This is a very important generalization to Banach spaces. It turns out you can define the
degree of I −F where F is a compact mapping. To recall what one of these is, here is the
definition.
Definition 21.9.1Let Ω be a bounded open set in X a Banach space and letF : Ω→ X be continuous. Then F is called compact if F
(B )
is precompact wheneverB is bounded. That is, if
{xn}
is a bounded sequence, then there is a subsequence
{xnk}
such that
{F (xnk)}
converges.
Theorem 21.9.2Let F : Ω→ X as above be compact. Then for eachε > 0, there exists F_{ε} : Ω→ X such that F has values in a finitedimensional subspace of X and sup_{x∈Ω}
∥Fε (x) − F (x)∥
< ε. In addition to this,
(I − F)
^{−1}
(compact set)
=compact set. (This is called “proper”.)
Proof: It is known that F
(Ω )
is compact. Therefore, there is an ε net for F
(Ω)
,
{Fxk}
_{k=1}^{n} satisfying
-----
F (Ω) ⊆ ∪kB (F xk,ε)
Now let
ϕ (Fx) ≡ (ε− ∥F x− F x ∥)+
k k
Thus this is equal to 0 if
∥Fx − F x∥
k
≥ ε and is positive if
∥Fx − Fx∥
k
< ε. Then
consider
∑n -ϕk(F-x)--
F ε(x ) ≡ F (xk) ∑ ϕi(Fx)
k=1 i
It clearly has values in span
n
({Fxk}k=1)
. How close is it to F
(x)
? Say Fx ∈ B
(F xk,ε)
.
Then for such x,
∥F (x)− F (xk)∥
< ε by definition. Hence
∑ ϕk(Fx)
∥F (x)− F ε(x )∥ = ∥F (xk) − F (x)∥∑-ϕ-(F-x)
k:∥F(x)−Fxk∥<ε i i
∑ -ϕk-(F-x)--
< ε ∑i ϕi(Fx) = ε
k
Of course x is arbitrary and so
sup∥Fε(x)− F (x)∥ < ε.
x∈Ω
Next consider the second claim. Let K be compact. Consider
− 1
{xk} ⊆ (I − F ) (K ).
It is necessary to show that it has a convergent subsequence. Then
{(I − F)(xk)}
is a
sequence in K and so it has a convergent subsequence still denoted with subscript k such
that
(I − F)
(xk)
→ y. The x_{k} are in a bounded set Ω and so, from compactness of F,
there is a further subsequence, still denoted with subscript k such that Fx_{k}→ z. It
follows that x_{k}→ y − z and hence every sequence in
(I − F)
^{−1}
(K)
has a convergent
subsequence. ■
Corollary 21.9.3Let F : Ω→ X as above be compact. Then for eachε > 0, there exists F_{ε} : Ω→ X such that F has values in a finitedimensional subspace of X and sup_{x∈Ω}
∥Fε(x)− F (x)∥
< ε. . In addition to this,
(I − F)
^{−1}
(compact set)
=compact set. (This is called “proper”.) If Ω is symmetricand F is odd (F
(− x )
= −F
(x)
) then one can also assume F_{ε}is also odd.
Proof: Suppose Ω is symmetric in that x ∈ Ω iff −x ∈ Ω. Suppose also that F is odd.
Thus F
(Ω)
is also symmetric. Thus F
(Ω)
is compact and symmetric. If y ∈ F
(Ω)
, then
y = Fx and so −y = −F
(x)
= F
(− x)
∈ F
(Ω)
. Choose the ε net to be symmetric. That
is, you have
(Fx)
_{k} in the net if and only if −
(F x)
_{k} is in the net. Just add
them in if needed. Therefore, there is an ε net for F
that is, ϕ_{−k} is centered at −Fx_{k} while ϕ_{k} is centered at Fx_{k}, each function equal to 0 off
B
(Fx ,ε)
k
and is positive on B
(F x ,ε)
k
. Then consider
m
F (x) ≡ ∑ε F (x )∑ϕk-(F-x)--
ε k iϕi(Fx)
k= −mε
m∑ ε --ϕk(F-(−-x))- m∑ε -ϕ−-k(F (x))-
Fε (− x) = F (xk)∑ ϕi(F (− x)) = F (xk)∑ ϕ−i(F (x ))
k=−mε i k=− mε i
m∑ ε -ϕ−k-(F-(x))-- m∑ε -ϕ−-k(F (x))-
= − F (− xk) ∑i ϕ−i(F (x)) = − F (x −k)∑i ϕ−i(F (x ))
k=m−mε k=−m ε
= − ∑ ε F (x )∑ϕk-(F (x)) = − F (x)
k= −m k iϕi(F (x )) ε
ε
The rest of the argument is the same. ■
Now let F : Ω→ X be compact and consider I −F. Is
(I − F )
(∂Ω)
closed? Suppose
(I − F)
x_{k}→ y. Then K ≡ y ∪
{(I − F) xk}
_{k=1}^{∞} is a compact set because if you have
any open cover, one of the open sets contains y and hence it contains all
(I − F)
x_{k}
except for finitely many which can then be covered by finitely many open sets in the open
cover. Hence, since
(I − F)
is proper,
(I − F)
^{−1}
(K )
is compact. It follows that there is
a subsequence, still called x_{k} such that x_{k}→ x ∈
(I − F)
^{−1}
(K)
. Then by continuity of
F,
(I − F)(xk) → (I − F )(x)
(I − F)(xk) → y
It follows y =
(I − F )
x and so in fact
(I − F )
(∂Ω)
is closed.
Lemma 21.9.4If F : Ω→ X is compact and Ω is a bounded open set in X, then
(I − F)
(∂Ω )
is closed.
Justification for definition of Leray Schauder Degree
Now let y
∕∈
(I − F )
(∂Ω)
, a closed set. Hence dist
(y,(I − F )(∂Ω))
> 4δ > 0. Now
let F_{k} be a sequence of approximations to F which have values in an increasing
sequence of finite dimensional subsets V_{k} each of which contains y. Thus lim_{k→∞}
sup_{
x∈Ω}
∥F (x)− Fk(x)∥
= 0. Consider
d(I − F | ,Ω ∩ V ,y)
k Vk k
Each of these is a well defined integer according to Theorem 21.8.3. For all k large
enough,
sxu∈pΩ∥(I − F )(x)− (I − Fk)(x)∥ < δ
Hence, for all such k,
B (y,3δ)∩ (I − Fk)(∂Ω) = ∅, that is dist(y,(I − Fk )(∂Ω )) > 3δ (21.9.11)
(21.9.11)
Note that this implies
dist(y,(I − Fk )(∂ (Ω∩ V ))) > 3δ
for any subspace V . If k < l are two such indices, then consider
d(I − Fk |Vk,Ω ∩Vk,y),d(I − Fl|Vl,Ω ∩ Vl,y)
Are they equal? Let V = V_{k} + V_{l}. Then by Theorem 21.8.4,
d (I − Fl|Vl,Ω ∩ Vl,y) = d(I − Fl|V ,Ω∩ V,y)
d(I − F | ,Ω ∩ V ,y) = d(I − F | ,Ω ∩ V,y)
k Vk k kV
So what about d
(I − Fl|V,Ω ∩V, y)
,d
(I − Fk|V ,Ω∩ V,y)
? Are these equal?
sup-∥Fl(x)− Fk (x)∥ ≤ sup-∥Fl(x)− F (x)∥+ sup--∥F (x)− Fk (x )∥ < 2δ
x∈Ω∩V x∈Ω∩V x∈Ω∩V
This implies for
h (x,t) = t(I − Fl)(x)+ (1− t)(I − Fk)(x),
and x ∈Ω ∩ V,y
∕∈
h
(∂ (Ω ∩ V),t)
for all t ∈
[0,1]
. To see this, let x ∈ ∂Ω
∥t(I − Fl)(x)+ (1− t)(I − Fk)(x)− y∥
= ∥t(I − Fk)(x)+ t(Fkx − Flx) +(1 − t)(I − Fk)(x)− y∥
= ∥(I − F )(x)+ t(F x − F x)∥ ≥ 3δ− t2δ ≥ δ
k k l
Hence
d (I − F |,Ω ∩ V,y) = d(I − F | ,Ω ∩ V,y)
lV kV
and so
lim d(I − Fk|Vk,Ω ∩ Vk,y)
k→ ∞
exists. A similar argument shows that this limit is independent of the sequence
{Fk}
of
approximating functions having values in a finite dimensional space. Thus we have the
following definition of the Leray Schauder degree.
Definition 21.9.5Let X be a Banach space and let F : X → X be compact. Thatis, F
(Ω )
is precompact whenever Ω is bounded. Let Ω be a bounded open setin X and let y
∕∈
(I − F)
(∂Ω )
. Let F_{k}be a sequence of operators which havevalues in finite dimensional spaces V_{k}such that V_{k}⊆ V_{k+1}
⋅⋅⋅
,y ∈ V_{k}, and
lim_{k→∞}sup_{
x∈Ω}
∥F (x)− Fk (x)∥
= 0. Then
D (I − F,Ω, y) ≡ lim d(I − Fk |Vk,Ω ∩Vk,y)
k→∞
In fact, the sequence on the right is eventually constant. So
D (I − F,Ω, y) ≡ d(I − Fk |Vk,Ω ∩Vk,y)
for all k sufficiently large.
The main properties of the Leray Schauder degree follow from the corresponding
properties of Brouwer degree.
Theorem 21.9.6Let D be the Leray Schauder degree just defined and let Ω be abounded open set y
∕∈
(I − F)
(∂ Ω)
where F is always a compact mapping. Then thefollowing properties hold:
D
(I,Ω,y)
= 1
If Ω_{i}⊆ Ω where Ω_{i}is open, Ω_{1}∩ Ω_{2} = ∅, and y
∈∕
Ω∖
(Ω1 ∪ Ω2)
then
D (I − F,Ω,y) = D (I − F, Ω1,y)+ D (I − F,Ω2,y)
If t → y
(t)
is continuous h : Ω×
[0,1]
→ X is continuous,
(x,t)
→ h
(x,t)
iscompact, (It takes bounded subsets of Ω×
[0,1]
to precompact sets in X) andif y
(t)
∕∈
(I − h)
(∂Ω,t)
for all t, then t → D
((I − h)(⋅,t),Ω,y (t))
isconstant.
Proof:The mapping x → 0 is clearly compact. Then an approximating sequence is
F_{k},F_{k}x = 0 for all k. Then
D (I,Ω,y) = lim d(I|Vk,Ω ∩ Vk,y) = 1
k→∞
For the second part, let k be large enough that for U = Ω,Ω_{1},Ω_{2},
D (I − F,U, y) = d(I − F | ,U ∩ V ,y)
k Vk k
where F_{k} is the sequence of approximating functions having finite dimensional range.
Then the result follows from the Brouwer degree. In fact,
D (I − F, Ω,y) = d(I − Fk|Vk,Ω∩ Vk,y)
= d(I − Fk|Vk,Ω1 ∩ Vk,y)+ d(I − Fk|Vk,Ω2 ∩ Vk,y)
= D (I − F,Ω1,y)+ D (I − F,Ω2,y)
this does the second claim of the theorem. Now consider the third one about homotopy
invariance.