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Boundary Value Problems
Volume 2007, Article ID 78029, 17 pages
doi:10.1155/2007/78029
Research Article
A Boundary Harnack Principle for Infinity-Laplacian and
Some Related Results
Tilak Bhattacharya
Received 27 June 2006; Revised 27 October 2006; Accepted 27 October 2006
Recommended by Jos
´
e Miguel Urbano
We prove a boundary comparison principle for positive infinity-harmonic functions for
smooth boundaries. As consequences, we obtain (a) a doubling property for certain pos-
itive infinity-harmonic functions in smooth bounded domains and the half-space, and
(b) the optimality of blowup rates of Aronsson’s examples of singular solutions in cones.
Copyright © 2007 Tilak Bhattacharya. This is an open access article distributed under
the Creative Commons Attribution License, which permits unrestricted use, distribution,
and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In this work, one of our main efforts is to prove a boundary Harnack principle for positive
infinity-harmonic functions on domains with smooth boundaries. This will generalize
the result in [1] proven for flat boundaries. In this connection, also see [2–5]. This result
will also be applied to study some special positive infinity-harmonic functions defined
on such domains. One could refer to these as infinity-harmonic measures, however, b eing
solutions to a nonlinear equation, these are not true measures. We derive some properties
of these functions and among these would be the doubling property. A decay rate and a
halving property for such functions on the half-space will also be presented. Another
application will be to show optimality of Aronsson’s singular examples in cones, thus
generalizing the result in [6, 7].
We now introduce notations for describing our results. Let Ω

⊂ R
n
, n ≥ 2, be a domain
with boundary ∂Ω.Wesayu is infinity-harmonic in Ω if u solves in the sense of viscosity
Δ
∞
u =
n

i, j=1
D
i
u(x)D
j
u(x)D
ij
u(x) =0, x ∈ Ω. (1.1)
For more discussion, see [8, 1, 9]. For a motivation for these problems, see [8, 10]. For
2 Boundary Value Problems
r>0andx
∈ R
n
, B
r
(x) will be the open ball centered at x and has radius r.Let

A denote
the closure of the set A and let χ
A
denote its characteristic function. Define Ω

r
(x) = Ω ∩
B
r
(x), P
r
(x) = ∂Ω ∩B
r
(x). We will assume throughout this work that ∂Ω ∈ C
2
.More
precisely,wefirstdefineforeveryx
∈ ∂Ω R
x
to be the radius of the largest interior ball
tangential to Ω at x. We will assume that R
y
> 0foreveryy ∈ ∂Ω and R
x
≥ R
y
/2 > 0,
x
∈ P
δ
y
(y), for some δ
y
> 0. For every x ∈ ∂Ω,setν
x

to be the inner unit normal at x
and x
s
= x + sν
x
, s>0. We will now state Theorem 1.1 which is the result about boundary
Harnack principle [2, 1, 3, 4].
Theorem 1.1 (Boundary Harnack Principle). Let Ω be a domain in
R
n
, n ≥ 2,with∂Ω
satisfying the interior ball condition as stated above. Let u and v be infinity-harmonic in Ω.
Suppose that y
∈ ∂Ω, 0 < 4δ ≤inf(δ
y
, R
y
/2),andu,v>0 in Ω
δ
y
(y).Supposethatu,andv
vanish continuously on P
δ
y
(y), then there exist positive constants C, C
1
, C
2
independent of
u, v,andδ, such that for every z

∈ Ω
δ
(y),
(i) u(z)
≤ cu(y
δ
),
(ii) c
1
u(y
δ
)/v(y
δ
) ≤ u(z)/v(z) ≤c
2
u(y
δ
)/v(y
δ
).
Inequality (i) is often referred to as the Carleson inequality. A proof is provided in
Section 2. At this time, we are unable to determine if Theorem 1.1 also holds when Ω
has Lipschitz continuous boundary. We will apply Theorem 1.1 to prove (a) the doubling
property of solutions of (1.2), and (b) the optimality of blowup rates of the Aronsson
singular functions in cones [6]. Let Ω be a bounded domain. Fix y
∈ ∂Ω;foreveryr>0,
define Q
r
(y) =∂Ω \


P
r
(y). Consider the problem
Δ
∞
u(x) =0, x ∈ Ω, u(x) = 1, x ∈ P
r
(y), u(x) = 0, x ∈ Q
r
(y). (1.2)
By a solution u of (1.2), we mean that (i) u is infinity-harmonic, in the viscosity sense, in
Ω, and (ii) u assumes the values 1 and 0 continuously on P
r
(y)andQ
r
(y). More precisely,
if x
∈ P
r
(y)andz → x, z ∈ Ω,thenu(z) → 1, and analogously for Q
r
(y). We show the ex-
istence of bounded solutions of (1.2)inLemma 3.1.Onecouldrefertou as the nonlinear
infinity-harmonic measure in Ω (although we have not shown uniqueness). Clearly, it is
not a true measure. Our motivation for studying such quantities arises from the works
[2–5]. In the context of boundary behavior, for instance the Fatou theorem, the works
[4, 5] have studied such solutions for the linearized version of the p-Laplacian for finite
p. We will show that requiring boundedness implies the maximum principle and com-
parison, see Lemma 3.1.LetH
={x ∈R

n
: x
n
> 0} denote the half-space in R
n
.Sete
n
to
be the unit vector along the positive x
n
-axis. Set T ={x ∈R
n
: x
n
= 0};fory ∈T,define
P
r
(y) =T ∩B
r
(y), Q
r
(y) =T \

P
r
(y), and M
u
y
(ρ) = sup
∂B

ρ
(y)∩H
u. Define a solution u of
Δ
∞
u(z) = 0, z ∈ H, u|
P
r
(y)
= 1, u|
Q
r
(y)
= 0, (1.3)
to be infinity-harmonic in Ω, in the sense of viscosity, 0
≤ u ≤ 1, continuous up to P
r
(y)
and Q
r
(y), and lim sup
ρ→∞
M
u
y
(ρ) = 0. We will address the existence and uniqueness of
such solutions in Lemma 3.4. We now state a result about the doubling property of solu-
tions of (1.2)and(1.3). For r>0, set o
3r
= 3re

n
.
Tilak Bhattacharya 3
Theorem 1.2. (a) Let Ω
⊂ R
n
be a bounded domain. For y ∈ ∂Ω, assume that P
r
(y) lies on
a connected component of ∂Ω.Letu
r
be a bounded solution of (1.2)inΩ and let r be small.
Then there are positive constants c, C independent of r, such that u
r
(y
r
) ≥ c and
u
2r
(z) ≤ Cu
r
(z), z ∈Ω \B
3r
(y). (1.4)
(b) Let H be the half-space in
R
n
.Letu
r
o

be the unique infinity-harmonic measure in H.
Then there exist universal constants C
1
> 0 and 0 <C
2
< 1 such that
u
2r
o
(z) ≤ C
1
u
r
o
(z), z ∈H \B
3r
(o), u
r
o

o
s

≤
C
2
u
2r
o


o
s

, s ≥ 3r. (1.5)
Estimates in Theorem 1.2 are well known for linear equations [3] and also for the
linearized version for the p-Laplacian [4, 5]. While we are able to prove the doubling
property for any C
2
domain (see Lemma 3.3), it is unclear how a halving property (i.e.,
f (t)
≤ cf(2t), f positive, increasing, and c<1) may be proven if tr ue. In particular, it
would be interesting to know if this is true when Ω is the unit ball. We now introduce
notations for Theorem 1.3.Forα>0, let C
α
stand for the interior of the half-infinite
cone in H,withapexato,thex
n
-axis as the axis of symmetry, and aperture 2α.For
r>0, let M
u
(r) = sup
z∈∂B
r
(o)∩C
α
u(z). We extend the result in [7] to show optimality of
the Aronsson singular examples [6].
Theorem 1.3. For α>0,letC
α
be as described above. Let u, v be positive infinity-harmonic

functions in C
α
. Assume that (i) both u and v vanish continuously on ∂C
α
\{o},
(ii) sup
r>0
M
u
(r) =∞, sup
r>0
M
v
(r) =∞,and(iii) lim
r→∞
M
u
(r) = lim
r→∞
M
v
(r) = 0.
Then there exists a constant C, depending on α, u,andv such that
1
C
≤
u(z)
v(z)
≤ C, z ∈ C
α

. (1.6)
Moreover, for e very m
= 1,2, 3, , if α = π/2m and ω is a direction in C
π/2m
,thenforan
appropriate

C =

C(ω),
1

C|z|
m
2
/(2m+1)
≤ u(z) ≤

C
|z|
m
2
/(2m+1)
, z ∈ C
π/2m
with z =|z|ω. (1.7)
ThelastconclusioninTheorem 1.3 will follow from the works [6, 7]. While Theorem
1.3 applies to special situations, the main purpose is to understand better the blowup
rates of singular solutions, and in some situations decay rates.
We now state some well-known results that will be used in this work. Let u>0be

infinity-harmonic in a domain Ω, suppose that a,b
∈ Ω such that the segment ab is at
least η>0awayfrom∂Ω, then the following Harnack inequality holds:
u(a)e
|a−b|/η
≥ u(b). (1.8)
4 Boundary Value Problems
Let B
r
(a) ⊂ Ω,ifω is a unit vector and 0 ≤t ≤ s<r,then
u(a + tω)
r −t
≤
u(a + sω)
r −s
, u(a + sω)(r
−s) ≤ u(a + tω)(r −t). (1.9)
We wil l re fer to (1.9) as the monotonicity property of u.For(1.8)and(1.9), see [8, 1, 11,
7, 12, 13]. Moreover, u is locally Lipschitz (C
1
if n = 2[14]) and satisfies the comparison
principle [15].
Finally, we mention that it is unclear if a boundary Holder continuity of the quotient
of two infinity-harmonic functions holds for smooth domains. Such a result for general
Lipschitz domains would undoubtedly be quite useful. For p-harmonic functions (finite
p), we direct the reader to the recent work by John Lewis and Kaj Nystrom “Boundary
Behaviour for p Harmonic Functions in Lipschitz and Starlike Lipschitz Ring Domains.”
We thank John Lewis for sending us this work.
2. Proof of Theorem 1.1
Our proof is an adaptation of the methods developed in [2, 1, 3]. Since Δ

∞
is t ranslation
and rotation invariant, we may assume that the origin o
∈ ∂Ω.Setosc
A
u = sup
z∈A
u(z) −
inf
z∈A
u(z) to be the oscillation function of u on the set A.RecallthatΩ
r
(y) = Ω ∩
B
r
(y), y ∈∂Ω.
Step 1 (oscillation estimate near the boundary). Let u>0 be infinity-harmonic in Ω
and vanishing on a neighborhood of o,in∂Ω.LetM
u
(r) = sup
z∈Ω
r
(o)
u(z). By the max-
imum principle, M
u
(r) > 0andu(z) ≤ M
u
(r), z ∈ Ω
r

(o). For 0 <α≤ β, consider the
function w(z)
= M
u
(α)+[M
u
(β) −M
u
(α)](|z|−α)/(β −α), z ∈Ω
β
\Ω
α
.Clearly,u ≤ w
on ∂(Ω
β
\Ω
α
). Thus u ≤ w in Ω
β
\Ω
α
.Thus
M
u
(γ) ≤M
u
(α)+

M
u

(β) −M
u
(α)

γ −α
β −α
, α
≤ γ ≤ β. (2.1)
This implies that osc
Ω
r
(o)
u = M
u
(r)isconvexinr.Sinceu(o) = 0, it follows that 2osc
Ω
r
(o)
u
≤ osc
Ω
2r
(o)
u.
Step 2 (Carleson inequality). We now use the interior ball condition. Since ∂Ω
∈ C
2
, R
x
≥

R
o
/2, x ∈ P
4δ
(o), with 4δ<inf(δ
y
,R
o
/2). For every x ∈ ∂Ω,letν
x
denote the unit inner
normal at x,andsetx
t
= x + tν
x
,0≤t ≤ R
x
.Wewillprovethatu(z) ≤Cu(o
δ
), z ∈ Ω
δ
(o).
We will adapt a device, based on the Harnack inequality, from [3]. For z
∈ Ω
δ
,definex
z
∈
∂Ω to be the point nearest to z.Alsosetd(z) =|x
z

−z|.Thenz = x
z
+ d(z)ν
x
z
= (x
z
)
d(z)
;
set z
s
= x
z
+2
s−1
d(z)ν
x
z
, s =1,2,3, By the Harnack inequality (1.8), for z ∈ Ω
3δ
(o),
u(z)
≤
⎧
⎪
⎪
⎪
⎨
⎪

⎪
⎪
⎩
Mu

z
2

:0<d(z) <
3δ
2
,
Mu

o
δ

: δ<d(z) < 3δ.
(2.2)
Tilak Bhattacharya 5
We take M
= e
8
. We now make an observation which will be used repeatedly in what
follows. If d(z)
≥ δ/2
s
,then
u(z)
≤ Mu


z
2

≤···≤
M
s
u

z
s

≤
M
s+1
u

o
δ

. (2.3)
Suppose now that there is a ξ
0
∈ Ω
δ
(o)suchthatu(ξ
0
) ≥ M
l+2
u(o

δ
), where l = l(δ)islarge
and will be determined later. Using the aforementioned observation, we obtain
dist

ξ
0
,∂Ω

≤
δ
2
l
. (2.4)
Let p
0
∈ ∂Ω be the nearest point to ξ
0
.Clearly,ξ
0
∈ Ω
2
−l
δ
(p
0
) ⊂ Ω
2δ
(o). Thus, osc
Ω

δ2
−l
(p
0
)
u
≥ u(ξ
0
); thus by Step 1,form = 1,2,3 ,
osc
Ω
δ2
−l+m
(p
0
)
u ≥ 2
m
osc
Ω
δ2
−l
(p
0
)
u ≥ 2
m
u

ξ

0

, (2.5)
where 2
m
≥ M
3
= e
24
. Select m = 60; thus osc
Ω
δ2
−l+m
(p
0
)
u ≥ 2
m
u(ξ
0
) ≥ M
l+5
u(o
δ
). Thus
there is a ξ
1
∈ Ω
δ2
−l+m

(p
0
)suchthatu(ξ
1
) ≥ M
l+5
u(o
δ
). Arguing as done in (2.4), we see
dist(ξ
1
,∂Ω) ≤ δ2
−l−3
. Letting p
1
∈ ∂Ω to be closest to ξ
1
,weseethatp
1
∈ Ω
2δ
(o). Repeat-
ing our previous argument,
osc
Ω
δ2
−l−3+m
(p
1
)

u ≥ 2
m
osc
Ω
δ2
−l−3
(p
1
)
u ≥ 2
m
u

ξ
1

≥
M
l+8
u

o
δ

. (2.6)
Thus we may find a ξ
2
∈ Ω
δ2
−l−3+m

(p
1
)suchthatu(ξ
2
) ≥ M
l+8
u(o
δ
), and dist(ξ
2
,∂Ω) ≤
δ2
−l−6
. Thus we obtain a sequence of points ξ
k
∈ Ω and p
k
∈ ∂Ω, k =1,2,3 ,suchthat
u

ξ
k

≥
M
l+2+3k
u(o
δ
), dist


ξ
k
,∂Ω

≤
δ2
−l−3k
, ξ
k
∈ Ω
δ2
−l−3(k−1)+m

p
k−1

. (2.7)
Note that


ξ
k
−o


≤
k−1

i=1



ξ
i+1
−ξ
i


+


ξ
0
−o


≤
δ

1+2
k−1

i=0
2
−l−3i+m

. (2.8)
Choose l
≥ 70, then |ξ
k
−o|≤2δ. Noting that u vanishes continuously on ∂Ω and letting

k
→∞in (2.7) result in a contradiction. Thus the Carleson inequality in Theorem 1.1
follows.
Step 3 (bounds near the boundary). We first derive a lower bound in terms of the distance
to the boundary. For every z
∈ Ω
δ
(o), let x
z
and d(z)beasinStep 2.Notethatd(z) ≤|z −
o|≤δ.Thusx
z
∈ Ω
2δ
(o). Call ζ
z
= x
z
+ δν
x
z
,observethatζ
z
∈ Ω
3δ
(o). By monotonicity
(1.9) and the interior ball condition, we have
u(z)
d(z)
≥

u

ζ
z

δ
≥ e
−6
u

o
δ

δ
, (2.9)
since
|ζ
z
−o
δ
|≤|x
z
+ δν
x
z
−δν
o
|≤4δ.
Let z
∈ Ω

δ
(o). As noted previously, x
z
∈ Ω
2δ
(o)andΩ
δ
(x
z
) ⊂ Ω
3δ
(o). Note that z ∈
Ω
δ
(x
z
). Set μ
z
= sup
Ω
δ
(x
z
)
u, then by comparison u(ξ) ≤ μ
z
|ξ −x
z
|/δ, ξ ∈ Ω
δ

(x
z
). Thus
6 Boundary Value Problems
u(z)
≤ μ
z
d(z)/δ. By the Carleson inequality, μ
z
≤ Cu(ζ
z
). Note that |x
δ
− o
δ
|=|x
z
+
δν
x
−δν
o
|≤4δ. By the Harnack inequality, u(ζ
z
) ≤ e
4
u(o
δ
). Thus there is universal


C,
such that
u(z)
d(z)
≤

C
u

o
δ

δ
, z
∈ Ω
δ
(o). (2.10)
If u, v are two positive infinity-harmonic functions in Ω
4δ
(o), then by (2.9)and(2.10),
there exist universal constants C
1
and C
2
such that
C
1
u

o

δ

v

o
δ

≤
u(z)
v(z)
≤ C
2
u

o
δ

v

o
δ

, z ∈ Ω
δ
(o). (2.11)
This proves Theorem 1.1.
Remark 2.1. We comment that the distance function d(z)
= dist(z,∂Ω), z ∈ Ω,isC
2
and

infinity-harmonic near ∂Ω. Also the oscillation estimate in Step 1 continues to hold for
Lipschitz boundaries. One could show a Carleson inequality by following the ideas in [2].
3. Proof of Theorem 1.2
In this section, we will assume that Ω is a bounded C
2
domain. For y ∈∂Ω and r>0, re-
call the definitions of P
r
(y)andQ
r
(y). Note that both P
r
(y)andQ
r
(y) are relatively open
in ∂Ω.Letu be a solution of (1.2). As in Section 2,forx
∈ ∂Ω, ν
x
and x
t
= x + tν
x
, t>0,
are as defined in Section 2. We will assume that Ω is bounded but we can extend our
arguments to the case of the half-space H.Wewillalwaystakeu to be bounded in this
section. This w ill imply the maximum principle. At this time, it is not clear whether un-
bounded solutions to (1.2) exist. Let C
y
be the connected component of ∂Ω that contains
y.InLemma 3.1, we assume that B

r
(y) ∩∂Ω = B
r
(y) ∩C
y
.
Lemma 3.1. Let Ω
∈ C
2
be a bounded domain. Let y ∈∂Ω and r>0. The following holds.
(i) There exists a solution u of the problem in (1.2) such that 0 <u<1 in Ω.
(ii) If v is any bounded solution of (1.2), then 0 <v<1 in Ω.
(iii) There are a maximal solution u
r
y
and a minimal solution u
r
y
,inΩ such that if v is
any bounded solution of (1.2), then
u
r
y
≤ v ≤ u
r
y
.
(iv) If t<r, then u
t
y

≤ lim
ρ↑r
u
ρ
y
=

u
r
y
≤ u
r
y
= lim
r↓r
u
r
y
.
Moreover, u
r
y
satisfies the following compar ison principle: if ω, w ∈ C(

Ω) are infinity-
harmonic, and ω
≤ u
r
y
≤ w on ∂Ω, then ω ≤ u

r
y
≤ w in Ω.
Proof. Fix y
∈ ∂Ω and r>0. We have broken up our proof into five steps. We first start
with the existence of bounded solutions.
Step 1 (existence). We use the existence results proven in [8, 15], for Lipschitz bound-
ary data. Let η>0besmall.SetI
r
(y) = ∂B
r
(y) ∩ ∂Ω,andfort>0, set S
t
= P
r
(y) ∪
(

x∈I
r
(y)
B
t
(x) ∩ ∂Ω). The set S
t
is obtained by appending a t-band to P
r
(y). For l =
1,2,3, ,let f
l

be such that
(i) f
l
∈ C(∂Ω),
(ii) f
l
(x) = 1, x ∈ P
r
(y),
Tilak Bhattacharya 7
(iii) f
l
(x) = 0, x ∈ ∂Ω \S
η/l
,
(iv) f
l
(x) = (η/l) −dist(x,P
r
(y))/(η/l), x ∈ S
η/l
.
Now let u
l
∈ C(

Ω) be the unique viscosity solution of the problem
Δ
∞
u

l
(z) = 0, z ∈ Ω, u
l
|
∂Ω
= f
l
. (3.1)
Clearly, 0 <u
l
< 1inΩ.Since f
l
≥ f
l+1
, by comparison, there is a function u
η
such that
u
l
↓ u
η
in Ω. We first show that if x ∈ P
r
(y)andz → x ∈ P
r
(y), z ∈ Ω,thenu
η
(z) → 1.
Consider the set Ω
δ

(x), where δ = inf
ξ∈Q
r
(y)
|x −ξ|/2. For z ∈Ω
δ
(x), set w(z) = 1 −|z −
x|/δ. By comparison, for every l, w ≤u
l
≤ 1inΩ
δ
(x). Thus 1 −|z −x|/δ ≤ u
η
(z) ≤ 1, z ∈
Ω
δ
(x). We see that lim
z→x
u
η
(z) = 1. For x ∈ Q
r
(y)andδ = inf
ξ∈P
r
(y)
|ξ −x|/2, it is clear
that for l large, u
l
(z) ≤|z −x|/δ, z ∈ B

δ
(x) . Thus u
η
(z) → 0asz → x. Moreover, the limit
function u
η
does not depend on the width η of the appended band S
η
.Anargument
based on comparison shows easily that for any η
1
, η
2
> 0, u
η
1
= u
η
2
.Setu = u
η
.Our
next step is to show that u is a viscosity solution in Ω. We first show that u is locally
Lipschitz in Ω. To see this, take x
1
∈ Ω and t>0suchthatB
4t
(x
1
) ⊂ Ω. Select x

2
∈ B
t
(x
1
);
set μ
l
= sup
B
4t
(x
1
)
u
l
. Applying monontonicity (1.9)inB
t
(x
1
), we have for every l,(μ
l
−
u
l
(x
1
))/t ≤ (μ
l
−u

l
(x
2
))/(t −|x
1
−x
2
|). Rearranging terms (see [1, Lemma 3.6], also see
[12]), noting that u
l
(x
1
), u
l
(x
2
) ≥ 0andμ
l
≤ μ
1
≤ 1, we obtain |u
l
(x
2
) −u
l
(x
1
)|/|x
1

−
x
2
|≤1/t. Fixing x
1
, x
2
and letting l →∞,weobtainthatu is locally Lipschitz. Fix ξ ∈ Ω
and for 0
≤ t<dist(ξ, pΩ), set M
l
(t) = sup
B
t
(ξ)
u
l
, m
l
= inf
B
t
(ξ)
u
l
, M(t) = sup
B
t
(ξ)
u,and

m(t)
= inf
B
t
(ξ)
u. Using that (i) u
k
≤ u
j
≤ u
l
when l<j<k, (ii) M
l
is convex and m
l
is
concave in t, it follows that for a<c<band z
∈ ∂B
c
(ξ),
b
−c
b −a
m
k
(a)+
c
−a
b −a
m

k
(b) ≤ m
k
(c) ≤ u
j
(z) ≤ M
l
(c) ≤
b −c
b −a
M
l
(a)+
c
−a
b −a
M
l
(b).
(3.2)
Now in (3.2)firstlettingk
→∞,replacingu
j
(z)byu(z), and then letting l →∞,weobtain
that M(t)isconvexandm(t) is concave. This implies that u is a viscosity solution [8]. Part
(i) now follows. A proof also could be worked by showing cone comparison.
Throughout the rest of the proof, u will stand for the solution constructed in Step 1.
Step 2 (comparison). We now prove an easy comparison result for u.Let f
∈ C(∂Ω)and
let u

f
∈ C(

Ω) be the unique infinity-harmonic function with boundary values f .Let
f
≤ χ
P
r
(y)
. Using comparison, we see that for every l, u
f
≤ u
l
in Ω.Thusu
f
≤ u in Ω.
Now let f
≥ χ
P
r
(y)
,setε>0. Since f ≥ 1inP
r
(y), there exists a δ>0suchthat f + ε ≥ 1
in B
δ
(x) ∩∂Ω,foreveryx ∈ ∂Ω ∩∂B
r
(y). Take l large so that η/l ≤ δ/2. By comparison,
u

l
≤ u
f
+ ε in Ω.Thuswehaveu ≤u
f
in Ω.
Step 3 (maximum principle). We now prove part (ii). Let v be any bounded solution of
(1.2). We will adapt an argument used in [11]. We observe that there is an R
0
> 0such
that for x
∈ P
2r
(y), R
x
≥ R
0
, and consequently,

x∈P
2r
(y)
B
R
0
/4
(x
R
0
/4

) ⊂ Ω.Inwhatfollows
we take the quantities σ, η<R
0
/10. We exploit the special geometry of P
r
(y) to achieve
our proof.
Set I
r
(y) = ∂Ω ∩∂B
r
(y); for every x ∈ I
r
(y)andσ>0, define m
x
(σ) = inf
∂B
σ
(x)∩Ω
v
and M
x
(σ) = sup
∂B
σ
(x)∩Ω
v.Clearly,m
x
(σ) ≤ 0andM
x

(σ) ≥ 1. We claim that M
x
is convex
8 Boundary Value Problems
and m
x
is concave in σ. To see this, take z ∈ Ω with 0 <a≤|z −x|≤b.Setw(z) = m
x
(a)+
[m
x
(b) −m
x
(a)](|z −x|−a)/(b − a). Clearly, w ≤ 0. By comparison, w ≤ v in (B
b
(x) \
B
a
(x)) ∩ Ω.Thusm
x
(σ)isconcaveinσ, and one can show analogously that M
x
(σ)is
convex. Define m
y
(σ) = inf
x∈I
r
(y)
m

x
(σ)andM
y
(σ) = sup
x∈I
r
(y)
M
x
(σ), then for σ>0,
(i) M
y
(σ) ≥ 1isconvex, m
y
(σ) ≤ 0isconcaveinσ,
(ii) m
y
(σ) ≤ v(z) ≤M
y
(σ), z ∈ Ω \∪
x∈I
r
(y)
B
σ
(x),
(iii) M
y
(σ) ↑, m
y

(σ) ↓ as σ ↓0.
(3.3)
Note that v
= 0or1on∂Ω \

x∈I
r
(y)
B
σ
(x). Thus (3.3)(i) follows easily. Now using (3.3)(i)
and comparison in the set Ω
\

x∈I
r
(y)
B
σ
(x)yields(3.3)(ii). Clearly, M
y
(σ)(m
y
(σ)) is
the supremum (infimum) of v on Ω
\

x∈I
r
(y)

B
σ
(x). The conclusion in (3.3)(iii) follows
by observing that

x∈I
r
(y)
B
σ
1
(x) ⊂

x∈I
r
(y)
B
σ
2
(x), when σ
1
>σ
2
.By(3.3), the quanti-
ties M(0)
= lim
σ→0
M
y
(σ)andm(0) = lim

σ→0
m
y
(σ) exist. By our assumptions, −∞ <
m(0)
≤ v ≤ M(0) < ∞. We show that m(0) = 0. Assume instead that m(0) < 0. Recall
that v is continuous up to Q
r
(y)andP
r
(y). For x ∈ ∂Ω,letρ(x) = dist(x,P
r
(y)) and
ρ(x) = dist(x,Q
r
(y)). For x ∈ Q
r
(y), define w
x
(z) = m(0)|z −x|/ρ(x) in the set Ω
ρ(x)
(x).
By comparison w
x
≤ v in Ω
ρ(x)
(x), and v ≥ m(0)/2, in Ω
ρ(x)/2
(x). For x ∈ P
r

(y), define
ω
x
(z) = 1+(m(0) −1)|z −x|/ ρ(x)inΩ
ρ(x)
(x). Then v ≥ ω
x
in Ω
ρ(x)
(x)andv ≥m(0)/2,
in Ω
ρ(x)/2
(x). Let η>0besmall.SetA
η
={x ∈ ∂Ω : ρ(x) ≥ η} and B
η
={x ∈ P
r
(y):
ρ(x) ≥ η}. We now apply the above observations to obtain
v(z)
≥
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪

⎪
⎪
⎩
m(0)
2
: z
∈ Ω
ρ(x)/2
(x), x ∈A
η
,
m(0)
2
: z
∈ Ω
ρ(x)/2
(x), x ∈B
η
.
(3.4)
Set S
=

η>0

x∈A
η
Ω
ρ(x)/2
(x)andT =


η>0

x∈B
η
Ω
ρ(x)/2
(x), and call G
y
= Ω \(S ∪T).
For l
= 1,2,3 ,letz
l
∈ Ω be such that v(z
l
) ≤ 7m(0)/8andv(z
l
) → m(0), as l →∞.By
(3.4), z
l
∈ Ω \G
y
, and by the maximum principle, dist(z
l
,I
r
(y)) →0.
In the discussion that follows, we will assume that n>2. Recalling that I
r
(y)= ∂B

r
(y) ∩
∂Ω, it follows that I
r
(y) is smooth. For every l,letx
l
∈ I
r
(y) be the closest point to
z
l
and d
l
=|x
l
−z
l
|. Note that the seg ment x
l
z
l
is normal to I
r
(y). Since x
l
∈ ∂B
r
(y),
yx
l

⊥ ∂B
r
(y), and so yx
l
⊥ I
r
(y). Let T
l
be the hyperplane tangential to ∂Ω at x
l
,andlet
Π
l
be the 2-dimensional plane containing the segments yx
l
and yz
l
.ThusΠ
l
⊥ I
r
(y)atx
l
and ν
x
l
lies in Π
l
.NotethatΠ
l

⊥ T
l
and I
r
(y)istangentialtoT
l
at x
l
.CallJ
l
= ∂Ω ∩Π
l
,
observe that the curve J
l
⊥ I
r
(y)atx
l
. It is easy to see that if x ∈ J
l
is close to x
l
, then (i)
ρ(x)
=|x −x
l
| if x ∈ P
r
(y), and (ii) ρ(x) =|x −x

l
| if x ∈ Q
r
(y). Now consider the set
C
l
= Π
l
∩∂B
d
l
(x
l
) \G
y
. As noted above z
l
∈ C
l
, moreover one can find α
l
∈ C
l
such that
v(α
l
) = 3m(0)/4. We will apply the Harnack inequality in C
l
to obtain a contradiction.
In (3.4), take η

= d
l
and we observe the following. Since ∂Ω ∈ C
2
and x
l
’s lie in a com-
pact set, it follows that for q
∈ C
l
, dist(q,∂Ω) ≈ dist(q,T
l
) = O(d
l
), as d
l
→ 0. In other
words, dist(q,∂Ω) has a lower bound of the order of d
l
. We show this as follows. First
note that since ∂Ω
∈ C
2
, it permits a local parametrization near x
l
,wherex
n
= ν
x
l

, x
n
= 0
is T
l
,andx
n
= φ(x
1
, ,x
n−1
) describes ∂Ω. Clearly, dist(q,∂Ω) ≤|q −x
l
|=d
l
.Wewill
Tilak Bhattacharya 9
show that (a) dist(q,∂Ω)
≥ dist(q,T
l
)+O(d
2
l
) and (b) dist(q,T
l
) ≈ O(d
l
), uniformly in
l. (a) Let (i) q
∂Ω

be the point on ∂Ω closest to q, (ii) let q
T
l
be the point on T
l
closest
to q, (iii) q
int
the point of intersection of the line, containing the seg ment qq
∂Ω
,andT
l
,
and (iv) let q
T
l
∂Ω
be the point on T
L
closest to q
∂Ω
.Clearly,|q −q
T
l
|≤|q −x
l
|=d
l
and
q

∂Ω
∈ B
2d
l
(x
l
). Since ∂Ω ∈ C
2
, |q
∂Ω
−q
T
l
∂Ω
|=O(d
2
l
). If |q −q
∂Ω
|≥|q −q
int
|,then|q −
q
∂Ω
|≥|q − q
int
|+ dist(q
∂Ω
,T
l

) =|q − q
int
|+ O(d
2
l
) ≥|q − q
T
l
|+ O(d
2
l
). Let |q − q
∂Ω
| <
|q −q
int
|.If|q −q
∂Ω
|≥|q −q
T
l
|, then we are done. Otherwise, |q −q
∂Ω
|+ |q
∂Ω
−q
T
l
∂Ω
|≥

|
q −q
T
l
∂Ω
|≥|q −q
T
l
|.Thus|q −q
∂Ω
|≥|q −q
T
l
|+ O(d
2
l
).
(b) We now estimate
|q − q
T
l
|.Letp
l
= J
l
∩ ∂B
d
l
(x
l

), then |q − p
l
|≥d
l
/2. See the
paragraph preceding proof of (a). Note that dist(p
l
,T
l
) = O(d
2
l
), since ∂Ω ∈ C
2
.Ifp
l
−
x
l
,ν
x
l
≥0, then dist(q,T
l
) ≥ d
l
/3. If p
l
−x
l

,ν
x
l
 < 0, it again follows that dist(q,T
l
) ≥
d
l
/3.
We now apply the Harnack inequality, employing the above estimate for (q, ∂Ω), to see
that for some c>0 independent of d
l
,

v

z
l

−
m(0)

≥
e
−c

v

α
l


−
m(0)

≥
e
−c
|m(0)|
4
. (3.5)
Letting l
→∞,weget0≥|m(0)|/4. Thus m(0) = 0. To show that M(0) =1, we work with
function 1
−u and in place of m(0), we take 1 −M(0). Arguing analogously, one may now
show that M(0)
= 1. When n = 2, I
r
(y) reduces to two points and one may again adapt
the above argument to obtain part (ii).
Step 4 (maximal solution u
r
y
). Our next goal is to show that u ≥ v,wherev is any
bounded solution of (1.2). Recall that for x
∈ ∂Ω, ν
x
is the unit inner normal to ∂Ω at
x and x
s
= x + sν

x
.Since∂Ω ∈ C
2
and is bounded, there exists a δ>0suchthatforevery
x
∈ ∂Ω, R
x
≥ δ.Letε>0, small, with ε ≤ min(1/10
4
,δ
2
/10
4
,r
2
/10
4
). For every x ∈ ∂Ω,
set Ω
ε
={x ∈Ω : dist(x,∂Ω) ≥ ε}.Then∂Ω
ε
={x
ε
: x ∈ ∂Ω}. We w ill estimate u and v on
∂Ω
ε
. To this end, set P
ε
={x

ε
: x ∈ P
r
(y)} and Q
ε
={x
ε
: x ∈ Q
r
(y)}.Notethat
Q
ε
= ∂Ω
ε
\

P
ε
, dist(∂Ω,∂Ω
ε
) = ε, Ω
ε
↑ Ω,asε ↓ 0. (3.6)
For z
∈ ∂Ω
ε
,letz
ε
be the nearest point on ∂Ω.Clearly,z = (z
ε

)
ε
.Ifz ∈ Q
ε
,thenu(z
ε
) = 0,
and if z
∈ P
ε
,thenu(z
ε
) = 1. Set
N
ε
=

z ∈ Q
ε
: dist

z
ε
,P
r
(y)

≥
√
ε


, O
ε
=

z ∈ P
ε
: dist

z
ε
,Q
r
(y)

≥
√
ε

. (3.7)
For v, we use comparison as follows. For x
∈ Q
r
(y) with dist(x,P
r
(y)) ≥
√
ε,
v(ξ)
≤

|
ξ −x|
√
ε
, ξ
∈ B
√
ε
(x) ∩Ω, implying that 0 <v(x
ε
) ≤
√
ε. (3.8)
Similarly, for x
∈ P
r
(y) with dist(x,Q
r
(y)) ≥
√
ε,
1
−v(ξ) ≤
|
ξ −x|
√
ε
, ξ
∈ B
√

ε
(x) ∩Ω, implying that 1 −
√
ε ≤ v(x
ε
) ≤ 1. (3.9)
10 Boundary Value Problems
Thus from (3.8)and(3.9), we obtain that
0 <v
≤
√
ε on N
ε
,1−
√
ε ≤ v<1onO
ε
. (3.10)
Note that (3.10) is satisfied by any solution of (1.2), and in particular holds also for
u. However, we will work with u
l
instead. Fix η>0, and recall from Step 1 that for
l
= 1,2, 3, ,
f
l
(x) =
(η/l) −dist

x, P

r
(y)

(η/l)
, x
∈ S
η/l
, f
l
(x) = 0, x ∈ ∂Ω \S
η/l
. (3.11)
For ease of presentation, set j
= 4l/η. We will work with l’s such that j
√
ε<1. For x ∈∂Ω
with dist(x,P
r
(y)) ≤3
√
ε,weseethat
u
l
(x) = 1, x ∈ P
r
(y), u
l
(x) ≥
(η/l) −3
√

ε
(η/l)
≥ 1 − j
√
ε, x ∈ P
r
(y). (3.12)
We now use comparison in B
√
ε
(x) ∩Ω, with dist(x,P
r
(y)) ≤2
√
ε.Setw
x
(z) = j
√
ε +(1−
j
√
ε)|z −x|/
√
ε.Clearly,w
x
≥ 1 −u
l
in B
√
ε

(x) ∩Ω. Using (3.9) and noting that u
l
≥ u,we
have for x
∈ ∂Ω,
(i) u
l
(x
ε
) ≥ 1 −
√
ε, x ∈ P
r
(y), with dist(x,Q
r
(y)) ≥
√
ε,
(ii) u
l
(x
ε
) ≥ (1 − j
√
ε)(1 −
√
ε), with dist(x,P
r
(y)) ≤2
√

ε.
Call J
ε
={x
ε
: x ∈ ∂Ω and dist(x,P
r
(y)) ≤ 2
√
ε}.From(3.7), J
ε
⊃ O
ε
, J
ε
∩N
ε
= ∅,and
u
l
(x
ε
) ≥ (1 − j
√
ε)(1 −
√
ε), x ∈ J
ε
. Using (3.8)and(3.9), we see that u
l

+2j
√
ε ≥ v on
∂Ω
ε
. By comparison, u
l
+2j
√
ε ≥ v in Ω
ε
. Letting ε → 0, we obtain u
l
≥ v in Ω.Now
letting l
→∞,weseethatu ≥v in Ω.
From here on, we call u
r
y
= u and refer to it as the maximal solution of (1.2); clearly,
v
≤ u
r
y
.
Step 5 (minimal solution
u
r
y
). It is clear from Step 1 that for r

1
<r
2
, u
r
1
y
≤ u
r
2
y
(working
with the corresponding u
l
’s). Note that u
r
y
is locally Lipschitz but uniformly so in r.Set
u
r
y
= sup
t<r
u
t
y
. Using Step 1, u
r
y
is a solution of (1.2)andu

r
y
≤ u
r
y
. The comparison prin-
ciple in Step 2 also holds. We now show that
u
r
y
≤ v,wherev is any solution of (1.2).
We do this by showing that u
t
y
≤ v whenever t<r.Fixt<r; we proceed as in Step 4.Let
δ>0beasinStep 4.Letd>0, small, such that 0 <d
≤ min (δ
2
/10
4
,r
2
/10
4
,(r −t)
2
/100);
set Ω
d
={z ∈ Ω : dist(z,∂Ω) ≥ d}.AsinStep 4,defineP

d,s
={x
d
: x ∈ P
s
(y)},wheres
is either r or t.NowdefineQ
d,s
analogously. Then (3.6)holds.Forz ∈ ∂Ω
d
,recallthat
z
d
∈ ∂Ω is such that |z −z
d
|=d.Nowforeachs =r,t, define the sets N
d,s
and O
d,s
, both
subsets of Ω
d
, along the lines of (3.7). Using (3.8), (3.9), and (3.10), we obtain
(i) 0 <u
t
y
(ξ) ≤
√
d, ξ ∈ N
d,t

,
(ii) 1
−
√
d ≤ u
t
y
(ξ) ≤1, ξ ∈ O
d,t
,
(iii) 0 <v(ξ)
≤
√
d, ξ ∈ N
d,r
,
(iv) 1
−
√
d ≤ v(x
d
) ≤ 1, x ∈ O
d,r
.
Clearly, O
d,r
⊃ O
d,t
, N
d,t

⊃ N
d,r
,andforsmalld, N
d,t
∩O
d,r
= ∅.Thusv +2
√
d ≥ u
t
y
on
∂Ω
d
. Using comparison in Ω
d
and taking d → 0, we obtain v ≥ u
t
y
. The claim now follows.
Call
u
r
y
the minimal solution. By Step 4 and arguments presented here, we have the first
Tilak Bhattacharya 11
part of part (iii), namely,
u
t
y

≤ u
r
y
≤ v ≤ u
r
y
,whenevert<r. (3.13)
Since u
r
y
≥ u
r
y
, r>r.Settingu = lim
r↓r
u
r
y
≥ u
r
y
, imitating Step 1, one can show that u
solves (1.2). Thus by Step 4,lim
r↓r
u
r
y
= u
r
y

.Part(iv)isalsoproven.

Remark 3.2. Ifonecouldprovethatlim
t↑r
u
t
y
= u
r
y
, then uniqueness would follow, how-
ever, equality here would be a stronger result. A proof of this is shown for the half-space
in Lemma 3.4.
Let v
r
y
> 0 satisfy the fol lowing:
Δ
∞
v
r
y
(z) = 0, z ∈ Ω, v
r
y
|
Q
r
(y)
= 0, v ∈ C



Ω \Ω
r
(y)

. (3.14)
Lemma 3.3. For y
∈ ∂Ω, r>0,letu
r
be a bounded solution of (1.2)andletv
r
y
be as in
(3.14). Assume that r is small. Then u
r
(y
r
) ≥ c, for some universal constant c>0.Moreover,
there exist universal constants C>0 and
C>0 such that
(i) u
2r
(z) ≤ Cu
r
(z), z ∈ Ω \Ω
3r
(y),
(ii) u
r

(z)/C ≤v
r
y
(z)/v
r
y
(y
r
) ≤ Cu
r
(z), z ∈ Ω \Ω
2r
(y).
Proof. By the maximum principle, 0 <u
r
< 1inΩ.Letw(z) = (r −|z − y|)/r, z ∈ Ω
r
(y).
Then u
r
≥ w on ∂B
r
(y) ∩Ω,andu
r
≥ w on P
r
(y). By comparison, u
r
≥ w in Ω
r

(y). Thus
u
r
(y
r/2
) ≥ w(y
r/2
) = 1/2. We may now use the Harnack inequalit y to conclude that
u
r

y
r

≥
e
−2
u
r

y
r/2

≥
1
2e
2
. (3.15)
We now prove part (i), the “doubling” property of u
r

in Ω \ Ω
3r
(y). We w ill use the
boundary Harnack inequalit y and comparison. Note that u
2r
= u
r
= 0, x ∈ ∂Ω \P
5r/2
(y).
We consider Ω
r/4
(x), x ∈ ∂B
5r/2
(y) ∩∂Ω.ByTheorem 1.1(ii),
u
r
(z)
u
2r
(z)
≥ C
1
u
r

x
r/4

u

2r

x
r/4

, z ∈ ∂B
5r/2
(y) ∩Ω
r/4
(x) . (3.16)
We now use the Harnack inequality and (3.15) to conclude that there are universal con-
stants C
2
, C
3
,andC
4
such that
u
r
(z)
u
2r
(z)
≥ C
2
u
r

y

5r/2

u
2r

y
5r/2

≥
C
3
u
r

y
r

u
2r

y
r

≥
C
4
, z ∈ ∂B
5r/2
(y) ∩Ω. (3.17)
We now use comparison in Ω

\Ω
5r/2
(y) to conclude part (i). We now prove part (ii). For
every x
∈ ∂Ω ∩∂B
2r
(y), we have by Theorem 1.1(ii) that
C
1
u
r

x
r/2

v
r
y

x
r/2

≤
u
r
(z)
v
r
y
(z)

≤ C
2
u
r

x
r/2

v
r
y

x
r/2

, z ∈ ∂B
2r
(y) ∩Ω
r/2
(x) . (3.18)
12 Boundary Value Problems
As done before, we may use the Harnack inequality to conclude that
C
3
u
r

y
r


v
r
y

y
r

≤
u
r
(z)
v
r
y
(z)
≤ C
4
u
r

y
r

v
r

y
r

, z ∈ ∂B

2r
(y) ∩Ω. (3.19)
Thus using (3.15), we obtain
v
r
y

y
r

u
r
(z)
C
≤ v
r
y
(z) ≤ Cv
r
y

y
r

u
r
(z), z ∈∂B
2r
(y) ∩Ω. (3.20)
The claim follows by comparison.


We now look at the case of the half-space H ={x∈R
n
: x
n
>0}.SetT ={x∈R
n
: x
n
=0};
for y
∈ T,defineP
r
(y) = H ∩B
r
(y), Q
r
(y) = T \

P
r
(y), and M
u
y
(ρ) = sup
∂B
ρ
(y)∩H
u.De-
fine a solution u of

Δ
∞
u(z) = 0, z ∈ H, u|
P
r
(y)
= 1, u|
Q
r
(y)
= 0, (3.21)
to satisfy the equation in the sense of viscosity, 0
≤ u ≤ 1, u is continuous up to P
r
(y)and
Q
r
(y), and limsup
ρ→∞
M
u
y
(ρ) = 0. Set L
y
={y + se
n
: s ∈ R}.IntheproofofLemma 3.4,
we make use of an example of a positive singular infinity-harmonic function in the half-
space [6, 11]. We utilize the definition in Step 2 of Theorem 1.3 as appears in Section 4.
For Lemma 3.4,wedefineφ(x)

= f (θ)/|x|
1/3
,whereθ is the conical angle at y and f (θ)
is the function f
m
(θ)whenm = 1. Then φ(x) blows up at y, vanishes elsewhere on T,and
decays to zero at inifinity. In what follows, we make frequent use of the results in [7].
Lemma 3.4 (Half-Space). Let H
={x ∈R
n
: x
n
> 0}. Then there exists a unique solution u
r
y
of the problem in (3.21) such that 0 <u
r
y
< 1.Moreover,ifσ
2
=

n−1
i
=1
(z
i
− y
i
)

2
, then u
r
y
(z)
is symmetric about the line L
y
,andu
r
y
(z) = u
r
y
(σ,z
n
) is decreasing in σ.
Proof. Let 0 <r<ρand set H
ρ
= H ∩B
ρ
(y). For l = 1,2,3, ,andη>0, define u
r,ρ
l
to be
the unique solution of the problem
Δ
∞
u
r,ρ
l

(z) = 0, z ∈ H ∩B
ρ
(y), u
ρ,r
l
|
T
= f
l
, (3.22)
where f
l
is the function as defined in Step 1 of Lemma 3.1.Clearly,0<u
r,ρ
l
< 1andby
comparison, u
r,ρ
l
↑ u
r
l
as ρ ↑∞. Arguing as in Step 1 of Lemma 3.1, we may show that
u
r
l
is infinity-har monic in H and u
r
l
|

T
= f
l
.SetM
l
(s) = sup
∂B
s
(y)∩H
u
r
l
; we now show
that M
l
(s) → 0ass →∞. The following are tr ue: (i) 0 <M
l
(s) ≤ 1, s>0, (ii) M
l
(s) = 1,
0 <s
≤ r, and (iii) M
l
(s) < 1 and is convex in s,whenevers>r+(η/l). Thus M
l
(s)isnon-
increasing. Let φ>0 be the Aronsson singular solution on H as described above. Adapting
the argument in Step 2 of Theorem 1.2 (this follows below), one may show that for some
C>0 depending only on φ(y
2r

), u
r,ρ
l
(z) ≤ Cφ(z), z ∈ ∂B
2r
(y) ∩H,and0= u
r,ρ
l
<φon
∂B
ρ
(y) ∩H.Thusu
r,ρ
l
(z) ≤ Cφ(z), z ∈ H
ρ
\H
2r
.Clearly,u
r
l
≤ Cφ,inH,andM
l
(s) → 0
as s
→∞.Henceu
r
l
solves (3.21) with modified boundary data f
l

.Sinceu
r
l
decreases in
l, it follows that u
r
l
→ u
r
y
,wherenowu
r
y
solves (3.21). Let v be any solution of (3.21).
We show that u
r
y
≥ v in H.Letε>0besmall.WeadaptStep 4 in Lemma 3.1 and use
Tilak Bhattacharya 13
thecomparisonlemma[7, Lemma 2.2]. We work with u
r
l
and estimate both u
r
l
and v on
the set
{x ∈ H : x
n
= ε}.Letr

0
> 100
√
ε be such that 0 ≤ sup(M
v
y
(ρ),M
r
l
) ≤ ε, ρ ≥ r
0
.We
work in S
r
0
ε
={x ∈H : x
n
≥ ε,and|x|≤r
0
}. By letting r
0
→∞and, as done in Step 4 of
Lemma 3.1, then letting ε
→ 0, one can show that u
r
l
≥ v in H for any l.Thusu
r
y

≥ v.Also
we may show by adapting Step 5 that u
t
y
≤ v ≤ u
r
y
, t<r.Callu
r
y
the maximal solution.
Note that u
r
y
is symmetric about L
y
, since by reflection, u
r,ρ
l
is symmetric about L
y
.Writ-
ing u
r,ρ
l
(z) = u
r,ρ
l
(σ,z
n

) and using reflection and comparison (see [7, Lemma 2.6]), we see
that u
r,ρ
l
(σ,z
n
)decreasesinσ. Thus the same holds for u
r
y
.
We now s cale as follows. For θ>0, set z
θ
= y + θ(z − y), z ∈ H,thenμ
θ
(z) = u
r
y
(z
θ
)
solves (3.21)withP
r
(y)replacedbyP
r/θ
(y). Clearly, by the maximality of u
r/θ
y
,
μ
θ

(z) ≤ u
r/θ
y
(z), implying that u
r
y
(z
θ
) ≤ u
r/θ
y
(z), z ∈ H. (3.23)
Using θ>1 and that u
r
y
is maximal, we see that if v is any solution of (3.21), then (3.23)
implies that u
r
y
(z
θ
) ≤ u
r/θ
y
(z) ≤ v(z) ≤ u
r
y
(z), z ∈ H. Letting θ ↑ 1 and using continuity,
we obtain uniqueness of u
r

y
. 
Proof of Theorem 1.2. Set T
s
={x
n
= s}, s>0andM(s) = sup
T
s
u.Letu
r
= u
r
o
solve (3.21).
We will assume that lim
ρ→∞
sup
∂B
ρ
(o)∩H
u
r
= 0. Our proof will use and adapt results from
[7].
Step 1. By Lemma 3.4, u
r
is unique. To show the doubling property of u
r
,weuseLemma

3.3(ii), the comparison result in [7, Lemma 2.2], and (3.15), that is, u
r
(o
r
) ≥ c>0. We
now focus on the halv ing property. By Lemma 3.4, u
r
is unique and u
2r
(x) = u
r
(x/2),
x
∈ H. Thus our goal is to show that
u
r

o
4r

≤
αu
2r

o
4r

=
αu
r


o
2r

, (3.24)
for some universal 0 <α<1. We now make an observation. By Lemma 3.4,fort>1,
u
r
(x) ≤ u
rt
(x) = u
r
(x/t). If ν is a unit vector with ν,e
n
≥0, then u
r
(sν), s>0, is a de-
creasing function of s. In particular, writing a point on the x
n
-axis as (0,x
n
), u
r
(0,x
n
)
decreases in x
n
.ByLemma 3.4, u(0,s) = M(s), s>0, and M(s) is decreasing. To see that
M(s)isconvexins,for0<s<t, consider the set H

s,t
={x ∈H : s<x
n
<t}. The function
w(x)
= M(s)+[M(t) − M(s)][x
n
−s]/(t −s) is infinity-harmonic in H
s,t
,andbycom-
parison, u
r
≤ w in H
s,t
. The claim follows. Since u
r
(o
2r
) = M(2r), u(o
4r
) = M(4r), and
lim
s→∞
M(s) = 0, by convexity it is clear that u(o
4r
)/u(o
2r
) = α<1. Our goal is to show
that α is independent of r.
Step 2 (decay estimate). We show that u

r
(x)decayslike|x|
−1/3
.Weusethework[7]. Let
v(x)
= f (θ)|x|
−1/3
,whereθ = θ(x) = cos
−1
x
n
/|x|, be the Aronsson example of a singular
solution in the half-space H (see Section 4). Consider the set A
t
= H ∩∂B
t
(o), t>0. Em-
ploying Theorem 1.1(ii), the Harnack inequality, and following the proof of Theorem 1.1
in [7], we see that there are universal constants C
1
, C
2
such that
C
1
u
r
(x)
v(x)
≤

u
r

o
t

v

o
t

≤
C
2
u
r
(x)
v(x)
, x
∈ A
t
, t ≥ 2r. (3.25)
14 Boundary Value Problems
Set Γ(t)
= sup
A
t
u
r
/v, γ(t) = inf

A
t
u
r
/v, t ≥ 2r. We now proceed as in [7, Corollary 2.3] to
see that there is a universal constant C
3
such that for t ≥2r and 2r ≤ t
1
≤ t
2
,
γ(t)
≤ Γ(t), Γ(t) ≤ C
3
γ(t), Γ

t
2

≤
Γ

t
1

, γ

t
2


≥
γ

t
1

. (3.26)
For generality, let λ
≥ 2andv(x) = v
λr
(x) = f (θ)(λr/|x|)
1/3
u
r
(o
λr
)/f(0). Then v(o
λr
) =
u
r
(o
λr
). Note that f (0) = (16)
−1/3
,andby(3.15) and the Harnack inequality, u
r
(o
λr

)
≥ e
1−2λ
u
r
(o
r/2
) ≥ e
1−2λ
/2. Using (3.25)and(3.26), we see that there are universal con-
stants C
4
and C
5
such that
C
5
≤ γ(λr) ≤ γ(t) ≤
u
r
(x)
v
λr
(x)
≤ Γ(t) ≤ Γ(λr) ≤ C
4
, x ∈ A
t
, t ≥ λr. (3.27)
Thus C

5
v
λr
(x) ≤ u
r
(x) ≤ C
4
v
λr
(x), x ∈ H \B
t
(o). Thus
C
5
f (θ)
f (0)

λr
|x|

1/3
≤
u
r
(x)
u
r

o
λr


≤
C
4
f (θ)
f (0)

λr
|x|

1/3
, |x|≥λr. (3.28)
Step 3. Using (3.28)andStep 1, it follows that for κ>1and
|x|=κλr, there are universal
constants C
6
and C
7
such that
C
6
κ
1/3
≤
u

o
κλr

u


o
2r

≤
C
7
κ
1/3
=⇒
C
6
κ
1/3
≤
M(κλr)
M(λr)
≤
C
7
κ
1/3
. (3.29)
Choose l
= sup(3,3/C
7
)andsetκ =(lC
7
)
3

> 3.
M(2λr)
≤
κ −2
κ −1
M(λr)+
1
κ −1
M(κλr)
≤
1
κ −1

κ −2+
1
l

M(λr) ≤
κ −5/3
κ −1
M(λr).
(3.30)
Clearly, α<1in(3.24) and is universal.

4. Proof of Theorem 1.3
In this section, we will present another application of Theorem 1.1. We show that any
two positive singular infinity-harmonic singular functions, defined in a cone, are compa-
rable. As a consequence, we will show the optimality of the blowup rates of the Aronsson
examples [6]. This will extend the results in [7]. First we prove a version of monotonicity
that holds in a cone.

Lemma 4.1 (Monotonicity). For 0 <α
≤ π/2,letC
α
denote the interior of the cone {x : x
n
>
0, x
n
=xcosα}.Letu>0 be ∞-harmonic in C
α
.Supposethatν isaunitvectorthatlies
in C
α
, that is, ν,e
n
≥cos α and let θ = α −cos
−1
e
n
,ν,thenfor0 <t<s, u(tν)/t
sinθ
≥
u(sν)/s
sinθ
,andu(tν)t
sinθ
≤ u(sν)s
sinθ
.
Tilak Bhattacharya 15

Proof. We use the version of the Harnack inequality proved in [7, Lemma 2.1]. Then
σ(τ)
= (t + τ(s −t))ν,0≤τ ≤ 1, while d(τ) = σ(τ)sinθ.Thus
u(tν)
≥ u(sν)exp

−

0
1
s
−t
sinθ

t + τ(s −t)

dτ

=
u(sν)exp

−
log(s/t)
sinθ

=
u(sν)

t
s


1/ sinθ
.
(4.1)
Thus u(tν)/t
1/ sinθ
≥ u(sν)/s
1/ sinθ
. Switching tν by sν yields the second inequality. 
Proof of Theorem 1.3. Our proof will be an adaptation of the proof of Theorem 1.1 in
[7]. First note that M
u
(r) is convex. By using comparison, we see that u(x) ≤ M
u
(t)+
[M
u
(s) −M
u
(t)](|x|−t)/(s −t)intheannulusC
α
∩(B
s
(o) \B
t
(o)), 0 <t<s.ThusM
u
(r)
is decreasing, lim
r→0

M
u
(r) =∞,andlim
r→∞
M
u
(r) =0.
Step 1. We will prove that any two positive solutions u and v arecomparableinC
α
.Now
consider the set C
α,r
= C
α
∩B
r
(o). Then (i) for x ∈ ∂C
α
∩∂B
r
(o), R
x
= r tanα, and (ii) for
y
∈ ∂C
α
∩B
r/4
(x), R
y

≥ (3r/4)tanα.InTheorem 1.1,wemaytakeδ = (r/8)tanα.Thus
there are universal constants C
1
and C
2
such that
C
1
u(z)
v(z)
≤
u

x
δ

v

x
δ

≤
C
2
u(z)
v(z)
, z
∈ C
α
∩B

(r tanα)/4
(x) . (4.2)
Set p
r
= re
n
;letS
r,x
be the great circle centered at o, has radius r, and passing through p
r
and x. Using the Harnack inequality, we may conclude that for ξ ∈ (C
α
∩ S
r,x
) \
B
(r tanα)/4
(x), there are constants A
1
= A
1
(α)andA
2
= A
2
(α)suchthatA
1
u(p
r
) ≤ u(ξ) ≤

A
2
u(p
r
). This holds for every x ∈ ∂C
α
∩∂B
r
(o). Using (4.2), we obtain that there are con-
stants C
3
= C
3
(α)andC
4
(α)suchthat
C
3
u(z)
v(z)
≤
u

p
r

v

p
r


≤
C
4
u(z)
v(z)
, z
∈ C
α,r
. (4.3)
It is clear that (4.3)holdsforeveryr>0. Define Γ(r)
= sup
z∈C
α,r
u(z)/v(z)andγ(r) =
inf
z∈C
α,r
u(z)/v(z). Thus from (4.3), there is a constant C
5
= C
5
(α)suchthatγ(r) ≤Γ(r) ≤
C
5
γ(r), r>0. By comparison, Γ(r) is decreasing and γ(r) is increasing in r,see[7, Lemma
2.2 and Corollary 2.3]. Thus using the above inequality, we see that 0 <γ(0)
≤ γ(r) ≤
Γ(r) ≤ Γ(0) ≤ C
5

γ(0) < ∞, r>0. Thus there is a constant C such that v(x)/C ≤ u(x) ≤
Cv(x), x ∈ C
α
.
Step 2. We now show the optimality of the Aronsson examples. We first observe that by
arguing as in [7, Lemmas 2.5 and 2.6], any solution u is axially sy mmetric in C
α
.Ifweset
ρ
2
=

n
i
=1
x
2
i
and θ =cos
−1
x, e
n
/|x|,thenu(x) =u(ρ,θ), x ∈ C
α
,and
Δ
∞
u = u
2
ρ

u
ρρ
+
2u
ρ
u
θ
u
ρθ
ρ
2
+
u
2
θ
u
θθ
ρ
4
−
u
ρ
u
2
θθ
ρ
3
= 0. (4.4)
Note that there is no explicit dependence on the dimension n.Foreachm
= 1,2, 3, ,set

16 Boundary Value Problems
α
=π/2m. The Aronsson example in the planar cone C
π/2m
is given by w
m
(x)=w
m
(|x|, θ) =
f
m
(θ)/|x|
m
2
/(2m+1)
,where
f
m
(θ) =




1 −
cos
2
t
k





(k−1)/2
cos t, θ =

t
0
sin
2
s
k −cos
2
s
ds, k
=−
m
2
2m +1
. (4.5)
Note that θ
= t − (1 + 1/m)arctan(mtant/(m + 1)). From above w
m
is symmetric in θ
and reinterpreting the polar angle θ to be the conical angle, we obtain an example in
higher dimensions. This continues to be a viscosity solution in C
π/2m
,seetheappendixin
[11, 7]. Note that w
m
(|x|, θ) > 0, −π/2m ≤ θ ≤π/2m,andw

m
(±π/2m) = 0. We now have
the desired conclusion by using Step 1.

Acknowledgments
We thank the referees for careful reading of the manuscript. Their various suggestions
have improved the presentation and the clarity of this work.
References
[1] T. Bhattacharya, “On the properties of ∞-harmonic functions and an application to capacitary
convex rings,” Electronic Journal of Differential Equations, vol. 2002, no. 101, pp. 1–22, 2002.
[2] P. Bauman, “Positive solutions of elliptic equations in nondivergence form and their adjoints,”
Arkiv f
¨
or Matematik, vol. 22, no. 2, pp. 153–173, 1984.
[3] L. Caffarelli, E. Fabes, S. Mortola, and S. Salsa, “Boundary behavior of nonnegative solutions of
elliptic operators in divergence form,” Indiana University Mathematics Journal,vol.30,no.4,pp.
621–640, 1981.
[4] E. Fabes, N. Garofalo, S. Mar
´
ın-Malave, and S. Salsa, “Fatou theorems for some nonlinear elliptic
equations,” Revista Matem
´
atica Iberoamericana, vol. 4, no. 2, pp. 227–251, 1988.
[5] J. J. Manfredi and A. Weitsman, “On the Fatou theorem for p-harmonic functions,” Communi-
cations in Partial Differential Equations, vol. 13, no. 6, pp. 651–668, 1988.
[6] G. Aronsson, “Construction of singular solutions to the p-harmonic equation and its limit equa-
tion for p
=∞,” Manuscripta Mathematica, vol. 56, no. 2, pp. 135–158, 1986.
[7] T. Bhattacharya, “A note on non-negative singular infinity-harmonic functions in the half-
space,” Revista Matem

´
atica Complutense, vol. 18, no. 2, pp. 377–385, 2005.
[8] G. Aronsson, M. G. Crandall, and P. Juutinen, “A tour of the theory of absolutely minimizing
functions,” Bulletin of the American Mathematical Society. New Series, vol. 41, no. 4, pp. 439–505,
2004.
[9] M. G. Crandall, L. C. Evans, and R. F. Gariepy, “Optimal Lipschitz extensions and the infinity
Laplacian,” Calculus of Variations and Partial Differential Equations, vol. 13, no. 2, pp. 123–139,
2001.
[10] T. Bhattacharya, E. DiBenedetto, and J. J. Manfredi, “Limits as p
→∞of Δ
p
u
p
= f and related
extremal problems. Some topics in nonlinear PDEs (Turin, 1989),” Rendiconti del Seminario
Matematico. Universit
`
a e Politecnico di Torino, pp. 15–68 (1991), 1989, special issue.
[11] T. Bhattacharya, “On the behaviour of
∞-harmonic functions near isolated points,” Nonlinear
Analysis. Theory, Methods & Applications, vol. 58, no. 3-4, pp. 333–349, 2004.
[12] T. Bhattacharya, “On the behaviour of
∞-harmonic functions on some special unbounded do-
mains,” Pacific Journal of Mathematics, vol. 219, no. 2, pp. 237–253, 2005.
Tilak Bhattacharya 17
[13] P. Lindqvist and J. J. Manfredi, “The Harnack inequality for ∞-harmonic functions,” Electronic
Journal of Differential Equations, vol. 1995, no. 4, pp. 1–5, 1995.
[14] O. Savin, “C
1
regularity for infinity harmonic functions in two dimensions,” Archive for Rational

Mechanics and Analysis, vol. 176, no. 3, pp. 351–361, 2005.
[15] G. Barles and J. Busca, “Existence and comparison results for fully nonlinear degenerate elliptic
equations without zeroth-order term,” Communications in Partial Differential Equations, vol. 26,
no. 11-12, pp. 2323–2337, 2001.
Tilak Bhattacharya: Department of Mathematics, Western Kentucky University, Bowling Green,
KY 42101, USA
Email address: