1 Introduction and Statement of Main results
The classical Trudinger–Moser inequality for bounded domain Ω⊂ℝn (see [24, 31]) asserts that there exists a constant C=C(n)>0 such that
(1.1)
sup
u
∈
W
0
1
,
n
(
Ω
)
,
|
∇
u
|
L
n
(
Ω
)
≤
1
∫
Ω
e
α
u
n
n
-
1
𝑑
x
≤
C
⇔
α
≤
α
n
=
n
ω
n
-
1
1
n
-
1
,
where ωn-1 stands for the (n-1)-dimensional surface measure of the (n-1)-sphere. For unbounded domains, and in particular for the whole euclidean space ℝn, several Trudinger–Moser type inequalities have been established. We can, for example, cite [1, 7, 22, 29]. Subsequently, Adimurthi and Sandeep in [2] extended the Trudinger–Moser inequality (1.1) for singular weights. More precisely, they proved the following result.
Theorem 1.1 ([2, Theorem 2.1]).
Let Ω⊂Rn(n≥2) be a smooth bounded domain. Then, for any α>0 and σ∈[0,n), we have
∫
Ω
e
α
|
u
|
n
n
-
1
|
x
|
σ
𝑑
x
<
+
∞
.
Moreover,
(1.2)
sup
{
∫
Ω
e
α
|
u
|
n
n
-
1
|
x
|
σ
d
x
,
u
∈
W
0
1
,
n
(
Ω
)
,
|
∇
u
|
L
n
(
Ω
)
≤
1
}
<
+
∞
⇔
α
α
n
+
σ
n
≤
1
.
That last inequality has been generalized to the whole space ℝn, n≥2, by Adimurthi and Yang in [3] as follows.
Theorem 1.2 ([3, Theorem 1.1]).
For all α>0 and σ∈[0,n), and u∈W1,n(Rn), we have
∫
ℝ
n
e
α
|
u
|
n
′
-
S
n
-
2
(
α
,
u
)
|
x
|
σ
𝑑
x
<
+
∞
,
where n′=nn-1 and Sn-2(α,u)=∑m=0n-2αmm!|u|mn′. Furthermore, for all α≤(1-σn)αn and τ>0, we have
sup
{
∫
ℝ
n
e
α
|
u
|
n
′
-
S
n
-
2
(
α
,
u
)
|
x
|
σ
d
x
,
∥
u
∥
1
,
τ
≤
1
}
<
+
∞
,
where
∥
u
∥
1
,
τ
n
=
∫
ℝ
n
(
|
∇
u
|
n
+
τ
|
u
|
n
)
𝑑
x
.
The question of the existence of extremal functions for that singular inequality has been discussed in [21]. We also have to mention the work [20] of Li, which can be regarded as an improvement of Theorem 1.2. Concerning the singular Trudinger–Moser inequalities, we can also mention the recent work of Nguyen and Takahashi [27]. Now, concerning Trudinger–Moser inequalities defined on weighted Sobolev spaces, we can, for example, cite [5, 4, 6, 8, 10, 11, 9, 14, 15, 17, 18, 26]. The majority of those works considered the restriction to radial functions, and in [18], although the weight is not necessarily radial but its growth permits to pass to the radial case through some radial rearrangement. This interest to reduce the inequality to the radial functions is mainly motivated by their capacity to increase the maximal growth of the integrability. Recently, Calanchi and Ruf studied in [11] the case of weighted Sobolev norms with weights of logarithmic type defined on the unit open ball ℬ of ℝn. In fact, they considered the subspace W0,rad1,n(ℬ,w1) defined as the radial functions of the completion of C0∞(ℬ) with respect to the norm
∥
u
∥
w
1
n
=
∫
ℬ
w
1
(
x
)
|
∇
u
|
n
𝑑
x
,
where w1(x)=(log1|x|)β(n-1) or w1(x)=(loge|x|)β(n-1), β≥0, x∈ℬ.
First, Calanchi and Ruf proved the following result.
Theorem 1.3 ([11, Theorem 1]).
Let 0<β<1. Then, for u∈W0,rad1,n(B,w1), we have
(1.3)
∫
ℬ
e
|
u
|
γ
𝑑
x
<
+
∞
⇔
γ
≤
γ
n
,
β
=
n
(
n
-
1
)
(
1
-
β
)
=
n
′
1
-
β
.
Moreover,
(1.4)
sup
{
∫
ℬ
e
α
|
u
|
γ
n
,
β
d
x
,
∥
u
∥
w
1
≤
1
}
<
+
∞
⇔
α
≤
α
n
,
β
=
n
[
ω
n
-
1
1
n
-
1
(
1
-
β
)
]
1
1
-
β
.
Clearly, if β=0, we obtain the classical Trudinger–Moser inequality. Secondly, Calanchi and Ruf considered the limiting case where β=1 and w1(x)=(loge|x|)n-1, x∈ℬ. In this case, the specific behavior of the weight function has an impact on the corresponding embeddings. In fact, the maximal growth e|s|nn-1 proved in the classical Trudinger–Moser inequality significantly increased so that a doubly exponential growth is now permitted. More precisely, they proved th following theorem.
Theorem 1.4 ([11, Theorem 4]).
We have
(1.5)
∫
ℬ
e
e
|
u
|
n
′
𝑑
x
<
+
∞
for all
u
∈
W
0
,
rad
1
,
n
(
ℬ
,
w
1
)
,
and
(1.6)
sup
{
∫
ℬ
e
a
e
w
n
-
1
1
n
-
1
|
u
|
n
′
d
x
,
u
∈
W
0
,
rad
1
,
n
(
ℬ
,
w
1
)
,
∥
u
∥
w
1
≤
1
}
<
+
∞
⇔
a
≤
n
.
Profiting of this new Trudinger–Moser inequality, Calanchi, Ruf and Sani proved in [12] the existence of a nontrivial radial solution for an elliptic problem defined on the unit ball of ℝ2, with nonlinearities having a doubly exponential growth at infinity.
In this paper, in a first step, we extend the inequalities proved in Theorem 1.3 and Theorem 1.4 when a singular term (as in Theorem 1.1) appears. In a second step, we generalize the result to the whole space ℝn. This generalization (which is not trivial at all), even in the absence of the singular term, is completely new and interesting in itself. Finally, an application is given by treating an elliptic equation defined on ℝn and whose nonlinearities enjoy a doubly exponential growth at infinity.
Throughout this work, we consider the standard weighted Sobolev space defined as the completion of C0∞(ℬ) with respect to the norm
∥
u
∥
w
n
=
∫
ℬ
w
(
x
)
|
∇
u
|
n
𝑑
x
,
and denote by W0,rad1,n(ℬ,w) the corresponding subspace of radial functions, where
(1.7)
w
(
x
)
=
|
log
(
e
|
x
|
)
|
β
(
n
-
1
)
.
The first result in the present work is the following extension of Theorem 1.3.
Theorem 1.5.
Let β∈[0,1), let w be defined by (1.7) and let u∈W0,rad1,n(B,w), n≥2. Then, for every α>0 and σ∈[0,n), we have
(1.8)
∫
ℬ
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
<
+
∞
.
Moreover,
(1.9)
sup
{
∫
ℬ
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
d
x
,
u
∈
W
0
,
rad
1
,
n
(
ℬ
,
w
)
,
∥
u
∥
w
≤
1
}
<
+
∞
⇔
α
α
n
,
β
+
σ
n
≤
1
,
with αn,β=n[ωn-11n-1(1-β)]11-β.
One can easily see that the value β=1 is a kind of second order limiting case. In reality, for that “limit value”, we find that the maximal growth of integrability is of double exponential type.
Theorem 1.6.
Let n≥2, let w be defined by (1.7) with β=1 and u∈W0,rad1,n(B,w). Then, for every σ∈[0,n),
(1.10)
∫
ℬ
e
e
|
u
|
n
′
|
x
|
σ
𝑑
x
<
+
∞
.
Moreover, we have
(1.11)
sup
u
∈
W
0
,
rad
1
,
n
(
ℬ
,
w
)
,
∥
u
∥
w
≤
1
∫
ℬ
e
a
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
|
x
|
σ
𝑑
x
<
+
∞
⇔
a
+
σ
≤
n
.
Remark 1.7.
Obviously, if β=0, then (1.9) turns out to be the classical singular sharp Trudinger–Moser inequality stated in (1.2). Also, if σ=0, then inequalities (1.9) and (1.11) are nothing else than (1.4) and (1.6), respectively.
In the second part of this work, we are interested in the case of an unbounded domain of ℝn. More precisely, we consider a radial weight
(1.12)
w
(
x
)
=
{
(
log
(
e
|
x
|
)
)
β
(
n
-
1
)
if
|
x
|
<
1
,
χ
(
|
x
|
)
if
|
x
|
≥
1
,
where, 0<β<1 and χ:[1,+∞[→]0,+∞[ is a continuous function such that χ(1)=1 and inft∈[1,+∞[χ(t)>0. Denote by E the weighted Sobolev space
E
=
{
u
∈
W
rad
1
,
n
(
ℝ
n
)
:
∫
ℝ
n
w
(
x
)
|
∇
u
|
n
𝑑
x
<
+
∞
}
.
We equip E with the standard Sobolev norm
∥
u
∥
n
=
∫
ℝ
n
|
∇
u
|
n
w
(
x
)
d
x
+
∫
ℝ
n
|
u
|
n
d
x
.
Throughout this part of the work, we use the notation
S
n
-
2
(
α
,
u
)
=
∑
k
=
0
n
-
2
α
k
k
!
|
u
|
k
.
We have the following first extension of Theorem 1.5.
Theorem 1.8.
Let n≥2, and let w be defined by (1.12). For all α>0 and u∈E, we have
(1.13)
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
.
Moreover, if ααn,β+σn<1, then
(1.14)
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
.
If ααn,β+σn>1, then
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
=
+
∞
.
The value ααn,β+σn=1 does not necessarily belong to the range of the positive values α for which the supremum in (1.14) is finite. However, the sharpness of the Trudinger–Moser inequality can be recovered when we consider a different functional space. More precisely, for 0<β<1 , define Eβ as the space of all the radial functions of the completion of C0∞(ℝn) with respect to the norm
∥
u
∥
β
=
|
∇
u
|
L
n
(
ℝ
n
,
w
)
+
|
u
|
L
d
β
(
ℝ
n
)
=
(
∫
ℝ
n
w
(
x
)
|
∇
u
|
n
d
x
)
1
n
+
(
∫
ℝ
n
|
u
|
d
β
d
x
)
1
d
β
,
where
d
β
=
n
′
(
1
-
β
)
n
′
-
1
+
β
.
For this space, we obtain the following sharp singular Trudinger–Moser inequality, which can be considered as a second extension of Theorem 1.5.
Theorem 1.9.
Let 0<β<1, 0<σ<n, and let w be defined by (1.12). For all u∈Eβ, we have
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
for all
α
>
0
.
Moreover, if β≤1n′+1, then
sup
u
∈
E
β
,
∥
u
∥
β
≤
1
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
⇔
α
α
n
,
β
+
σ
n
≤
1
.
Finally, we provide the following extension of Theorem 1.6.
Theorem 1.10.
For all α>0, σ∈[0,n) and u∈E, we have
(1.15)
∫
ℝ
n
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
<
+
∞
.
If a≤(n-σ)e(infs≥1χ(s))1n(n-1), then
(1.16)
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
<
+
∞
.
If a>(n-σ)exp(1n-1∫0+∞logn(1+t)e-nt𝑑t), then
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
=
+
∞
.
In the last part of this work, we prove the existence of at least two solutions for the singular elliptic equation
-
div
(
w
(
x
)
|
∇
u
|
n
-
2
∇
u
)
+
|
u
|
n
-
2
u
=
f
(
x
,
u
)
|
x
|
σ
+
ϵ
h
in
ℝ
n
,
h
∈
E
*
∖
{
0
}
,
where w is defined by (1.12) with β=1 and f:ℝn×ℝ→ℝ is a Carathéodory function enjoying a doubly exponential growth governed by the Trudinger–Moser inequality (1.16). See Section 7 for more details.
2 Proof of Theorem 1.5
Using Hölder’s inequality and (1.3), we obtain (1.8). Indeed, if t>1 is chosen so that σt<n, then
∫
ℬ
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
≤
(
∫
ℬ
e
α
t
(
t
-
1
)
|
u
|
n
′
1
-
β
𝑑
x
)
t
-
1
t
(
∫
ℬ
d
x
|
x
|
t
σ
)
1
t
<
∞
.
Now, in order to prove (1.9), we start with the case when ααn,β+σn<1. Choosing t>1 such that
α
α
n
,
β
+
σ
t
n
=
1
and using Hölder’s inequality, from (1.4), it follows that
sup
∥
u
∥
w
≤
1
∫
ℬ
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
d
x
≤
sup
∥
u
∥
w
≤
1
(
∫
ℬ
e
α
n
,
β
|
u
|
n
′
1
-
β
)
α
α
n
,
β
(
∫
ℬ
d
x
|
x
|
n
t
)
σ
t
n
<
∞
.
It remains to prove (1.9) in the case when ααn,β+σn=1. Let γ=ααn,β. Then σn=1-γ. Thus, it suffices to show that
sup
∥
u
∥
w
≤
1
∫
ℬ
e
γ
α
n
,
β
|
u
|
n
′
1
-
β
|
x
|
(
1
-
γ
)
n
𝑑
x
<
∞
.
For u∈W0,rad1,n(ℬ,w), we have
(2.1)
∫
ℬ
w
(
x
)
|
∇
u
|
n
𝑑
x
=
ω
n
-
1
∫
0
1
w
(
r
)
|
u
′
(
r
)
|
n
r
n
-
1
𝑑
r
and
(2.2)
∫
ℬ
e
γ
α
n
,
β
|
u
|
n
′
1
-
β
|
x
|
(
1
-
γ
)
n
𝑑
x
=
ω
n
-
1
∫
0
1
e
γ
α
n
,
β
|
u
(
r
)
|
n
′
1
-
β
r
γ
n
-
1
𝑑
r
.
Since C0,rad∞(ℬ) is dense in W0,rad1,n(ℬ,w), we may assume that u∈C0∞(ℬ) and u≥0 (otherwise, replace u by |u|). Since the problem is radially symmetric, we make the change of variable
|
x
|
=
e
1
-
t
γ
n
,
and set
(2.3)
ψ
(
t
)
=
γ
1
-
β
n
′
ω
n
-
1
1
n
n
(
n
-
1
)
(
1
-
β
)
n
(
1
-
β
)
n
-
1
n
u
(
e
1
-
t
γ
n
)
.
By (2.1) and (2.2), we have
(2.4)
∫
ℬ
ω
(
x
)
|
∇
u
|
n
𝑑
x
=
1
(
1
-
β
)
n
-
1
∫
γ
n
+
∞
(
ψ
′
(
t
)
)
n
t
β
(
n
-
1
)
𝑑
t
,
and the exponential integral
n
e
-
γ
n
ω
n
-
1
∫
ℬ
e
γ
α
n
,
β
|
u
|
n
′
1
-
β
|
x
|
(
1
-
γ
)
n
𝑑
x
=
1
γ
∫
γ
n
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
.
It suffices to show that there exists a constant C>0 such that
∫
γ
n
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
≤
C
for all
ψ
with
∥
ψ
∥
n
=
1
(
1
-
β
)
n
-
1
∫
γ
n
+
∞
(
ψ
′
(
t
)
)
n
t
β
(
n
-
1
)
𝑑
t
≤
1
.
Hence, the problem is reduced to the estimate of a one-dimensional calculus inequality whose proof is mainly based on a result due to Leckband (see [19]). To this end, the following definitions are needed (see [19, 25]).
Definition 2.1.
A continuous function ρ:[0,+∞)→[0,+∞) is called C*-function if there is a constant Cρ>0 such that for every 0<d<+∞, there exists a constant D=D(d)>0 with
ρ
(
(
l
+
d
)
s
)
≤
C
ρ
ρ
(
(
l
-
1
)
s
)
for all
l
≥
D
and
0
<
s
<
+
∞
.
Definition 2.2.
A C1-function N:[0,+∞)→[0,+∞) is called C*-convex function if N(0)=0, N is convex, N∈C1([0,+∞)), and the function ρ defined by the differential equation ρ(N(t))=N′(t) is a C*-function.
Now, we state a general functional inequality by Leckband [19], which is a generalization of a previous result by Neugebauer [25], which in turn is a generalization of Moser’s integral inequality [24], see also [23].
Theorem 2.3 (Leckband’s Inequality [19]).
Let f∈Ln([0,+∞)) be such that ∥f∥Ln([0,+∞))=1, let φ:R+→R+, with φ≥0, be locally integrable, and set
G
(
x
)
=
(
φ
n
n
-
1
(
y
)
d
y
)
n
-
1
n
𝑎𝑛𝑑
F
(
x
)
=
∫
0
x
f
(
y
)
φ
(
y
)
𝑑
y
.
Let ϕ≥0 be a nonincreasing function on [0,+∞) and N(t) a C*-convex function. Then there exists a constant C>0 such that
∫
0
+
∞
ϕ
(
N
(
G
(
t
)
)
-
N
(
F
(
t
)
)
)
𝑑
N
(
G
(
t
)
)
≤
C
∥
ϕ
∥
L
1
(
[
0
,
+
∞
)
)
.
Now, we are ready to complete the proof of Theorem 1.5 by applying the Leckband inequality to
f
(
t
)
=
ψ
′
(
t
)
(
t
β
1
-
β
)
n
-
1
n
𝟙
[
γ
n
,
+
∞
)
(
t
)
,
φ
(
t
)
=
(
t
β
1
-
β
)
1
-
n
n
,
where ψ(t)∈C1([0,+∞)). First case. For ∥ψ∥=1(1-β)n-1∫γn+∞(ψ′(t))ntβ(n-1)𝑑t=1, we have
∫
0
+
∞
|
f
(
t
)
|
n
d
t
=
∫
γ
n
+
∞
|
ψ
′
(
t
)
|
n
t
β
(
n
-
1
)
(
1
-
β
)
n
-
1
d
t
=
1
.
Furthermore, with this choice of φ, one has
G
(
x
)
=
(
∫
0
x
φ
n
n
-
1
(
y
)
𝑑
y
)
n
-
1
n
=
(
∫
0
x
1
-
β
t
β
𝑑
t
)
n
-
1
n
=
x
(
1
-
β
)
n
-
1
n
and
F
(
x
)
=
∫
0
x
f
(
y
)
φ
(
y
)
𝑑
y
=
∫
0
x
ψ
′
(
y
)
𝟙
[
γ
n
,
+
∞
)
(
y
)
𝑑
y
=
{
ψ
(
x
)
-
ψ
(
γ
n
)
=
ψ
(
x
)
if
x
≥
γ
n
,
0
if
x
<
γ
n
.
Finally, set
N
(
s
)
=
s
n
(
n
-
1
)
(
1
-
β
)
,
ϕ
(
s
)
=
e
-
s
.
Then
N
(
G
(
t
)
)
=
t
and
N
(
F
(
t
)
)
=
{
ψ
n
′
1
-
β
(
t
)
if
t
≥
γ
n
,
0
if
t
<
γ
n
.
By Leckband’s inequality, we obtain
∫
γ
n
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
≤
∫
0
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
≤
C
∫
0
+
∞
e
-
s
𝑑
s
=
C
<
+
∞
.
Second case. For 0<∥ψ∥<1, taking ψ1=ψ∥ψ∥, according to the first case, we infer
∫
γ
n
+
∞
e
(
ψ
1
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
≤
C
.
Therefore,
∫
γ
n
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
≤
∫
γ
n
+
∞
e
(
ψ
1
(
t
)
)
n
′
1
-
β
-
t
𝑑
t
≤
C
.
Now, we prove that inequality (1.9) is sharp in the sense that if ααn,β+σn=γ+σn>1, then the supremum in (1.9) is infinite. We have
∫
ℬ
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
=
ω
n
-
1
e
n
-
σ
γ
n
∫
γ
n
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
e
(
σ
-
n
)
t
γ
n
𝑑
t
.
It is sufficient to test
∫
γ
n
+
∞
e
(
ψ
(
t
)
)
n
′
1
-
β
e
(
σ
-
n
)
t
γ
n
𝑑
t
on the family of functions
ξ
k
(
t
)
=
{
t
1
-
β
-
(
γ
n
)
1
-
β
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
n
,
γ
n
≤
t
≤
k
+
γ
n
,
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
n
-
1
n
,
t
≥
k
+
γ
n
.
We can easily see that
∫
γ
n
+
∞
|
ξ
k
′
(
t
)
|
n
t
β
(
n
-
1
)
(
1
-
β
)
n
-
1
𝑑
t
=
1
.
Then
sup
∥
u
∥
w
≤
1
∫
ℬ
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
≥
ω
n
-
1
e
n
-
σ
γ
n
∫
γ
n
+
∞
e
|
ξ
k
|
n
′
1
-
β
e
(
σ
-
n
)
t
n
γ
𝑑
t
≥
ω
n
-
1
e
n
-
σ
γ
n
e
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
1
-
β
∫
k
+
γ
n
+
∞
e
(
σ
-
n
)
t
n
γ
𝑑
t
=
ω
n
-
1
e
n
-
σ
n
-
σ
e
(
k
+
γ
n
)
(
[
1
-
γ
n
k
+
γ
n
]
1
1
-
β
+
(
σ
-
n
)
n
γ
)
→
∞
as
k
→
+
∞
.
3 Proof of Theorem 1.6
Since σ<n, we may choose t>1 such that σt<n. By Hölder’s inequality, it follows from (1.5) that
∫
ℬ
e
e
|
u
|
n
′
|
x
|
σ
𝑑
x
≤
(
∫
ℬ
e
t
t
-
1
e
|
u
|
n
′
𝑑
x
)
t
-
1
t
(
∫
ℬ
d
x
|
x
|
σ
t
)
1
t
<
+
∞
.
Now, assume that a+σ<n and choose t>1 such that
a
n
+
σ
t
n
=
1
.
Using again Hölder’s inequality and taking (1.6) into account, we infer
sup
∥
u
∥
w
≤
1
∫
ℬ
e
a
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
|
x
|
σ
d
x
≤
sup
∥
u
∥
w
≤
1
(
∫
ℬ
e
n
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
d
x
)
a
n
(
∫
ℬ
d
x
|
x
|
n
t
)
σ
t
n
<
+
∞
.
We establish (1.11) in the remaining case a+σ=n. To this end, we make the change of variable
|
x
|
=
e
-
t
a
,
and set
(3.1)
ψ
(
t
)
=
a
1
n
′
ω
n
-
1
1
n
u
(
e
-
t
a
)
.
Hence,
(3.2)
∫
ℬ
w
(
x
)
|
∇
u
|
n
d
x
=
∫
0
+
∞
|
ψ
′
(
t
)
|
n
(
1
+
t
a
)
n
-
1
d
t
and
∫
ℬ
e
a
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
|
x
|
n
-
a
𝑑
x
=
ω
n
-
1
a
∫
0
+
∞
e
a
e
a
-
1
|
ψ
(
t
)
|
n
′
e
-
t
𝑑
t
.
We claim that there exists a positive constant C>0 such that
(3.3)
∫
0
+
∞
e
a
e
a
-
1
|
ψ
(
t
)
|
n
′
e
-
t
d
t
≤
C
for all
ψ
with
∫
0
+
∞
|
ψ
′
(
t
)
|
n
(
1
+
t
a
)
n
-
1
d
t
≤
1
.
This result is achieved by applying Leckband’s inequality stated in Theorem 2.3 to
f
(
t
)
=
ψ
′
(
t
)
(
1
+
t
a
)
1
n
′
,
φ
(
t
)
=
(
1
+
t
a
)
1
-
n
n
.
Then
∫
0
+
∞
|
f
(
t
)
|
n
d
t
=
∫
0
+
∞
|
ψ
′
(
t
)
|
n
(
1
+
t
a
)
n
-
1
≤
1
.
Furthermore, with this choice of φ, one has
G
(
x
)
=
(
∫
0
x
φ
n
n
-
1
(
y
)
𝑑
y
)
n
-
1
n
=
(
∫
0
x
d
y
1
+
y
a
)
n
-
1
n
=
a
n
-
1
n
(
log
(
1
+
x
a
)
)
n
-
1
n
and
F
(
x
)
=
∫
0
x
f
(
y
)
φ
(
y
)
𝑑
y
=
∫
0
x
ψ
′
(
y
)
𝑑
y
=
ψ
(
x
)
.
Finally, set
N
(
s
)
=
e
1
a
s
n
n
-
1
-
1
,
ϕ
(
s
)
=
e
-
a
s
.
Then, we get
N
(
G
(
t
)
)
=
t
a
and
N
(
F
(
t
)
)
=
e
1
a
(
ψ
(
t
)
)
n
′
-
1
.
With these particular choices, Leckband’s inequality implies that (3.3) holds true. In the next step of the proof of Theorem 1.6, we show that if a+σ>n, then
sup
∥
u
∥
w
≤
1
∫
ℬ
e
a
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
|
x
|
σ
𝑑
x
=
+
∞
.
We have
∫
ℬ
e
a
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
|
x
|
σ
𝑑
x
=
ω
n
-
1
a
∫
0
+
∞
e
a
e
a
-
1
|
ψ
(
t
)
|
n
′
e
(
σ
-
n
)
t
a
𝑑
t
.
It is sufficient to test
∫
0
+
∞
e
a
e
a
-
1
|
ψ
(
t
)
|
n
′
e
(
σ
-
n
)
t
a
𝑑
t
on the family of functions
ψ
k
(
t
)
=
{
a
1
n
′
log
(
1
+
t
a
)
log
1
n
(
1
+
k
)
,
0
≤
t
≤
a
k
,
a
1
n
′
log
n
-
1
n
(
1
+
k
)
,
t
≥
a
k
.
We have
∫
0
+
∞
|
ψ
k
′
(
t
)
|
n
(
1
+
t
a
)
n
-
1
d
t
=
1
and
sup
∥
ψ
∥
≤
1
∫
0
+
∞
e
a
e
|
ψ
|
n
′
e
(
σ
-
n
)
t
a
𝑑
t
≥
∫
a
k
+
∞
e
a
e
a
-
1
|
ψ
k
(
t
)
|
n
′
e
(
σ
-
n
)
t
a
𝑑
t
≥
e
a
+
a
k
∫
a
k
+
∞
e
(
σ
-
n
)
t
a
𝑑
t
≥
a
e
a
n
-
σ
e
(
a
+
(
σ
-
n
)
)
k
→
+
∞
as
k
→
+
∞
,
where
∥
ψ
∥
=
∫
0
+
∞
|
ψ
′
(
t
)
|
n
(
1
+
t
a
)
n
-
1
d
t
.
This ends the proof of Theorem 1.6.
4 Proof of Theorem 1.8
In order to prove Theorem 1.8, we need the following radial lemma, for which, for the convenience of the reader, we provide a detailed proof.
Lemma 4.1.
Let u∈E. Then there exists a constant c such that
|
u
(
x
)
|
≤
c
|
u
|
L
n
(
ℝ
n
)
n
-
1
|
∇
u
|
L
n
(
ℝ
n
,
w
)
|
x
|
(
1
-
n
)
n
,
x
≠
0
.
Proof.
By the definition of the functional space E, it suffices to show the inequality above for u∈C0,rad∞(ℝn). For such a function, we denote φ(s)=u(x), |x|=s. For r≥1, one has
|
φ
(
r
)
|
n
≤
n
∫
r
+
∞
|
(
φ
′
(
s
)
)
(
φ
(
s
)
)
n
-
1
s
n
-
1
n
s
1
-
n
n
|
d
s
≤
n
(
∫
r
+
∞
|
φ
′
(
s
)
|
n
s
n
-
1
d
s
)
1
n
(
∫
r
+
∞
|
φ
(
s
)
|
n
s
n
s
n
-
1
d
s
)
n
-
1
n
=
n
ω
n
-
1
r
n
-
1
(
w
n
-
1
∫
r
+
∞
|
φ
′
(
s
)
|
n
s
n
-
1
d
s
)
1
n
(
ω
n
-
1
∫
r
+
∞
|
φ
(
s
)
|
n
s
n
-
1
d
s
)
n
-
1
n
≤
n
ω
n
-
1
r
n
-
1
(
1
inf
s
≥
1
χ
(
s
)
)
1
n
(
∫
|
x
|
≥
r
w
(
x
)
|
∇
u
|
n
d
x
)
1
n
(
∫
|
x
|
≥
r
|
u
|
n
d
x
)
n
-
1
n
.
The proof is complete. ∎
For u∈E, we have
(4.1)
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
=
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
+
∫
|
x
|
≥
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
.
On the one hand,
∫
|
x
|
≥
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
d
x
≤
∑
k
=
n
-
1
+
∞
α
k
k
!
∫
|
x
|
≥
1
|
u
|
k
n
′
1
-
β
d
x
=
α
n
-
1
(
n
-
1
)
!
∫
|
x
|
≥
1
|
u
|
n
1
-
β
d
x
+
∑
k
=
n
+
∞
α
k
k
!
∫
|
x
|
≥
1
|
u
|
k
n
′
1
-
β
d
x
.
For k≥n, by Lemma 4.1 we have
∫
|
x
|
≥
1
|
u
|
k
n
′
1
-
β
𝑑
x
≤
(
c
∥
u
∥
)
k
n
′
1
-
β
∫
|
x
|
≥
1
d
x
|
x
|
k
1
-
β
=
ω
n
-
1
(
c
∥
u
∥
)
k
n
′
1
-
β
∫
1
+
∞
r
n
-
1
-
k
1
-
β
𝑑
r
(4.2)
=
ω
n
-
1
(
c
∥
u
∥
)
k
n
′
1
-
β
1
k
1
-
β
-
n
≤
(
c
∥
u
∥
)
k
n
′
1
-
β
ω
n
-
1
(
1
-
β
)
n
β
.
For k=n-1, it suffices to remind that E↪W1,n(ℝn)↪Ln1-β(ℝn), with the embeddings being continuous, and as a consequence, there exists a positive constant C(β)>0 such that
∫
|
x
|
≥
1
|
u
|
n
1
-
β
d
x
≤
C
(
β
)
∥
u
∥
n
1
-
β
.
Using that last inequality with (4.2), we deduce
(4.3)
∫
|
x
|
≥
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
≤
α
n
-
1
(
n
-
1
)
!
C
(
β
)
∥
u
∥
n
1
-
β
+
ω
n
-
1
(
1
-
β
)
n
β
e
α
(
c
∥
u
∥
)
n
′
1
-
β
<
+
∞
.
Now, in order to estimate the first integral in (4.1), set
v
(
x
)
=
{
u
(
x
)
-
u
(
e
1
)
if
|
x
|
<
1
,
0
if
|
x
|
>
1
,
where e1=(1,0,…,0) is the first vector in the canonical basis of ℝn. Notice that v∈W0,rad1,n(ℬ,w). An elementary calculus gives the following inequality: for all ϵ>0, we have
(4.4)
(
a
+
b
)
n
′
1
-
β
≤
(
1
+
ϵ
)
a
n
′
1
-
β
+
1
+
ϵ
(
(
1
+
ϵ
)
1
-
β
n
′
-
1
+
β
-
1
)
n
′
-
1
+
β
1
-
β
b
n
′
1
-
β
.
Discarding the nonnegative term Sn-2(α,|u|n′1-β), by (4.4), Hölder’s and Young’s inequalities, we have
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
≤
(
∫
|
x
|
<
1
d
x
|
x
|
σ
p
′
)
1
p
′
(
∫
|
x
|
<
1
e
α
p
|
u
|
n
′
1
-
β
𝑑
x
)
1
p
≤
C
(
∫
|
x
|
<
1
e
α
p
|
u
|
n
′
1
-
β
𝑑
x
)
1
p
(4.5)
≤
C
exp
(
1
+
ϵ
(
(
1
+
ϵ
)
1
-
β
n
′
-
1
+
β
-
1
)
n
′
-
1
+
β
1
-
β
|
u
(
e
1
)
|
n
′
1
-
β
)
(
∫
|
x
|
<
1
e
α
p
(
1
+
ϵ
)
|
v
|
n
′
1
-
β
𝑑
x
)
1
p
,
where p>1 is chosen such that σp′<n. Thanks to Theorem 1.3, we obtain
(4.6)
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
.
Combining (4.3) with (4.6), we deduce from (4.1) that
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
.
This ends the proof of (1.13). Now, in order to prove (1.14), which consists of the Trudinger–Moser inequality itself, observe that (4.3) leads to
sup
u
∈
E
,
∥
u
∥
≤
1
∫
|
x
|
≥
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
≤
sup
u
∈
E
,
∥
u
∥
≤
1
[
α
n
-
1
(
n
-
1
)
!
C
(
β
)
∥
u
∥
n
1
-
β
+
ω
n
-
1
(
1
-
β
)
n
β
exp
(
α
(
c
∥
u
∥
)
n
′
1
-
β
)
]
(4.7)
≤
α
n
-
1
(
n
-
1
)
!
C
(
β
)
+
ω
n
-
1
(
1
-
β
)
n
β
exp
(
α
c
n
′
1
-
β
)
.
On the other hand, inequality (4.5) gives
(4.8)
sup
u
∈
E
,
∥
u
∥
≤
1
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
d
x
≤
C
sup
u
∈
E
,
∥
u
∥
≤
1
(
∫
|
x
|
<
1
e
α
p
(
1
+
ϵ
)
|
v
|
n
′
1
-
β
d
x
)
1
p
.
Let ααn,β+σn<1. Clearly, there exists ϵ>0 such that αp(1+ϵ)αn,β+σn<1. Having in mind that
∥
v
∥
w
n
=
∫
ℬ
w
(
x
)
|
∇
u
|
n
𝑑
x
≤
∥
u
∥
n
≤
1
,
from (1.9) and (4.8), it follows that
(4.9)
sup
u
∈
E
,
∥
u
∥
≤
1
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
.
Combining (4.7) with (4.9), we obtain
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
for
α
α
n
,
β
+
σ
n
<
1
.
In the next step of the proof of Theorem 1.8, we show that if ααn,β+σn=γ+σn>1, then
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
=
+
∞
.
As in the proof of Theorem 1.5, we make use of the change of variables
|
x
|
=
e
1
-
t
γ
n
,
and let ψ be given by (2.3). From (2.4), a direct computation gives
(4.10)
∫
|
x
|
≥
1
|
∇
u
|
n
χ
(
|
x
|
)
d
x
=
γ
β
(
n
-
1
)
n
β
(
n
-
1
)
(
1
-
β
)
n
-
1
∫
-
∞
γ
n
|
ψ
′
(
t
)
|
n
χ
(
e
1
-
t
γ
n
)
d
t
,
(4.11)
∫
ℝ
n
|
u
|
n
d
x
=
γ
(
β
-
1
)
(
n
-
1
)
-
1
e
n
(
1
-
β
)
n
-
1
∫
ℝ
(
ψ
(
t
)
)
n
e
-
t
γ
d
t
and
(4.12)
∫
ℝ
n
e
γ
α
n
,
β
|
u
|
n
′
1
-
β
-
S
n
-
2
(
γ
α
n
,
β
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
=
ω
n
-
1
e
n
-
σ
γ
n
∫
ℝ
[
e
(
ψ
(
t
)
)
n
′
1
-
β
-
S
n
-
2
(
1
,
(
ψ
(
t
)
)
n
′
1
-
β
)
]
e
(
σ
-
n
γ
n
)
t
𝑑
t
.
Now, consider the following family of functions:
ξ
k
(
t
)
=
{
θ
(
t
1
-
β
-
(
γ
n
)
1
-
β
)
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
n
,
γ
n
≤
t
≤
k
+
γ
n
,
θ
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
n
-
1
n
,
k
+
γ
n
≤
t
<
+
∞
,
0
,
t
≤
γ
n
,
where 0<θ<1 is a positive real number to be determined later. We have
∫
ℝ
(
exp
(
|
ξ
k
(
t
)
|
n
′
1
-
β
)
-
S
n
-
2
(
1
,
|
ξ
k
(
t
)
|
n
′
1
-
β
)
)
e
(
σ
-
n
γ
n
)
t
𝑑
t
≥
∫
k
+
γ
n
+
∞
(
exp
(
θ
n
′
1
-
β
(
k
+
γ
n
)
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
1
-
β
)
-
S
n
-
2
(
1
,
θ
n
′
1
-
β
(
k
+
γ
n
)
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
1
-
β
)
)
e
(
σ
-
n
γ
n
)
t
d
t
≥
S
n
-
2
(
1
,
θ
n
′
1
-
β
(
k
+
γ
n
)
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
1
-
β
)
γ
n
σ
-
n
e
(
σ
-
n
γ
n
)
(
k
+
γ
n
)
+
∫
k
+
γ
n
+
∞
exp
(
θ
n
′
1
-
β
(
k
+
γ
n
)
[
1
-
γ
n
k
+
γ
n
]
1
1
-
β
)
e
(
σ
-
n
γ
n
)
t
𝑑
t
=
S
n
-
2
(
1
,
θ
n
′
1
-
β
(
k
+
γ
n
)
(
(
k
+
γ
n
)
1
-
β
-
(
γ
n
)
1
-
β
)
1
1
-
β
)
γ
n
n
-
σ
e
(
σ
-
n
γ
n
)
(
k
+
γ
n
)
(4.13)
+
γ
n
σ
-
n
exp
(
θ
n
′
1
-
β
(
k
+
γ
n
)
[
(
1
-
γ
n
k
+
γ
n
)
1
1
-
β
+
σ
-
n
γ
n
]
)
→
+
∞
if
θ
n
′
1
-
β
>
n
-
σ
γ
n
.
If γ+σn>1, then one can easily find 0<θ<1 such that
n
-
σ
n
γ
<
θ
n
′
1
-
β
<
1
.
Now, for k≥1, consider the function uk∈E such that
ξ
k
(
t
)
=
γ
1
-
β
n
′
ω
n
-
1
1
n
n
(
n
-
1
)
(
1
-
β
)
n
(
1
-
β
)
n
-
1
n
u
k
(
x
)
.
By (2.4), (4.10) and (4.11), we have
∥
u
k
∥
n
=
∫
|
x
|
<
1
|
∇
u
k
|
n
w
(
x
)
d
x
+
∫
|
x
|
<
1
|
u
k
|
n
d
x
=
θ
n
+
o
k
(
1
)
.
Consequently, there exists k0 sufficiently large such that ∥uk∥≤1 for all k≥k0. Finally, in view of (4.13) and (4.12), we deduce that if γ+σn>1, then
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
exp
(
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
≥
∫
ℝ
n
exp
(
α
|
u
k
|
n
′
1
-
β
)
-
S
n
-
2
(
α
,
|
u
k
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
→
+
∞
as
k
→
+
∞
.
This ends the proof of Theorem 1.8.
5 Proof of Theorem 1.9
We start by proving a radial lemma for the space Eβ, similar to the one established in Lemma 4.1.
Lemma 5.1.
There exists a positive constant c>0 such that for all u∈Eβ, we have
|
u
(
x
)
|
≤
c
|
x
|
n
′
-
1
+
β
n
′
|
∇
u
|
L
n
(
ℝ
n
,
w
)
n
′
-
1
+
β
n
′
|
u
|
L
d
β
(
ℝ
n
)
1
-
β
n
′
for all
|
x
|
≥
1
.
Proof.
By the definition of Eβ, it suffices to show the inequality for u∈C0,rad∞(ℝn). As in the proof of Lemma 4.1, denote φ(s)=u(x), |x|=s. For r≥1, we have
|
φ
(
r
)
|
n
′
n
′
-
1
+
β
≤
(
n
′
n
′
-
1
+
β
)
∫
r
+
∞
|
φ
′
(
s
)
|
⋅
|
φ
(
s
)
|
1
-
β
n
′
-
1
+
β
s
n
-
1
n
s
1
-
n
n
d
s
≤
(
n
′
n
′
-
1
+
β
)
(
∫
r
+
∞
|
φ
′
(
s
)
|
n
s
n
-
1
d
s
)
1
n
(
∫
r
+
∞
|
φ
(
s
)
|
d
β
s
-
1
d
s
)
1
n
′
≤
(
n
′
(
n
′
-
1
+
β
)
r
n
-
1
)
(
∫
r
+
∞
|
φ
′
(
s
)
|
n
s
n
-
1
d
s
)
1
n
(
∫
r
+
∞
|
φ
(
s
)
|
d
β
s
-
1
d
s
)
1
n
′
≤
(
n
′
(
n
′
-
1
+
β
)
ω
n
-
1
r
n
-
1
)
(
1
inf
s
≥
1
χ
(
s
)
)
1
n
(
∫
|
x
|
≥
1
|
∇
u
|
n
w
(
x
)
d
x
)
1
n
(
∫
|
x
|
≥
1
|
u
|
d
β
d
x
)
1
n
′
.
The proof is complete. ∎
In order to prove that
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
for all
u
∈
E
β
,
it suffices to argue exactly as in (4.2) and (4.6). In fact, for k≥n, by Lemma 5.1, we have
∫
|
x
|
≥
1
|
u
|
k
n
′
1
-
β
d
x
≤
ω
n
-
1
(
c
∥
u
∥
)
k
n
′
1
-
β
∫
1
+
∞
r
n
-
1
r
k
(
n
′
-
1
+
β
1
-
β
)
d
r
≤
ω
n
-
1
(
c
∥
u
∥
)
k
n
′
1
-
β
1
k
(
n
′
-
1
+
β
1
-
β
)
-
n
≤
ω
n
-
1
(
c
∥
u
∥
)
k
n
′
1
-
β
1
-
β
n
(
n
′
-
1
+
β
)
-
n
(
1
-
β
)
.
Now, observe that, since n1-β>dβ, again by Lemma 5.1, we infer
∫
|
x
|
≥
1
|
u
|
n
1
-
β
d
x
=
∫
|
x
|
≥
1
|
u
|
d
β
|
u
|
n
1
-
β
-
d
β
d
x
≤
(
c
∥
u
∥
)
n
1
-
β
-
d
β
∫
ℝ
n
|
u
|
d
β
d
x
.
Proceeding as in (4.7), we can easily see that
sup
u
∈
E
β
,
∥
u
∥
β
≤
1
∫
|
x
|
≥
1
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
<
+
∞
for all
α
>
0
.
Next, we claim that if β≤1n′+1, then
(5.1)
sup
u
∈
E
β
,
∥
u
∥
β
≤
1
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
<
+
∞
for all
α
≤
α
n
,
β
(
1
-
σ
n
)
.
Let u∈Eβ∖{0} be such that ∥u∥β≤1 and choose ϵ>0 such that
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
=
(
∫
|
x
|
<
1
|
∇
u
|
n
w
(
x
)
d
x
)
-
1
.
Using inequality (4.4), it follows that
(5.2)
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
≤
exp
(
α
(
1
+
ϵ
)
|
u
(
e
1
)
|
n
′
1
-
β
(
(
1
+
ϵ
)
1
-
β
n
′
-
1
+
β
-
1
)
n
′
-
1
+
β
1
-
β
)
∫
|
x
|
<
1
e
α
(
1
+
ϵ
)
|
v
|
n
′
1
-
β
|
x
|
σ
𝑑
x
,
where v is given by
v
(
x
)
=
{
u
(
x
)
-
u
(
e
1
)
if
|
x
|
<
1
,
0
if
|
x
|
≥
1
.
We have
∫
|
x
|
<
1
|
∇
(
(
1
+
ϵ
)
1
-
β
n
′
v
)
|
n
w
(
x
)
d
x
=
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
∫
|
x
|
<
1
|
∇
u
|
n
w
(
x
)
d
x
=
1
.
By Theorem 1.5, for α≤αn,β(1-σn), there exists a constant Cβ>0 such that
∫
|
x
|
<
1
e
α
(
1
+
ϵ
)
|
v
|
n
′
1
-
β
|
x
|
σ
𝑑
x
≤
C
β
.
On the other hand, by the proof of Lemma 5.1 and using Young’s inequality, it follows that
|
u
(
e
1
)
|
n
′
n
′
-
1
+
β
≤
c
1
(
∫
|
x
|
≥
1
|
∇
u
|
n
w
(
x
)
d
x
+
∫
|
x
|
≥
1
|
u
|
d
β
d
x
)
≤
c
1
(
∫
ℝ
n
|
∇
u
|
n
w
(
x
)
d
x
+
∫
ℝ
n
|
u
|
d
β
d
x
-
∫
|
x
|
<
1
|
∇
u
|
n
w
(
x
)
d
x
)
=
c
1
(
|
∇
u
|
L
n
(
ℝ
n
,
w
)
n
+
|
u
|
L
d
β
(
ℝ
n
)
d
β
-
1
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
)
≤
c
1
(
|
∇
u
|
L
n
(
ℝ
n
,
w
)
+
|
u
|
L
d
β
(
ℝ
n
)
-
1
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
)
≤
c
1
(
1
-
1
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
)
,
where we used the fact that ∥u∥β≤1 and dβ≥1 (because β≤1n′+1). Hence,
|
u
(
e
1
)
|
n
′
1
-
β
≤
c
2
(
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
-
1
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
)
n
′
-
1
+
β
1
-
β
.
Putting the last inequality in (5.2), we deduce that
(5.3)
∫
|
x
|
<
1
e
α
|
u
|
n
′
1
-
β
|
x
|
σ
𝑑
x
≤
C
β
exp
(
c
2
(
1
+
ϵ
)
(
(
1
+
ϵ
)
(
1
-
β
)
(
n
-
1
)
-
1
)
n
′
-
1
+
β
1
-
β
(
(
1
+
ϵ
)
1
-
β
n
′
-
1
+
β
-
1
)
n
′
-
1
+
β
1
-
β
(
1
+
ϵ
)
(
n
′
-
1
+
β
)
(
n
-
1
)
)
.
Observe that the function defined on ]1,+∞[ by
x
↦
(
x
(
1
-
β
)
(
n
-
1
)
-
1
)
n
′
-
1
+
β
1
-
β
(
x
1
-
β
n
′
-
1
+
β
-
1
)
n
′
-
1
+
β
1
-
β
x
(
n
′
-
1
+
β
)
(
n
-
1
)
-
1
is bounded. In view of (5.3), we can easily conclude that (5.1) follows. Finally, proceeding exactly as in the last part of the proof of Theorem 1.8, we can prove that if ααn,β+σn>1, then
sup
u
∈
E
β
,
∥
u
∥
β
≤
1
∫
ℝ
n
e
α
|
u
|
n
′
1
-
β
-
S
n
-
2
(
α
,
|
u
|
n
′
1
-
β
)
|
x
|
σ
𝑑
x
=
+
∞
.
This completes the proof of Theorem 1.9.
6 Proof of Theorem 1.10
We start by proving the first conclusion of Theorem 1.10. For u∈E, we have
∫
ℝ
n
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
(6.1)
=
∫
|
x
|
<
1
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
+
∫
|
x
|
≥
1
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
.
The second integral in (6.1) can be estimated as follows:
∫
|
x
|
≥
1
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
∑
k
=
n
-
1
+
∞
α
k
k
!
∫
|
x
|
≥
1
(
e
|
u
|
n
′
-
1
)
k
𝑑
x
.
Having in mind that the function s↦es-1s is increasing on [0,+∞[ and using Lemma 4.1, we get
e
|
u
(
x
)
|
n
′
-
1
≤
e
c
n
′
∥
u
∥
n
′
-
1
c
n
′
∥
u
∥
n
′
|
u
(
x
)
|
n
′
for all
x
∈
ℝ
n
,
|
x
|
≥
1
.
Thus,
(6.2)
∫
|
x
|
≥
1
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
d
x
≤
∑
k
=
n
-
1
+
∞
α
k
k
!
(
e
c
n
′
∥
u
∥
n
′
-
1
c
n
′
∥
u
∥
n
′
)
k
∫
|
x
|
≥
1
|
u
|
n
′
k
d
x
.
By Lemma 4.1, for k≥n+1, we infer
∫
|
x
|
≥
1
|
u
|
n
′
k
d
x
=
ω
n
-
1
∫
1
+
∞
|
u
(
r
)
|
n
′
k
r
n
-
1
d
r
≤
ω
n
-
1
(
c
∥
u
∥
)
n
′
k
∫
|
x
|
≥
1
r
n
-
k
-
1
d
r
≤
(
c
∥
u
∥
)
n
′
k
ω
n
-
1
n
-
k
≤
ω
n
-
1
(
c
∥
u
∥
)
n
′
k
.
For k=n-1 (resp. k=n), by the continuous embedding E↪Ln(ℝn) (resp. E↪Ln2n-1(ℝn)), there exists a positive constant C1>0 such that
(6.3)
∫
|
x
|
≥
1
|
u
|
n
(
x
)
d
x
≤
C
1
∥
u
∥
n
(resp. C2>0 such that
(6.4)
∫
|
x
|
≥
1
|
u
|
n
2
n
-
1
d
x
≤
C
2
∥
u
∥
n
2
n
-
1
)
.
Combining (6.2), (6.3) and (6.4), we obtain
∫
|
x
|
≥
1
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
ω
n
-
1
∑
k
=
n
+
1
+
∞
α
k
k
!
(
c
n
′
∥
u
∥
n
′
e
c
n
′
∥
u
∥
n
′
-
1
c
n
′
∥
u
∥
n
′
)
k
+
C
1
α
n
-
1
∥
u
∥
n
+
C
2
α
n
∥
u
∥
n
2
n
-
1
(6.5)
≤
ω
n
-
1
e
α
(
e
c
n
′
∥
u
∥
n
′
-
1
)
+
C
1
α
n
-
1
∥
u
∥
n
+
C
2
α
n
∥
u
∥
n
2
n
-
1
<
+
∞
.
Now, in order to estimate the first integral in (6.1), recall the elementary inequality
(6.6)
(
a
+
b
)
n
′
≤
(
1
+
ϵ
)
a
n
′
+
1
+
ϵ
(
(
1
+
ϵ
)
1
n
′
-
1
-
1
)
n
′
-
1
b
n
′
for all
a
,
b
≥
0
and all
ϵ
>
0
.
Set
(6.7)
v
(
x
)
=
{
u
(
x
)
-
u
(
e
1
)
if
|
x
|
<
1
,
0
if
|
x
|
≥
1
.
We have v∈W0,rad1,n(ℬ,w) and
∫
ℬ
|
∇
v
|
n
w
(
x
)
d
x
=
∫
ℬ
|
∇
u
|
n
w
(
x
)
d
x
≤
∥
u
∥
<
+
∞
.
By (6.6), we have
|
u
|
n
′
≤
(
1
+
ϵ
)
|
v
|
n
′
+
1
+
ϵ
(
(
1
+
ϵ
)
1
n
′
-
1
-
1
)
n
′
-
1
|
u
(
e
1
)
|
n
′
.
Thus,
∫
|
x
|
<
1
e
α
e
|
u
|
n
′
|
x
|
σ
𝑑
x
≤
∫
|
x
|
<
1
exp
(
α
exp
(
1
+
ϵ
(
(
1
+
ϵ
)
1
n
′
-
1
-
1
)
n
′
-
1
|
u
(
e
1
)
|
n
′
)
)
exp
(
α
e
(
1
+
ϵ
)
|
v
|
n
′
)
|
x
|
σ
𝑑
x
.
From (1.10), it follows that
(6.8)
∫
|
x
|
<
1
e
α
(
e
|
u
|
n
′
-
1
)
-
S
n
-
2
(
α
,
e
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
<
+
∞
.
Combining (6.5) and (6.8), we deduce that (1.15) holds true. Next, in view of inequality (6.5), one can easily see that
∫
|
x
|
≥
1
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
ω
n
-
1
exp
(
a
(
e
c
n
′
ω
n
-
1
1
n
-
1
-
1
)
)
+
C
3
.
Thus,
(6.9)
sup
u
∈
E
,
∥
u
∥
≤
1
∫
|
x
|
≥
1
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
ω
n
-
1
exp
(
a
(
e
c
n
′
ω
n
-
1
1
n
-
1
-
1
)
)
+
C
3
.
On the other hand, let u∈E be such that ∥u∥≤1. For v defined by (6.7), from (6.6), we have
(6.10)
∫
|
x
|
<
1
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
∫
|
x
|
<
1
e
a
e
ω
n
-
1
1
n
-
1
1
+
ϵ
(
(
1
+
ϵ
)
1
n
′
-
1
-
1
)
n
′
-
1
|
u
(
e
1
)
|
n
′
e
ω
n
-
1
1
n
-
1
(
1
+
ϵ
)
|
v
|
n
′
|
x
|
σ
𝑑
x
.
Choose ϵ>0 such that
(
1
+
ϵ
)
1
-
n
=
∫
|
x
|
<
1
|
∇
u
|
n
w
(
x
)
d
x
.
By the proof of Lemma 4.1, one can easily note that
|
u
(
e
1
)
|
n
≤
n
ω
n
-
1
(
1
inf
s
≥
1
χ
(
s
)
)
1
n
(
∫
|
x
|
≥
1
|
∇
u
|
n
w
(
x
)
d
x
)
1
n
(
∫
|
x
|
≥
1
|
u
|
n
d
x
)
n
-
1
n
≤
1
ω
n
-
1
(
1
inf
s
≥
1
χ
(
s
)
)
1
n
(
∫
|
x
|
≥
1
|
∇
u
|
n
w
(
x
)
d
x
+
∫
|
x
|
≥
1
|
u
|
n
d
x
)
≤
1
ω
n
-
1
(
1
inf
s
≥
1
χ
(
s
)
)
1
n
(
1
-
∫
|
x
|
<
1
|
∇
u
|
n
w
(
x
)
d
x
)
.
Hence,
ω
n
-
1
1
n
-
1
1
+
ϵ
(
(
1
+
ϵ
)
1
n
′
-
1
-
1
)
n
′
-
1
|
u
(
e
1
)
|
n
′
≤
(
inf
s
≥
1
χ
(
s
)
)
-
1
n
(
n
-
1
)
.
Putting that last inequality in (6.10), we obtain
(6.11)
∫
|
x
|
<
1
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
∫
|
x
|
<
1
e
a
e
(
inf
s
≥
1
χ
(
s
)
)
-
1
n
(
n
-
1
)
e
ω
n
-
1
1
n
-
1
(
1
+
ϵ
)
|
v
|
n
′
|
x
|
σ
𝑑
x
.
Set ξ=(1+ϵ)n-1nv. Since ξ∈W0,rad1,n(ℬ,w) and ∥ξ∥w≤1, for a≤(n-σ)e(infs≥1χ(s))1n(n-1), it follows from (6.11) that
sup
u
∈
E
,
∥
u
∥
≤
1
∫
|
x
|
<
1
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
(6.12)
≤
sup
u
∈
W
0
,
rad
1
,
n
(
ℬ
,
w
)
,
∥
u
∥
w
≤
1
∫
|
x
|
<
1
e
(
n
-
σ
)
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
|
x
|
σ
𝑑
x
<
+
∞
.
Combining (6.9) with (6.12), we conclude that
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
<
+
∞
.
Now, we prove that (1.16) holds true. For u∈E, let ψ be defined by (3.1) with |x|=e-ta. Then
∫
|
x
|
≥
1
|
∇
u
|
n
χ
(
|
x
|
)
d
x
=
∫
-
∞
0
|
ψ
′
(
t
)
|
n
χ
(
e
-
t
a
)
d
t
and
∫
ℝ
n
|
u
|
n
d
x
=
1
a
n
∫
ℝ
|
ψ
(
t
)
|
n
e
-
n
t
a
d
t
.
Hence, having (3.2) in mind,
∥
u
∥
n
=
∫
ℝ
n
|
∇
u
|
n
w
(
x
)
d
x
+
∫
ℝ
n
|
u
|
n
d
x
=
∫
0
+
∞
|
ψ
′
(
t
)
|
n
(
1
+
t
a
)
n
-
1
d
t
+
∫
-
∞
0
|
ψ
′
(
t
)
|
n
χ
(
e
-
t
a
)
d
t
+
1
a
n
∫
ℝ
|
ψ
(
t
)
|
n
e
-
n
t
a
d
t
.
In a similar way, we have
∫
ℝ
n
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
=
ω
n
-
1
a
∫
-
∞
+
∞
[
e
a
(
e
a
-
1
|
ψ
(
t
)
|
n
′
-
1
)
-
S
n
-
2
(
1
,
e
a
-
1
|
ψ
(
t
)
|
n
′
-
1
)
]
e
(
σ
-
n
a
)
t
𝑑
t
.
For k>0, set
ψ
k
(
t
)
=
{
a
1
n
′
log
(
1
+
t
a
)
log
1
n
(
1
+
k
)
,
0
≤
t
≤
a
k
,
a
1
n
′
log
n
-
1
n
(
1
+
k
)
,
t
≥
a
k
,
0
,
t
≤
0
.
We have
∫
0
+
∞
(
ψ
k
′
(
t
)
)
n
(
1
+
t
a
)
n
-
1
𝑑
t
=
1
and
1
a
n
∫
ℝ
(
ψ
k
(
t
)
)
n
e
-
n
t
a
𝑑
t
=
1
n
log
n
-
1
(
1
+
k
)
e
-
n
k
+
1
a
log
(
1
+
k
)
∫
0
a
k
log
n
(
1
+
t
a
)
e
-
n
t
a
𝑑
t
=
1
n
log
n
-
1
(
1
+
k
)
e
-
n
k
+
1
log
(
1
+
k
)
∫
0
k
log
n
(
1
+
t
)
e
-
n
t
𝑑
t
.
Thus,
∥
ψ
k
∥
n
=
1
+
ϵ
k
,
where
ϵ
k
=
1
n
log
n
-
1
(
1
+
k
)
e
-
n
k
+
1
log
(
1
+
k
)
∫
0
k
log
n
(
1
+
t
)
e
-
n
t
𝑑
t
.
It is easy to see that
(6.13)
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
k
→
exp
(
-
1
n
-
1
∫
0
+
∞
log
n
(
1
+
t
)
e
-
n
t
𝑑
t
)
as
k
→
+
∞
.
We have
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≥
∫
a
k
+
∞
[
e
a
e
a
-
1
(
ψ
k
(
t
)
)
n
′
∥
ψ
k
∥
n
′
-
1
-
S
n
-
2
(
a
,
e
a
-
1
(
ψ
k
(
t
)
)
n
′
∥
ψ
k
∥
n
′
-
1
)
]
e
(
σ
-
n
a
)
t
𝑑
t
≥
a
σ
-
n
e
(
σ
-
n
)
k
∑
m
=
0
n
-
2
a
m
m
!
(
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
-
1
)
m
+
a
n
-
σ
exp
(
a
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
)
e
(
σ
-
n
)
k
(6.14)
≥
a
σ
-
n
e
(
σ
-
n
)
k
∑
m
=
0
n
-
2
a
m
m
!
(
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
-
1
)
m
+
a
n
-
σ
exp
(
k
(
(
σ
-
n
)
+
a
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
k
)
)
.
By (6.13), it follows that
(
σ
-
n
)
+
a
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
k
→
(
σ
-
n
)
+
a
exp
(
-
1
n
-
1
∫
0
+
∞
log
n
(
1
+
t
)
e
-
n
t
𝑑
t
)
as
k
→
+
∞
.
Hence, if
a
>
(
n
-
σ
)
exp
(
1
n
-
1
∫
0
+
∞
log
n
(
1
+
t
)
e
-
n
t
𝑑
t
)
,
then
exp
(
k
(
(
σ
-
n
)
+
a
exp
(
log
(
1
+
k
)
(
1
+
ϵ
k
)
1
n
-
1
)
k
)
)
→
+
∞
as
k
→
+
∞
.
Finally, from inequality (6.14), we deduce that
sup
u
∈
E
,
∥
u
∥
≤
1
∫
ℝ
n
e
a
(
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
-
S
n
-
2
(
a
,
e
ω
n
-
1
1
n
-
1
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
=
+
∞
.
This completes the proof of Theorem 1.10.
Corollary 6.1.
For all t>0, if u∈E is such that ∥u∥≤(ωn-11n-1t)1n′, then there exists a positive constant C=C(t)>0 such that
∫
ℝ
n
e
(
n
-
σ
)
e
(
inf
s
≥
1
χ
(
s
)
)
1
n
(
n
-
1
)
(
e
t
|
u
|
n
′
-
1
)
-
S
n
-
2
(
(
n
-
σ
)
e
(
inf
s
≥
1
χ
(
s
)
)
1
n
(
n
-
1
)
,
e
t
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
C
.
7 Application to a Singular Elliptic Equation
In this section, we consider the following singular elliptic equation:
(7.1)
-
div
(
w
(
x
)
|
∇
u
|
n
-
2
∇
u
)
+
|
u
|
n
-
2
u
=
f
(
x
,
u
)
|
x
|
σ
+
ϵ
h
in
ℝ
n
,
h
∈
E
*
∖
{
0
}
,
where w is defined by (1.12), with β=1, and f:ℝn×ℝ→ℝ is a Carathéodory function. Moreover, we assume the following:
There exist constants b1, b2, α>n, β>0 and 0<γ<(n-σ)e(infs≥1χ(s))1n(n-1) such that
|
f
(
x
,
s
)
|
≤
b
1
|
s
|
n
-
1
+
b
2
|
s
|
α
-
1
(
e
γ
(
e
β
|
s
|
n
′
-
1
)
-
S
n
-
2
(
γ
,
e
β
|
s
|
n
′
-
1
)
)
for all
s
∈
ℝ
and a.e.
x
∈
ℝ
n
.
There exists θ>n such that
θ
F
(
x
,
s
)
=
θ
∫
0
s
f
(
x
,
t
)
𝑑
t
≤
s
f
(
x
,
s
)
for all
s
∈
ℝ
and a.e.
x
∈
ℝ
n
.
There exist A>0 and r>n such that
F
(
x
,
s
)
≥
A
s
r
for all
s
≥
0
and a.e.
x
∈
ℝ
n
.
Concerning the functional space E, observe that since infx∈ℝnw(x)>0, the following embeddings are continuous:
E
↪
W
1
,
n
(
ℝ
n
)
↪
L
t
(
ℝ
n
)
for all
t
≥
n
.
Moreover, for t>n, the last embeddings are compact. In fact, we can apply Strauss’ theorem [30] to obtain that Wrad1,n(ℝn) is compactly embedded into the Lebesgue space Lq(ℝn), for all q>n. On the other hand, for μ≥n and u∈E, we have
∫
ℝ
n
|
u
|
μ
|
x
|
σ
𝑑
x
<
+
∞
,
and there exists C>0 such that
(7.2)
|
u
|
x
|
σ
μ
|
L
μ
(
ℝ
n
)
≤
C
∥
u
∥
.
In fact, by Hölder’s inequality, for t>1 and t′>1 such that 1t+1t′=1, and 0<σt′<n, we have
∫
ℝ
n
|
u
|
μ
|
x
|
σ
𝑑
x
=
∫
|
x
|
≥
1
|
u
|
μ
|
x
|
σ
𝑑
x
+
∫
|
x
|
<
1
|
u
|
μ
|
x
|
σ
𝑑
x
≤
∫
|
x
|
≥
1
|
u
|
μ
d
x
+
(
∫
|
x
|
<
1
|
u
|
μ
t
d
x
)
1
t
(
∫
|
x
|
<
1
d
x
|
x
|
σ
t
′
)
1
t
′
≤
c
∥
u
∥
μ
+
c
′
(
∫
|
x
|
<
1
|
u
|
μ
t
d
x
)
1
t
≤
C
∥
u
∥
μ
,
where we have used the continuous embeddings
E
↪
W
1
,
n
(
ℝ
n
)
↪
L
μ
(
ℝ
n
)
and
E
↪
L
μ
t
(
ℝ
n
)
.
In particular,
λ
σ
=
inf
u
∈
E
∖
{
0
}
∥
u
∥
n
∫
ℝ
n
|
u
|
n
|
x
|
σ
𝑑
x
>
0
.
Clearly, a weak solution of equation (7.1) is a function u∈E such that
∫
ℝ
n
w
(
x
)
|
∇
u
|
n
-
2
∇
u
∇
v
d
x
+
∫
ℝ
n
|
u
|
n
-
2
u
v
d
x
=
∫
ℝ
n
f
(
x
,
u
)
v
|
x
|
σ
d
x
+
ϵ
〈
h
,
v
〉
for all
v
∈
E
.
Here, we state our existence result.
Theorem 7.1.
Assume that (H1)–(H3) hold. Furthermore, assume that
lim sup
s
→
0
F
(
x
,
s
)
|
s
|
n
<
λ
σ
, uniformly with respect to
x
∈
ℝ
n
.
Then there exist A0>0 large enough and ϵ0>0 small enough such that if A>A0 and 0<ϵ<ϵ0, then problem (7.1) has at least two weak solutions.
7.1 The Variational Formulation
From (H1), there exist b3 and b4 such that
(7.3)
|
F
(
x
,
s
)
|
≤
b
3
|
s
|
n
+
b
4
|
s
|
α
(
e
γ
(
e
β
|
s
|
n
′
-
1
)
-
S
n
-
2
(
γ
,
e
β
|
s
|
n
′
-
1
)
)
for all
s
∈
ℝ
and a.e.
x
∈
ℝ
n
.
Given u∈E, by (7.3),
(7.4)
∫
ℝ
n
|
F
(
x
,
u
)
|
|
x
|
σ
𝑑
x
≤
b
3
∫
ℝ
n
|
u
|
n
|
x
|
σ
𝑑
x
+
b
4
∫
ℝ
n
|
u
|
α
(
e
γ
(
e
β
|
u
|
n
′
-
1
)
-
S
n
-
2
(
γ
,
e
β
|
u
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
.
By Hölder’s inequality, we get
∫
ℝ
n
|
u
|
α
(
e
γ
(
e
β
|
u
|
n
′
-
1
)
-
S
n
-
2
(
γ
,
e
β
|
u
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
(7.5)
≤
(
∫
ℝ
n
|
u
|
α
+
1
|
x
|
σ
𝑑
x
)
α
α
+
1
(
∫
ℝ
n
e
γ
(
α
+
1
)
(
e
β
|
u
|
n
′
-
1
)
-
S
n
-
2
(
γ
(
α
+
1
)
,
e
β
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
)
1
α
+
1
,
where we have used the elementary inequality
(
e
s
-
∑
j
=
0
n
-
2
s
j
j
!
)
r
≤
e
r
s
-
∑
j
=
0
n
-
2
(
r
s
)
j
j
!
,
for all r≥1, s≥0. By (7.5), (7.2) and Theorem 1.10, we can immediately deduce from (7.4) that
∫
ℝ
n
|
F
(
x
,
u
)
|
|
x
|
σ
𝑑
x
<
+
∞
.
Consequently, the energy functional Iϵ:E→ℝ, given by
I
ϵ
(
u
)
=
∥
u
∥
n
n
-
∫
ℝ
n
F
(
x
,
u
)
|
x
|
σ
𝑑
x
-
ϵ
〈
h
,
u
〉
,
is well defined. Furthermore, using standard arguments, we can easily show that Iϵ∈C1(E,ℝ) and that the critical points of the functional Iϵ are precisely the weak solutions of problem (7.1).
7.2 Proof of Theorem 7.1
The proof is done in two steps. First, we prove the existence of a solution with negative energy using the variational principle of Ekeland. Next, applying the mountain pass theorem without the Palais–Smale condition, we establish the existence of a second solution with positive energy.
Lemma 7.2.
If f satisfies (H1) and (H4), then there exist ϵ1>0, ρ>0 and r>0 such that
(7.6)
I
ϵ
(
u
)
≥
ρ
for all
u
∈
E
,
∥
u
∥
=
r
,
provided that 0<ϵ<ϵ1.
Proof.
By (H4), there exist τ>0 and δ>0 such that
F
(
x
,
s
)
≤
λ
σ
-
τ
n
|
s
|
n
for all
|
s
|
≤
δ
and a.e.
x
∈
ℝ
n
.
This inequality, together with (H1), implies
F
(
x
,
s
)
≤
λ
σ
-
τ
n
|
s
|
n
+
b
5
|
s
|
α
(
e
γ
(
e
β
|
s
|
n
′
-
1
)
-
S
n
-
2
(
γ
,
e
β
|
s
|
n
′
-
1
)
)
for all
s
∈
ℝ
and a.e.
x
∈
ℝ
n
.
Thus, for all u∈E,
∫
ℝ
n
F
(
x
,
u
)
|
x
|
σ
𝑑
x
≤
λ
σ
-
τ
n
∫
ℝ
n
|
u
|
n
|
x
|
σ
𝑑
x
+
b
5
∫
ℝ
n
|
u
|
α
(
e
γ
(
e
β
|
u
|
n
′
-
1
)
-
S
n
-
2
(
γ
,
e
β
|
u
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
(7.7)
≤
λ
σ
-
τ
n
∫
ℝ
n
|
u
|
n
|
x
|
σ
𝑑
x
+
b
5
(
∫
ℝ
n
|
u
|
α
t
|
x
|
σ
𝑑
x
)
1
t
(
∫
ℝ
n
(
e
γ
t
′
(
e
β
|
u
|
n
′
-
1
)
-
S
n
-
2
(
γ
t
′
,
e
β
|
u
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
)
1
t
′
,
where t>1 is such that γt′<(n-σ)e(infs≥1χ(s))1n(n-1). By Corollary 6.1, for u∈E, with ∥u∥=r≤(ωn-11n-1β)1n′, we have
∫
ℝ
n
e
γ
t
′
(
e
β
|
u
|
n
′
-
1
)
-
S
n
-
2
(
γ
t
′
,
e
β
|
u
|
n
′
-
1
)
|
x
|
σ
𝑑
x
≤
b
6
.
Putting this inequality in (7.7) and having in mind that E is continuously embedded into Lα+1(ℝn), it follows that
∫
ℝ
n
F
(
x
,
u
)
|
x
|
σ
𝑑
x
≤
λ
σ
-
τ
n
∥
u
∥
n
λ
σ
+
b
7
∥
u
∥
α
for all
u
∈
E
,
∥
u
∥
=
r
.
Hence,
I
ϵ
(
u
)
≥
τ
n
λ
σ
∥
u
∥
n
-
b
7
∥
u
∥
α
-
b
8
ϵ
|
h
|
E
*
∥
u
∥
.
Clearly, we could choose r>0 sufficiently small such that
τ
n
λ
σ
r
n
-
1
-
b
7
r
α
-
1
>
0
.
By this choice, it suffices to take ϵ<ϵ1=τnλσrn-1-b7rα-1b8|h|E*. The proof of Lemma 7.2 can be concluded by choosing ρ=τnλσrn-b7rα-b8ϵ|h|E*r. ∎
Lemma 7.3.
Assume that f satisfies assumptions (H1) and (H2). Let (uk)⊂E be a sequence of Palais–Smale of Iϵ such that
∥
u
k
∥
n
′
≤
ω
n
-
1
1
n
-
1
β
for all
k
.
Then (uk) has a strongly convergent subsequence in E.
Proof.
Since E is reflexive, there exists u∈E such that, up to a subsequence, uk⇀u weakly in E. Moreover, without loss of generality, we can assume that uk(x)→u(x) a.e., x∈ℝn. We claim that
(7.8)
lim
k
→
+
∞
∫
ℝ
n
f
(
x
,
u
k
)
(
u
k
-
u
)
|
x
|
σ
𝑑
x
=
0
.
For R>0, we have
∫
ℝ
n
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
=
∫
|
x
|
≥
R
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
+
∫
|
x
|
<
R
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
.
By Hölder’s inequality, we have
(7.9)
∫
|
x
|
≥
R
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
≤
(
∫
|
x
|
≥
R
|
f
(
x
,
u
k
)
|
n
′
|
x
|
σ
𝑑
x
)
1
n
′
(
∫
|
x
|
≥
R
|
u
k
-
u
|
n
|
x
|
σ
𝑑
x
)
1
n
.
Clearly, in view of the radial Lemma 4.1, on can easily see that the sequence
(
∫
|
x
|
≥
R
|
f
(
x
,
u
k
)
|
n
′
|
x
|
σ
𝑑
x
)
k
is bounded in ℝ. Next, observe that
∫
|
x
|
≥
R
|
u
k
-
u
|
n
|
x
|
σ
d
x
≤
1
R
σ
∫
ℝ
n
|
u
k
-
u
|
n
d
x
≤
b
9
R
σ
for all
k
.
This fact together with (7.9) imply
sup
k
∫
|
x
|
≥
R
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
≤
b
10
R
σ
→
0
,
R
→
+
∞
.
Thus, for any ϵ>0, there exists Rϵ>0 large enough such that
(7.10)
sup
k
∫
|
x
|
≥
R
ϵ
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
<
ϵ
2
.
On the other hand, by Hölder’s inequality, we have
(7.11)
∫
|
x
|
<
R
ϵ
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
≤
(
∫
|
x
|
<
R
ϵ
|
u
k
-
u
|
t
′
|
x
|
σ
𝑑
x
)
1
t
′
(
∫
|
x
|
<
R
ϵ
|
f
(
x
,
u
k
)
|
t
|
x
|
σ
𝑑
x
)
1
t
,
where t is any positive real number in ]1,+∞[. By (H1), we have
∫
|
x
|
<
R
ϵ
|
f
(
x
,
u
k
)
|
t
|
x
|
σ
𝑑
x
≤
b
11
∫
|
x
|
<
R
ϵ
|
u
k
|
t
(
n
-
1
)
|
x
|
σ
𝑑
x
+
b
11
∫
|
x
|
<
R
ϵ
|
u
k
|
(
α
-
1
)
t
(
e
γ
t
(
e
β
|
u
k
|
n
′
-
1
)
-
S
n
-
2
(
γ
t
,
e
β
|
u
k
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
.
Since {x∈ℝn, |x|<Rϵ} is bounded, it is immediate that the sequence of integrals
(
∫
|
x
|
<
R
ϵ
|
u
k
|
t
(
n
-
1
)
|
x
|
σ
𝑑
x
)
k
is bounded in ℝ. On the other hand, again by Hölder’s inequality, we obtain
∫
|
x
|
<
R
ϵ
|
u
k
|
(
α
-
1
)
t
(
e
γ
t
(
e
β
|
u
k
|
n
′
-
1
)
-
S
n
-
2
(
γ
t
,
e
β
|
u
k
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
(7.12)
≤
(
∫
|
x
|
<
R
ϵ
|
u
k
|
p
t
(
α
-
1
)
|
x
|
σ
𝑑
x
)
1
p
(
∫
|
x
|
<
R
ϵ
e
γ
t
p
′
(
e
β
|
u
k
|
n
′
-
1
)
-
S
n
-
2
(
γ
t
p
′
,
e
β
|
u
k
|
n
′
-
1
)
|
x
|
σ
𝑑
x
)
1
p
′
,
where p∈]1,+∞[. Clearly, we can choose t>1 and p>1 such that γtp′<(n-σ)e(infs≥1χ(s))1n(n-1). By Corollary 6.1, from (7.12), we can deduce that the sequence of integrals
(
∫
|
x
|
<
R
ϵ
|
u
k
|
(
α
-
1
)
t
(
e
γ
t
(
e
β
|
u
k
|
n
′
-
1
)
-
S
n
-
2
(
γ
t
,
e
β
|
u
k
|
n
′
-
1
)
)
|
x
|
σ
𝑑
x
)
k
is also bounded in ℝ. We have
(7.13)
∫
|
x
|
<
R
ϵ
|
u
k
-
u
|
t
′
|
x
|
σ
d
x
≤
(
∫
|
x
|
<
R
ϵ
|
u
k
-
u
|
t
′
q
′
d
x
)
1
q
′
(
∫
|
x
|
<
R
ϵ
d
x
|
x
|
σ
q
)
1
q
,
where q>1 is chosen so that σq<n. By virtue of the Rellich–Kondrachov compact embeddings theorem, we infer
∫
|
x
|
<
R
ϵ
|
u
k
-
u
|
t
′
q
′
d
x
→
0
,
k
→
+
∞
.
Hence, inequality (7.13) implies that
∫
|
x
|
<
R
ϵ
|
u
k
-
u
|
t
′
|
x
|
σ
𝑑
x
→
0
,
k
→
+
∞
.
Using this fact, we deduce from (7.11) that
∫
|
x
|
<
R
ϵ
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
→
0
,
k
→
+
∞
.
Thus, there exists k0(ϵ)≥1 large enough such that
(7.14)
∫
|
x
|
<
R
ϵ
|
f
(
x
,
u
k
)
(
u
k
-
u
)
|
|
x
|
σ
𝑑
x
<
ϵ
2
for all
k
≥
k
0
(
ϵ
)
.
The claim (7.8) follows by combining (7.10) and (7.14).
Since 〈Iϵ′(uk),uk-u〉=ok(1), by (7.8), we have
(7.15)
∫
ℝ
n
w
(
x
)
|
∇
u
k
|
n
-
2
∇
u
k
∇
(
u
k
-
u
)
d
x
+
∫
ℝ
n
|
u
k
|
n
-
2
u
k
(
u
k
-
u
)
𝑑
x
=
o
k
(
1
)
.
Using the weak convergence of (uk) to u in E, we infer
∫
ℝ
n
w
(
x
)
|
∇
u
|
n
-
2
∇
u
∇
(
u
k
-
u
)
d
x
+
∫
ℝ
n
|
u
|
n
-
2
u
(
u
k
-
u
)
d
x
=
o
k
(
1
)
.
Putting that identity in (7.15), it follows that
∫
ℝ
n
w
(
x
)
(
|
∇
u
k
|
n
-
2
∇
u
k
-
|
∇
u
|
n
-
2
∇
u
)
∇
(
u
k
-
u
)
d
x
+
∫
ℝ
n
(
|
u
k
|
n
-
2
u
k
-
|
u
|
n
-
2
u
)
(
u
k
-
u
)
𝑑
x
=
o
k
(
1
)
.
Therefore, uk→u strongly in E. This completes the proof of Lemma 7.3. ∎
7.2.1 Existence of a First Solution
We will apply the Ekeland variational principle to prove the existence of a weak solution for (7.1) with negative energy. In (7.6), we could choose r small enough such that
r
≤
(
ω
n
-
1
1
n
-
1
β
)
1
n
′
.
Set Br={u∈E;∥u∥≤r}. Since h≠0, one can easily see that
c
r
=
inf
u
∈
B
r
I
ϵ
(
u
)
<
0
.
On the other hand, by (7.6), for 0<ϵ<ϵ1, we have
inf
∥
u
∥
=
r
I
ϵ
(
u
)
>
inf
∥
u
∥
≤
r
I
ϵ
(
u
)
.
From the variational principle of Ekeland (see [16]), there exists a Palais–Smale sequence (uk)⊂Br of Iϵ at the level cr. By Lemma 7.4, there exists u∈E such that, always up to a subsequence, uk→u strongly in E. Therefore, u is a critical point of the functional Iϵ and Iϵ(u)=cr<0.
7.2.2 Existence of a Second Solution
In this subsection, we check that the functional Iϵ has the mountain pass geometry. First, observe that by (H2), there exist two positive constant a and b such that
F
(
x
,
s
)
≥
a
|
s
|
θ
-
b
for all
s
∈
ℝ
and a.e.
x
∈
ℝ
n
.
Consequently, for all u∈C0,rad∞(ℝn)∖{0},
I
ϵ
(
t
u
)
≤
t
n
n
∥
u
∥
n
-
a
t
θ
∫
ℝ
n
|
u
|
θ
|
x
|
σ
𝑑
x
-
b
∫
supp
(
u
)
d
x
|
x
|
σ
-
ϵ
t
〈
h
,
u
〉
for all
t
≥
0
,
which implies that Iϵ(tu)→-∞ as t→+∞, since θ>n. By this fact together with Lemma 7.2, we can apply the mountain pass theorem without the Palais–Smale condition (see [28]) to obtain a Palais–Smale sequence (vk) in E at the level c, given by
c
=
inf
γ
∈
Γ
max
t
∈
[
0
,
1
]
I
ϵ
(
γ
(
t
)
)
,
where Γ={γ∈C([0,1],E),Iϵ(γ(0))=0,and I(γ(1))<0}.
Furthermore, we can easily see that the mountain pass level c depends on the parameter A appearing in (H3). More precisely, we have the following result.
Lemma 7.4.
Let 0<ϵ<ϵ1. For all η>0, there exists Aη>0 such that c<η for all A>Aη.
Proof.
Let φ∈E∖{0} be such that φ≥0 and 〈h,φ〉>0. We have
c
≤
sup
0
≤
t
≤
1
I
ϵ
(
γ
(
t
)
)
≤
sup
t
≥
0
I
ϵ
(
t
φ
)
.
By (H3), it follows that
c
≤
sup
t
≥
0
(
∥
φ
∥
n
n
t
n
-
A
|
φ
|
x
|
σ
r
|
L
r
(
ℝ
n
)
r
t
r
+
ϵ
1
t
〈
h
,
φ
〉
)
.
Let ψ(t)=α1tn-α2tr+α3t, with
α
1
=
∥
φ
∥
n
n
,
α
2
=
A
|
φ
|
x
|
σ
r
|
L
r
(
ℝ
n
)
r
and
α
3
=
ϵ
1
〈
h
,
φ
〉
.
Then
c
≤
sup
t
≥
0
ψ
(
t
)
≤
max
{
sup
0
≤
t
≤
1
(
α
1
t
n
-
α
2
t
r
+
α
3
t
)
,
sup
t
≥
1
(
α
1
t
n
-
α
2
t
r
+
α
3
t
)
}
(7.16)
≤
max
{
sup
0
≤
t
≤
1
(
(
α
1
+
α
3
)
t
-
α
2
t
r
)
,
sup
t
≥
1
(
(
α
1
-
α
2
)
t
r
+
α
3
t
)
}
.
Clearly, we have α1-α2+α3<0 and, as a consequence, supt≥1ψ(t)<0, provided that A is chosen large enough. For 0≤t≤1, define B(t)=(α1+α3)t-α2tr. Obviously, this function reaches its maximum at the point tm=((α1+α3)rα2)1r-1<1. By (7.16), it follows that
c
≤
(
α
1
+
α
3
)
t
m
-
α
2
t
m
r
=
(
α
1
+
α
3
)
(
(
α
1
+
α
3
)
r
α
2
)
1
r
-
1
-
α
2
(
(
α
1
+
α
3
)
r
α
2
)
r
r
-
1
=
(
(
α
1
+
α
3
)
r
α
2
)
1
r
-
1
[
α
1
+
α
3
-
α
1
+
α
3
r
]
=
(
(
α
1
+
α
3
)
r
α
2
)
1
r
-
1
(
α
1
+
α
3
)
(
1
-
1
r
)
=
(
(
α
1
+
α
3
)
r
)
r
r
-
1
(
r
-
1
)
(
1
α
2
)
1
r
-
1
.
This allows us to conclude the proof of Lemma 7.4. ∎
In the rest of the proof, we will try to control the growth of the norm ∥vk∥ in order to be able to apply Lemma 7.3 and consequently obtain the compactness required for the final passage to the limit.
Lemma 7.5.
Under assumptions (H1)–(H3), there exist A0>0 and 0<ϵ0<ϵ1 such that
lim
n
→
+
∞
sup
∥
v
k
∥
≤
(
ω
n
-
1
1
n
-
1
β
)
1
n
′
,
provided that A>A0 and ϵ<ϵ0.
Proof.
By (H2), it follows that
(7.17)
(
1
n
-
1
θ
)
∥
v
k
∥
n
-
b
12
ϵ
∥
v
k
∥
≤
I
ϵ
(
v
k
)
-
1
θ
〈
I
ϵ
′
(
v
k
)
,
v
k
〉
≤
c
+
∘
k
(
1
)
(
1
+
∥
v
k
∥
)
for all
k
∈
ℕ
.
Denote X=lim supn→+∞∥vk∥. Passing to the upper limit in (7.17), we obtain
(
1
n
-
1
θ
)
X
n
≤
b
12
ϵ
X
+
c
.
If X≥1, we have
X
≤
X
n
≤
c
1
n
-
1
θ
+
b
12
ϵ
X
1
n
-
1
θ
,
which implies that
X
(
1
-
b
12
ϵ
1
n
-
1
θ
)
≤
c
1
n
-
1
θ
.
If X≤1, we have
X
n
≤
c
1
n
-
1
θ
+
b
12
ϵ
X
1
n
-
1
θ
≤
c
1
n
-
1
θ
+
b
12
ϵ
1
n
-
1
θ
.
Thus,
X
≤
(
c
1
n
-
1
θ
+
b
12
ϵ
1
n
-
1
θ
)
1
n
.
By Lemma 7.4, we can find A0>0 and ϵ0<ϵ1 such that, for A>A0 and ϵ<ϵ0, one has
X
≤
(
ω
n
-
1
1
n
-
1
β
)
1
n
′
.
The proof is complete. ∎
By virtue of Lemma 7.3, we deduce that, up to a subsequence, there exists v∈E such that vk→v strongly in E. Therefore, v is a critical point of Iϵ such that Iϵ(v)=c≥ρ>0. The proof of Theorem 7.1 is now complete.
