14.3 Something Which Is Not A Retract
The next lemma is a fundamental result which will be used to develop the Brouwer
degree. It will also be used to give a short proof of the Brouwer fixed point
theorem in the exercises. This major fixed point theorem is probably the most
fundamental theorem in nonlinear analysis. The proof outlined in the exercises is from
[? ] .
Lemma 14.3.1 Let g : U → ℝ n be C 2 where U is an open subset of ℝ n .
Then
where here
ij ≡ g i,j ≡ . Also, cof ij =
.
Proof: From the cofactor expansion theorem,
∑n
det(Dg ) = gi,j cof (Dg)ij
i=1
and so
∂det(Dg-)
∂gi,j = cof (Dg )ij (14.3.2)
(14.3.2)
which shows the last claim of the lemma. Also
∑
δkjdet(Dg ) = gi,k(cof (Dg ))ij (14.3.3)
i
(14.3.3)
because if k ≠ j this is just the cofactor expansion of the determinant of a matrix in which
the k th and j th columns are equal. Differentiate 14.3.3 with respect to x j and sum on j .
This yields
∑ ∑ ∑
δkj∂-(detDg-)gr,sj = gi,kj(cof (Dg )) + gi,kcof (Dg ) .
r,s,j ∂gr,s ij ij ij ij,j
Hence, using δ kj = 0 if j ≠ k and 14.3.2 ,
∑ ∑ ∑
(cof (Dg ))rs gr,sk = gr,ks(cof (Dg))rs + gi,kcof (Dg)ij,j.
rs rs ij
Subtracting the first sum on the right from both sides and using the equality of mixed
partials,
∑ ( ∑ )
gi,k( (cof (Dg ))ij,j) = 0.
i j
If det
≠ 0 so that
is invertible, this shows
∑
j ij,j = 0. If
det
= 0, let
where ε k → 0 and det
≡ det
≠ 0. Then
∑ (cof (Dg )) = lim ∑ (cof (Dgk )) = 0 ■
j ij,j k→∞ j ij,j
Definition 14.3.2 Let h be a function defined on an open set, U ⊆ ℝ n . Then
h ∈ C k
if there exists a function g defined on an open set, W containng U such
that g =
h on U and g is C k .
Lemma 14.3.3 There does not exist h ∈ C 2
such that h :
B →
∂B which also has the property that h =
x for all x ∈ ∂B . That is,
there is no retraction of B to ∂B .
Proof: Suppose such an h exists. Let λ ∈
and let
p λ ≡ x +
λ .
This function,
p λ is a homotopy of the identity map and the retraction,
h .
Let
∫
I(λ) ≡ -----det(Dp λ(x))dx.
B(0,R)
Then using the dominated convergence theorem,
′ ∫ ∑ ∂det(Dp-λ(x))∂pλij-(x)-
I (λ) = B(0,R) ∂pλi,j ∂ λ
∫ i.j
= ∑ ∑ ∂det(Dp-λ(x))(hi(x)− xi) dx
B(0,R) i j ∂pλi,j ,j
∫ ∑ ∑
= ----- cof (Dpλ (x ))ij(hi(x) − xi),j dx
B(0,R) i j
Now by assumption,
h i =
x i on
∂ B and so one can integrate by parts and
write
∫
I′(λ) = − ∑ ∑ cof (Dp (x)) (h (x)− x )dx = 0.
i B(0,R) j λ ij,j i i
Therefore, I
equals a constant. However,
( )
I (0) = mn B-(0,R-) > 0
but
∫ ∫
I(1) = det(Dh (x))dmn = # (y)dmn = 0
B(0,R) ∂B(0,R)
because from polar coordinates or other elementary reasoning, m n
= 0
.
■
The last formula uses the change of variables formula for functions which are not one
to one. In this formula, #
equals the number of
x such that
h =
y . To see this is
so in case you have not seen this, note that
h is
C 1 and so the inverse function theorem
from advanced calculus applies. Thus
∫ ∫
----- det (Dh (x))dmn = det(Dh (x))dmn
B (0,R ) [de∫t(Dh (x))>0]
+ det(Dh (x ))dmn
[det(Dh(x))<0]
Thus
h is locally one to one on the two open sets
, .
Now use inverse function theorem and change of variables for one to one
h to verify that
both of these integrals equal 0. You cover
with countably many balls
on which
h is one to one and then use change of variables for each of these integrals over
intersected with this ball.
The following is the Brouwer fixed point theorem for C 2 maps.
Lemma 14.3.4 If h ∈ C 2
and h :
B → B , then h has a
fixed point, x such that h =
x .
Proof: Suppose the lemma is not true. Then for all x ,
≠ 0
. Then
define
g (x) = h(x)+ -x-− h-(x)t(x)
|x − h (x)|
where t
is nonnegative and is chosen such that
g ∈ ∂B . This mapping is
illustrated in the following picture.
If x → t
is
C 2 near
B , it will follow
g is a
C 2 retraction onto
∂B
contrary to Lemma
14.3.3 . Thus
t is the nonnegative solution to
( )
H (x,t) ≡ |h(x)|2 +2 h (x ), x-−-h(x) t+ t2 = R2 (14.3.4)
|x − h(x)|
(14.3.4)
Then
( )
Ht (x,t) = 2 h (x), x-−-h(x)- + 2t.
|x − h(x)|
If this is nonzero for all x near B , it follows from the implicit function theorem
that t is a C 2 function of x . Then from 14.3.4
( )
x-−-h(x)-
2t = − 2 h (x ),|x − h(x)|
∘ --(---------------)2---(-----------)-
± 4 h (x), x-−-h(x)- − 4 |h (x)|2 − R2
|x − h(x)|
and so
( )
H (x,t) = 2t +2 h (x ), x-−-h(x)
t |x − h(x)|
∘ -(-----------)----(---------------)2-
= ± 4 R2 − |h (x)|2 +4 h (x ), x-−-h(x)
|x − h(x)|
If
< R, this is nonzero. If
=
R, then it is still nonzero unless
But this cannot happen because the angle between h
and
x − h cannot be
π∕ 2
.
Alternatively, if the above equals zero, you would need
(h (x ),x) = |h(x)|2 = R2
which cannot happen unless x = h
which is assumed not to happen. Therefore,
x → t is
C 2 near
B and so
g given above contradicts Lemma
14.3.3 .
■
Then the Brouwer fixed point theorem is as follows.
Theorem 14.3.5 Let f : B → B be continuous, this being a ball in
ℝ p . Then it has a fixed point x ∈ B such that f
=
x .
Proof: You can extend f to assume it is defined on all of ℝ p , f
⊆ B ,
the convex hull of
B . Then letting
be a mollifier, let
f n ≡ f ∗ ψ n .
Thus
|∫ | ∫ ∫
|fn(x)| = || f (t)ψn(x − t)dt||≤ |f (t)|ψn (x− t)dt ≤ R ψn (x− t)dt = R
| ℝp | ℝp ℝp
and so the restriction of f n to B is C 2
. Therefore, there exists
x n ∈ B such that
f n =
x n . The functions
f n converge uniformly to
f on
B .
|∫ |
|| ||
|f (x) − fn(x)| = || B(0,1)(f (x)− f (x − t))ψn (t)dt||
∫ n
≤ |f (x) − f (x − t)|ψn (t) dt < ε
B(0,n1)
provided
n is large enough, this for every
x ∈ B , this by uniform continuity of
f on
B . There exists a subsequence, still called
which converges to
x ∈ B . Then using the uniform convergence of
f n to
f ,
f (x) = lnim→∞ f (xn) = nl→im∞ fn (xn ) = nli→m∞ xn = x ■
Definition 14.3.6 A nonempty topological space A is said to have the fixed point
property if every continuous mapping f : A → A has a fixed point.