1 Introduction
In the following, we investigate an ordered pair of functions v,w:ℝN→ℝ which are the sub- and supersolution of the equation
(1.1)
(
-
Δ
)
p
s
u
+
q
(
x
)
|
u
|
p
-
2
u
=
g
in
D
,
where s∈(0,1), p>1, q∈L∞(D) is a nonnegative function, g∈Lp′(D) with p′=pp-1 being the conjugate of p, and (-Δ)ps is the s-fractional p-Laplacian (up to a constant). Recall that for suitable (s,p) and some smoothness conditions on u we may write (see [11, Proposition 2.12])
(
-
Δ
)
p
s
u
(
x
)
=
lim
ϵ
→
0
+
∫
ℝ
N
∖
B
ϵ
(
x
)
|
u
(
x
)
-
u
(
y
)
|
p
-
2
(
u
(
x
)
-
u
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
y
,
x
∈
ℝ
N
.
In order to derive a strong comparison principle for the fractional p-Laplacian we use a weak setting. We denote by Ws,p(ℝN) as usual the fractional Sobolev space of order (s,p) given by
W
s
,
p
(
ℝ
N
)
=
{
u
∈
L
p
(
ℝ
N
)
:
∫
ℝ
N
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
<
∞
}
,
and for an open set D⊂ℝN we denote
(1.2)
𝒲
0
s
,
p
(
D
)
:=
{
u
∈
W
s
,
p
(
ℝ
N
)
:
u
≡
0
on
ℝ
N
∖
D
}
.
For an introduction into fractional Sobolev spaces, we refer to [6]. Finally, we also use the space
(1.3)
W
~
s
,
p
(
D
)
:=
{
u
∈
L
loc
p
(
ℝ
N
)
:
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
<
∞
}
to admit functions with a certain growth at infinity. Given an open set D⊂ℝN, q∈L∞(D), a function v∈W~s,p(D) is called a supersolution of (1.1) if for all nonnegative φ∈𝒲0s,p(D) with compact support in ℝN we have
(1.4)
∫
ℝ
N
∫
ℝ
N
|
v
(
x
)
-
v
(
y
)
|
p
-
2
(
v
(
x
)
-
v
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
+
∫
D
q
|
v
|
p
-
2
v
φ
𝑑
x
≥
∫
D
g
φ
𝑑
x
.
Similarly, we call v a subsolution of (1.1) if -v is a supersolution of (1.1). If v is a sub- and a supersolution of (1.1) andv∈𝒲0s,p(D), then we call v a solution of (1.1). We note that indeed the left-hand side in (1.4) is well-defined as is shown in Lemma 2.4 below.
Equations involving the fractional Laplacian, that is, the case of p=2, have been studied extensively in recent years (see, e.g., [1] and the references therein), whilst for its nonlinear counterpart there are still several unanswered questions. Existence of solutions and their regularity has been treated in [5, 15, 10, 11]. In particular, the question of existence of nontrivial solutions to problem (1.1) in the case q=0 with nontrivial outside data has been studied in [5, 15]. Let us also mention [17], where the Rayleigh quotient associated to (-Δ)ps has been studied, and [14], which analyzes the obstacle problem associated with the fractional p-Laplacian. In this work, we prove a strong comparison principle for equations of type (1.1) and apply this to equations in starshaped rings and in the half space.
Theorem 1.1 (Strong Comparison Principle).
Let s∈(0,1), let p>1, let D⊂RN be an open set, let q∈L∞(D), q≥0, let g∈Lp′(D), where p′=pp-1, and let v,w∈W~s,p(D) be such that v is a supersolution and w a subsolution of (1.1) with v≥w. Assume one of the following conditions holds:
1
1
-
s
<
p
≤
2
, and
v
∈
L
∞
(
ℝ
N
)
or
w
∈
L
∞
(
ℝ
N
)
.
p
≥
2
and for some
α
∈
(
0
,
1
]
with
α
(
p
-
2
)
>
s
p
-
1
we have
v
∈
C
loc
α
(
D
)
∩
L
∞
(
ℝ
N
)
or
w
∈
C
loc
α
(
D
)
∩
L
∞
(
ℝ
N
)
.
Then either v=w a.e. in RN or
essinf
K
(
v
-
w
)
>
0
for all
K
⋐
D
.
The weak comparison principle for the fractional p-Laplacian with q=0 goes back to [17] (see also [11, 15]). However, the validity of a strong comparison principle is already a delicate question in the case s=1, i.e. the case of the classical p-Laplacian. We refer here to the works [18, 19]. Note that in the above nonlocal case neither v nor w need to be solutions, and indeed to achieve such a statement we strongly use the nonlocal structure of the fractional p-Laplacian. In the case p=2, of course, the strong comparison principle follows from the strong maximum principle by linearity (see, e.g., [7]). But in general, when p≠2, the strong maximum principle for the fractional p-Laplacian does not imply the strong comparison principle due to the nonlinear structure of the operator. For the strong maximum principle and a Hopf-type lemma for the fractional p-Laplacian, we refer to the recent work [4].
For an application of Theorem 1.1, we investigate bounded nonnegative solutions of (1.1) in starshaped rings, i.e. we analyze
(1.5)
{
(
-
Δ
)
p
s
u
+
q
(
x
)
|
u
|
p
-
2
u
=
0
in
D
=
D
0
∖
D
¯
1
,
u
=
0
on
ℝ
N
∖
D
0
,
u
=
1
on
D
1
,
where D0,D1⊂ℝN are open sets with 0∈D1¯⊂D0. For our main statement, we recall that a subset A of ℝN is said starshaped with respect to the pointx¯∈A if for every x∈A the segment (1-s)x¯+sx, s∈[0,1], is contained in A. If x¯=0 (as we can always assume up to a translation), we simply say that A is starshaped, meaning that for every x∈A we have sx∈A for s∈[0,1], or equivalently
A
is starshaped if
s
A
⊆
A
for every
s
∈
[
0
,
1
]
.
The set A is said strictly starshaped if 0 is in the interior of A and any ray starting from 0 intersects the boundary of A in only one point.
By U(ℓ), ℓ∈ℝ, we denote the superlevel sets of a function u:ℝN→ℝ:
U
(
ℓ
)
:=
{
u
≥
ℓ
}
=
{
x
∈
ℝ
N
:
u
(
x
)
≥
ℓ
}
.
Theorem 1.2.
Let s∈(0,1), let p>1, and let D=D0∖D¯1, where D0,D1⊂RN are open bounded sets such that 0∈D1 and D¯1⊂D0. Let q:D→[0,∞) such that the following conditions hold:
q
is a bounded Borel-function.
For all
t
≥
1
and
x
∈
ℝ
N
such that
t
x
∈
D
, we have
t
s
p
q
(
t
x
)
≥
q
(
x
)
.
Moreover, let u be a continuous bounded weak solution of (1.5) such that 0≤u≤1, and assume D0 and D1 are starshaped. Then the superlevel sets U(ℓ) of u are starshaped for ℓ∈(0,1).
If in addition D0 and D1 are strictly starshaped sets and
p
≥
2
and
u
∈
C
loc
α
(
D
)
for some
α
∈
(
0
,
1
]
with
α
(
p
-
2
)
>
s
p
-
1
,
then the superlevel sets U(ℓ) of u are strictly starshaped for ℓ∈(0,1).
Remark 1.3.
The starshapedness of superlevel sets is indeed a consequence of the weak comparison principle, hence the assumptions are rather general in this case. To prove the strict starshapedness of superlevel sets, however, we need the strong comparison principle and hence stronger assumptions on u, s, and p in view of Theorem 1.1. We note that in the case q≡0, existence and local Hölder regularity of solutions of (1.5) has been discussed in [5], so Theorem 1.2 can be applied for p∈(0,1), s<p-1p, and for any p≥2, s<1p.
In the case p=2, neither the bounds on s nor the regularity assumption on u are necessary (see [12]).
Let us close this introduction with the following further result in half spaces (see also [8, 3, 2] for similar results).
Theorem 1.4.
Let s∈(0,1), p>1 and set R+N:={x∈RN:x1>0}. Moreover, let q∈L∞(R+N), q≥0, and let u∈W0s,p(R+N)∩L∞(R+N) be a nonnegative continuous function which satisfies
(
-
Δ
)
p
s
u
+
q
(
x
)
|
u
|
p
-
2
u
=
0
in
ℝ
+
N
,
lim
|
x
|
→
∞
u
(
x
)
=
0
.
If q is increasing in the direction of x1, i.e. q(x+te1)≥q(x) for all x∈Ω, t≥0, and
p
≥
2
and
u
∈
C
loc
α
(
ℝ
+
N
)
for some
α
∈
(
0
,
1
]
with
α
(
p
-
2
)
>
s
p
-
1
,
then u≡0 on RN.
The article is organized as follows: In Section 2, we give some basic properties on the involved function spaces and useful elementary inequalities. In Section 3, we give the proof of a variant of a weak comparison principle and then prove Theorem 1.1. The proofs of Theorem 1.2 and 1.4 are given in Section 4.
2 Preliminaries and Notation
We will use the following notation: For subsets D,U⊂ℝN, we denote by Dc:=ℝN∖D the complement of D in ℝN, and we write dist(D,U):=inf{|x-y|:x∈D,y∈U}. If D={x} is a singleton, we write dist(x,U) in place of dist({x},U). The notation U⋐D means that U¯ is compact and contained in D. For U⊂ℝN and r>0, we consider Br(U):={x∈ℝN:dist(x,U)<r}, and we let, as usual Br(x)=Br({x}) be the open ball in ℝN centered at x∈ℝN with radius r>0. For any subset M⊂ℝN, we denote by 1M:ℝN→ℝ the characteristic function of M, and by diam(M) the diameter of M. If M is measurable, |M| denotes the Lebesgue measure of M. For the unit ball, we will use in particular ωN:=|B1(0)|. Moreover, if w:M→ℝ is a function, we let w+=max{w,0} and w-=-min{w,0} denote the positive and negative part of w, respectively.
2.1 Some Elementary Inequalities
We use the notation a∗q:=|a|q-1a for any a∈ℝ, q>0. Note that for a≥0 we have a∗q=aq, and for a<0 we have a∗q=-|a|q. Moreover, we have the following elementary inequalities of this function.
Lemma 2.1 (see [11, Section 2.2]).
For all b≥0, q>0,
(2.1)
(
a
+
b
)
q
≤
max
{
1
,
2
q
-
1
}
(
a
q
+
b
q
)
if
a
≥
0
.
(2.2)
(
a
+
b
)
∗
q
-
a
∗
q
≥
2
1
-
q
b
q
if
a
∈
ℝ
,
q
≥
1
.
Lemma 2.2.
Let M,q>0. Then there are CM,1,CM,2>0 such that for all a∈[-M,M], b≥0,
(2.3)
a
∗
q
-
(
a
-
b
)
∗
q
≤
C
M
,
1
max
{
b
,
b
q
}
,
(2.4)
(
a
+
b
)
∗
q
-
a
∗
q
≥
C
M
,
2
min
{
b
,
b
q
}
.
Proof.
Inequality (2.3) is shown in [11, Section 2.2]. Moreover, if q≥1, then (2.4) follows from (2.2) and indeed no bound on a is needed. For q∈(0,1), fix M>0 as stated and let a∈[-M,M]. Note that the map t↦(t+a)∗q-t∗q satisfies, for t∈[-1,1],
(
t
+
1
)
∗
q
-
t
∗
q
=
q
∫
t
t
+
1
|
v
|
q
-
1
d
v
≥
q
max
{
|
t
|
,
|
t
+
1
|
}
q
-
1
≥
q
2
q
-
1
since q<1. Hence for b>max{0,|a|} we have
(
a
+
b
)
∗
q
-
a
∗
q
b
q
=
(
a
b
+
1
)
∗
q
-
(
a
b
)
∗
q
≥
q
2
q
-
1
.
And for 0≤b≤|a|, using again that q<1, we have
(
a
+
b
)
∗
q
-
a
∗
q
=
q
b
∫
0
1
|
a
+
v
b
|
q
-
1
𝑑
v
≥
q
(
|
a
|
+
|
b
|
)
q
-
1
b
≥
q
2
q
-
1
M
q
-
1
b
.
∎
Lemma 2.3 (see [16, Lemma 2]).
For all q∈(0,1], there is C>0 such that
(2.5)
|
(
a
+
b
)
∗
q
-
a
∗
q
|
≤
C
|
b
|
q
for all
a
,
b
∈
ℝ
.
2.2 Function Spaces and Their Properties
In the following, we let s∈(0,1), p>1, and D⊂ℝN open. We let 𝒲0s,p(D) and W~s,p(D) as introduced in (1.2) and (1.3), respectively. Moreover, we set formally for functions u,v:ℝN→ℝ,
〈
u
,
v
〉
s
,
p
:=
∫
ℝ
N
∫
ℝ
N
(
u
(
x
)
-
u
(
y
)
)
∗
(
p
-
1
)
(
v
(
x
)
-
v
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
.
Recall that 𝒲0s,p(D) is a Banach space with the norm
∥
u
∥
W
s
,
p
:=
(
∥
u
∥
L
p
(
D
)
p
+
〈
u
,
u
〉
s
,
p
)
1
p
,
and that it corresponds to the completion of Cc∞(D) with respect to this norm (see, e.g., [9]).
Lemma 2.4.
The map 〈⋅,⋅〉s,p:W~s,p(D)×W0s,p(D)→R is well-defined.
Proof.
We have by Hölder’s inequality with q=pp-1,
|
〈
u
,
v
〉
s
,
p
|
≤
∫
ℝ
N
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
-
1
|
v
(
x
)
-
v
(
y
)
|
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
=
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
-
1
|
v
(
x
)
-
v
(
y
)
|
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
+
∫
ℝ
N
∖
D
∫
D
|
u
(
x
)
-
u
(
y
)
|
p
-
1
|
v
(
x
)
|
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
≤
(
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
[
N
q
+
s
(
p
-
1
)
]
q
𝑑
x
𝑑
y
)
1
q
(
∫
D
∫
ℝ
N
|
v
(
x
)
-
v
(
y
)
|
p
|
x
-
y
|
[
N
p
+
s
]
p
𝑑
x
𝑑
y
)
1
p
+
(
∫
ℝ
N
∖
D
∫
D
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
[
N
q
+
s
(
p
-
1
)
]
q
𝑑
x
𝑑
y
)
1
q
(
∫
ℝ
N
∖
D
∫
D
|
v
(
x
)
|
p
|
x
-
y
|
[
N
p
+
s
]
p
𝑑
x
𝑑
y
)
1
p
≤
2
(
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
)
1
q
∥
v
∥
W
s
,
p
(
ℝ
N
)
<
∞
.
∎
Lemma 2.5.
We have that W~s,p(D) is a vector space with the following properties:
C
c
0
,
1
(
ℝ
N
)
⊂
W
s
,
p
(
ℝ
N
)
⊂
W
~
s
,
p
(
D
)
.
If
u
∈
W
~
s
,
p
(
D
)
, then
u
±
∈
W
~
s
,
p
(
D
)
.
Proof.
The fact that W~s,p(D) is a vector space follows from (2.1). Moreover, we have Cc0,1(ℝN)⊂Ws,p(ℝN) (see, e.g., [9]), and u∈Ws,p(ℝN)⊂W~s,p(D) is trivial. For the second statement, note that we have |u|∈W~s,p(D) since
|
u
(
x
)
-
u
(
y
)
|
≥
|
|
u
|
(
x
)
-
|
u
|
(
y
)
|
for all
x
,
y
∈
ℝ
N
.
Hence 2u±=|u|±u∈W~s,p(D). ∎
Lemma 2.6.
Let D be bounded and u∈W~s,p(D) with u=0 on RN∖D. Then u∈W0s,p(D).
Proof.
Since D is bounded, we have u∈Lp(D). Moreover,
∫
ℝ
N
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
=
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
+
∫
ℝ
N
∖
D
∫
D
|
u
(
x
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
≤
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
+
∫
ℝ
N
∫
D
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
=
2
∫
D
∫
ℝ
N
|
u
(
x
)
-
u
(
y
)
|
p
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
,
which is bounded by assumption. ∎
An immediate consequence of Lemmas 2.6 and 2.5 is the following corollary.
Corollary 2.7.
Let D be bounded and u∈W~s,p(D) with u≥0 on RN∖D. Then u-∈W0s,p(D).
In the following, we also say that v∈W~s,p(D) satisfies (in weak sense)
(
-
Δ
)
p
s
v
≥
g
in
D
for g∈[𝒲0s,p(D)]′, the dual of 𝒲0s,p(D), if for all nonnegative φ∈𝒲0s,p(D) with compact support in ℝN we have
〈
u
,
φ
〉
s
,
p
≥
∫
Ω
g
(
x
)
φ
(
x
)
𝑑
x
.
Similarly, we use “≤” and “=”.
Lemma 2.8 (cf. [11, Lemma 2.9]).
Let t>0 and let u∈W~s,p(D) satisfy (-Δ)psu=g in D for some g∈[W0s,p(D)]′. Then the function v:RN→R, v(x)=u(tx) satisfies v∈W~s,p(t-1D) and
(
-
Δ
)
p
s
v
=
t
s
p
g
(
t
⋅
)
on
t
-
1
D
.
Proof.
Let φ∈𝒲0s,p(t-1D). Then clearly φ(⋅t)∈𝒲0s,p(D) and
〈
v
,
φ
〉
s
,
p
=
∫
ℝ
N
∫
ℝ
N
(
u
(
x
)
-
u
(
y
)
)
∗
(
p
-
1
)
(
φ
(
x
t
)
-
φ
(
y
t
)
)
|
x
t
-
y
t
|
N
+
s
p
𝑑
x
𝑑
y
=
t
-
N
+
s
p
∫
ℝ
N
∫
ℝ
N
(
u
(
x
)
-
u
(
y
)
)
∗
(
p
-
1
)
(
φ
(
x
t
)
-
φ
(
y
t
)
)
|
x
-
y
|
N
+
s
p
t
-
2
N
𝑑
x
𝑑
y
=
t
-
N
+
s
p
∫
D
g
(
x
)
φ
(
x
t
)
𝑑
x
=
∫
t
-
1
D
t
s
p
g
(
t
x
)
φ
(
x
)
𝑑
x
.
∎
3 Comparison Principles
The following is a slight variant of the weak maximum principle presented in [11, Proposition 2.10], [15, Lemma 6], and [17, Lemma 9].
Lemma 3.1.
Let D⊂RN be an open set, let q∈L∞(D), q≥0, and g∈Lp′(D), where p′=pp-1 . If v,w∈W~s,p(D) are super- and subsolution, respectively, of (-Δ)psu+q(x)u∗(p-1)=g such that v≥w in RN∖D and
lim inf
|
x
|
→
∞
(
v
(
x
)
-
w
(
x
)
)
≥
0
,
then v≥w a.e. in RN.
Proof.
First assume D is bounded and set u(x)=v(x)-w(x) and
Q
(
x
,
y
)
:=
(
p
-
1
)
∫
0
1
|
w
(
x
)
-
w
(
y
)
+
t
(
u
(
x
)
-
u
(
y
)
)
|
p
-
2
𝑑
t
,
P
(
x
)
=
(
p
-
1
)
∫
0
1
|
w
(
x
)
+
t
u
(
x
)
|
p
-
2
𝑑
t
.
Note that P≥0 on ℝN, and Q(x,y)=Q(y,x)≥0 for (x,y)∈ℝN×ℝN. Moreover, if P(x)=0, then w(x)=0=u(x), and if Q(x,y)=0, then w(x)=w(y) and u(x)=u(y). Note that u≥0 on ℝN∖D, and hence u-∈𝒲0s,p(D) due to Corollary 2.7. Note that for any φ∈𝒲0s,p(D), φ≥0, we have
-
∫
D
q
(
x
)
P
(
x
)
u
(
x
)
φ
(
x
)
𝑑
x
=
-
∫
D
q
(
x
)
v
∗
(
p
-
1
)
(
x
)
φ
(
x
)
+
q
(
x
)
w
∗
(
p
-
1
)
(
x
)
φ
(
x
)
d
x
≤
〈
v
,
φ
〉
s
,
p
-
〈
w
,
φ
〉
s
,
p
=
∫
ℝ
N
∫
ℝ
N
[
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
-
(
w
(
x
)
-
w
(
y
)
)
∗
(
p
-
1
)
]
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
=
∫
ℝ
N
∫
ℝ
N
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
(
u
(
x
)
-
u
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
𝑑
x
𝑑
y
.
Since D is bounded, we have u-∈𝒲0s,p(D). Hence, since
[
u
(
x
)
-
u
(
y
)
]
[
u
-
(
x
)
-
u
-
(
y
)
]
=
-
2
u
+
(
x
)
u
-
(
y
)
-
(
u
-
(
x
)
-
u
-
(
y
)
)
2
≤
0
for x,y∈ℝN, we have with φ=u-,
-
∥
q
-
∥
L
∞
(
D
)
∫
D
u
-
(
x
)
2
P
(
x
)
𝑑
x
≤
-
∫
ℝ
N
∫
ℝ
N
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
(
u
-
(
x
)
-
u
-
(
y
)
)
2
𝑑
x
𝑑
y
≤
-
∫
D
u
-
(
x
)
2
∫
ℝ
N
∖
D
2
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
𝑑
y
𝑑
x
≤
0
.
Since q≥0, this implies
0
=
∫
D
u
-
(
x
)
2
∫
ℝ
N
∖
D
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
𝑑
y
𝑑
x
,
which by an argumentation as in [17, Lemma 9] is only possible if u-=0 a.e. Indeed, we have for a.e. (x,y)∈D×(ℝN∖D) either u-(x)=0 or Q(x,y)=0, but in the latter case we have u(x)=u(y) as mentioned above. Since x∈suppu- and y∈ℝN∖D, we have u(x)≤0≤u(y), so that u(x)=0=u(y). Hence also in the latter case it follows that u-(x)=0. Hence the claim follows for bounded sets.
If D is unbounded, then since
lim inf
|
x
|
→
∞
(
v
(
x
)
-
w
(
x
)
)
≥
0
,
we find for every ϵ>0 a number R>0 such that vϵ≥w in ℝN∖DR, where vϵ:=v+ϵ and DR:=BR(0)∩D. Moreover, for φ∈𝒲0s,p(D), φ≥0, with compact support in ℝN, using that q≥0 and t↦t∗(p-1) is increasing, we have
〈
v
ϵ
,
φ
〉
s
,
p
+
∫
D
q
(
x
)
v
ϵ
∗
(
p
-
1
)
(
x
)
φ
(
x
)
𝑑
x
≥
∫
ℝ
N
∫
ℝ
N
(
v
(
x
)
+
ϵ
-
(
v
(
y
)
+
ϵ
)
)
∗
(
p
-
1
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
+
∫
D
q
(
x
)
v
∗
(
p
-
1
)
(
x
)
φ
(
x
)
𝑑
x
=
〈
v
,
φ
〉
s
,
p
+
∫
D
q
(
x
)
v
∗
(
p
-
1
)
(
x
)
φ
(
x
)
𝑑
x
≥
∫
D
g
(
x
)
φ
(
x
)
𝑑
x
,
hence vϵ is a supersolution of (-Δ)psu+q(x)u∗(p-1)=g in DR, while w is a subsolution of this equation in DR. The first part gives v+ϵ=vϵ≥w in ℝN, and since ϵ>0 is arbitrary, this finishes the proof. ∎
Remark 3.2.
We note that almost with the same proof it is possible to show the following: Let D⊂ℝN be an open bounded set, let q∈L∞(D), q≥0, and assume v,w∈W~s,p(D) satisfy in weak sense
(
-
Δ
)
p
s
v
-
(
-
Δ
)
p
s
w
≥
-
q
(
x
)
(
v
-
w
)
∗
(
p
-
1
)
in
D
,
v
≥
w
in
ℝ
N
∖
D
.
Then v≥w a.e. in ℝN.
Lemma 3.3 (see [11, Lemma 2.8]).
Let D⊂RN be an open bounded set, let K⋐RN∖D, and let h∈Lloc1(RN). Then for any v∈W~s,p(D) we have in weak sense
(
-
Δ
)
p
s
(
v
+
h
1
K
)
=
(
-
Δ
)
p
s
v
+
H
in
D
,
with
H
(
x
)
=
2
∫
K
(
(
v
(
x
)
-
v
(
y
)
)
-
h
(
y
)
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
|
x
-
y
|
N
+
s
p
𝑑
y
for a.e. x∈D.
Lemma 3.4.
Let D⊂RN be an open bounded set and let K⋐RN∖D with |K|>0.
If
p
≥
2
, then there is
C
>
0
such that for any
v
∈
W
~
s
,
p
(
D
)
and any
δ
∈
(
0
,
1
]
we have in weak sense
(3.1)
(
-
Δ
)
p
s
(
v
-
δ
1
K
)
≥
(
-
Δ
)
p
s
v
+
C
δ
p
-
1
,
(
-
Δ
)
p
s
(
v
+
δ
1
K
)
≤
(
-
Δ
)
p
s
v
-
C
δ
p
-
1
.
If
p
∈
(
1
,
2
)
, then for any
M
>
0
there is
C
>
0
such that for any
v
∈
W
~
s
,
p
(
D
)
with
∥
v
∥
L
∞
(
ℝ
N
)
≤
M
and any
δ
∈
(
0
,
1
]
we have in weak sense
(3.2)
(
-
Δ
)
p
s
(
v
-
δ
1
K
)
≥
(
-
Δ
)
p
s
v
+
C
δ
,
(
-
Δ
)
p
s
(
v
+
δ
1
K
)
≤
(
-
Δ
)
p
s
v
-
C
δ
.
Proof.
By Lemma 3.3, we have in weak sense
(
-
Δ
)
p
s
(
v
-
δ
1
K
)
=
(
-
Δ
)
p
s
v
+
2
∫
K
(
v
(
x
)
-
v
(
y
)
+
δ
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
|
x
-
y
|
N
+
s
p
𝑑
y
.
We will start by showing the first inequalities in (3.1) and (3.2).
Case 1: p≥2. Then by inequality (2.2) we have
2
∫
K
(
v
(
x
)
-
v
(
y
)
+
δ
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
|
x
-
y
|
N
+
s
p
𝑑
y
≥
2
3
-
p
δ
∗
(
p
-
1
)
∫
K
1
|
x
-
y
|
N
+
s
p
𝑑
y
.
Hence the first inequality in (3.1) holds with C1=23-p|K|supx∈D,y∈K|x-y|-N-sp.
Case 2: p∈(1,2). Then with (2.4) we have
2
∫
K
(
v
(
x
)
-
v
(
y
)
+
δ
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
|
x
-
y
|
N
+
s
p
𝑑
y
≥
C
M
,
2
δ
∫
K
1
|
x
-
y
|
N
+
s
p
𝑑
y
.
Hence the first inequality in (3.2) holds with C2=CM,2|K|supx∈D,y∈K|x-y|-N-sp.
For the second inequalities in (3.1) and (3.2), note that
(
-
Δ
)
p
s
(
v
-
δ
1
K
)
=
(
-
Δ
)
p
s
v
+
2
∫
K
(
v
(
x
)
-
v
(
y
)
-
δ
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
|
x
-
y
|
N
+
s
p
𝑑
y
=
(
-
Δ
)
p
s
v
-
2
∫
K
(
v
(
x
)
-
v
(
y
)
-
δ
+
δ
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
-
δ
)
∗
(
p
-
1
)
|
x
-
y
|
N
+
s
p
𝑑
y
.
Hence this part follows similarly. ∎
Lemma 3.5.
Let D⊂RN be an open bounded set and let s∈(0,1), p>1, and f∈Cc2(D).
If
1
1
-
s
<
p
≤
2
, then for any
v
∈
W
~
s
,
p
(
D
)
there is
C
>
0
such that for all
a
∈
[
-
1
,
1
]
in weak sense
(3.3)
|
(
-
Δ
)
p
s
(
v
-
a
f
)
-
(
-
Δ
)
p
s
v
|
≤
|
a
|
p
-
1
C
in
supp
f
.
If
p
≥
2
, then for any
v
∈
W
~
s
,
p
(
D
)
∩
C
loc
α
(
D
)
∩
L
∞
(
ℝ
N
)
, α∈(0,1] with α(p-2)>sp-1, there is C>0 such that for all a∈[-1,1] in weak sense
(3.4)
|
(
-
Δ
)
p
s
(
v
-
a
f
)
-
(
-
Δ
)
p
s
v
|
≤
|
a
|
C
in
supp
f
.
Proof.
Let s∈(0,1) and a∈[-1,1], fix f∈Cc2(D), R=2diam(D)+1, U:=suppf, and
K
:=
∑
|
α
|
≤
2
∥
∂
α
f
∥
L
∞
(
ℝ
N
)
.
Moreover, let v∈W~s,p(D). If p∈(11-s,2], then we have for any φ∈𝒲0s,p(U), φ≥0, and a constant C1>0 given by (2.5),
|
∫
ℝ
N
∫
ℝ
N
(
(
v
(
x
)
-
v
(
y
)
-
a
(
f
(
x
)
-
f
(
y
)
)
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
|
≤
2
C
1
|
a
|
p
-
1
∫
ℝ
N
φ
(
x
)
∫
ℝ
N
|
f
(
x
)
-
f
(
x
+
y
)
|
p
-
1
|
y
|
N
+
s
p
𝑑
x
𝑑
y
≤
2
C
1
|
a
|
p
-
1
K
p
-
1
∫
U
φ
(
x
)
(
∫
B
R
(
0
)
|
y
|
-
N
+
(
1
-
s
)
p
-
1
𝑑
y
+
∫
ℝ
N
∖
B
R
(
0
)
|
y
|
-
N
-
s
p
𝑑
y
)
𝑑
x
≤
2
C
1
N
ω
N
|
a
|
p
-
1
K
p
-
1
∫
U
φ
(
x
)
(
∫
0
R
r
(
1
-
s
)
p
-
2
𝑑
r
+
∫
R
∞
r
-
1
-
s
p
𝑑
r
)
𝑑
x
≤
2
|
a
|
p
-
1
C
1
K
p
-
1
N
ω
N
(
1
(
1
-
s
)
p
-
1
+
2
s
p
)
∫
U
φ
(
x
)
𝑑
x
,
where ωN=|B1(0)|. Hence (3.3) holds with
C
=
2
C
1
K
p
-
1
N
ω
N
(
1
(
1
-
s
)
p
-
1
+
2
s
p
)
.
Next let p≥2, assume additionally that v∈L∞(ℝN)∩Clocα(D) for some α∈(0,1) with α(p-2)>sp-1, and set for x,y∈ℝN,
Q
(
x
,
y
)
:=
(
p
-
1
)
∫
0
1
|
v
(
x
)
-
v
(
y
)
-
a
t
(
f
(
x
)
-
f
(
y
)
)
|
p
-
2
𝑑
t
.
Note that Q(x,y)=Q(y,x)≥0 for any x,y∈ℝN. Moreover, there is
C
2
=
C
2
(
p
,
∥
v
∥
L
∞
(
ℝ
N
)
,
f
)
and
C
3
=
C
3
(
p
,
∥
v
∥
C
s
+
ϵ
(
D
)
,
f
)
such that
(3.5)
Q
(
x
,
y
)
≤
C
2
for
x
,
y
∈
ℝ
N
,
(3.6)
Q
(
x
,
y
)
≤
C
3
|
x
-
y
|
α
(
p
-
2
)
for
x
,
y
∈
U
,
where we used that U¯⊂D. Moreover, we have dist(suppf,Dc)=δ>0. Fix φ∈𝒲0s,p(U), φ≥0. Then, since
(
v
(
x
)
-
v
(
y
)
-
a
(
f
(
x
)
-
f
(
y
)
)
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
=
-
a
Q
(
x
,
y
)
(
f
(
x
)
-
f
(
y
)
)
,
we have
|
∫
ℝ
N
∫
ℝ
N
(
(
v
(
x
)
-
v
(
y
)
-
a
(
f
(
x
)
-
f
(
y
)
)
)
∗
(
p
-
1
)
-
(
v
(
x
)
-
v
(
y
)
)
∗
(
p
-
1
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
s
p
𝑑
x
𝑑
y
|
=
|
a
|
|
∫
ℝ
N
∫
ℝ
N
(
f
(
x
)
-
f
(
y
)
)
(
φ
(
x
)
-
φ
(
y
)
)
|
x
-
y
|
N
+
s
p
Q
(
x
,
y
)
𝑑
x
𝑑
y
|
≤
2
|
a
|
∫
U
φ
(
x
)
|
lim
ϵ
→
0
∫
B
ϵ
c
(
x
)
(
f
(
x
)
-
f
(
y
)
)
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
𝑑
y
|
𝑑
x
.
We now follow closely the lines of the proof of [13, Lemma 3.8] to show that
(3.7)
sup
x
∈
D
|
lim
ϵ
→
0
∫
B
ϵ
c
(
x
)
(
f
(
x
)
-
f
(
y
)
)
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
𝑑
y
|
≤
C
4
for a constant C4=C4(D,N,p,f,∥v∥Cα(D),∥v∥L∞(ℝN))>0. Clearly, once (3.7) is shown, this finishes the proof of (3.4). To see (3.7), fix x∈U and note that Bδ(x)⊂D. Let ϵ∈(0,δ). Moreover, since f∈Cc2(D), we have
|
2
f
(
x
)
-
f
(
x
-
y
)
-
f
(
x
+
y
)
|
≤
K
|
y
|
2
for all
x
,
y
∈
ℝ
N
,
and from (3.6) and (2.1) there is K2>0 such that
|
Q
(
x
,
x
-
y
)
-
Q
(
x
,
x
+
y
)
|
≤
K
2
|
y
|
α
(
p
-
2
)
for all
x
∈
supp
f
,
y
∈
B
δ
(
0
)
.
Hence with (3.5) and (3.6), we have
|
∫
B
ϵ
c
(
x
)
(
f
(
x
)
-
f
(
y
)
)
Q
(
x
,
y
)
|
x
-
y
|
N
+
s
p
𝑑
y
|
=
|
∫
B
ϵ
c
(
0
)
(
f
(
x
)
-
f
(
x
±
y
)
)
Q
(
x
,
x
±
y
)
|
y
|
N
+
s
p
𝑑
y
|
=
|
1
2
∫
B
ϵ
c
(
0
)
(
2
f
(
x
)
-
f
(
x
+
y
)
-
f
(
x
-
y
)
)
Q
(
x
,
x
+
y
)
|
y
|
N
+
s
p
d
y
+
1
2
∫
B
ϵ
c
(
0
)
(
f
(
x
)
-
f
(
x
-
y
)
)
(
Q
(
x
,
x
+
y
)
-
Q
(
x
,
x
-
y
)
)
|
y
|
-
N
-
s
p
d
y
|
≤
C
3
K
2
∫
B
δ
(
0
)
∖
B
ϵ
(
0
)
|
y
|
2
-
N
-
s
p
+
α
(
p
-
2
)
𝑑
y
+
C
2
4
K
∫
B
δ
c
(
0
)
|
y
|
-
N
-
s
p
𝑑
y
+
K
2
K
2
∫
B
δ
(
0
)
∖
B
ϵ
(
0
)
|
y
|
1
+
α
(
p
-
2
)
-
N
-
s
p
𝑑
y
≤
C
3
2
N
K
1
ω
N
∫
0
δ
ρ
1
-
s
p
+
α
(
p
-
2
)
𝑑
ρ
+
C
2
K
4
N
ω
N
∫
δ
∞
ρ
-
1
-
s
p
𝑑
ρ
+
K
2
K
2
N
ω
N
∫
0
δ
ρ
α
(
p
-
2
)
-
s
p
𝑑
ρ
=
C
3
K
N
ω
N
δ
2
-
s
p
+
α
(
p
-
2
)
2
(
2
-
s
p
+
α
(
p
-
2
)
)
+
C
2
K
4
s
p
N
ω
N
δ
-
s
p
+
K
2
K
N
ω
N
δ
1
-
2
s
+
α
(
p
-
2
)
2
(
1
-
s
p
+
α
(
p
-
2
)
)
=
:
C
4
<
∞
,
where we have used α(p-2)>sp-1. ∎
Lemma 3.6.
Let D⊂RN be an open bounded set, let K⋐RN∖D with |K|>0, δ∈(0,1], s∈(0,1), and p>1. Moreover, fix f∈Cc2(D) with 0≤f≤1.
If
1
1
-
s
<
p
≤
2
, then for any
v
∈
W
~
s
,
p
(
D
)
∩
L
∞
(
ℝ
N
)
there is
a
0
,
b
>
0
such that in weak sense for all
a
∈
(
0
,
a
0
]
,
(3.8)
(
-
Δ
)
p
s
(
v
-
a
f
-
δ
1
K
)
≥
(
-
Δ
)
p
s
v
+
b
in
supp
f
,
(
-
Δ
)
p
s
v
-
b
≥
(
-
Δ
)
p
s
(
v
+
a
f
+
δ
1
K
)
in
supp
f
.
If
p
≥
2
, then for any
v
∈
W
~
s
,
p
(
D
)
∩
C
α
(
D
)
∩
L
∞
(
ℝ
N
)
, α∈(0,1] with α(p-2)>sp-1, there is a0,b>0 such that (3.8) holds in weak sense for all a∈(0,a0].
Proof.
Fix v, s, p as stated. By Lemma 3.4, we have in all cases
(
-
Δ
)
p
s
(
v
-
a
f
-
δ
1
K
)
≥
(
-
Δ
)
p
s
(
v
-
a
f
)
+
C
min
{
δ
,
δ
p
-
1
}
for some C>0. Moreover, by Lemma 3.5 we have for some C~>0 in weak sense
(
-
Δ
)
p
s
(
v
-
a
f
)
≥
(
-
Δ
)
p
s
v
-
max
{
a
,
a
p
-
1
}
C
~
,
a
∈
(
0
,
1
]
.
Hence we may fix a0=a0(δ)∈(0,1] such that b=-max{a0,a0p-1}C~+Cmin{δ,δp-1}>0. This shows the first inequality in (3.8). The second inequality in (3.8) follows similarly. ∎
Proof of Theorem 1.1.
We start with the case v∈L∞(ℝN). Assume there is M⋐ℝN with 0<|M|<|D| and δ:=essinfM(v-w)>0. Without loss of generality, we may assume δ≤1. Fix K⋐D∖M and let f∈Cc2(D∖M) be given with f≡1 in K and 0≤f≤1. Let a0,b>0 be given by Lemma 3.6 and fix a∈(0,a0]. Furthermore, let ua:=v-af-δ1M and note that u∈W~s,p(suppf). Then, assuming for (ii) in addition v∈Clocα(D) for α∈(0,1] as stated, we have in weak sense in suppf by Lemma 3.6,
(
-
Δ
)
p
s
u
a
≥
(
-
Δ
)
p
s
v
+
b
≥
-
q
(
x
)
v
∗
(
p
-
1
)
+
b
≥
-
q
(
x
)
u
a
∗
(
p
-
1
)
+
b
+
q
(
x
)
(
(
v
-
a
f
)
∗
(
p
-
1
)
-
v
∗
(
p
-
1
)
)
≥
-
q
(
x
)
u
a
∗
(
p
-
1
)
+
b
-
∥
q
∥
L
∞
(
D
)
(
v
∗
(
p
-
1
)
-
(
v
-
a
)
∗
(
p
-
1
)
)
.
By (2.5) if p≤2, or by (2.3) with M=∥v∥L∞(ℝN), there is C>0 depending only on p, and if p>2 on M, such that
(
v
∗
(
p
-
1
)
-
(
v
-
a
)
∗
(
p
-
1
)
)
≤
C
max
{
a
,
a
p
-
1
}
.
It follows that
b
-
∥
q
∥
L
∞
(
D
)
(
v
∗
(
p
-
1
)
-
(
v
-
a
)
∗
(
p
-
1
)
)
≥
b
-
∥
q
∥
L
∞
(
D
)
C
max
{
a
,
a
p
-
1
}
.
Hence we may choose a1∈(0,a0] such that b-∥q∥L∞(D)Cmax{a1,a1p-1}>0. Then ua1 satisfies
(
-
Δ
)
p
s
u
a
1
+
q
(
x
)
u
a
1
∗
(
p
-
1
)
≥
g
in
supp
f
,
and ua1≥w in ℝN∖suppf. Lemma 3.1 implies ua1≥w a.e. in suppf, and hence v≥w+a1f in suppf. In particular,
v
≥
w
+
a
1
in
K
.
Since K and f were chosen arbitrarily, this shows
essinf
K
(
v
-
w
)
>
0
for all
K
⋐
D
∖
M
.
Since 0<|M|<|D|, we may fix M~⋐D∖M with δ~:=essinfM~(v-w)>0. By repeating the above argument now with D∖M~ in place of D∖M, this shows the claim for case (i) and for case (ii) with v∈Clocα(D)∩L∞(ℝN).
If w∈Clocα(D)∩L∞(ℝN), we note that Lemma 3.6 implies also the existence of a0,b>0 such that
(
-
Δ
)
p
s
(
w
+
a
f
+
δ
1
K
)
≤
(
-
Δ
)
p
s
w
-
b
for all a∈(0,a0]. Proceeding as in the first case, we find v≥w+af in suppf, and since f was chosen arbitrarily, this finishes the proof similarly to the first case. ∎
Remark 3.7.
We note that, as for the weak maximum principle, it is also with almost the same proof possible to show the following: Let D⊂ℝN be an open set, let q∈L∞(D), q≥0, s∈(0,1), p>1, and assume that v,w∈W~s,p(D) satisfy in weak sense
(
-
Δ
)
p
s
v
-
(
-
Δ
)
p
s
w
≥
-
q
(
x
)
(
v
-
w
)
∗
(
p
-
1
)
in
D
,
v
≥
w
in
ℝ
N
.
If one of the two conditions
1
1
-
s
<
p
≤
2
and v∈L∞(ℝN) or w∈L∞(ℝN), or
p
≥
2
and for some α∈(0,1] with α(p-2)>sp-1 we have v∈Clocα(D)∩L∞(ℝN) or w∈Clocs(D)∩L∞(ℝN),
holds, then either v≡w a.e. in ℝN or essinfK(v-w)>0 for all K⋐D.
Corollary 3.8 (Strong Maximum Principle).
Let D⊂RN be an open set and let f:D×R→[0,∞) be such that f(x,0)=0 for all x∈D and
∫
K
f
(
x
,
v
(
x
)
)
φ
(
x
)
𝑑
x
<
∞
for all
K
⋐
D
and
v
,
φ
∈
L
p
(
K
)
,
v
,
φ
≥
0
.
Moreover, let v∈W~s,p(D) be a nonnegative function satisfying in weak sense
(
-
Δ
)
p
s
v
≥
f
(
x
,
v
)
in
D
.
If p>11-s, then either v≡0 in RN or essinfKv>0 for all K⋐Ω.
Proof.
Since 0∈Ws,p(ℝN)∩Cc0,1(ℝN) satisfies (-Δ)ps0=0≤f(x,v)≤(-Δ)psv, an application of Theorem 1.1 proves the claim. ∎
4 Starshaped Superlevel Sets
As in [12], we use the following observation.
Lemma 4.1 (see [12, Lemma 3.1]).
Let u:RN→R such that M=maxRNu=u(0). Then the superlevel sets U(ℓ), ℓ∈R, of u are all (strictly) starshaped if and only if u(tx)≤u(x) (u(tx)<u(x)) for every x∈RN and every t>1.
Theorem 4.2.
Let D=D0∖D¯1 with D0,D1⊂RN open bounded sets such that 0∈D1 and D¯1⊂D0. Moreover, let b0,b1∈L∞(RN)∩W~s,p(RN) such that b0≡0 on ∂D0 and b1≡1 on ∂D1, let q∈L∞(D) be a nonnegative function, and let g∈Lp′(D) with p′=pp-1 such that both functions satisfy (A2), i.e. the following condition holds:
For all
t
≥
1
and
x
∈
ℝ
N
such that
t
x
∈
D
0
∖
D
¯
1
, we have
t
s
p
q
(
t
x
)
≥
q
(
x
)
and
t
s
p
g
(
t
x
)
≥
g
(
x
)
.
Let u∈W~s,p(RN) be a continuous solution of
{
(
-
Δ
)
p
s
u
+
q
(
x
)
|
u
|
p
-
2
u
=
-
g
in
D
,
u
=
b
0
on
D
0
c
,
u
=
b
1
on
D
1
,
such that 0≤u≤1 in D. If b0|D0c and b1|D1 have starshaped superlevel sets, then the following holds:
If
D
0
and
D
1
are starshaped sets, then the superlevel sets
U
(
ℓ
)
of
u
are starshaped for
ℓ
∈
(
0
,
1
)
.
If
D
0
and
D
1
are strictly starshaped sets,
0
<
u
<
1
in
D
and
p
≥
2
and
u
∈
C
loc
α
(
D
)
for some
α
∈
(
0
,
1
]
with
α
(
p
-
2
)
>
s
p
-
1
,
then the superlevel sets
U
(
ℓ
)
of
u
are strictly starshaped for
ℓ
∈
(
0
,
1
)
.
Proof.
We proceed as in the proof of [12, Theorem 1.8]. Let D0, D1, b0, b1, and u be given as for (i). Note that by assumption it follows that u∈L∞(ℝN). Denote for any t>1 and any function v:ℝN→ℝ,
v
t
(
x
)
=
v
(
t
x
)
x
∈
ℝ
N
.
Thanks to Lemma 4.1, the starshapedness of the level sets of u is equivalent to
u
≥
u
t
in
ℝ
N
for
t
>
1
.
Observe that since the superlevel sets of b0 and b1 are starshaped and 0≤u≤1 in D, we have u≥ut in ℝN∖D0 and in t-1D¯1, and
u
(
x
)
≥
u
t
(
x
)
for
x
∈
D
0
∖
(
t
-
1
D
0
)
and
x
∈
D
¯
1
∖
(
t
-
1
D
¯
1
)
.
Put Dt=(t-1D0)∖D¯1. It remains to investigate ut in Dt. Note that since D0 is bounded, Dt is empty for t large enough. By Lemma 2.8, we get (in weak sense)
(
-
Δ
)
p
s
u
t
=
t
s
p
[
(
-
Δ
)
p
s
u
]
t
=
-
t
s
p
q
(
t
x
)
u
t
∗
(
p
-
1
)
-
t
s
p
g
(
t
x
)
≤
-
q
(
x
)
u
t
∗
(
p
-
1
)
+
g
(
x
)
in
D
t
,
where we used that ut≥0 in ℝN. Hence Lemma 3.1 implies u≥ut in ℝN. This proves (i).
If in addition u, D0, and D1 satisfy the assumptions of (ii), then observe that with the same argument as above Theorem 1.1 yields either ut≡u in ℝN or u>ut in Dt. Since u≡ut in ℝN is not possible for t>1 due to the strict inequality 0<u<1 in D, we must have u>ut in Dt for all t>1. This proves (ii). ∎
Proof of Theorem 1.2.
If D0 and D1 are starshaped, then the result follows immediately from Theorem 4.2 (i). Hence let D0 and D1 be strictly starshaped. Since the functions v≡1 satisfies (-Δ)psv+q(x)|v|p-2v=q(x)≥0 in D and v≥u, v≢u in Dc, Theorem 1.1 implies u<1 in D. In a similar way, the function w≡0 satisfies (-Δ)psw+q(x)|w|p-2w=0 in D and u≥w, u≢w in Dc, Theorem 1.1 implies u>0 in D. Hence the claim follows from Theorem 4.2 (ii) with b0≡0 and b1≡1. ∎
Proof of Theorem 1.4.
Let q, u be as stated and assume u≢0 on ℝN. Then by the strong comparison principle we must have u>0 on ℝ+N since 0 is solution of (-Δ)psu+q(x)u∗(p-1)=0 and u≥0. For t≥0, denote ut(x):=u(x+te1) for x∈ℝN and Ht:={x∈ℝ+N:x1>t}. Then we have in weak sense for all t≥0,
(
-
Δ
)
p
s
u
+
q
(
x
)
u
∗
(
p
-
1
)
≥
(
-
Δ
)
p
s
u
t
(
x
)
+
q
(
x
)
u
t
∗
(
p
-
1
)
on
H
t
by the assumptions on q. Hence for all t≥0, since lim|x|→∞(u(x)-ut(x))=0, Lemma 3.1 implies ut≤u on Ωt. We state the following claim:
(4.1)
for all
t
>
0
,
we have
u
t
<
u
on
H
t
.
Fix t>0 and assume by contradiction that ut≡u on ℝN. But then u≡0 on {x∈ℝN:0<x1<t}, which is a contradiction to the fact that u>0 on ℝ+N. Hence ut≢u on ℝN, and Theorem 1.1 implies (4.1) since t>0 is arbitrary.
Note that (4.1) implies that u is strictly decreasing in x1, but since u≥0, u=0 on (ℝ+N)c and u is continuous, this is a contradiction, and hence we must have u≡0 on ℝN as claimed. ∎
