In this section, is a more general version of the mountain pass theorem. It is generalized
in two ways. First, the space is not a Hilbert space and second, the derivative of the
functional is not assumed to be Lipschitz. Instead of using I^{′} one uses the pseudogradient
in an appropriate differential equation. This is a significant generalization because there
is no convenient projection map from X^{′} to X like there is in Hilbert space. This is why
the use of the psedogradient is so interesting. For many more considerations
of this sort of thing, see [?]. First is a deformation theorem. Here I will be
defined on a Banach space X and I^{′}
(x)
∈ X^{′}. First recall the Palais Smale
conditions.
Definition 22.1.12A functional I satisfies the Palais Smale conditions meansthat if
{I (uk)}
is a bounded sequence and I^{′}
(uk )
→ 0, then
{uk}
is precompact.That is, it has a subsequence which converges.
Here is a picture which illustrates the main conclusion of the following theorem. The
idea is that you modify the functional on some set making it smaller and leaving it
unchanged off that set.
PICT
Theorem 22.1.13Let I be C^{1}, I is non constant, satisfy the Palais Smale condition, and I^{′}is bounded on bounded sets. Also suppose that c ∈ ℝ is such thateither I^{−1}
([c − δ,c+ δ])
= ∅ for some δ > 0 or I^{−1}
([c − δ,c+ δ])
≠∅ for all δ > 0
and IF I
(u)
= c, then I^{′}
(u)
≠0. Then for each sufficiently small ε > 0, there is aconstant δ ∈
(0,ε)
and a function η :
[0,1]
× X → X such that
η
(0,u)
= u
η
(1,u)
= u on I^{−1}
(X ∖(c− ε,c+ ε))
I
(η(t,u))
≤ I
(u)
η
( )
1,I−1(− ∞, c+ δ]
⊆ I^{−1}(−∞,c−δ], so I
(η(1,u))
≤ c−δ if I
(u)
≤ c + δ.
The main part of this conclusion is the statement about u → η
(1,u)
contained in
parts 2. and 4. The other two parts are there to facilitate these two although they are
certainly interesting for their own sake.
Proof:Suppose I^{−1}
([ ˆ ˆ])
c− δ,c+ δ
= ∅ for some
ˆ
δ
> 0. Then
( ) ( )
ˆδ ˆδ
I− 1 (− ∞, c+ 2] ⊆ I−1 (− ∞, c− 2]
and you could take ε = δ and let η
(t,u)
= u. The conclusion holds with δ =
ˆδ
∕2.
Therefore, assume I^{−1}
([c − δ,c+ δ])
≠∅ for all δ > 0. Since I is nonconstant, ε > 0
can be chosen small enough that
I−1(X ∖(c− ε,c+ ε)) ⁄= ∅.
Always let ε be this small. Note that I nonconstant is part of the assumptions.
Claim 1:For all small enough ε > 0, if u ∈ I^{−1}
([c− ε,c+ ε])
, then I^{′}
(u)
≠0 and in
fact, for such ε, there exists σ
(ε)
> 0, such that σ
(ε)
< min
(ε,1)
,
′
∥I(u)∥
> σ
(ε)
for all
u ∈ I^{−1}
([c − ε,c+ ε])
.
Proof of Claim 1:If the claim is not so, then there is
{uk }
,ε_{k},σ_{k}→ 0,
′
∥I (uk)∥
_{X′}< σ_{k},
and I
(uk)
∈
[c− εk,c+ εk]
but
′
∥I(uk)∥
_{X′}≤ σ_{k}. However, from the Palais Smale
condition, there is a subsequence, still denoted as u_{k} which converges to some u.
Now I
(uk)
∈
[c− εk,c+ εk]
and so I
(u)
= c while I^{′}
(u )
= 0 contrary to the
hypothesis. This proves Claim 1. From now on, ε will be sufficiently small that this
holds.
Now define for δ < ε (The precise description of small δ will be described later.
However, it will be δ < σ
(ε)
∕2, but this exact description is only used at the end.)
A ≡ I−1(X ∖ (c − ε,c +ε))
−1
B ≡ I ([c− δ,c + δ])
Thus A and B are disjoint closed sets. Recall that it is assumed that B≠∅ since otherwise,
there is nothing to prove. Also it is assumed throughout that ε > 0 is such that A≠∅
thanks to I not being constant. Thus these are nonempty sets and we do not have to fuss
with worrying about meaning when one is empty.
Claim 2: For any u,dist
(u,A )
+ dist
(u,B )
> 0.
This is so because if not, then both summands would be zero and this requires that
u ∈ A ∩ B since these sets are closed. But A ∩ B = ∅.
this last inequality from the inequalities of 22.1.5, and so this implies the third condition
since it says that the function t → I
(η (t,u))
is decreasing.
It remains to consider the last condition. This involves an appropriate choice of small
δ. It was chosen small and now it will be seen how small. It is desired to verify
that
( −1 ) −1
η 1,I ((− ∞, c+ δ]) ⊆ I ((− ∞, c− δ])
Suppose it is not so. Then there exists u such that I
(u)
∈ (c − δ,c + δ] but
I
(η (1,u))
> c−δ. We can assume that I
(u)
∈ (c−δ,c + δ] because if I
(u)
≤ c−δ, then
so is I
(η(1,u))
from what was just shown. Hence g
(u)
= 1. Then using the fact that
g
(u)
= 1,
∫ 1d
c− δ < I (η(1,u)) = I(u)+ dt (I(η))dt
0
∫
= I(u)− 1〈I′(η),h(∥V (η)∥)V (η)〉 dt
0
∫ 1
= I (u )+ − h (∥V (η)∥)〈I′(η),V (η)〉dt
0
∫
≤c+δ 1( ′ 2)
≤ I (u )+ 0 − h (∥V (η)∥)∥I (η)∥ dt
Then
∫ 1
c− δ+ h (∥V (η)∥)∥I′(η)∥2dt < I (u) ≤ c+ δ
0
Thus
∫ 1
c− 2δ+ h (∥V (η)∥)∥I′(η)∥2dt < c
0
Also, it is being assumed that I
(η(1,u))
> c − δ and so by the third conclusion
shown above, η
h (∥V (η(t,u))∥ ) ≤ h(∥I′(η(t,u))∥ ) ≤ h(σ (ε)) ≤-1--
X X′ σ(ε)
In fact, σ
(ε)
< 1 so h
(σ(ε))
= 1 so the above estimate, while correct is sloppy.
Hence
∫
1--1- 2
c− 2δ+ 0 σ (ε)σ(ε) dt < c
So far it was only assumed δ < ε. As indicated above, δ was chosen small enough that
−2δ + σ
(ε)
> 0. Hence this yields a contradiction. Thus the last conclusion is verified.
■
Imagine a valley surrounded by a ring of mountains. On the other side of this ring of
moutains, there is another low place. Then there must be some path from the valley to
the exterior low place which goes through a point where the gradient equals 0, the
gradient being the gradient of a function f which gives the altitude of the land. This is
the idea of the mountain pass theorem. The critical point where ∇f = 0 is the mountain
pass.
Theorem 22.1.14Let X be a Banach space and let I : X → ℝ be a C^{1}functionalhaving I^{′}bounded on bounded sets and such that I satisfies the Palais Smale condition.Suppose I
(0)
= 0 and I
(u)
≥ a > 0 for all
∥u∥
= r. Suppose also that there existsv,
∥v∥
> r such that I
(v)
≤ 0. Then define
Γ ≡ {g ∈ C ([0,1];X) : g(0) = 0,g(1) = v}
Let
c ≡ inf max I(g(t))
g∈Γ0≤t≤1
Then c is a critical value of I meaning that there exists u such that I
(u)
= c andI^{′}
(u)
= 0. In particular, there is u≠0 such that I^{′}
(u)
= 0.
Proof: First note that c ≥ a > 0. Suppose c is not a critical value. Then either
I^{−1}
((c− δ,c+ δ))
= ∅ for some δ > 0 in which case the conclusion of the deformation
theorem, (Theorem 22.1.13) holds, or for all δ > 0, I^{−1}
((c− δ,c+ δ))
≠∅ and if I
(u )
= c,
then I^{′}
(u)
≠0 in which case the deformation theorem also holds. Then by this theorem,
for ε > 0,ε sufficiently small, ε < c, there is η : X → X and a δ < ε small enough
that
η(I−1((− ∞, c+ δ])) ⊆ I−1((− ∞, c− δ])
and η leaves unchanged I^{−1}
(X ∖(c− ε,c+ ε))
. Then there is g ∈ Γ such that
tm∈a[0x,1]I(g(t)) < c+ δ
Then in particular, I
(g(t))
< c + δ for every t. Hence you look at η ∘ g. We
know that g
(0)
,g
(1)
are both in the set
[I(u)∈∕(c− ε,c+ ε)]
because they
are both 0 or less than 0 and so η leaves these unchanged. Hence η ∘ g ∈ Γ
and
I (η ∘g (t)) ≤ c− δ
for all t ∈
[0,1]
. Thus
c = gin∈fΓ0m≤atx≤1I (g (t)) ≤ mta∈x[0,1]I(η∘ g(t)) ≤ c − δ