YEVGENYA MOVSHOVICH
Received 15 July 2003
Using the notion of inferior mean due to M. Heins, we establish two inequalities for such a mean relative to a positive harmonic function defined on the open unit ball or half-space inRn+1.
1. Introduction
In connection withEPspaces, M. Heins proved the following PL-Lemma (unpublished).
Lemma1.1 (PL-Lemma). Ifuis a positive function on the annulus{R <|z|<1}with a subharmonic logarithm, andγ are rectifiable Jordan curves in{r <|z|<1}separating 0 from∞,then
lim
r→1infγ
γu(z)|dz| =limr→1
|z|=ru(z)|dz|. (1.1)
Wu showed in [4] that for a positive harmonic function in the unit disc, one has in most cases inequality, while equality occurs for functions whose boundary measures are absolutely continuous. She also showed that there exists a nonzero lower bound of the lim inf for this class of functions in the disc. The bound is achieved for functions whose boundary measures, for example, are purely singular. We generalize these results to higher dimensions.
LetΩbe the open unit ball or upper half-space inRn+1and letSdenote its boundary. Letube a positive harmonic function onΩ, which, by Riesz’s theorem, is given by a Borel measureµwith the total measureµonS.
Definition 1.2. LetΓbe a piecewiseC1-smooth hypersurface inA
δ= {q∈Ω:d(q,S)< δ}
separating the two boundaries ofAδ.Theinferior meanofuis defined by
IM(u)=lim
δ→0Γ⊂infAδ
Γu(q)dΓ. (1.2)
Letωnbe the volume of the unit sphere inRn+1, and letMn=ωn+1/πωn.In this paper,
we establish the following theorem.
Copyright©2005 Hindawi Publishing Corporation
Theorem1.3. For any positive harmonic functionuonΩwith boundary measureµ,there exists the following inequality:
IM(u)≤ µ. (1.3)
Equality occurs for thoseuwhose boundary measuresµare absolutely continuous, when the inferior mean is attained along boundaries ofAδnot equal toSasδ→0.
Theorem1.4. For any positive harmonic functionuonΩ,with boundary measureµ,there exists the following inequality:
IM(u)≥Mnµ. (1.4)
Equality occurs foruwith point-mass boundary measuresµconcentrated atp0,whenIM(u) is attained along the boundary of the set
˜
Ω=q∈Ω:d(q,S)< σ2\q∈Ω:q−p0< σ asσ→0. (1.5)
The proofs rely on Sard’s theorem (see [3]), and inequality (2.5) obtained below.
2. A surface measure lemma
Given spherical anglesφi∈[0,π],i < n,φn∈[0, 2π], we includeφ0=π/2 andφn+1= 0.For a point q∈Rn+1, the relation between its Cartesian (x1,...,x
n+1) and spherical (r,φ1,...,φn) coordinates is given by
xj=Xjcosφj, Xj=r j−1
i=0
sinφi,r= |q|. (2.1)
From [2, Section 676], we know that on a spherer=const, the Jacobian of this relation satisfies
In= D
x1,...,xn+1
Dr,φ1,...,φn
=XnIn−1= ··· =
n
k=1
Xk. (2.2)
IfSnis the unit sphere inRn+1anddSnis its volume element, then the volume element
onr=const equals
rndSn=Indφ1···dφn. (2.3)
We taker=1 in (2.2) and (2.3) to compute the constant
Mn=
Sn∩{0<φ1<π/2} 2 cosφ1
ωn dS
n=2ωn−1 nωn =
ωn+1
πωn. (2.4)
When a hypersurfaceΓis given byr=r(φ1,...,φn), then its volume element satisfies
dΓ≥rndSn. (2.5)
Lemma2.1. If Γis locally given byr=r(φ1,...,φn), then
dΓ=
1 +
n
k=1 rφ2k X2
k
rndSn. (2.6)
Proof. Assume thatΓis also given byxn+1=f(x1,...,xn).We know thatdΓ≥dRn,
be-cause in this case,
dΓ=1 +|gradf|2dx1···dx
n. (2.7)
We differentiatexn+1=f with respect toφ1,...,φn, solve the system by Cramer’s rule for
∂ f /∂xi, and substitute the result into (2.7), thus obtaining
dΓ=
n+1 i=1 J2
i(n+ 1)dφ1···dφn with Ji(n+ 1)=D
x1,...,xi−1,xi+1,...,xn+1
Dφ1,...,φn . (2.8)
Next, we show by induction onmthat
m
i=1 J2
i(m)=Im2−1
1 +
m−1
k=1 rφ2k X2
k
, m=1, 2,..., (2.9)
which is known to be true form=1, 2, 3.Assume that it is also true form=4,...,n.For JacobiansJi,i < n, we obtain a recurrence relation using the product rule
Ji(n+ 1)=D
..., cosφnXn, sinφnXn
Dφ1,...,φn
=0 + 0 +Xncos2φnJi(n) +Xnsin2φnJi(n)=XnJi(n).
(2.10)
In order to obtain a recurrence relation forJ2
n+Jn2+1, we use likewise the product rule in Jn,Jn+1.We also apply the chain rule to
Dx1,...,xn−1,Xn
Dφ1,...,φn =
Dx1,...,xn−1,Xn
Dr,φ1,...,φn−1
Dr,φ1,...,φn−1
Dφ1,...,φn , (2.11)
noting that this Jacobian depends onφnonly implicitly through the equation forr.Then
J2
n(n+ 1) +Jn2+1(n+ 1)=Xn2Jn2(n) +In2−1rφ2n. (2.12)
Applying (2.2) and the induction assumption to the sum withm=n, we obtain
n+1
i=1 J2
i(n+ 1)=Xn2 n
i=1 J2
i(n) +In2−1rφ2n=I 2
n
1 +
n−1
k=1 rφ2k X2
k
+I 2
nrφ2n X2
n
. (2.13)
3. Poisson kernel
We recall that a positive harmonic functionuonΩhas a representation via the Poisson-Stieltjes integralu(q)=SP(q,p)dµ(p).We write the kernel in the usual half-space coor-dinatesq=(y,s), withs∈Sso that y=dist(q,S)=dist(q,s).We have (cf. [1, pages 12, 127])
P(q,p)= 2y−κy2 ωn|q−p|n+1=
2y−κy2 ωn
y2+|s−p|2(1−κy)(n+1)/2, κ=
0 for half-space, 1 for ball.
(3.1)
By direct integration ofPoverΓδ=∂Aδ =Sin the half-space for all positiveδ, or by the
mean value property ofPin the unit ball as a harmonic function ofqforδ <1, we obtain
ΓδP(q,p)dΓδ=(1−κδ)
n. (3.2)
4. Proof ofTheorem 1.3
The upper bound ofIMfollows from (3.2) right away:
IM(u)≤lim
δ→0
ΓδudΓδ=limδ→0
S
ΓδP(q,p)dΓδdµ(p)= µ. (4.1)
To prove equality, letuhave absolutely continuous boundary measureµ. Sard’s theorem and (2.5) allow us to use, just as in [4], the existence of nonzerou∗to show thatIM≥ µ. LetΓjbe aC1-smooth hypersurface separating boundaries ofA1/ j, j=3, 4,....
Con-sider
Γ
j=
q∈Γj:y=yj(s) is defined in some neighborhood ofq
. (4.2)
By Sard’s theorem, the imageSjofΓjunder the mapq→shas full measure inS.For each points, we choose a preimage onΓjnearest toS, and denote this subset ofΓjbyΓj.It has the same image inSasΓj, moreover,Γj andSjare the coordinate charts related via the mapq=(y,s)→s. From (2.5) and (2.7), we havedΓj≥(1−κyj)ndS.Thus,
ΓjudΓj≥
Γ j
u(q)dΓj≥
Sj
uyj,s1−κyjndS=
Su
yj,s1−κyjndS. (4.3)
Then Fatou’s lemma and the existence of the nontangential limit ofua.e. yield
lim
j→∞
ΓjudΓj≥
Slim infj→∞ u
yj,s
1−κyj n
dS=
Su
5. Proof ofTheorem 1.4
We use local spherical coordinates with the origins atp∈Sand thex1-axis orthogonal to S.Thus, 0≤φ1< π/2, and the Poisson kernel
P(q,p)= 2y−κy2 ωn|q−p|n+1=
2 cosφ1−κr
ωnrn . (5.1)
Forδ∈(0, 1), letΓbe aC1-smooth hypersurface inA
δseparating boundaries ofAδ.We
may assume that everyq∈Γhas a neighborhood in whichr=r(φ1,...,φn) is defined (see
the argument using Sard’s theorem in the proof ofTheorem 1.3). Fubini’s theorem, (2.4), (2.5), and plane geometry yield
ΓudΓ=
S
Γ
2 cosφ1−κr
ωnrn dΓdµ(p)
>
S
{r<√2δ−δ2<cosφ1}
2 cosφ1−κr ωnrn r
ndSndµ(p)
= µMn
(1−δ)n−κ √
2δ−δ2 Mn
.
(5.2)
We obtain the lower bound forIMwhenδ→0.
To prove equality, assume thatuhas the boundary measureµthat is concentrated at pointp0∈S.Then,
u(q)=Pq,p0µp0, µp0= µ. (5.3)
Letσ∈(0,δ).Note that the boundary of ˜Ω= {d(q,S)< σ2} \ {|q−p0|< σ}is formed by two hypersurfaces:Γ1consisting of pointsqon a sphere|q−p0| =σwith the distancey toSlarger thanσ2; andΓ
2consisting of pointsq=(y,s) on a level hypersurface y=σ2 with|q−p0| ≥σ.
OnΓ1, we use spherical coordinates with the origin atp0.We see from (5.1) and (5.3) that
u(q)=2 cosφ1−κσ
ωnσn µ, (5.4)
and from (2.3) thatdΓ1=σndSn.We use these two facts and (2.4) to estimate the integral ofuoverΓ1as follows:
Γ1
udΓ1<µ
Sn∩{0<φ1<π/2} 2 cosφ1
ωnσn σ
ndSn=M
nµ. (5.5)
Once we show that
Γ2
udΓ2=O(σ), (5.6)
Letαbe the distance fromp0tosalong a geodesic inS.Then
α≥s−p0≥ 2
πα, (5.7)
and onΓ2,|s−p0| ≥σ= √
σ2−σ4, where our coordinates are (y,s)=(σ2,s).Also equal-ity (2.3) implies thatdΓ2=(1−κσ2)ndS.Hence, it follows that
Γ2
udΓ2=µ ωn
|s−p0|≥σ
2σ2−κσ4
σ4+s−p021−κσ2(n+1)/2
1−κσ2ndS
<µ ωn
|s−p0|≥σ 2σ2 |s−p0|n+1dS
<µ ωn 2σ
2ω
n−1
∞
σ
αn−1dα
(2/π)αn+1=O(σ).
(5.8)
References
[1] S. Axler, P. Bourdon, and W. Ramey,Harmonic Function Theory, Graduate Texts in Mathemat-ics, vol. 137, Springer-Verlag, New York, 1992.
[2] G. M. Fichtenholz,Differential- und Integralrechnung, vol. 3, VEB Deutsher Verlag der Wis-senschaften, Berlin, 1982.
[3] A. Sard,The measure of the critical values of differentiable maps, Bull. Amer. Math. Soc.48 (1942), 883–890.
[4] J. M. Wu,An integral problem for positive harmonic functions, Ph.D. thesis, University of Illinois, Illinois, 1974.
Yevgenya Movshovich: Department of Mathematics, University of Illinois at Urbana-Champaign (UIUC), 1409 W. Green Street, Urbana, IL 61801, USA
