# Multiplicity of Solutions for Neumann Problems for Semilinear Elliptic Equations

Authors: Yu-Cheng An; Hong-Min Suo
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

Using the minimax methods in critical point theory, we study the multiplicity of solutions for a class of Neumann problems in the case near resonance. The results improve and generalize some of the corresponding existing results.


---

## Body

## 1. Introduction

The aim of this paper is to study the following semilinear Neumann problem:(1)
-
Δ
u
+
β
(
x
)
u
=
λ
u
+
f
(
x
,
u
)
,
x
∈
Ω
,
∂
u
∂
n
=
0
,
x
∈
∂
Ω
.
Here Ω
⊂
R
N
(
N
≥
3
) is a bounded domain with a C
1 boundary ∂
Ω and n
(
·
) denotes the outward unit normal on ∂
Ω. Also β
∈
L
s
(
Ω
) with s
>
N
/
2 and it may change sign. The reaction f
(
x
,
u
) is a Carathéodory function and satisfies the following assumptions:(
f
0
)
for everyM
>
0, there exists a function L
M
∈
L
2
(
Ω
) such that |
f
(
x
,
t
)
|
≤
L
M
(
x
) for all |
t
|
≤
M and a.e. x
∈
Ω;
(
f
∞
)
lim
⁡
|
t
|
→
∞
(
f
(
x
,
t
)
/
t
)
=
0 uniformly for x
∈
Ω.Recently, there have been many papers concerned with the Neumann problems; see [1–5] and the references therein. Or, more specifically, in Li [1] and Qian [2], the left hand side differential operator is -
Δ
u
+
β
u, with β
∈
R, β
>
0. In Motreanu et al. [3], Tang and Wu [4], and Motreanu et al. [5], the differential operator is -
Δ
u (i.e., β
=
0). Semilinear Neumann problems with unbounded and indefinite potential, especially, were studied by Gasiński and Papageorgiou [6]. They obtained two multiplicity theorems. In addition, the same problems were studied by Papageorgiou and Rădulescu [7]. They dealt with equations in which the reaction f
(
x
,
u
) exhibits an asymmetric behavior at +
∞ and at -
∞ (jumping nonlinearity) and they proved multiplicity theorems providing sign information for all the solutions.On the other hand, for the perturbed problem, Mawhin and Schmitt [8] first considered the two-point boundary value problem
(2)
-
u
′′
-
λ
u
=
f
(
x
,
u
)
+
h
(
x
)
,
u
(
0
)
=
u
(
π
)
=
0
.
Under the assumption that f is bounded and satisfies a sign condition, if the parameter λ is sufficiently close to λ
1 from left, problem (2) has at least three solutions; if λ
1
≤
λ
<
λ
2, problem (2) has at least one solution, where λ
1, λ
2 are the first and second eigenvalues of the corresponding linear problem. Ma et al. [9] considered the boundary value problem Δ
u
+
λ
u
+
f
(
x
,
u
)
=
h
(
x
) defined on a bounded open set Ω
⊂
R
N, no matter whether the boundary conditions are Dirichlet or Neumann condition; as the parameter λ approaches λ
1 from left, there exist three solutions. Moreover, existence of three solutions was obtained for the quasilinear problem in bounded domains as the parameter λ approaches λ
1 from left. In [10, 11], these results were extended to the perturbed p-Laplacian equation in R
N. In [12], Ou and Tang extended above some results to some elliptic systems with the Dirichlet boundary conditions. de Paiva and Massa in [13], especially, studied the semilinear elliptic boundary value problem in any spatial dimension and using variational techniques; they showed that a suitable perturbation will turn the almost resonant situation (λ near to λ
k, i.e., near resonance with a nonprincipal eigenvalue) in a situation where the solutions are at least two. In [14], those results were extended to the degenerate elliptic equations in the bounded domain.Motivated by the above idea, we have the goal in this paper of extending these results in [6, 12–14] to some elliptic equations with the Neumann boundary conditions. Here, it is worth pointing out that (
f
∞
) is weaker than (
f
1
) in [13] (or (
A
) in [14]). More to the point, there are functions satisfying the assumptions of our main results in Section 2 and not satisfying the assumptions in [13, 14]. For example, let f
(
x
,
t
)
=
t
/
ln
⁡
⁡
(
1
+
|
t
|
). Then f satisfies the assumptions for our Theorem 5 in Section 2 and does not satisfy (
f
1
) in [13] (or (
A
) in [14]).The rest of our paper is organized as follows. In Section2 we give some preliminary lemmas and our main results. Section 3 gives the detailed proofs of our main results based on several estimates, whose proof will be presented in Sections 4 and 5.


## 2. Preliminaries and Main Results

Let the Sobolev spaceX
=
H
1
(
Ω
). Denote
(3)
∥
u
∥
1,2
=
(
∫
Ω
‍
(
|
∇
u
|
2
+
u
2
)
d
x
)
1
/
2
to be the norm of u in X and ∥
u
∥
p the norm of u in L
p
(
Ω
). The space X is a Hilbert space. For a discussion about the space setting, we refer to [15] and the references therein.Again, we recall the properties of the eigenvalue problem as follows (see [6]):
(4)
-
Δ
u
+
β
(
x
)
u
=
λ
u
,
x
∈
Ω
,
∂
u
∂
n
=
0
,
x
∈
∂
Ω
.
The eigenvalue problems in (4) have a sequence {
λ
k
}
k
≥
1 of eigenvalues, such that
(5)
-
∞
<
λ
1
<
λ
2
≤
λ
3
≤
⋯
≤
λ
k
≤
⋯
⟶
+
∞
i
i
≤
λ
3
≤
⋯
≤
λ
k
≤
⋯
⟶
+
∞
as
k
⟶
+
∞
.
There is also a corresponding sequence {
ϕ
k
}
k
≥
1
⊂
H
1
(
Ω
) of eigenfunctions which form an orthonormal basis of L
2
(
Ω
) and an orthogonal basis of H
1
(
Ω
). Moreover, we know that ϕ
k
∈
C
1
,
α
(
Ω
) for some α
∈
(
0,1
) and all k
≥
1.We denote byE
k the eigenspace corresponding to an eigenvalue λ
k, and we can decompose H
1
(
Ω
)
=
⨁
k
=
1
∞
E
k. We set
(6)
σ
(
u
)
=
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
)
d
x
.
The eigenvalues admit the following variational characterizations in terms of the Rayleigh quotient σ
(
u
)
/
∥
u
∥
2
2 for all u
∈
H
1
(
Ω
):
(7)
λ
1
=
inf
⁡
{
σ
(
u
)
∥
u
∥
2
2
:
u
∈
H
1
(
Ω
)
∖
{
0
}
}
,
(8)
λ
k
=
inf
⁡
{
σ
(
u
)
∥
u
∥
2
2
:
u
∈
⨁
i
=
k
∞
E
i
,
u
≠
0
}
=
sup
⁡
{
σ
(
u
)
∥
u
∥
2
2
:
u
∈
⨁
i
=
1
k
E
i
,
u
≠
0
}
.
In (7), the infimum is realized on E
1. Also, in (8), both the infimum and the supremum are realized on E
k. All the eigenspaces have the so-called unique continuation property. The first eigenvalue λ
1 is simple and it is clear from (7) that the corresponding eigenfunctions do not change sign. Namely, we can suppose that ϕ
1 is strictly positive on Ω. We mention that all the other eigenvalues have nodal eigenfunctions. For more properties to the eigenvalue problem (4), see [6, 7].By the presence of functionβ, weak solutions of (1) must be found in a suitable space. To this purpose, letting θ
>
max
⁡
{
-
λ
1
,
0
}, we introduce a new inner product on H
1
(
Ω
) by
(9)
〈
u
,
v
〉
=
∫
Ω
‍
(
∇
u
∇
v
+
β
(
x
)
u
v
+
θ
u
v
)
d
x
for u
,
v
∈
H
1
(
Ω
) and the associated norm
(10)
∥
u
∥
=
(
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
+
θ
u
2
)
d
x
)
1
/
2
,
u
∈
H
1
(
Ω
)
.Lemma 1.
Letβ
∈
L
s
(
Ω
) with s
>
N
/
2 and it may change sign, if θ
>
max
⁡
{
-
λ
1
,
0
}; then ∥
·
∥ is equivalent to the usual Sobolev norm ∥
·
∥
1,2.Proof.
By virtue of Ho
¨lder’s inequality, we have
(11)
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
+
θ
u
2
)
d
x
≤
∫
Ω
‍
(
|
∇
u
|
2
+
θ
u
2
)
d
x
+
∥
β
∥
s
∥
u
∥
2
s
′
2
≤
max
⁡
{
1
,
θ
}
∥
u
∥
1,2
2
+
∥
β
∥
s
∥
u
∥
2
s
′
2
≤
max
⁡
{
1
,
θ
}
∥
u
∥
1,2
2
+
K
2
∥
β
∥
s
∥
u
∥
1,2
2
≤
ξ
2
∥
u
∥
1,2
2
,
where K is the constant of Sobolev imbedding from H
1
(
Ω
)
↪
L
2
s
′
(
Ω
), s
′
=
s
/
(
s
-
1
), ξ
2
=
max
⁡
{
1
,
θ
,
K
2
∥
β
∥
s
}.
On the other hand, ifθ
>
max
⁡
{
-
λ
1
,
0
}, then there exists ξ
1
>
0 such that
(12)
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
+
θ
u
2
)
d
x
≥
ξ
1
∥
u
∥
1,2
2
.
If not, it is clear from (7) that
(13)
χ
(
u
)
=
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
+
θ
u
2
)
d
x
≥
0
.
Exploiting the 2-homogeneity of χ we can find a sequence {
u
n
}
n
≥
1
⊂
H
1
(
Ω
), such that
(14)
∥
u
n
∥
1,2
=
1
,
∀
n
≥
1
,
χ
(
u
n
)
⟶
0
+
.
Passing to a subsequence if necessary, we may assume that
(15)
u
n
⇀
u
weakly
in
H
1
(
Ω
)
,
(16)
u
n
⟶
u
strongly
in
L
2
s
′
(
Ω
)
.
The sequential weak lower semicontinuity of σ and (15) imply that
(17)
σ
(
u
)
≤
-
∫
Ω
‍
θ
u
2
d
x
≤
λ
1
∥
u
∥
2
2
.
So σ
(
u
)
=
λ
1
∥
u
∥
2
2 and thus u
=
γ
ϕ
1, with some γ
∈
R.
Ifγ
=
0, then u
n
→
0 in H
1
(
Ω
), which contradicts the fact that ∥
u
n
∥
1,2
=
1 for all n
≥
1. If γ
≠
0, then, from (17), we have σ
(
ϕ
1
)
<
λ
1
∥
ϕ
1
∥
2
2, which contradicts (7).
Combining (11) and (12), we have
(18)
ξ
1
∥
u
∥
1,2
2
≤
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
+
θ
u
2
)
d
x
≤
ξ
2
∥
u
∥
1,2
2
.
Namely, ∥
·
∥ is equivalent to the usual Sobolev norm ∥
·
∥
1,2.From now on we take(
H
1
(
Ω
)
,
〈
·
〉
,
∥
·
∥
) as our working space, θ
>
max
⁡
{
-
λ
1
,
0
}, ∥
ϕ
1
∥
=
1. In view of Lemma 1 and the Rellich-Kondrachov Compactness theorem (see [15, Theorem  1]), we directly get the following lemma.Lemma 2.
Letβ
∈
L
s
(
Ω
) with s
>
N
/
2 and it may change sign. Then the space H
1
(
Ω
) is compactly embedded in L
p
(
Ω
) for 1
≤
p
<
2
* and continuously embedded in L
2
*
(
Ω
); hence there exists S
>
0 such that
(19)
∥
u
∥
p
≤
S
∥
u
∥
,
∀
u
∈
H
1
(
Ω
)
.In addition, we also need the following lemmas.Lemma 3 (from Lemma  4.6 of [13]).
LetX be a Hilbert space with orthonormal direct sum splitting X
=
V
⊕
Z
⊕
W. Moreover, let dim
⁡
(
V
⊕
Z
)
<
∞. For ρ
>
R
>
0, set
(20)
A
=
{
u
∈
W
:
∥
u
∥
≥
R
}
∪
{
u
∈
Z
⊕
W
:
∥
u
∥
=
R
}
,
B
=
{
u
∈
V
⊕
Z
:
∥
u
∥
=
ρ
}
.
Then A and B link.Lemma 4 (from Theorem  8.1 of [16]).
LetX
=
X
1
⊕
X
2 be a Hilbert space where X
1 has finite dimension and J
∈
C
1
(
X
,
R
) satisfying the (
P
.
S
.
) condition and such that, for given ρ
1
,
ρ
2
>
0,
(21)
sup
⁡
u
∈
ρ
1
S
1
J
(
u
)
<
a
=
inf
⁡
u
∈
ρ
2
B
2
J
(
u
)
≤
b
=
sup
⁡
u
∈
ρ
1
B
1
J
(
u
)
<
inf
⁡
u
∈
ρ
2
S
2
J
(
u
)
,
where B
i and S
i represent the unit ball and the unit sphere in X
i
:
i
=
1,2. Then there exists a critical point u
0 such that J
(
u
0
)
∈
[
a
,
b
].Next, letF
(
x
,
t
)
=
∫
0
t
‍
f
(
x
,
s
)
d
s; in order to state our main results, we introduce the following assumptions on the nonlinear term:(
f
1
)
lim
⁡
|
t
|
→
∞
(
f
(
x
,
t
)
t
/
|
t
|
)
=
+
∞ uniformly with respect to x
∈
Ω;
(
f
2
)
lim
⁡
|
t
|
→
∞
F
(
x
,
t
)
=
+
∞ uniformly with respect to x
∈
Ω;
(
f
3
)
lim
⁡
|
t
|
→
∞
(
f
(
x
,
t
)
t
/
|
t
|
)
=
-
∞ uniformly with respect to x
∈
Ω;
(
f
4
)
lim
⁡
|
t
|
→
∞
F
(
x
,
t
)
=
-
∞ uniformly with respect to x
∈
Ω.Our main results are given by the following theorems.Theorem 5.
Letλ
k  (k
≥
2) be an eigenvalue of multiplicity m and β
∈
L
s
(
Ω
) with s
>
N
/
2 and it may change sign. Suppose that the conditions (
f
0
) and (
f
∞
) hold and one of the sets of hypotheses (
f
1
) and (
f
2
). Then there exists δ
0
>
0 such that for λ
∈
(
λ
k
-
δ
0
,
λ
k
) problem (1) has at least two solutions.Theorem 6.
Letλ
k  (k
≥
2) be an eigenvalue of multiplicity m and β
∈
L
s
(
Ω
) with s
>
N
/
2 and it may change sign. Suppose that the conditions (
f
0
) and (
f
∞
) hold and one of the sets of hypotheses (
f
3
) and (
f
4
). Then there exists δ
1
>
0 such that for λ
∈
(
λ
k
,
λ
k
+
δ
1
) problem (1) has at least two solutions.Theorem 7.
Letβ
∈
L
s
(
Ω
) with s
>
N
/
2 and it may change sign. Suppose that the conditions (
f
0
) and (
f
∞
) hold, and the nonlinearity f satisfies (
f
2
). Then, for λ
<
λ
1 sufficiently close to λ
1, problem (1) has at least three solutions.


## 3. Proof of Theorems

The associated functional of problem (1) is
(22)
J
(
u
)
=
1
2
∫
Ω
‍
(
|
∇
u
|
2
+
β
(
x
)
u
2
)
d
x
-
λ
2
∫
Ω
‍
u
2
d
x
-
∫
Ω
‍
F
(
x
,
u
)
d
x
=
1
2
∥
u
∥
2
-
λ
+
θ
2
∫
Ω
‍
u
2
d
x
-
∫
Ω
‍
F
(
x
,
u
)
d
x
for u
∈
H
1
(
Ω
). Under the conditions (
f
0
) and (
f
∞
), it is easy to verify that, for every λ
∈
R, J
∈
C
1
(
H
1
(
Ω
)
,
R
) and
(23)
〈
J
′
(
u
)
,
v
〉
=
∫
Ω
‍
(
∇
u
∇
v
+
β
(
x
)
u
v
)
d
x
-
λ
∫
Ω
‍
u
v
d
x
-
∫
Ω
‍
f
(
x
,
u
)
v
d
x
,
for u
,
v
∈
H
1
(
Ω
). Moreover, critical points of J are exactly weak solutions of problem (1).It follows from(
f
0
) and (
f
∞
) that for every ε
>
0 there exist M
~
ε
>
0 and L
M
~
ε
∈
L
2
(
Ω
) such that
(24)
|
f
(
x
,
t
)
|
≤
ε
|
t
|
+
|
L
M
~
ε
(
x
)
|
,
for all t
∈
R and a.e. x
∈
Ω, which implies that
(25)
|
F
(
x
,
t
)
|
≤
ε
2
|
t
|
2
+
|
L
M
~
ε
(
x
)
|
|
t
|
,
for all t
∈
R and a.e. x
∈
Ω. From this and Hölder’s inequality, we have
(26)
|
∫
Ω
‍
F
(
x
,
u
)
d
x
|
≤
ε
2
S
2
∥
u
∥
2
+
L
~
ε
∥
u
∥
,
where L
~
ε
=
S
∥
L
M
~
ε
∥
2 and S is the best embedding constant.In addition, we will use several times the estimates below.From (7) and (8), we have
(27)
∥
u
∥
2
≥
(
θ
+
λ
1
)
∫
Ω
‍
u
2
d
x
,
∀
u
∈
H
1
(
Ω
)
,
(28)
∥
u
∥
2
≤
(
θ
+
λ
k
)
∫
Ω
‍
u
2
d
x
,
∀
u
∈
span
⁡
{
ϕ
1
,
ϕ
2
,
…
,
ϕ
k
}
,
(29)
∥
u
∥
2
≥
(
θ
+
λ
k
+
1
)
∫
Ω
‍
u
2
d
x
,
∀
u
∈
(
span
⁡
{
ϕ
1
,
ϕ
2
,
…
,
ϕ
k
}
)
⊥
.Proposition 8.
Assume that(
f
0
) and (
f
∞
) hold, and suppose that λ
≠
λ
k for any k
∈
N
+. Then J satisfies the (
P
.
S
.
) condition.Proof.
For any sequence{
u
n
}
⊂
H
1
(
Ω
) such that
(30)
|
J
(
u
n
)
|
<
∞
∀
n
,
J
′
(
u
n
)
⟶
0
as
n
⟶
∞
,
we need to prove that {
u
n
} has a convergent subsequence. By the standard argument, it suffices to show that {
u
n
} is bounded in H
1
(
Ω
). Suppose by contradiction that ∥
u
n
∥ as n
→
∞. Let w
n
=
u
n
/
∥
u
n
∥. Then ∥
w
n
∥
=
1, so we may suppose that there is w
∈
H
1
(
Ω
) such that
(31)
w
n
⇀
w
weakly
in
H
1
(
Ω
)
,
w
n
⟶
w
strongly
in
L
p
(
Ω
)
,
where
p
∈
[
1
,
2
*
)
,
as n
→
∞. Then, for any v
∈
H
1
(
Ω
), from (24) and (27), we have
(32)
1
∥
u
n
∥
∫
Ω
‍
f
(
x
,
u
n
)
v
d
x
≤
1
∥
u
n
∥
∫
Ω
‍
(
ε
|
u
n
|
+
|
L
M
~
ε
(
x
)
|
)
|
v
|
d
x
≤
1
∥
u
n
∥
(
ε
∥
u
n
∥
2
∥
v
∥
2
+
∥
L
M
~
ε
∥
2
∥
v
∥
2
1
1
)
≤
ε
θ
+
λ
1
∥
v
∥
2
+
∥
L
M
~
ε
∥
2
∥
v
∥
2
∥
u
n
∥
which shows that
(33)
lim
⁡
n
→
∞
1
∥
u
n
∥
∫
Ω
‍
f
(
x
,
u
n
)
v
d
x
≤
ε
θ
+
λ
1
∥
v
∥
2
.
By the arbitrariness of ε, one has
(34)
lim
⁡
n
→
∞
1
∥
u
n
∥
∫
Ω
‍
f
(
x
,
u
n
)
v
d
x
=
0
.
Like in the proof of (34), from (26), it follows that
(35)
lim
⁡
n
→
∞
1
∥
u
n
∥
2
∫
Ω
‍
F
(
x
,
u
n
)
v
d
x
=
0
.
Thus, for any v
∈
H
1
(
Ω
), by (30), (31), and (34), passing to the limit in
(36)
〈
J
′
(
u
n
)
,
v
〉
∥
u
n
∥
=
∫
Ω
‍
(
∇
w
n
∇
v
+
β
(
x
)
w
n
v
)
d
x
-
λ
∫
Ω
‍
w
n
v
d
x
-
∫
Ω
‍
f
(
x
,
u
n
)
v
d
x
gives
(37)
∫
Ω
‍
(
∇
w
n
∇
v
d
x
+
β
(
x
)
w
n
v
)
d
x
-
λ
∫
Ω
‍
w
n
v
d
x
=
0
,
which implies that
(38)
w
∈
Ker
⁡
(
-
Δ
+
β
-
λ
)
.
In addition, by (30), (31), and (35), passing to the limit in
(39)
J
(
u
n
)
∥
u
n
∥
2
=
1
2
∫
Ω
‍
(
|
∇
w
n
|
2
+
β
(
x
)
w
n
2
+
θ
w
n
2
)
d
x
-
θ
+
λ
2
∫
Ω
‍
w
n
2
d
x
-
1
∥
u
n
∥
2
∫
Ω
‍
F
(
x
,
u
n
)
d
x
gives
(40)
1
2
-
θ
+
λ
2
∫
Ω
‍
w
2
d
x
=
0
.
Ifθ
+
λ
=
0, then, from (40), we have 1
/
2
=
0, which is a contradiction.
Ifθ
+
λ
≠
0, then, from (40), we have w
≠
0. From this and (38) it follows that λ is an eigenvalue of operator -
Δ
+
β, a contradiction; the proof is completed.Hereafter, we set(41)
V
=
span
⁡
{
ϕ
1
,
…
,
ϕ
k
-
1
}
,
Z
=
span
⁡
{
ϕ
k
,
…
,
ϕ
k
+
m
-
1
}
=
E
k
,
W
=
(
V
⊕
Z
)
⊥
,
and we define
(42)
B
V
=
{
u
∈
V
:
∥
u
∥
≤
1
}
,
B
V
Z
=
{
u
∈
V
⊕
Z
:
∥
u
∥
≤
1
}
,
B
Z
W
=
{
u
∈
Z
⊕
W
:
∥
u
∥
≤
1
}
,
and S
V, S
V
Z, and S
Z
W, respectively, are their relative boundaries.Theorems5 and 6 will be a consequence of the geometry in Propositions 9 and 10 stated below, whose proofs will be postponed to Sections 4 and 5.Proposition 9.
Ifλ
∈
(
λ
k
-
1
,
λ
k
) and hypotheses (
f
0
) and (
f
∞
) are satisfied, then there exist constants D
λ and ρ
λ such that
(43)
J
(
u
)
≥
D
λ
,
f
o
r
u
∈
Z
⊕
W
,
(44)
J
(
u
)
<
D
λ
,
f
o
r
u
∈
ρ
λ
S
V
.
Moreover, if one of the sets of hypotheses (
f
1
) and (
f
2
) is satisfied, then there exists δ
0 such that for λ
∈
(
λ
k
-
δ
0
,
λ
k
) there exist D
W
,
D
λ
∈
R, ρ
λ
>
R
1
>
0 such that, in addition to (43) and (44),
(45)
J
(
u
)
≥
D
W
,
f
o
r
u
∈
W
,
(46)
J
(
u
)
<
D
W
,
f
o
r
u
∈
R
1
S
V
Z
,
(47)
J
(
u
)
<
D
W
,
f
o
r
u
∈
V
,
∥
z
∥
≥
R
1
.
(The values with index λ depend on λ; the others may be fixed uniformly.)Proposition 10.
Ifλ
∈
(
λ
k
,
λ
k
+
m
) and hypotheses (
f
0
) and (
f
∞
) are satisfied, then there exist constants K
λ and β
λ such that
(48)
J
(
z
)
≥
K
λ
,
f
o
r
z
∈
W
,
(49)
J
(
z
)
<
K
λ
,
f
o
r
z
∈
β
λ
S
V
Z
.
Moreover, if one of the sets of hypotheses (
f
3
) and (
f
4
) is satisfied, then there exists δ
1 such that for λ
∈
(
λ
k
,
λ
k
+
δ
1
) there exist K
λ
,
K
V
,
E
∈
R, β
λ
>
R
2
>
0, and ξ
>
0 such that, in addition to (48) and (49),
(50)
J
(
u
)
<
K
V
,
f
o
r
u
∈
V
,
(51)
J
(
u
)
>
K
V
,
f
o
r
u
∈
R
2
S
Z
W
,
(52)
J
(
u
)
>
K
V
,
f
o
r
u
∈
W
,
∥
u
∥
≥
R
2
,
(53)
J
(
u
)
>
E
,
f
o
r
u
∈
R
2
B
Z
W
,
(54)
J
(
u
)
<
E
,
f
o
r
u
∈
ξ
S
V
.
(The values with index λ depend on λ; the others may be fixed uniformly.)Proof of Theorem5.
Since the functionalJ satisfies the (
P
.
S
.
) condition, we can apply two times the saddle point theorem (see, e.g., [17]); let
(55)
Γ
k
-
1
=
{
γ
∈
C
0
(
ρ
λ
B
V
;
E
)
s.t.
γ
|
ρ
λ
S
V
=
id
}
,
Γ
k
=
{
γ
∈
C
0
(
R
1
B
V
Z
;
E
)
s.t.
γ
|
R
1
S
V
Z
=
id
}
.
The first solution, which we denote byu
k
-
1 and may be obtained for any λ
∈
(
λ
k
-
1
,
λ
k
) with just hypotheses (
f
0
) and (
f
∞
), corresponds to a critical point at the level
(56)
c
k
-
1
=
inf
⁡
γ
∈
Γ
k
-
1
sup
⁡
w
∈
ρ
λ
B
V
J
(
γ
(
w
)
)
.
The criticality of this level is guaranteed by the estimates (43) and (44), since ρ
λ
S
V and Z
⊕
W link; that is, the image of any map in Γ
k
-
1 intersects Z
⊕
W.
The second solution, which we denote byu
k, corresponds to a critical point at the critical level
(57)
c
k
=
inf
⁡
γ
∈
Γ
k
sup
⁡
w
∈
R
1
B
V
Z
J
(
γ
(
w
)
)
.
Actually, this is a critical level because of the estimates (45) and (46), since R
1
S
V
Z and W link.
To conclude the proof, we need to show that these two solutions are distinct.
We observe first that by estimate (45) we have that c
k
≥
D
W, then we observe that we may build a map γ
0
∈
Γ
k
-
1 in such a way that its image is the union between the annulus {
u
∈
V
:
∥
u
∥
∈
[
R
1
,
ρ
λ
]
} and the image of a (
k
-
1
)-dimensional ball in R
1
S
V
Z whose boundary is R
1
S
V. By the estimates (46) and (47), we deduce that sup
⁡
w
∈
ρ
λ
B
V
J
(
γ
0
(
w
)
)
<
D
W, and as a consequence c
k
-
1
<
D
W, proving that the two solutions are distinct, for being at different critical levels.Proof of Theorem6.
Since the functionalJ satisfies the (
P.S.
) condition, we can apply the saddle point theorem and Lemma 4.
The first solution, which we denote byw
k and may be obtained for any λ
∈
(
λ
k
,
λ
k
+
m
) with just hypotheses (
f
0
) and (
f
∞
), is again obtained through the saddle point theorem and corresponds to a critical point at the critical level
(58)
d
k
=
inf
⁡
γ
∈
Γ
k
sup
⁡
w
∈
β
λ
B
V
Z
J
(
γ
(
w
)
)
,
where now
(59)
Γ
k
=
{
γ
∈
C
0
(
β
λ
B
V
Z
;
E
)
s.t.
γ
|
β
λ
S
V
Z
=
id
}
.
The criticality is guaranteed by estimates (48) and (49), since β
λ
S
V
Z and W link.
The second solution, which we denote byw
k
-
1, comes from Lemma 3, where we set X
1
=
V and X
2
=
Z
⊕
W; actually we have the structure
(60)
sup
⁡
ξ
S
V
J
(
u
)
<
E
=
inf
⁡
R
2
B
Z
W
J
(
u
)
≤
sup
⁡
ξ
B
V
J
(
u
)
<
K
V
<
inf
⁡
R
2
S
Z
W
J
(
u
)
and then we have a critical point w
k
-
1 at the level d
k
-
1
≤
K
V.
Finally, in order to prove that these two solutions are distinct, we need a sharper estimate ford
k than that given by (49). For this we use Lemma 3 to guarantee that, for any map γ
∈
Γ
k, since β
λ
>
R
2, one has that the image of γ either intersects R
2
S
Z
W or has a point u
∈
W with ∥
u
∥
≥
R
2. This implies that
(61)
sup
⁡
w
∈
β
λ
B
V
Z
J
(
γ
(
w
)
)
>
K
V
,
by estimates (51) and (52), and then d
k
>
K
V proving that the two solutions are distinct, for being at different critical levels.Proof of Theorem7.
The proof will be divided into four steps.
Step  1. For λ
<
λ
1, from the definition of λ
1, (26), and (27), we get
(62)
J
(
u
)
≥
1
2
(
λ
1
-
λ
θ
+
λ
1
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Letting ε
=
(
λ
1
-
λ
)
/
2
S
2
(
θ
+
λ
1
), it follows that J is coercive in H
1
(
Ω
).
Similarly, from (26) and (29), we obtain
(63)
J
(
w
)
≥
1
2
(
λ
2
-
λ
θ
+
λ
2
-
ε
S
2
)
∥
w
∥
2
-
L
~
ε
∥
w
∥
≥
1
2
(
λ
2
-
λ
1
θ
+
λ
2
-
ε
S
2
)
∥
w
∥
2
-
L
~
ε
∥
w
∥
,
for all w
∈
(
span
⁡
{
ϕ
1
}
)
⊥. Let ε
=
(
λ
2
-
λ
1
)
/
2
S
2
(
θ
+
λ
2
); hence J is coercive in (
span
⁡
{
ϕ
1
}
)
⊥ and J is bounded from below on (
span
⁡
{
ϕ
1
}
)
⊥, and, moreover, there is a constant M, independent of λ, such that inf
⁡
(
span
⁡
{
ϕ
1
}
)
⊥
J
≥
M.
Step  2. If λ
<
λ
1 is sufficiently close to λ
1, we have t
-
<
0
<
t
+ such that J
(
t
±
ϕ
1
)
<
M. In fact, for λ
<
λ
1, we have
(64)
J
(
t
ϕ
1
)
=
λ
1
-
λ
2
(
θ
+
λ
1
)
∥
t
ϕ
1
∥
2
-
∫
Ω
‍
F
(
x
,
t
ϕ
1
)
d
x
.
For any fixed u
~
∈
R with |
u
~
|
=
1, from (
f
2
), we get lim
⁡
|
t
|
→
∞
F
(
x
,
t
u
~
)
=
+
∞ uniformly in x
∈
Ω. From Fatou’s lemma and ϕ
1
>
0 in Ω, it follows that
(65)
lim
⁡
t
→
+
∞
∫
Ω
‍
F
(
x
,
t
ϕ
1
)
d
x
≥
∫
Ω
‍
lim
⁡
t
→
+
∞
F
(
x
,
t
ϕ
1
)
d
x
=
+
∞
,
and hence, taking t
+
(
>
0
) large enough, we get
(66)
∫
Ω
‍
F
(
x
,
t
+
ϕ
1
)
d
x
>
-
M
+
1
.
For λ
1
-
2
(
θ
+
λ
1
)
/
(
t
+
)
2
<
λ
<
λ
1, combining (64) and (66) yields J
(
t
+
ϕ
1
)
<
M. A similar conclusion holds for some t
-
<
0.
Step  3. If λ
<
λ
1, let
(67)
Σ
±
=
{
z
∈
H
1
(
Ω
)
:
z
=
±
t
ϕ
1
+
w
with
t
>
0
,
m
m
m
m
m
m
x
c
m
w
∈
(
span
⁡
{
ϕ
1
}
)
⊥
}
.
The functional J satisfies the (
P.S.
)
c
,
Σ
+ and (
P.S.
)
c
,
Σ
- condition for all c
<
M.
In fact, let{
z
n
}
⊂
∑
+
‍ satisfy J
(
z
n
)
→
c
<
M and J
′
(
z
n
)
→
0 as n
→
∞. From Proposition 8, there is z
∈
H
1
(
Ω
) such that z
n
→
z strongly in H
1
(
Ω
). If z
∈
∂
Σ
+
=
(
span
⁡
{
ϕ
1
}
)
⊥, from the second conclusion of Step 1, we get J
(
z
n
)
→
c
≥
M, which is impossible. Hence, c
∈
Σ
+ and J satisfies the (
P.S.
)
c
,
Σ
+. Similarly we have that (
P.S.
)
c
,
Σ
- holds for all c
<
M.
Step  4. Three solutions are obtained.
Ifλ
<
λ
1 is sufficiently close to λ
1, from Steps 1 and 2, we get -
∞
<
inf
⁡
Σ
±
J
<
M, which implies that J is bounded below in Σ
+. Consequently, from Ekeland’s variational principle, there exists {
z
n
}
⊂
Σ
+ such that J
(
z
n
)
→
inf
⁡
Σ
+
J and J
′
(
z
n
)
→
0 as n
→
∞. Since J satisfies (
P.S.
)
c
,
Σ
+ for all c
<
M, there is z
+
∈
Σ
+ such that J
(
z
+
)
=
inf
⁡
Σ
+
J; that is, the infimum is attained in Σ
+. A similar conclusion holds in Σ
-. So J has two distinct critical points, denoted by z
+, z
-.
Suppose thatJ
(
z
+
)
≥
J
(
z
-
). Letting
(68)
ψ
(
z
)
=
J
(
z
+
z
+
)
-
J
(
z
+
)
,
e
=
z
-
-
z
+
,
we have ψ
(
0
)
=
0, ψ
(
e
)
≤
0 and since z
+ is the local minimum of J, there are r
,
ρ
>
0 such that ψ
(
z
)
≥
ρ for ∥
z
∥
=
r. Since ψ
′
=
J
′ and ψ satisfies the (
P.S.
) condition, from the mountain pass theorem, the number
(69)
c
=
inf
⁡
h
∈
Γ
max
⁡
t
∈
[
0,1
]
J
(
h
(
t
)
)
,
where Γ
=
{
h
∈
C
(
[
0,1
]
,
H
1
(
Ω
)
)
:
h
(
0
)
=
z
+
,
h
(
1
)
=
z
-
}, is a critical value of J. From the definition of c, we have c
≥
M and obtain a third critical point of J. Hence, the proof is completed.


## 4. Proof of Estimates

In this section we will prove all the estimates in Propositions9 and 10.


### 4.1. Estimates of the Saddle Geometry

Lemma 11.
Under hypotheses(
f
0
) and (
f
∞
), one gets the following:(i) for λ
∈
(
λ
k
-
1
,
λ
k
), there exists D
λ satisfying (43) and D
W
∈
R satisfying (45);
(ii) for λ
∈
(
λ
k
,
λ
k
+
m
) one has the following:
(a) there exists K
λ
∈
R satisfying (48),
(b) for a given R
2
>
0, there exists E
∈
R satisfying (53).Proof.
Letu
∈
W; using estimates (26) and (29) we get
(70)
J
(
u
)
≥
1
2
(
λ
k
+
m
-
λ
θ
+
λ
k
+
m
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Forλ
∈
(
λ
k
-
1
,
λ
k
), letting ε
<
(
λ
k
+
m
-
λ
k
)
/
S
2
(
θ
+
λ
k
+
m
)
<
(
λ
k
+
m
-
λ
)
/
S
2
(
θ
+
λ
k
+
m
), it follows that J is bounded below in W; that is, there exists a D
W as in (45).
Forλ
∈
(
λ
k
,
λ
k
+
m
), then the same estimate holds but the constant cannot be made independent of λ, giving (48).
In the same way, letu
∈
Z
⊕
W and set δ
=
λ
k
-
λ
>
0; we get
(71)
J
(
u
)
≥
1
2
(
λ
k
-
λ
θ
+
λ
k
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
≥
1
2
(
δ
θ
+
λ
k
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Letting ε
<
δ
/
S
2
(
θ
+
λ
k
), it follows that J is bounded below in Z
⊕
W; that is, there exists a D
λ such that for all u
∈
Z
⊕
W we have (43), where again the constant D
λ depends on δ, that is, on λ.
Finally, (71) with λ
∈
(
λ
k
,
λ
k
+
m
) implies
(72)
J
(
u
)
≥
1
2
(
λ
k
-
λ
k
+
m
θ
+
λ
k
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Then, no matter the value of λ, J is bounded from below in any bounded subset of Z
⊕
W, giving (53) for a suitable value of E.Lemma 12.
Under hypotheses(
f
0
) and (
f
∞
), one gets the following:(i) for λ
∈
(
λ
k
-
1
,
λ
k
), given the constant D
λ
∈
R, there exists ρ
λ
>
0 satisfying (44);
(ii) for λ
∈
(
λ
k
,
λ
k
+
m
) one has the following:
(a) there exists K
V
∈
R satisfying (50),
(b) for a given K
λ
∈
R, there exists β
λ
>
0 satisfying (49),
(c) for a given E
∈
R, there exists ξ
>
0 satisfying (54).Moreover, given the valuesR
1, R
2, one may always choose ρ
λ
>
R
1, β
λ
>
R
2 as claimed in Propositions 9 and 10.Proof.
Letu
∈
V, by estimates (26) and (28); we get
(73)
J
(
u
)
≤
1
2
(
λ
k
-
1
-
λ
θ
+
λ
k
-
1
+
ε
S
2
)
∥
u
∥
2
+
L
~
ε
∥
u
∥
.
Forλ
∈
(
λ
k
-
1
,
λ
k
), letting ε
<
(
λ
-
λ
k
-
1
)
/
S
2
(
θ
+
λ
k
-
1
), then one obtains (44) for suitably large ρ
λ
>
R
1.
Forλ
∈
(
λ
k
,
λ
k
+
m
), letting ε
<
(
λ
k
-
λ
k
-
1
)
/
S
2
(
θ
+
λ
k
-
1
), one obtains, for suitable K
V and ξ
>
0, (50) and (54).
Finally, lettingu
∈
V
⊕
Z and setting δ
=
λ
-
λ
k
>
0, we get
(74)
J
(
u
)
≤
1
2
(
λ
k
-
λ
θ
+
λ
k
+
ε
S
2
)
∥
u
∥
2
+
L
~
ε
∥
u
∥
≤
1
2
(
-
δ
θ
+
λ
k
+
ε
S
2
)
∥
u
∥
2
+
L
~
ε
∥
u
∥
.
Letting ε
<
δ
/
S
2
λ
k, it is clear that (once δ is fixed) this goes to -
∞ and then we may find the claimed β
λ
>
R
2 such that (49) holds.
Observe thatK
V and E can be chosen uniformly for λ
∈
(
λ
k
,
λ
k
+
m
), while ρ
λ, β
λ will in fact depend on λ.


### 4.2. Estimating the Effect of the Nontrivial Perturbation

In this section we will prove the remaining inequalities in Propositions9 and 10, those which rely on the hypothesis (
f
1
) or (
f
2
) or (
f
3
) or (
f
4
), which, roughly speaking, say that the perturbation F is nontrivial in such a way that a new solution arises when λ is sufficiently near to the eigenvalue λ
k. The proof is simpler for Theorem 5, since we need to estimate the functional in the compact set S
V
Z, while for Theorem 6 the same kind of estimate is required in the noncompact set S
Z
W.


#### 4.2.1. EstimatingJ in S
V
Z

For the next estimates, we will need the following lemma.Lemma 13.
Hypothesis(
f
2
) implies that there exists a nondecreasing function D
:
(
0
,
+
∞
)
→
R such that
(75)
lim
⁡
R
→
+
∞
D
(
R
)
=
+
∞
,
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
>
D
(
R
)
.Proof.
First we claim that there exists a constantη
>
0 such that the sets Ω
u
=
{
x
∈
Ω
:
|
u
(
x
)
|
>
η
} have measure |
Ω
u
|
>
η, for all u
∈
S
V
Z.
Actually,V
⊕
Z is a finite-dimensional subspace and the functions u
∈
S
V
Z are smooth; they are uniformly bounded; that is, there exists M
>
0 such that |
u
(
x
)
|
≤
M for all x
∈
Ω. Suppose that for η
n
→
0
(
η
n
<
1
) there exists {
u
n
}
⊂
S
V
Z such that |
Ω
u
n
|
≤
η
n.
On one hand, by (28), one obtains
(76)
1
θ
+
λ
k
≤
∫
Ω
‍
u
2
d
x
.
On the other hand,(77)
∫
Ω
‍
u
2
d
x
=
∫
Ω
u
n
|
u
n
|
2
d
x
+
∫
Ω
-
Ω
u
n
|
u
n
|
2
d
x
≤
M
2
|
Ω
u
n
|
+
η
n
2
|
Ω
-
Ω
u
n
|
≤
η
n
(
M
2
+
|
Ω
|
)
⟶
0
.
That is a contradiction.
Now, for anyH
>
0, we will show that we can find a R
~ large enough so that ∫
Ω
‍
F
(
x
,
R
u
)
d
x
≥
H for any u
∈
S
V
Z and R
≥
R
~; this means that
(78)
lim
⁡
R
→
∞
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
=
+
∞
.
Actually, it follows from(
f
2
) that for any M
>
0 there exists t
0
(
>
0
) such that F
(
x
,
t
)
>
M for |
t
|
>
t
0. For R
>
t
0
/
η, one has Ω
u
⊆
{
x
∈
Ω
:
|
R
u
(
x
)
|
>
t
0
}, and then one gets
(79)
∫
|
R
u
|
≥
t
0
‍
F
(
x
,
R
u
)
d
x
≥
M
η
.
ForR
≤
t
0
/
η, by (
f
0
) and (
f
2
), there exist C
~
>
0 and L
C
~
∈
L
2
(
Ω
) such that F
(
x
,
t
)
≥
-
C
~
(
1
+
L
C
~
(
x
)
), for t
∈
R and a.e. x
∈
Ω.
LetM
=
(
H
+
C
~
|
Ω
|
+
C
~
|
Ω
|
1
/
2
∥
L
C
~
∥
2
)
η
-
1; one finally obtains
(80)
∫
Ω
‍
F
(
x
,
R
u
)
d
x
=
∫
|
R
u
|
≥
t
0
‍
F
(
x
,
R
u
)
d
x
+
∫
|
R
u
|
≤
t
0
‍
F
(
x
,
R
u
)
d
x
≥
M
η
-
C
~
|
Ω
|
-
C
~
|
Ω
|
1
/
2
∥
L
C
~
∥
2
=
H
.
It is elementary that
(81)
D
(
R
)
=
inf
⁡
ρ
≥
R
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
is well defined and satisfies the claim.Now we may prove the following.Lemma 14.
Consider Theorem5 with one of the sets of hypotheses (
f
1
) and (
f
2
). Given the constant D
W
∈
R, there exist R
1, δ
0
>
0 such that, for any λ
∈
(
λ
k
-
δ
0
,
λ
k
), (46) and (47) hold.Proof.
We consider the two sets of hypotheses separately.
(i) In case(
f
1
), it follows from (
f
0
) and (
f
1
) that for any M
>
0 there exist C
M and L
M
∈
L
2
(
Ω
) such that
(82)
F
(
x
,
t
)
≥
M
|
t
|
-
C
M
(
1
+
L
M
(
x
)
)
,
for t
∈
R and x
∈
Ω; in particular we set M
=
1. Let u
∈
R
S
V
Z; for being in a finite-dimensional subspace, all the norms are equivalent, so that (set δ
=
λ
k
-
λ
>
0 and use estimates (28) and (82))
(83)
J
(
u
)
≤
λ
k
-
λ
2
(
θ
+
λ
k
)
∥
u
∥
2
-
∥
u
∥
+
C
0
≤
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
-
∥
u
∥
+
C
0
≤
δ
2
(
θ
+
λ
k
)
R
2
-
R
+
C
0
,
where C
0
=
C
1
|
Ω
|
1
/
2
∥
L
1
∥
2
+
C
1
|
Ω
|.
(ii) In case(
f
2
), let D
(
R
) be as in Lemma 13, for ∥
u
∥
=
R; let u
=
w
+
ϕ with w
∈
V and ϕ
∈
Z
=
E
k:
(84)
J
(
u
)
≤
λ
k
-
1
-
λ
2
(
θ
+
λ
k
-
1
)
∥
w
∥
2
+
λ
k
-
λ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
≤
λ
k
-
1
-
λ
k
+
δ
2
(
θ
+
λ
k
-
1
)
∥
w
∥
2
+
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
≤
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
λ
k
-
λ
k
-
1
4
(
θ
+
λ
k
-
1
)
∥
w
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
.
We assume that δ
≤
(
λ
k
-
λ
k
-
1
)
/
2; it is easy to see that (
λ
k
-
1
-
λ
k
)
∥
w
∥
2
/
4
(
θ
+
λ
k
-
1
)
≤
C for some constant C; we estimate
(85)
J
(
z
)
≤
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
C
-
D
(
R
)
≤
δ
2
(
θ
+
λ
k
)
R
2
-
D
(
R
)
+
C
.
Considering (83) and (85), we see that since lim
⁡
R
→
∞
⁡
D
(
R
)
=
+
∞ by Lemma 13, we may fix R
1 so that C
-
D
(
R
1
)
<
D
W
-
1 (or C
0
-
R
1
<
D
W
-
1 for the case (
f
1
)) and then for 0
<
δ
<
min
⁡
{
2
(
θ
+
λ
k
)
/
R
1
2
,
(
λ
k
-
λ
k
-
1
)
/
2
} one gets (46).
To obtain (47), we observe that (since λ
>
λ
k
-
1) if ϕ
=
0, that is, if u
∈
V, then in estimates (83) and (85) we may avoid the term δ
R
2
/
2
(
θ
+
λ
k
) so that (remember that D
(
R
) is nondecreasing) J
(
u
)
<
D
W
-
1 for ∥
u
∥
>
R
1.


#### 4.2.2. EstimatingJ in S
Z
W

We consider the corresponding estimates of the previous lemma, for Theorem6.Lemma 15.
Consider Theorem6 with one of the sets of hypotheses (
f
3
) and (
f
4
). Given the constant K
V
∈
R, there exist R
2, δ
1
>
0 such that, for any λ
∈
(
λ
k
,
λ
k
+
δ
1
), (51) and (52) hold.Proof.
Lettingλ
=
λ
k
+
δ, we see from (70) that property (52) will be satisfied provided that R
2 is large enough (say R
2
>
R
~); observe that this value can be made independent from λ once δ is small enough.
Now we consider the two sets of hypotheses separately.
(i) In case(
f
3
), suppose u
∈
E
k
⊕
W; we can assume that u
=
w
+
ϕ, with w
∈
W and ϕ
∈
E
k. Since E
k is a finite dimension subspace, all the norms are equivalent, so that there exists K
>
0 such that for all ϕ
∈
E
k we have ∥
ϕ
∥
≤
K
∥
ϕ
∥
1. In addition, by (
f
0
) and (
f
3
), there exist C
2 and L
2
∈
L
2
(
Ω
) such that
(86)
-
F
(
x
,
s
)
≥
K
|
s
|
-
C
2
(
1
+
L
2
(
x
)
)
,
uniformly in x
∈
Ω. So by (29) and (86),
(87)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
K
∥
w
+
ϕ
∥
1
-
C
4
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
K
∥
ϕ
∥
1
-
K
∥
-
w
∥
1
-
C
4
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∥
ϕ
∥
-
C
3
∥
w
∥
-
C
4
,
where C
3
=
|
Ω
|
1
/
2
S
K, C
4
=
C
2
|
Ω
|
1
/
2
∥
L
2
∥
2
+
C
2
|
Ω
|. Since
(88)
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
≤
∥
w
∥
+
∥
ϕ
∥
-
δ
2
(
θ
+
λ
k
)
(
∥
ϕ
∥
2
+
∥
w
∥
2
)
≤
(
1
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
)
∥
ϕ
∥
+
∥
w
∥
,
supposing δ
≤
(
λ
k
+
m
-
λ
k
)
/
2, (87) becomes
(89)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
C
3
∥
w
∥
-
C
4
-
∥
w
∥
+
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
(
C
3
+
1
)
∥
w
∥
-
C
4
+
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
,
since (
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
>
0, so
(90)
h
(
w
)
=
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
(
C
3
+
1
)
∥
w
∥
-
C
4
is bounded below for all w
∈
W; that is, there exists C
5
∈
R such that h
(
w
)
≥
C
5; by (89) one gets
(91)
J
(
u
)
≥
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
+
C
5
=
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
+
∥
u
∥
+
C
5
.
(ii) In case(
f
4
), first we give some conclusions which are similar to Lemma 3 of [18]. Under the property of F, there exists a const C, and G
∈
C
(
R
,
R
) which is subadditive, that is,
(92)
G
(
s
+
t
)
≤
G
(
s
)
+
G
(
t
)
for all s
,
t
∈
R, and coercive, that is,
(93)
G
(
s
)
⟶
+
∞
as |
s
|
→
∞, and satisfies that
(94)
G
(
s
)
≤
|
s
|
+
4
for all s
∈
R, such that
(95)
-
F
(
x
,
s
)
≥
G
(
s
)
-
C
for all s
∈
R and x
∈
Ω
¯.
In fact, since-
F
(
x
,
s
)
→
+
∞ as |
s
|
→
∞ uniformly for all x
∈
Ω
¯, there exists a sequence of positive integers n
k with n
k
+
1
>
2
n
k for all positive integers k such that
(96)
-
F
(
x
,
s
)
≥
k
for all |
s
|
≥
n
k and all x
∈
Ω
¯. Let n
0
=
0 and define
(97)
G
(
s
)
=
k
+
2
+
|
s
|
-
n
k
-
1
n
k
-
n
k
-
1
for n
k
-
1
≤
|
s
|
<
n
k, where k
∈
N.
By the definition ofG we have
(98)
k
+
2
≤
G
(
s
)
≤
k
+
3
for all n
k
-
1
≤
|
s
|
<
n
k. By (
f
4
) and F
∈
C
1
(
Ω
¯
×
R
,
R
), there exists C
F
>
0 such that
(99)
-
F
(
x
,
s
)
≥
-
C
F
,
∀
(
x
,
s
)
∈
(
Ω
,
R
2
)
.
It follows that
(100)
-
F
(
x
,
s
)
≥
G
(
s
)
-
C
,
where C
=
C
F
+
4. In fact, when n
k
-
1
≤
|
s
|
<
n
k, for some k
≥
2, one has, by (96) and (98),
(101)
-
F
(
x
,
s
)
≥
k
-
1
≥
G
(
s
)
-
4
≥
G
(
s
)
-
C
for all x
∈
Ω
¯. When |
s
|
<
n
1, we have, by (98) and (99),
(102)
-
F
(
x
,
s
)
≥
-
C
F
=
4
-
C
≥
G
(
s
)
-
C
for all x
∈
Ω
¯.
It is obvious thatG is continuous and coercive. Moreover, one has
(103)
G
(
s
)
≤
|
s
|
+
4
for all s
∈
R. In fact, for every s
∈
R, there exists k
∈
N such that
(104)
n
k
-
1
≤
|
s
|
<
n
k
,
which implies that
(105)
G
(
s
)
≤
(
k
-
1
)
+
4
≤
n
k
-
1
+
4
≤
|
s
|
+
4
for all s
∈
R by (98) and the fact that n
k
≥
k for all integers k
≥
0.
Now we only need to prove the subadditivity ofG. Let
(106)
n
k
-
1
≤
|
s
|
<
n
k
,
n
j
-
1
≤
|
t
|
<
n
j
and m
=
max
⁡
{
k
,
j
}. Then we have
(107)
|
s
+
t
|
≤
|
s
|
+
|
t
|
<
n
k
+
n
j
≤
2
n
m
<
n
m
+
1
.
Hence we obtain, by (98),
(108)
G
(
s
+
t
)
≤
m
+
4
≤
k
+
2
+
j
+
2
≤
G
(
s
)
+
G
(
t
)
,
which shows that G is subadditive.
Foru
∈
E
k
⊕
W, we assume that u
=
w
+
ϕ, with w
∈
W and ϕ
∈
E
k; letting 0
<
δ
<
(
λ
k
+
m
-
λ
k
)
/
2, by (28), (92), (94), and (95), one gets
(109)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
+
w
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
∫
Ω
‍
G
(
-
w
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
∫
Ω
‍
(
|
w
|
+
4
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
z
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
S
|
Ω
|
∥
w
∥
-
C
1
=
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
,
where g
(
w
)
=
(
(
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
)
∥
w
∥
2
-
S
|
Ω
|
∥
w
∥
-
C
1, C
1
=
(
4
+
C
)
|
Ω
|. Since ϕ
∈
E
k, E
k is a finite-dimensional subspace, and G is coercive, from the proof of (75), one can get
(110)
lim
⁡
∥
ϕ
∥
→
∞
∫
Ω
‍
G
(
ϕ
)
d
x
=
+
∞
.
That is, ∫
Ω
‍
G
(
ϕ
)
d
x is coercive on E
k. Since (
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
>
0, so g
(
w
) is coercive on W, and ∫
Ω
‍
G
(
ϕ
)
d
x and g
(
w
) are bounded below, so it is obvious that
(111)
lim
⁡
∥
u
∥
→
∞
(
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
)
=
+
∞
for all u
∈
Z
⊕
W.
Considering (91), (109), and (111), we can choose R
2 large enough such that for all ∥
u
∥
≥
R
2 one gets
(112)
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
>
K
V
+
1
(or R
2
+
C
5
>
K
V
+
1 for the case (
f
3
)) and property (52) holds; then for 0
<
δ
<
min
⁡
{
2
(
θ
+
λ
k
)
/
R
2
2
,
(
λ
k
+
m
-
λ
k
)
/
2
}
=
δ
1 and u
∈
R
2
S
Z
W one gets J
(
u
)
>
K
V; that is, the property (51) holds.


## 4.1. Estimates of the Saddle Geometry

Lemma 11.
Under hypotheses(
f
0
) and (
f
∞
), one gets the following:(i) for λ
∈
(
λ
k
-
1
,
λ
k
), there exists D
λ satisfying (43) and D
W
∈
R satisfying (45);
(ii) for λ
∈
(
λ
k
,
λ
k
+
m
) one has the following:
(a) there exists K
λ
∈
R satisfying (48),
(b) for a given R
2
>
0, there exists E
∈
R satisfying (53).Proof.
Letu
∈
W; using estimates (26) and (29) we get
(70)
J
(
u
)
≥
1
2
(
λ
k
+
m
-
λ
θ
+
λ
k
+
m
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Forλ
∈
(
λ
k
-
1
,
λ
k
), letting ε
<
(
λ
k
+
m
-
λ
k
)
/
S
2
(
θ
+
λ
k
+
m
)
<
(
λ
k
+
m
-
λ
)
/
S
2
(
θ
+
λ
k
+
m
), it follows that J is bounded below in W; that is, there exists a D
W as in (45).
Forλ
∈
(
λ
k
,
λ
k
+
m
), then the same estimate holds but the constant cannot be made independent of λ, giving (48).
In the same way, letu
∈
Z
⊕
W and set δ
=
λ
k
-
λ
>
0; we get
(71)
J
(
u
)
≥
1
2
(
λ
k
-
λ
θ
+
λ
k
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
≥
1
2
(
δ
θ
+
λ
k
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Letting ε
<
δ
/
S
2
(
θ
+
λ
k
), it follows that J is bounded below in Z
⊕
W; that is, there exists a D
λ such that for all u
∈
Z
⊕
W we have (43), where again the constant D
λ depends on δ, that is, on λ.
Finally, (71) with λ
∈
(
λ
k
,
λ
k
+
m
) implies
(72)
J
(
u
)
≥
1
2
(
λ
k
-
λ
k
+
m
θ
+
λ
k
-
ε
S
2
)
∥
u
∥
2
-
L
~
ε
∥
u
∥
.
Then, no matter the value of λ, J is bounded from below in any bounded subset of Z
⊕
W, giving (53) for a suitable value of E.Lemma 12.
Under hypotheses(
f
0
) and (
f
∞
), one gets the following:(i) for λ
∈
(
λ
k
-
1
,
λ
k
), given the constant D
λ
∈
R, there exists ρ
λ
>
0 satisfying (44);
(ii) for λ
∈
(
λ
k
,
λ
k
+
m
) one has the following:
(a) there exists K
V
∈
R satisfying (50),
(b) for a given K
λ
∈
R, there exists β
λ
>
0 satisfying (49),
(c) for a given E
∈
R, there exists ξ
>
0 satisfying (54).Moreover, given the valuesR
1, R
2, one may always choose ρ
λ
>
R
1, β
λ
>
R
2 as claimed in Propositions 9 and 10.Proof.
Letu
∈
V, by estimates (26) and (28); we get
(73)
J
(
u
)
≤
1
2
(
λ
k
-
1
-
λ
θ
+
λ
k
-
1
+
ε
S
2
)
∥
u
∥
2
+
L
~
ε
∥
u
∥
.
Forλ
∈
(
λ
k
-
1
,
λ
k
), letting ε
<
(
λ
-
λ
k
-
1
)
/
S
2
(
θ
+
λ
k
-
1
), then one obtains (44) for suitably large ρ
λ
>
R
1.
Forλ
∈
(
λ
k
,
λ
k
+
m
), letting ε
<
(
λ
k
-
λ
k
-
1
)
/
S
2
(
θ
+
λ
k
-
1
), one obtains, for suitable K
V and ξ
>
0, (50) and (54).
Finally, lettingu
∈
V
⊕
Z and setting δ
=
λ
-
λ
k
>
0, we get
(74)
J
(
u
)
≤
1
2
(
λ
k
-
λ
θ
+
λ
k
+
ε
S
2
)
∥
u
∥
2
+
L
~
ε
∥
u
∥
≤
1
2
(
-
δ
θ
+
λ
k
+
ε
S
2
)
∥
u
∥
2
+
L
~
ε
∥
u
∥
.
Letting ε
<
δ
/
S
2
λ
k, it is clear that (once δ is fixed) this goes to -
∞ and then we may find the claimed β
λ
>
R
2 such that (49) holds.
Observe thatK
V and E can be chosen uniformly for λ
∈
(
λ
k
,
λ
k
+
m
), while ρ
λ, β
λ will in fact depend on λ.


## 4.2. Estimating the Effect of the Nontrivial Perturbation

In this section we will prove the remaining inequalities in Propositions9 and 10, those which rely on the hypothesis (
f
1
) or (
f
2
) or (
f
3
) or (
f
4
), which, roughly speaking, say that the perturbation F is nontrivial in such a way that a new solution arises when λ is sufficiently near to the eigenvalue λ
k. The proof is simpler for Theorem 5, since we need to estimate the functional in the compact set S
V
Z, while for Theorem 6 the same kind of estimate is required in the noncompact set S
Z
W.


### 4.2.1. EstimatingJ in S
V
Z

For the next estimates, we will need the following lemma.Lemma 13.
Hypothesis(
f
2
) implies that there exists a nondecreasing function D
:
(
0
,
+
∞
)
→
R such that
(75)
lim
⁡
R
→
+
∞
D
(
R
)
=
+
∞
,
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
>
D
(
R
)
.Proof.
First we claim that there exists a constantη
>
0 such that the sets Ω
u
=
{
x
∈
Ω
:
|
u
(
x
)
|
>
η
} have measure |
Ω
u
|
>
η, for all u
∈
S
V
Z.
Actually,V
⊕
Z is a finite-dimensional subspace and the functions u
∈
S
V
Z are smooth; they are uniformly bounded; that is, there exists M
>
0 such that |
u
(
x
)
|
≤
M for all x
∈
Ω. Suppose that for η
n
→
0
(
η
n
<
1
) there exists {
u
n
}
⊂
S
V
Z such that |
Ω
u
n
|
≤
η
n.
On one hand, by (28), one obtains
(76)
1
θ
+
λ
k
≤
∫
Ω
‍
u
2
d
x
.
On the other hand,(77)
∫
Ω
‍
u
2
d
x
=
∫
Ω
u
n
|
u
n
|
2
d
x
+
∫
Ω
-
Ω
u
n
|
u
n
|
2
d
x
≤
M
2
|
Ω
u
n
|
+
η
n
2
|
Ω
-
Ω
u
n
|
≤
η
n
(
M
2
+
|
Ω
|
)
⟶
0
.
That is a contradiction.
Now, for anyH
>
0, we will show that we can find a R
~ large enough so that ∫
Ω
‍
F
(
x
,
R
u
)
d
x
≥
H for any u
∈
S
V
Z and R
≥
R
~; this means that
(78)
lim
⁡
R
→
∞
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
=
+
∞
.
Actually, it follows from(
f
2
) that for any M
>
0 there exists t
0
(
>
0
) such that F
(
x
,
t
)
>
M for |
t
|
>
t
0. For R
>
t
0
/
η, one has Ω
u
⊆
{
x
∈
Ω
:
|
R
u
(
x
)
|
>
t
0
}, and then one gets
(79)
∫
|
R
u
|
≥
t
0
‍
F
(
x
,
R
u
)
d
x
≥
M
η
.
ForR
≤
t
0
/
η, by (
f
0
) and (
f
2
), there exist C
~
>
0 and L
C
~
∈
L
2
(
Ω
) such that F
(
x
,
t
)
≥
-
C
~
(
1
+
L
C
~
(
x
)
), for t
∈
R and a.e. x
∈
Ω.
LetM
=
(
H
+
C
~
|
Ω
|
+
C
~
|
Ω
|
1
/
2
∥
L
C
~
∥
2
)
η
-
1; one finally obtains
(80)
∫
Ω
‍
F
(
x
,
R
u
)
d
x
=
∫
|
R
u
|
≥
t
0
‍
F
(
x
,
R
u
)
d
x
+
∫
|
R
u
|
≤
t
0
‍
F
(
x
,
R
u
)
d
x
≥
M
η
-
C
~
|
Ω
|
-
C
~
|
Ω
|
1
/
2
∥
L
C
~
∥
2
=
H
.
It is elementary that
(81)
D
(
R
)
=
inf
⁡
ρ
≥
R
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
is well defined and satisfies the claim.Now we may prove the following.Lemma 14.
Consider Theorem5 with one of the sets of hypotheses (
f
1
) and (
f
2
). Given the constant D
W
∈
R, there exist R
1, δ
0
>
0 such that, for any λ
∈
(
λ
k
-
δ
0
,
λ
k
), (46) and (47) hold.Proof.
We consider the two sets of hypotheses separately.
(i) In case(
f
1
), it follows from (
f
0
) and (
f
1
) that for any M
>
0 there exist C
M and L
M
∈
L
2
(
Ω
) such that
(82)
F
(
x
,
t
)
≥
M
|
t
|
-
C
M
(
1
+
L
M
(
x
)
)
,
for t
∈
R and x
∈
Ω; in particular we set M
=
1. Let u
∈
R
S
V
Z; for being in a finite-dimensional subspace, all the norms are equivalent, so that (set δ
=
λ
k
-
λ
>
0 and use estimates (28) and (82))
(83)
J
(
u
)
≤
λ
k
-
λ
2
(
θ
+
λ
k
)
∥
u
∥
2
-
∥
u
∥
+
C
0
≤
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
-
∥
u
∥
+
C
0
≤
δ
2
(
θ
+
λ
k
)
R
2
-
R
+
C
0
,
where C
0
=
C
1
|
Ω
|
1
/
2
∥
L
1
∥
2
+
C
1
|
Ω
|.
(ii) In case(
f
2
), let D
(
R
) be as in Lemma 13, for ∥
u
∥
=
R; let u
=
w
+
ϕ with w
∈
V and ϕ
∈
Z
=
E
k:
(84)
J
(
u
)
≤
λ
k
-
1
-
λ
2
(
θ
+
λ
k
-
1
)
∥
w
∥
2
+
λ
k
-
λ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
≤
λ
k
-
1
-
λ
k
+
δ
2
(
θ
+
λ
k
-
1
)
∥
w
∥
2
+
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
≤
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
λ
k
-
λ
k
-
1
4
(
θ
+
λ
k
-
1
)
∥
w
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
.
We assume that δ
≤
(
λ
k
-
λ
k
-
1
)
/
2; it is easy to see that (
λ
k
-
1
-
λ
k
)
∥
w
∥
2
/
4
(
θ
+
λ
k
-
1
)
≤
C for some constant C; we estimate
(85)
J
(
z
)
≤
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
C
-
D
(
R
)
≤
δ
2
(
θ
+
λ
k
)
R
2
-
D
(
R
)
+
C
.
Considering (83) and (85), we see that since lim
⁡
R
→
∞
⁡
D
(
R
)
=
+
∞ by Lemma 13, we may fix R
1 so that C
-
D
(
R
1
)
<
D
W
-
1 (or C
0
-
R
1
<
D
W
-
1 for the case (
f
1
)) and then for 0
<
δ
<
min
⁡
{
2
(
θ
+
λ
k
)
/
R
1
2
,
(
λ
k
-
λ
k
-
1
)
/
2
} one gets (46).
To obtain (47), we observe that (since λ
>
λ
k
-
1) if ϕ
=
0, that is, if u
∈
V, then in estimates (83) and (85) we may avoid the term δ
R
2
/
2
(
θ
+
λ
k
) so that (remember that D
(
R
) is nondecreasing) J
(
u
)
<
D
W
-
1 for ∥
u
∥
>
R
1.


### 4.2.2. EstimatingJ in S
Z
W

We consider the corresponding estimates of the previous lemma, for Theorem6.Lemma 15.
Consider Theorem6 with one of the sets of hypotheses (
f
3
) and (
f
4
). Given the constant K
V
∈
R, there exist R
2, δ
1
>
0 such that, for any λ
∈
(
λ
k
,
λ
k
+
δ
1
), (51) and (52) hold.Proof.
Lettingλ
=
λ
k
+
δ, we see from (70) that property (52) will be satisfied provided that R
2 is large enough (say R
2
>
R
~); observe that this value can be made independent from λ once δ is small enough.
Now we consider the two sets of hypotheses separately.
(i) In case(
f
3
), suppose u
∈
E
k
⊕
W; we can assume that u
=
w
+
ϕ, with w
∈
W and ϕ
∈
E
k. Since E
k is a finite dimension subspace, all the norms are equivalent, so that there exists K
>
0 such that for all ϕ
∈
E
k we have ∥
ϕ
∥
≤
K
∥
ϕ
∥
1. In addition, by (
f
0
) and (
f
3
), there exist C
2 and L
2
∈
L
2
(
Ω
) such that
(86)
-
F
(
x
,
s
)
≥
K
|
s
|
-
C
2
(
1
+
L
2
(
x
)
)
,
uniformly in x
∈
Ω. So by (29) and (86),
(87)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
K
∥
w
+
ϕ
∥
1
-
C
4
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
K
∥
ϕ
∥
1
-
K
∥
-
w
∥
1
-
C
4
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∥
ϕ
∥
-
C
3
∥
w
∥
-
C
4
,
where C
3
=
|
Ω
|
1
/
2
S
K, C
4
=
C
2
|
Ω
|
1
/
2
∥
L
2
∥
2
+
C
2
|
Ω
|. Since
(88)
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
≤
∥
w
∥
+
∥
ϕ
∥
-
δ
2
(
θ
+
λ
k
)
(
∥
ϕ
∥
2
+
∥
w
∥
2
)
≤
(
1
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
)
∥
ϕ
∥
+
∥
w
∥
,
supposing δ
≤
(
λ
k
+
m
-
λ
k
)
/
2, (87) becomes
(89)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
C
3
∥
w
∥
-
C
4
-
∥
w
∥
+
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
(
C
3
+
1
)
∥
w
∥
-
C
4
+
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
,
since (
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
>
0, so
(90)
h
(
w
)
=
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
(
C
3
+
1
)
∥
w
∥
-
C
4
is bounded below for all w
∈
W; that is, there exists C
5
∈
R such that h
(
w
)
≥
C
5; by (89) one gets
(91)
J
(
u
)
≥
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
+
C
5
=
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
+
∥
u
∥
+
C
5
.
(ii) In case(
f
4
), first we give some conclusions which are similar to Lemma 3 of [18]. Under the property of F, there exists a const C, and G
∈
C
(
R
,
R
) which is subadditive, that is,
(92)
G
(
s
+
t
)
≤
G
(
s
)
+
G
(
t
)
for all s
,
t
∈
R, and coercive, that is,
(93)
G
(
s
)
⟶
+
∞
as |
s
|
→
∞, and satisfies that
(94)
G
(
s
)
≤
|
s
|
+
4
for all s
∈
R, such that
(95)
-
F
(
x
,
s
)
≥
G
(
s
)
-
C
for all s
∈
R and x
∈
Ω
¯.
In fact, since-
F
(
x
,
s
)
→
+
∞ as |
s
|
→
∞ uniformly for all x
∈
Ω
¯, there exists a sequence of positive integers n
k with n
k
+
1
>
2
n
k for all positive integers k such that
(96)
-
F
(
x
,
s
)
≥
k
for all |
s
|
≥
n
k and all x
∈
Ω
¯. Let n
0
=
0 and define
(97)
G
(
s
)
=
k
+
2
+
|
s
|
-
n
k
-
1
n
k
-
n
k
-
1
for n
k
-
1
≤
|
s
|
<
n
k, where k
∈
N.
By the definition ofG we have
(98)
k
+
2
≤
G
(
s
)
≤
k
+
3
for all n
k
-
1
≤
|
s
|
<
n
k. By (
f
4
) and F
∈
C
1
(
Ω
¯
×
R
,
R
), there exists C
F
>
0 such that
(99)
-
F
(
x
,
s
)
≥
-
C
F
,
∀
(
x
,
s
)
∈
(
Ω
,
R
2
)
.
It follows that
(100)
-
F
(
x
,
s
)
≥
G
(
s
)
-
C
,
where C
=
C
F
+
4. In fact, when n
k
-
1
≤
|
s
|
<
n
k, for some k
≥
2, one has, by (96) and (98),
(101)
-
F
(
x
,
s
)
≥
k
-
1
≥
G
(
s
)
-
4
≥
G
(
s
)
-
C
for all x
∈
Ω
¯. When |
s
|
<
n
1, we have, by (98) and (99),
(102)
-
F
(
x
,
s
)
≥
-
C
F
=
4
-
C
≥
G
(
s
)
-
C
for all x
∈
Ω
¯.
It is obvious thatG is continuous and coercive. Moreover, one has
(103)
G
(
s
)
≤
|
s
|
+
4
for all s
∈
R. In fact, for every s
∈
R, there exists k
∈
N such that
(104)
n
k
-
1
≤
|
s
|
<
n
k
,
which implies that
(105)
G
(
s
)
≤
(
k
-
1
)
+
4
≤
n
k
-
1
+
4
≤
|
s
|
+
4
for all s
∈
R by (98) and the fact that n
k
≥
k for all integers k
≥
0.
Now we only need to prove the subadditivity ofG. Let
(106)
n
k
-
1
≤
|
s
|
<
n
k
,
n
j
-
1
≤
|
t
|
<
n
j
and m
=
max
⁡
{
k
,
j
}. Then we have
(107)
|
s
+
t
|
≤
|
s
|
+
|
t
|
<
n
k
+
n
j
≤
2
n
m
<
n
m
+
1
.
Hence we obtain, by (98),
(108)
G
(
s
+
t
)
≤
m
+
4
≤
k
+
2
+
j
+
2
≤
G
(
s
)
+
G
(
t
)
,
which shows that G is subadditive.
Foru
∈
E
k
⊕
W, we assume that u
=
w
+
ϕ, with w
∈
W and ϕ
∈
E
k; letting 0
<
δ
<
(
λ
k
+
m
-
λ
k
)
/
2, by (28), (92), (94), and (95), one gets
(109)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
+
w
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
∫
Ω
‍
G
(
-
w
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
∫
Ω
‍
(
|
w
|
+
4
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
z
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
S
|
Ω
|
∥
w
∥
-
C
1
=
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
,
where g
(
w
)
=
(
(
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
)
∥
w
∥
2
-
S
|
Ω
|
∥
w
∥
-
C
1, C
1
=
(
4
+
C
)
|
Ω
|. Since ϕ
∈
E
k, E
k is a finite-dimensional subspace, and G is coercive, from the proof of (75), one can get
(110)
lim
⁡
∥
ϕ
∥
→
∞
∫
Ω
‍
G
(
ϕ
)
d
x
=
+
∞
.
That is, ∫
Ω
‍
G
(
ϕ
)
d
x is coercive on E
k. Since (
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
>
0, so g
(
w
) is coercive on W, and ∫
Ω
‍
G
(
ϕ
)
d
x and g
(
w
) are bounded below, so it is obvious that
(111)
lim
⁡
∥
u
∥
→
∞
(
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
)
=
+
∞
for all u
∈
Z
⊕
W.
Considering (91), (109), and (111), we can choose R
2 large enough such that for all ∥
u
∥
≥
R
2 one gets
(112)
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
>
K
V
+
1
(or R
2
+
C
5
>
K
V
+
1 for the case (
f
3
)) and property (52) holds; then for 0
<
δ
<
min
⁡
{
2
(
θ
+
λ
k
)
/
R
2
2
,
(
λ
k
+
m
-
λ
k
)
/
2
}
=
δ
1 and u
∈
R
2
S
Z
W one gets J
(
u
)
>
K
V; that is, the property (51) holds.


## 4.2.1. EstimatingJ in S
V
Z

For the next estimates, we will need the following lemma.Lemma 13.
Hypothesis(
f
2
) implies that there exists a nondecreasing function D
:
(
0
,
+
∞
)
→
R such that
(75)
lim
⁡
R
→
+
∞
D
(
R
)
=
+
∞
,
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
>
D
(
R
)
.Proof.
First we claim that there exists a constantη
>
0 such that the sets Ω
u
=
{
x
∈
Ω
:
|
u
(
x
)
|
>
η
} have measure |
Ω
u
|
>
η, for all u
∈
S
V
Z.
Actually,V
⊕
Z is a finite-dimensional subspace and the functions u
∈
S
V
Z are smooth; they are uniformly bounded; that is, there exists M
>
0 such that |
u
(
x
)
|
≤
M for all x
∈
Ω. Suppose that for η
n
→
0
(
η
n
<
1
) there exists {
u
n
}
⊂
S
V
Z such that |
Ω
u
n
|
≤
η
n.
On one hand, by (28), one obtains
(76)
1
θ
+
λ
k
≤
∫
Ω
‍
u
2
d
x
.
On the other hand,(77)
∫
Ω
‍
u
2
d
x
=
∫
Ω
u
n
|
u
n
|
2
d
x
+
∫
Ω
-
Ω
u
n
|
u
n
|
2
d
x
≤
M
2
|
Ω
u
n
|
+
η
n
2
|
Ω
-
Ω
u
n
|
≤
η
n
(
M
2
+
|
Ω
|
)
⟶
0
.
That is a contradiction.
Now, for anyH
>
0, we will show that we can find a R
~ large enough so that ∫
Ω
‍
F
(
x
,
R
u
)
d
x
≥
H for any u
∈
S
V
Z and R
≥
R
~; this means that
(78)
lim
⁡
R
→
∞
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
=
+
∞
.
Actually, it follows from(
f
2
) that for any M
>
0 there exists t
0
(
>
0
) such that F
(
x
,
t
)
>
M for |
t
|
>
t
0. For R
>
t
0
/
η, one has Ω
u
⊆
{
x
∈
Ω
:
|
R
u
(
x
)
|
>
t
0
}, and then one gets
(79)
∫
|
R
u
|
≥
t
0
‍
F
(
x
,
R
u
)
d
x
≥
M
η
.
ForR
≤
t
0
/
η, by (
f
0
) and (
f
2
), there exist C
~
>
0 and L
C
~
∈
L
2
(
Ω
) such that F
(
x
,
t
)
≥
-
C
~
(
1
+
L
C
~
(
x
)
), for t
∈
R and a.e. x
∈
Ω.
LetM
=
(
H
+
C
~
|
Ω
|
+
C
~
|
Ω
|
1
/
2
∥
L
C
~
∥
2
)
η
-
1; one finally obtains
(80)
∫
Ω
‍
F
(
x
,
R
u
)
d
x
=
∫
|
R
u
|
≥
t
0
‍
F
(
x
,
R
u
)
d
x
+
∫
|
R
u
|
≤
t
0
‍
F
(
x
,
R
u
)
d
x
≥
M
η
-
C
~
|
Ω
|
-
C
~
|
Ω
|
1
/
2
∥
L
C
~
∥
2
=
H
.
It is elementary that
(81)
D
(
R
)
=
inf
⁡
ρ
≥
R
inf
⁡
u
∈
R
S
V
Z
∫
Ω
‍
F
(
x
,
u
)
d
x
is well defined and satisfies the claim.Now we may prove the following.Lemma 14.
Consider Theorem5 with one of the sets of hypotheses (
f
1
) and (
f
2
). Given the constant D
W
∈
R, there exist R
1, δ
0
>
0 such that, for any λ
∈
(
λ
k
-
δ
0
,
λ
k
), (46) and (47) hold.Proof.
We consider the two sets of hypotheses separately.
(i) In case(
f
1
), it follows from (
f
0
) and (
f
1
) that for any M
>
0 there exist C
M and L
M
∈
L
2
(
Ω
) such that
(82)
F
(
x
,
t
)
≥
M
|
t
|
-
C
M
(
1
+
L
M
(
x
)
)
,
for t
∈
R and x
∈
Ω; in particular we set M
=
1. Let u
∈
R
S
V
Z; for being in a finite-dimensional subspace, all the norms are equivalent, so that (set δ
=
λ
k
-
λ
>
0 and use estimates (28) and (82))
(83)
J
(
u
)
≤
λ
k
-
λ
2
(
θ
+
λ
k
)
∥
u
∥
2
-
∥
u
∥
+
C
0
≤
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
-
∥
u
∥
+
C
0
≤
δ
2
(
θ
+
λ
k
)
R
2
-
R
+
C
0
,
where C
0
=
C
1
|
Ω
|
1
/
2
∥
L
1
∥
2
+
C
1
|
Ω
|.
(ii) In case(
f
2
), let D
(
R
) be as in Lemma 13, for ∥
u
∥
=
R; let u
=
w
+
ϕ with w
∈
V and ϕ
∈
Z
=
E
k:
(84)
J
(
u
)
≤
λ
k
-
1
-
λ
2
(
θ
+
λ
k
-
1
)
∥
w
∥
2
+
λ
k
-
λ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
≤
λ
k
-
1
-
λ
k
+
δ
2
(
θ
+
λ
k
-
1
)
∥
w
∥
2
+
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
≤
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
-
λ
k
-
λ
k
-
1
4
(
θ
+
λ
k
-
1
)
∥
w
∥
2
-
∫
Ω
‍
F
(
x
,
u
)
d
x
.
We assume that δ
≤
(
λ
k
-
λ
k
-
1
)
/
2; it is easy to see that (
λ
k
-
1
-
λ
k
)
∥
w
∥
2
/
4
(
θ
+
λ
k
-
1
)
≤
C for some constant C; we estimate
(85)
J
(
z
)
≤
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
C
-
D
(
R
)
≤
δ
2
(
θ
+
λ
k
)
R
2
-
D
(
R
)
+
C
.
Considering (83) and (85), we see that since lim
⁡
R
→
∞
⁡
D
(
R
)
=
+
∞ by Lemma 13, we may fix R
1 so that C
-
D
(
R
1
)
<
D
W
-
1 (or C
0
-
R
1
<
D
W
-
1 for the case (
f
1
)) and then for 0
<
δ
<
min
⁡
{
2
(
θ
+
λ
k
)
/
R
1
2
,
(
λ
k
-
λ
k
-
1
)
/
2
} one gets (46).
To obtain (47), we observe that (since λ
>
λ
k
-
1) if ϕ
=
0, that is, if u
∈
V, then in estimates (83) and (85) we may avoid the term δ
R
2
/
2
(
θ
+
λ
k
) so that (remember that D
(
R
) is nondecreasing) J
(
u
)
<
D
W
-
1 for ∥
u
∥
>
R
1.


## 4.2.2. EstimatingJ in S
Z
W

We consider the corresponding estimates of the previous lemma, for Theorem6.Lemma 15.
Consider Theorem6 with one of the sets of hypotheses (
f
3
) and (
f
4
). Given the constant K
V
∈
R, there exist R
2, δ
1
>
0 such that, for any λ
∈
(
λ
k
,
λ
k
+
δ
1
), (51) and (52) hold.Proof.
Lettingλ
=
λ
k
+
δ, we see from (70) that property (52) will be satisfied provided that R
2 is large enough (say R
2
>
R
~); observe that this value can be made independent from λ once δ is small enough.
Now we consider the two sets of hypotheses separately.
(i) In case(
f
3
), suppose u
∈
E
k
⊕
W; we can assume that u
=
w
+
ϕ, with w
∈
W and ϕ
∈
E
k. Since E
k is a finite dimension subspace, all the norms are equivalent, so that there exists K
>
0 such that for all ϕ
∈
E
k we have ∥
ϕ
∥
≤
K
∥
ϕ
∥
1. In addition, by (
f
0
) and (
f
3
), there exist C
2 and L
2
∈
L
2
(
Ω
) such that
(86)
-
F
(
x
,
s
)
≥
K
|
s
|
-
C
2
(
1
+
L
2
(
x
)
)
,
uniformly in x
∈
Ω. So by (29) and (86),
(87)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
K
∥
w
+
ϕ
∥
1
-
C
4
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
K
∥
ϕ
∥
1
-
K
∥
-
w
∥
1
-
C
4
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∥
ϕ
∥
-
C
3
∥
w
∥
-
C
4
,
where C
3
=
|
Ω
|
1
/
2
S
K, C
4
=
C
2
|
Ω
|
1
/
2
∥
L
2
∥
2
+
C
2
|
Ω
|. Since
(88)
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
≤
∥
w
∥
+
∥
ϕ
∥
-
δ
2
(
θ
+
λ
k
)
(
∥
ϕ
∥
2
+
∥
w
∥
2
)
≤
(
1
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
)
∥
ϕ
∥
+
∥
w
∥
,
supposing δ
≤
(
λ
k
+
m
-
λ
k
)
/
2, (87) becomes
(89)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
C
3
∥
w
∥
-
C
4
-
∥
w
∥
+
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
(
C
3
+
1
)
∥
w
∥
-
C
4
+
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
,
since (
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
>
0, so
(90)
h
(
w
)
=
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
(
C
3
+
1
)
∥
w
∥
-
C
4
is bounded below for all w
∈
W; that is, there exists C
5
∈
R such that h
(
w
)
≥
C
5; by (89) one gets
(91)
J
(
u
)
≥
(
1
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
)
∥
u
∥
+
C
5
=
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
+
∥
u
∥
+
C
5
.
(ii) In case(
f
4
), first we give some conclusions which are similar to Lemma 3 of [18]. Under the property of F, there exists a const C, and G
∈
C
(
R
,
R
) which is subadditive, that is,
(92)
G
(
s
+
t
)
≤
G
(
s
)
+
G
(
t
)
for all s
,
t
∈
R, and coercive, that is,
(93)
G
(
s
)
⟶
+
∞
as |
s
|
→
∞, and satisfies that
(94)
G
(
s
)
≤
|
s
|
+
4
for all s
∈
R, such that
(95)
-
F
(
x
,
s
)
≥
G
(
s
)
-
C
for all s
∈
R and x
∈
Ω
¯.
In fact, since-
F
(
x
,
s
)
→
+
∞ as |
s
|
→
∞ uniformly for all x
∈
Ω
¯, there exists a sequence of positive integers n
k with n
k
+
1
>
2
n
k for all positive integers k such that
(96)
-
F
(
x
,
s
)
≥
k
for all |
s
|
≥
n
k and all x
∈
Ω
¯. Let n
0
=
0 and define
(97)
G
(
s
)
=
k
+
2
+
|
s
|
-
n
k
-
1
n
k
-
n
k
-
1
for n
k
-
1
≤
|
s
|
<
n
k, where k
∈
N.
By the definition ofG we have
(98)
k
+
2
≤
G
(
s
)
≤
k
+
3
for all n
k
-
1
≤
|
s
|
<
n
k. By (
f
4
) and F
∈
C
1
(
Ω
¯
×
R
,
R
), there exists C
F
>
0 such that
(99)
-
F
(
x
,
s
)
≥
-
C
F
,
∀
(
x
,
s
)
∈
(
Ω
,
R
2
)
.
It follows that
(100)
-
F
(
x
,
s
)
≥
G
(
s
)
-
C
,
where C
=
C
F
+
4. In fact, when n
k
-
1
≤
|
s
|
<
n
k, for some k
≥
2, one has, by (96) and (98),
(101)
-
F
(
x
,
s
)
≥
k
-
1
≥
G
(
s
)
-
4
≥
G
(
s
)
-
C
for all x
∈
Ω
¯. When |
s
|
<
n
1, we have, by (98) and (99),
(102)
-
F
(
x
,
s
)
≥
-
C
F
=
4
-
C
≥
G
(
s
)
-
C
for all x
∈
Ω
¯.
It is obvious thatG is continuous and coercive. Moreover, one has
(103)
G
(
s
)
≤
|
s
|
+
4
for all s
∈
R. In fact, for every s
∈
R, there exists k
∈
N such that
(104)
n
k
-
1
≤
|
s
|
<
n
k
,
which implies that
(105)
G
(
s
)
≤
(
k
-
1
)
+
4
≤
n
k
-
1
+
4
≤
|
s
|
+
4
for all s
∈
R by (98) and the fact that n
k
≥
k for all integers k
≥
0.
Now we only need to prove the subadditivity ofG. Let
(106)
n
k
-
1
≤
|
s
|
<
n
k
,
n
j
-
1
≤
|
t
|
<
n
j
and m
=
max
⁡
{
k
,
j
}. Then we have
(107)
|
s
+
t
|
≤
|
s
|
+
|
t
|
<
n
k
+
n
j
≤
2
n
m
<
n
m
+
1
.
Hence we obtain, by (98),
(108)
G
(
s
+
t
)
≤
m
+
4
≤
k
+
2
+
j
+
2
≤
G
(
s
)
+
G
(
t
)
,
which shows that G is subadditive.
Foru
∈
E
k
⊕
W, we assume that u
=
w
+
ϕ, with w
∈
W and ϕ
∈
E
k; letting 0
<
δ
<
(
λ
k
+
m
-
λ
k
)
/
2, by (28), (92), (94), and (95), one gets
(109)
J
(
w
+
ϕ
)
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
+
w
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
(
λ
k
+
δ
)
2
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
∫
Ω
‍
G
(
-
w
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
ϕ
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
∫
Ω
‍
(
|
w
|
+
4
)
d
x
-
C
|
Ω
|
≥
λ
k
+
m
-
λ
k
4
(
θ
+
λ
k
+
m
)
∥
w
∥
2
-
δ
2
(
θ
+
λ
k
)
∥
z
∥
2
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
S
|
Ω
|
∥
w
∥
-
C
1
=
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
-
δ
2
(
θ
+
λ
k
)
∥
u
∥
2
,
where g
(
w
)
=
(
(
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
)
∥
w
∥
2
-
S
|
Ω
|
∥
w
∥
-
C
1, C
1
=
(
4
+
C
)
|
Ω
|. Since ϕ
∈
E
k, E
k is a finite-dimensional subspace, and G is coercive, from the proof of (75), one can get
(110)
lim
⁡
∥
ϕ
∥
→
∞
∫
Ω
‍
G
(
ϕ
)
d
x
=
+
∞
.
That is, ∫
Ω
‍
G
(
ϕ
)
d
x is coercive on E
k. Since (
λ
k
+
m
-
λ
k
)
/
4
(
θ
+
λ
k
+
m
)
>
0, so g
(
w
) is coercive on W, and ∫
Ω
‍
G
(
ϕ
)
d
x and g
(
w
) are bounded below, so it is obvious that
(111)
lim
⁡
∥
u
∥
→
∞
(
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
)
=
+
∞
for all u
∈
Z
⊕
W.
Considering (91), (109), and (111), we can choose R
2 large enough such that for all ∥
u
∥
≥
R
2 one gets
(112)
g
(
w
)
+
∫
Ω
‍
G
(
ϕ
)
d
x
>
K
V
+
1
(or R
2
+
C
5
>
K
V
+
1 for the case (
f
3
)) and property (52) holds; then for 0
<
δ
<
min
⁡
{
2
(
θ
+
λ
k
)
/
R
2
2
,
(
λ
k
+
m
-
λ
k
)
/
2
}
=
δ
1 and u
∈
R
2
S
Z
W one gets J
(
u
)
>
K
V; that is, the property (51) holds.


## 5. Proof of the Geometry in Propositions9 and 10

We finally give the proof of Propositions9 and 10, which is nothing but a resume of the lemmata above, verifying that all the constants can be chosen sequentially without contradictions.Proof of Proposition9.
Under hypotheses(
f
0
) and (
f
∞
), if we fix a value λ, then we obtain the constant D
λ from Lemma 11 and with this we get ρ
λ from Lemma 12. If we consider also one of the two sets of hypotheses (
f
1
) and (
f
2
), then we proceed as follows. First of all, we determine (once forever) the constant D
W from Lemma 11; with this we obtain from Lemma 14 the values R
1 and δ
0. Then, for any (now fixed) λ
∈
(
λ
k
-
δ
0
,
λ
k
), we obtain from Lemma 11 the value D
λ. Finally, we can get from Lemma 12 the corresponding value of ρ
λ
>
R
1.Proof of Proposition10.
Under hypotheses(
f
0
) and (
f
∞
), if we fix a value λ
∈
(
λ
k
,
λ
k
+
m
), then we obtain the constant K
λ from Lemma 11 and with this we get β
λ from Lemma 12. If we consider also one of the two sets of hypotheses (
f
3
) and (
f
4
), then we proceed as follows. First of all, we determine (once forever) the constant K
V from Lemma 12; with this we obtain from Lemma 15 the values R
2 and δ
1. Since we have R
2, we can get from Lemma 11 the constant E and with this obtain ξ from Lemma 12.
Finally, for any (now fixed)λ
∈
(
λ
k
,
λ
k
+
δ
1
), we obtain from Lemma 11 the constant K
λ and with this we get from Lemma 12 the corresponding value of β
λ
>
R
2.



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*Source: 360581-2014-05-29.xml*