This is from Evans [?]. It is an interesting theorem. See also [?] for more general
versions. It has to do with differentiable functions defined on a Hilbert space H.
Thus I : H → ℝ will be differentiable. Then the following is the Palais Smale
condition.
Definition 22.1.1A functional I satisfies the Palais Smale conditions means thatif
{I (uk)}
is a bounded sequence and I^{′}
(uk)
→ 0, then
{uk}
is precompact. Thatis, it has a subsequence which converges.
It will be assumed that I is C^{1}
(H;ℝ )
and also that I^{′} is Lipschitz on bounded sets.
By I^{′}
(u)
is meant the element of H such that
I (u + v) = I(u)+ (I′(u),v) + o(v)
H
Such exists because of the Riesz representation theorem. Note that, from the assumption
that I^{′} is Lipschitz continuous, it follows that I^{′} is bounded on every bounded
set.
First is a deformation theorem. The notation
[I (u) ∈ S ]
means
{u : I(u) ∈ S}
.
Theorem 22.1.2Let I be C^{1}, I is non constant, satisfy the Palais Smalecondition, and I^{′}is Lipschitz continuous on bounded sets. Also suppose thatc ∈ ℝ is such that either
[I(u) ∈ [c− δ,c+ δ]]
= ∅ for some δ > 0 or
[I(u) ∈ [c− δ,c+ δ]]
≠∅ for all δ > 0 and IF I
(u)
= c, then I^{′}
(u)
≠0. Thenfor each sufficiently small ε > 0, there is a constant δ ∈
(0,ε)
and a functionη :
[0,1]
× H → H such that
η
(0,u)
= u
η
(1,u)
= u on
[I(u) ∕∈ (c− ε,c + ε)]
I
(η(t,u))
≤ I
(u)
η
(1,[I (u ) ≤ c+ δ])
⊆
[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(u) ∈ [c − δ,c+ δ]]
= ∅ for some δ > 0. Then
[I(u) ≤ c+ δ∕2]
⊆
[I (u ) ≤ c− δ∕2]
and you could take ε = δ and let η
(t,u)
= u.
Therefore, assume
[I(u) ∈ [c− δ,c +δ]]
≠∅ for all δ > 0. Since I is nonconstant, ε > 0 can
be chosen small enough that
[I(u) ∕∈ (c− ε,c+ ε)]
≠∅. Always let ε be this
small.
Claim 1:For all small enough ε > 0, if u ∈
[I(u) ∈ [c− ε,c+ ε]]
,I^{′}
(u)
≠0 and in
fact, for such ε, there exists σ
(ε)
> 0 such that σ
(ε)
< ε,
′
∥I (u)∥
> σ
(ε)
for all u ∈
[I(u) ∈ [c− ε,c+ ε]]
.
Proof of Claim 1:If claim is not so, then there is
{uk}
,ε_{k},σ_{k}→ 0,
′
∥I(uk)∥
< σ_{k},
and I
(uk)
∈
[c− εk,c+ εk]
but
′
∥I (uk)∥
≤ 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.
Now define for δ < ε (The description of small δ will be described later.)
A ≡ [I(u) ∕∈ (c− ε,c+ ε)]
B ≡ [I(u) ∈ [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 would be zero and this requires that u ∈ A ∩ B
since these sets are closed. But A ∩ B = ∅.
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 choosing δ still smaller if
necessary. It is desired to verify that
η(1,[I (u) ≤ c+ δ]) ⊆ [I(u) ≤ c− δ]
Suppose it is not so. Then there exists u ∈
[I (u) ≤ c+ δ]
but I
(η (1,u))
> c − δ.
∫ 1
c− δ < I (η (1,u)) = I (u) − (I′(η),g(u)h(∥I′(η)∥)I′(η))dt
∫ 1 0
= I (u) − g (u ) h(∥I′(η)∥)∥I′(η(t,u))∥2dt
0
∫ 1 ′ ′ 2
< c+ δ − g(u) h(∥I (η)∥)∥I (η(t,u ))∥ dt
0
Then
∫
1 ′ ′ 2
c − 2δ + g(u) 0 h(∥I (η(t,u))∥)∥I (η (t,u))∥ dt < c
If I
(u)
≤ c−δ, there is nothing to show because in this case I
(η (1,u))
≤ I
(u)
≤ c−δ.
Hence we can assume that I
(u)
> c − δ and also that I
(u )
≤ c + δ. Thus u ∈ B and so
g
(u)
= 1. Thus
∫ 1
c− 2δ+ h(∥I′(η(t,u))∥)∥I′(η (t,u))∥2dt < c
0
Also, it is being assumed that I
(η(1,u))
> c − δ and so by the third conclusion
shown above, η
(t,u)
∈ B for t ∈
[0,1]
. We also know that for such values of
η
(t,u)
,
∥I′(η(t,u))∥
≥ σ
(ε)
from Claim 1. If
∥I′(η(t,u))∥
> 1, the integrand equals
∥I′(η(t,u))∥
≥ σ
(ε)
. if
∥I′(η(t,u))∥
≤ 1, the integrand is
∥I′(η(t,u))∥
^{2}≥ σ
(ε)
^{2}.
Thus
∫
1 ( 2)
c− 2δ+ 0 min σ(ε),σ(ε) dt < c
and the only restriction on δ was that it should be smaller than ε. Although it was not
mentioned above, δ was chosen so small that −2δ + min
( σ(ε),σ(ε)2)
> 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 22.1.3Let H be a Hilbert space and let I : H → ℝ be a C^{1}functional havingI^{′}Lipschitz continuous and such that I satisfies the Palais Smale condition. SupposeI
(0)
= 0 and I
(u)
≥ a > 0 for all
∥u∥
= r. Suppose also that there exists v,
∥v∥
> r suchthat I
(v)
≤ 0. Then define
Γ ≡ {g ∈ C ([0,1];H) : g(0) = 0,g(1) = v}
Let
c ≡ ingf∈Γ m0a≤xt≤1 I(g(t))
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 by the
deformation theorem, for ε > 0,ε sufficiently small, there is η : H → H and a δ < ε small
enough that
η([I (u ) ≤ c+ δ]) ⊆ [I (u) ≤ c− δ]
and η leaves unchanged
[I (u ) ∕∈ (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 and so η
leaves these unchanged. Hence η ∘ g ∈ Γ and
I (η ∘g (t)) ≤ c− δ
for all t ∈
[0,1]
. Thus
c = inf max I (g (t)) ≤ max I(η∘ g(t)) ≤ c − δ
g∈Γ0≤t≤1 t∈[0,1]
which is clearly a contradiction. ■
The Palais Smale conditions are pretty restrictive. For example, let I
(x)
= cosx.
Thus I : ℝ → ℝ. Then let u_{k} = kπ. Clearly I
(uk)
is bounded and lim_{k→∞}I
(uk)
= 0 but
{uk}
is not precompact. However, here is a simple case which does satisfy the Palais
Smale conditions.
Example 22.1.4Let I : ℝ^{d}→ ℝ satisfy lim_{}
|x|
→∞I
(x)
= ∞. Then I satisfies thePalais Smale conditions.
The growth condition implies that if I
(xk)
is bounded, then so is
{xk }
and so this
sequence is precompact. Nothing needs to be said about I^{′}