|n
∂u(x)
∂r
=
n
P
i=1
xi ∂u
r ∂xi .
Proof Applying the second Green’s formula to the harmonic functions u(x) and v(x) =
xn
+
n
−R
n in the resulting sphere B (r, R) = B(r, R) ∩ R+ , we obtain
Z
∂v(x)
∂u(x)
− v(x)
u(x)
dσ(x) = 0,
∂n
∂n
∂B + (r,R)
where ∂/∂n is the differentiation along the inward normal onto B + (r, R).
Since the function v(x) is harmonic in Rn+ \ {0}, the equations
∂v(x)
nxn
= n+1 , x = (x′ , xn )
∂n
R
hold on the half sphere {x : |x| = R, xn ≥ 0}, and the equations
∂v(x)
|x|n n − 1
1
v(x) = 0,
=−
+
∂n
r
rn
Rn
v(x) = 0,
hold on {x : |x| = r, xn > 0}. Moreover, the equations
v(x) = 0,
∂v(x)
1
1
= ′n− n
∂n
|x |
R
are true on {x = (x′ , 0) : r < |x′ | < R}. Thus
Z
∂v(x)
∂u(x)
0=
u(x)
− v(x)
dσ(x)
+
∂n
∂n
∂Br,R
Z
Z
∂u(x)
∂v(x′ ) ′
∂v(x)
=
u(x)
− v(x)
dσ(x) +
u(x′ )
dx
+
+
∂n
∂n
∂n
{x∈∂Br,R
, xn >0}
{x∈∂Br,R
, xn =0}
Z
Z
nxn
1
1
′
′
=
u(x) n+1 dx +
u(x )
− n dx′
R
|x′ |n
R
{x: |x|=R; xn >0}
{x: r<|x′ |<R, xn =0}
Z
Z
nxn
xn
xn ∂u(x)
−
u(x) n+1 dx′ −
− n
dσ(x).
n
r
|x|
R
∂r
{x: |x|=r; xn >0}
{x: |x′ |=r, xn =0}
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ACTA MATHEMATICA SCIENTIA
The remainder term
Z
−
{x: |x|=r; xn >0}
nxn
u(x)dσ(x) −
rn+1
Z
{x: r<|x|<R, xn =0}
Vol.32 Ser.B
xn
xn
− n
|x|n
R
∂u(x)
dσ(x)
∂r
c1 (r)
= n+1 − c2 (r)
R
is denoted by Au (r, R), a function depending on r and R. This completes the proof of Lemma
1.
2
Next, we will give a precise Nevanlinna’s representation of harmonic functions in the half
sphere by using Hörmander’s theorem [5].
Lemma 2 (Nevanlinna’s representation of harmonic functions in the half sphere) Let
+
u(x) be a harmonic function in Rn+ . Then, on the closed half sphere B R = B R ∩ Rn+ with
B R = {x : |x| = R}, we have
u(x) =
Z
{y∈Rn
+ : |y|=R, yn >0}
2xn
+
ωn
Z
R2 − |x|2
ωn R
′
{y∈Rn
+ : r<|y |<R, yn =0}
1
1
−
|y − x|n
|y − x∗ |n
u(y)dσ(y)
1
Rn
1
−
′
n
n
′
|y − x|
|x| |y − x
e|n
u(y ′ )dy ′ ,
where the reflection x
e of x in ∂BR is defined by x
e = R2 x/|x|2 , its direction corresponds with
x, and |x||e
x| = R 2 .
Proof Note that x ∈ Rn \{0}, the inversion x → x
e is the identity on ∂BR . It is also an
e
′
involution, that is, (e
x) = x for every x 6= 0. If x, x and y, y ′ are two pairs of corresponding
points, then the equation |x||x′ | = |y||y ′ | shows that the triangles with vertices 0, x, y and
0, x′ , y ′ are similar, so that we may interchange x and x′ to obtain
|y|/|x′ | = |x|/|y ′ | = |x − y|/|x′ − y ′ |,
(4)
|y|/|x| = |x′ |/|y ′ | = |x′ − y|/|x − y ′ |.
(5)
p
If |y| = R, then y ′ = y and (4) gives |x − y|/|x′ − y| = |x|/|x′ | = |x|/R, which means
that the sphere is harmonic with respect to x and x′ .
Let
E(x) = −
|x|2−n
,
(n − 2)ωn
n > 2,
where ωn is the area of the unit sphere in Rn . We know E is locally integrable in Rn . Now we
define the following Green’s function
|x|
|y|
GR (x, y) = E(x − y) − E (e
x − y)
= E(x − y) − E (x − ye)
R
R
for x, y ∈ B̄R and x 6= y, where the second equality follows from (5). The first expression shows
that GR is a harmonic function of y for fixed x 6= y, while the second expression shows that
GR is a harmonic function of x for fixed y 6= x. Clearly, we have
GR (x, y) = 0 if |x| = R
or |y| = R, and GR (x, y) ≤ 0.
No.4
Y.H. Zhang et al: ON THE LOWER BOUND FOR A CLASS OF HARMONIC FUNCTIONS
1491
For fixed x with |x| < R the inequality |x − y| < |x||x′ − y|/R is satisfied for all y in BR . Define
a Green’s function by
+
∗
∗
G+
R (x, y) = GR (x, y ) − GR (x, y) = GR (x, y) − GR (x , y), x, y ∈ BR , x 6= y
+
for the half ball BR
= BR ∩ Rn+ . Recall that ∗ denotes the reflection operation on the boundary
+
n
n
plane ∂R+ . It is clear that G+
R (x, y) = 0 if x or y is in ∂R+ , and GR (x, y) − E(x − y) is
harmonic in x and y. We see that the Poisson kernel
+
+
PR+ (x, y) = ∂G+
R (x, y)/∂n, x ∈ BR , y ∈ ∂BR
+
is positive, because G+
R (x, y) → −∞ as x → y, so that by the maximum principle GR (x, y) < 0
+
+
+
in BR × BR . If u(x) is harmonic near RR , Hörmander [2] proved that
Z
u(x) =
PR+ (x, y)u(y)dσ(y).
+
∂BR
Below we provide an explict calculation for PR+ (x, y).
When |y| = R,
∂GR (x, y)
∂GR (x, y)
1 R2 − |x|2
=
=
,
∂n
∂y
ωn R |y − x|n
∂GR (x, y ∗ )
1 R2 − |x|2
∂GR (x, y ∗ )
=
=
,
∗
∂n
∂y
ωn R |y ∗ − x|n
|y| = R,
|y ∗ | = R.
So
∂G+
∂GR (x, y) ∂GR (x∗ , y)
R (x, y)
=
−
∂n ∂n
∂n
R2 − |x|2
1
1
=
−
.
ωn R
|y − x|n
|y − x∗ |n
PR+ (x, y) =
When |y| < R,
Hence,
yn = 0,
∂GR (x, y) xn
1
1
Rn
∂GR (x, y)
= −
=
−
,
∂n
∂yn
ωn |y − x|n
|y − x
e|n |x|n
yn =0
∂GR (x∗ , y)
∂GR (x∗ , y) xn
1
1
Rn
=−
=−
−
.
f∗ |n |x|n
∂n
∂yn
ωn |y − x∗ |n
|y − x
yn =0
∂G+
∂GR (x, y) ∂GR (x∗ , y)
R (x, y)
=
−
∂n
∂n
∂n
2xn
1
1
Rn
=
−
,
ωn |y − x|n
|y − x
e|n |x|n
PR+ (x, y) =
u(x) =
Z
{y∈Rn
+ : |y|=R, yn >0}
2xn
+
ωn
Z
R2 − |x|2
ωn R
′
{y∈Rn
+ : r<|y |<R, yn =0}
The proof of Lemma 2 is completed.
1
1
−
n
|y − x|
|y − x∗ |n
u(y)dσ(y)
1
Rn
1
−
|y ′ − x|n
|x|n |y ′ − x
e|n
u(y ′ )dy ′ .
2
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ACTA MATHEMATICA SCIENTIA
Vol.32 Ser.B
Proof of Theorem 1
We apply Lemma 1 to the harmonic function u(x) to obtain
n
Rn+1
Z
+
=
R
Z
xn u− (x)dσ(x)
{x: |x|=R, xn >0}
1
1
u (x )
− n
′ |n
|x
R
′
{x: r<|x |<R, xn =0}
Z
Z
n
+
x
u
(x)dσ(x)
+
n
n+1
−
′
{x: |x|=R, xn >0}
dx′ + Au (r, R)
+
′
u (x )
{x: r<|x′ |<R, xn =0}
1
1
− n
′
n
|x |
R
dx′ . (6)
Here and in the sequel, u− = (−u)+ , i.e., u = u+ − u− , and the remainder term is
Au (r, R) =
c1 (r)
− c2 (r)
Rn+1
in which c1 , c2 are functions depending only on r.
The terms on the right-hand side of (6) can be estimated by using (1):
n
Rn+1
Z
xn u+ (x)dσ(x) ≤ nωn KRρ−1 ,
{x: |x|=R, xn >0}
Z
+
′
u (x )
{x: r<|x′ |<R, xn =0}
1
1
− n
|x′ |n
R
dx′ ≤
nωn−1 KRρ−1
.
(ρ − 1)(ρ + n − 1)
Thus, for R ≥ 2 and (6), we can obtain
n
Rn+1
Z
≤
Z
{x: r<|x′ |<R, xn =0}
Z
n
2
2n − 1
{x:
xn u− (x)dσ(x) ≤ cRρ−1 ,
(7)
{x: |x|=R, xn >0}
u− (x′ ) ′
dx
|x′ |n
r<|x′ |<R,
u− (x′ )
xn =0}
1
1
−
|x′ |n
(2R)n
dx′ ≤ cRρ−1 ,
(8)
where c is a constant independent of the variables.
Lemma 2 and estimates (7) and (8) allow us to find a lower bound for the function u(x).
We have
Z
R2 − |x|2
1
1
−u(x) =
−
(−u(y))dσ(y)
ωn R
|y − x|n
|y − x∗ |n
{y: |y|=R, yn >0}
Z
2xn
1
Rn
1
+
−
(−u(y ′ ))dy ′ .
(9)
ωn {y: r<|y′ |<R, yn =0} |y ′ − x|n
|x|n |y ′ − x
e|n
Both integral kernels are positive, since they are derivatives of the Green function with respect to
the outward normal. This permits us to replace −u by u− in the integrals and simultaneously
to transform identity (9) into the inequality. In the following we will estimate the kernels
appearing in the integrals.
No.4
Y.H. Zhang et al: ON THE LOWER BOUND FOR A CLASS OF HARMONIC FUNCTIONS
1493
Let r = |x| ≥ 2, R = 2r. Then, by the mean value theorem for derivatives, we obtain the
estimate of the kernel in the first integral
R2 − |x|2
1
1
−
ωn R
|y − x|n
|y − x∗ |n
R2 − r 2
1
1
=
−
ωn R
|R2 − 2Rr cos(θ − ϕ) + r2 |n/2
|R2 − 2Rr cos(θ + ϕ) + r2 |n/2
R2 − r2 2Rr sin θ sin ϕ
c sin θ
≤ n−1 ,
(10)
≤
ωn R
(R − r)n+2
R
where xn = |x| cos θ, yn = |y| cos ϕ, and c is a constant depending only on n. So
Z
R2 − |x|2
1
1
−
(−u(y))dσ(y)
ωn R
|y − x|n
|y − x∗ |n
{y: |y|=R, yn >0}
Z
1
1
R2 − |x|2
≤
−
u− (y)dσ(y)
≤
n
∗ |n
ω
R
|y
−
x|
|y
−
x
n
{y: |y|=R, yn >0}
Z
c
≤ n
u− (y)yn dσ(y) ≤ cRρ ,
R {y: |y|=R, yn >0}
where c is a constant independent of the variables.
Using the inequalities

n
n

 |y| sin ϕ, y ∈ Rn+ ,
n
|y − x| ≥
n

 |x| ,
y ∈ Rn+ ,
2n
yn = 0,
|y| ≥ 1,
yn = 0,
|y| < 1,
and setting r = |x| ≥ 2, R = 2r, we obtain the estimate of the kernel in the second integral:

xn
1


, y ∈ Rn+ , yn = 0, |y| ≥ 1,

n

ωn |y| sinn ϕ
xn
1
1
Rn
− ′
≤

ωn |y ′ − x|n
|y − x
e|n |x|n
xn 2n


,
y ∈ Rn+ , yn = 0, |y| < 1.

ωn |x|n
For the second integral, we have, for |y| ≥ 1,
Z
1
1
Rn
2xn
−
(−u(y ′ ))dy ′
|y ′ − x|n
|y ′ − x
e|n |x|n
{y: r<|y ′ |<R, yn =0} ωn
Z
2xn
1
1
Rn
=
−
u− (y ′ )dy ′
′ − x|n
′−x
n |x|n
ω
|y
|y
e
|
′
n
{y: r<|y |<R, yn =0}
1 + Rρ
≤ cK n−1 .
sin
ϕ
Substituting inequalities (10) and (11) into (9), we obtain
Z 1
1 + rρ
2n
−u(x) ≤ ARρ + cK n−1 +
u− (t) + u− (−t) dt.
n−1
sin
ϕ ωn r
−1
By using (2), the third integral is bounded uniformly with respect to x′ .
Therefore, for |x| ≥ 2, we get
u(x) ≥ −cK
Thus Theorem 1 is proved.
1 + rρ
.
sinn−1 ϕ
(11)
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Vol.32 Ser.B
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