Proof.
(A1) and (A2) are obvious. Let
n
>
k
>
2
; we assume that
σ
≤
δ
and define
(17)
β
k
(
σ
)
:
=
sup
u
∈
Z
k
,
∥
u
∥
=
1

u

σ
,
β
k
(
δ
)
:
=
sup
u
∈
Z
k
,
∥
u
∥
=
1

u

δ
.
Observe that
(18)

u

σ
≤
β
k
(
σ
)
∥
u
∥
,

u

δ
≤
β
k
(
δ
)
∥
u
∥
,
for any
u
∈
Z
k
. Note that
q
<
p
*
; there exists a constant
c
>
0
such that
(19)

u

q
≤
c
∥
u
∥
.
By the Sobolev trace imbedding inequality, we have
(20)

u

L
p
(
∂
Ω
)
p
≤
K
∥
u
∥
p
.
Then we take
Λ
*
=
1
/
4
K
such that, for all
η
<
Λ
*
,
(21)
η
p

u

L
p
(
∂
Ω
)
p
≤
1
4
p
∥
u
∥
p
.
By (F3), for any
ɛ
>
0
, there exists
c
ɛ
such that
(22)

G
(
x
,
u
)

≤
ɛ

u

p
+
c
ɛ

u

q
.
Then, by (F1)–(F3) and (18)–(21), we obtain
(23)
Φ
λ
(
u
)
=
1
p
∥
u
∥
p

∫
Ω
G
(
x
,
u
)
d
x

η
p
∫
∂
Ω

u

p
d
s

λ
∫
Ω
F
(
x
,
u
)
d
x
≥
3
4
p
∥
u
∥
p

ɛ

u

p
p

c
ɛ

u

q
q

c

u

σ
σ

c

u

δ
δ
≥
3
4
p
∥
u
∥
p

ɛ
c
∥
u
∥
p

c
∥
u
∥
q

c
β
k
(
σ
)
σ
∥
u
∥
σ

c
β
k
(
δ
)
δ
∥
u
∥
δ
.
Note that
p
<
q
; we may choose
ɛ
>
0
and
R
>
0
sufficiently small that
(24)
1
4
p
∥
u
∥
p

ɛ
c
∥
u
∥
p

c
∥
u
∥
q
≥
0
holds true for any
u
∈
W
1
,
p
(
Ω
)
with
∥
u
∥
≤
R
. So we have
(25)
Φ
λ
(
u
)
≥
1
2
p
∥
u
∥
p

c
β
k
(
σ
)
σ
∥
u
∥
σ

c
β
k
(
δ
)
δ
∥
u
∥
δ
,
for any
u
∈
Z
k
with
∥
u
∥
≤
R
. Choosing
(26)
ρ
k
:
=
(
4
p
c
β
k
(
σ
)
σ
+
4
p
c
β
k
(
δ
)
δ
)
1
/
(
p

δ
)
,
by Lemma 7, for
β
k
(
σ
)
→
0
,
β
k
(
δ
)
→
0
as
k
→
∞
, it follows that
ρ
k
→
0
as
k
→
∞
, so there exists
k
0
such that
ρ
k
≤
R
when
k
≥
k
0
. Thus, for
k
≥
k
0
,
u
∈
Z
k
, and
∥
u
∥
=
ρ
k
, we have
Φ
λ
(
u
)
≥
ρ
k
p
/
4
p
>
0
; then
a
k
(
λ
)
≥
0
for all
λ
∈
[
1,2
]
.
On the other hand, if
u
∈
Y
k
with
∥
u
∥
being small enough, since all the norms are equivalent on the finite dimensional space and
σ
<
p
, then
b
k
(
λ
)
<
0
for all
λ
∈
[
1,2
]
.
Furthermore, if
u
∈
Z
k
with
∥
u
∥
≤
ρ
k
,
k
≥
k
0
, we see that
(27)
Φ
λ
(
u
)
≥

c
β
k
(
σ
)
σ
ρ
k
σ

c
β
k
(
δ
)
δ
ρ
k
δ
⟶
0
as
k
⟶
∞
.
Therefore,
d
k
(
λ
)
→
0
as
k
→
∞
. Thus, (A3) holds.
Proof.
Since
Φ
λ
n
′

Y
n
(
u
(
λ
n
)
)
=
0
, then
(29)
1

η
∫
∂
Ω

u
(
λ
n
)

p
∥
u
(
λ
n
)
∥
p
d
s
=
∫
Ω
λ
n
f
(
x
,
u
(
λ
n
)
u
(
λ
n
)
)
+
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)
∥
u
(
λ
n
)
∥
p
d
x
.
We can choose
0
<
Λ
<
Λ
*
and if
η
<
Λ
such that
1

η
K
>
0
. If, up to a subsequence,
∥
u
(
λ
n
)
∥
→
∞
as
n
→
∞
, then, by (F2),
(30)
1
+

η

K
≥
∫
Ω
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)
∥
u
(
λ
n
)
∥
p
d
x
≥
1
2
(
1

η
K
)
,
for
n
is large enough. Obviously, it is a condition if (F4)(1) holds.
Otherwise, we set
w
n
=
u
(
λ
n
)
/
∥
u
(
λ
n
)
∥
; then, up to a subsequence,
(31)
w
n
⇀
w
in
E
,
w
n
⟶
w
in
L
t
(
Ω
)
for
1
≤
t
<
p
*
,
w
n
(
x
)
⟶
w
(
x
)
a
.
e
.
x
∈
Ω
.
If
w
≠
0
in
E
and
lim

u

→
∞
g
(
x
,
u
)
/

u

p

2
u
=

∞
in (F4)(2), then, for
n
is large enough, by Fatou’s Lemma, we have that
(32)

1
2
(
1

η
K
)
≥
∫
Ω

g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

u
(
λ
n
)

p

w
n

p
d
x
≥
c
+
∫
{
x
∈
Ω
:
w
≠
0
,

u
(
λ
n
)

≥
c
}

g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

u
(
λ
n
)

p

w
n

p
d
x
⟶
∞
;
this is a contradiction. It is similar if
lim

u

→
∞
g
(
x
,
u
)
/

u

p

2
u
=
∞
in (F4)(3). Thus,
w
=
0
.
Let
t
n
∈
[
0,1
]
such that
(33)
Φ
λ
n
(
t
n
u
(
λ
n
)
)
:
=
max
t
∈
[
0,1
]
Φ
λ
n
(
t
u
(
λ
n
)
)
.
For any
c
>
0
large enough, and
w
¯
n
:
=
(
2
p
c
)
1
/
p
w
n
, for
n
is large enough, we have that
(34)
Φ
λ
n
(
t
n
u
(
λ
n
)
)
≥
Φ
λ
n
(
w
¯
n
)
=
2
c

∫
Ω
G
(
x
,
w
¯
n
)
d
x

η
p
∫
∂
Ω

w
¯
n

p
d
s

λ
n
∫
Ω
F
(
x
,
w
¯
n
)
d
x
≥
c
,
which implies that
lim
n
→
∞
Φ
λ
n
(
t
n
u
(
λ
n
)
)
→
∞
. Obviously,
(35)
〈
Φ
λ
n
′
(
t
n
u
(
λ
n
)
)
,
t
n
u
(
λ
n
)
〉
=
0
.
It follows that
(36)
∞
=
lim
n
→
∞
(
Φ
λ
n
(
t
n
u
(
λ
n
)
)

1
p
〈
Φ
′
λ
n
(
t
n
u
(
λ
n
)
)
,
t
n
u
(
λ
n
)
〉
)
≤
lim
n
→
∞
λ
n
∫
Ω
(
1
p
f
(
x
,
t
n
u
(
λ
n
)
)
t
n
u
(
λ
n
)

F
(
x
,
t
n
u
(
λ
n
)
)
1
p
)
d
x
+
∫
Ω
(
1
p
g
(
x
,
t
n
u
(
λ
n
)
)
t
n
u
(
λ
n
)

G
(
x
,
t
n
u
(
λ
n
)
)
)
d
x
.
If (F4)(2) holds, we have that
(
1
/
p
)
f
(
x
,
u
)
u

F
(
x
,
u
)
and
(
1
/
p
)
g
(
x
,
u
)
u

G
(
x
,
u
)
are decreasing in
u
for
u
is large enough. Therefore,
(37)
1
p
f
(
x
,
s
u
)
s
u

F
(
x
,
s
u
)
+
1
p
g
(
x
,
s
u
)
s
u

G
(
x
,
s
u
)
≤
c
for all
s
>
0
and
u
∈
ℝ
; it is a contradiction.
If (F4)(3) holds, then we have that
(38)
∞
≤
c
∫
Ω

u
(
λ
n
)

σ
d
x
+
∫
Ω
(
1
p
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

G
(
x
,
u
(
λ
n
)
)
)
d
x
,
which implies
(39)
∫
Ω
(
1
p
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

G
(
x
,
u
(
λ
n
)
)
)
d
x
⟶
∞
.
On the other hand, by the property of
u
(
λ
n
)
, for
n
is large enough, since
α
>
max
{
δ
,
σ
}
, we have that
(40)
b
k
(
1
)
≥
λ
n
∫
Ω
(
1
p
f
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

F
(
x
,
u
(
λ
n
)
)
)
d
x
+
∫
Ω
(
1
p
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

G
(
x
,
u
(
λ
n
)
)
)
d
x
≥
1
2
∫
Ω
(
1
p
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

G
(
x
,
u
(
λ
n
)
)
)
d
x
+
1
2
c
∫
Ω

u
(
λ
n
)

α
d
x

1
2
c
∫
Ω

u
(
λ
n
)

δ
d
x

1
2
c
∫
Ω

u
(
λ
n
)

σ
d
x
≥
c
∫
Ω
(
1
p
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

G
(
x
,
u
(
λ
n
)
)
)
d
x

c
;
this implies that
∫
Ω
(
(
1
/
p
)
g
(
x
,
u
(
λ
n
)
)
u
(
λ
n
)

G
(
x
,
u
(
λ
n
)
)
)
d
x
is bounded, which contradicts (39).
By the above arguments, we have that
{
u
(
λ
n
)
}
is bounded.