14. THE BROUWER THEOREM FOR THE DISK
Notation: Throughout this handout
D  {x R
2
:
by subscripts
x
A x , Pu
D
its boundary.
 1}
2
D  {x R
:
 1}
x
is the closed unit disk and
is oriented counter clockwise. Partial derivatives are denoted
, etc. A function of several variables is called continuously differentiable iff all of
its first order partial derivatives exist and are continuous, and is twice continuously differentiable iff all of
its second order partial derivatives exist and are continuous.
Lemma (Weierstrass Theorem): If
that
  ψ
Lemma: If
g:D  D
Pf: Write
 ε
u
is continuous and
 g(x)
f(x)
 ε
2
 ( a(x), b(x) )
f(x)
 ε /(2
u
2
f(x)  h(x)
h(x)
2

because
2
2)

f(x)
g(x)
f(x)  g(x)
u

1

ε  0
for all
ψ
in two variables so
then there is a twice continuously differentiable
x D
α(x) and β(x)
h(x)  ( α(x), β(x) )
.
2
a(x)  α(x)
2
2
then there is a polynomial
.
. The coordinate functions a and b are continuous. By the Weierstrass
Theorem there are polynomials
b β
ε  0
and
.
f :D  D
with
  C(D)

2
a  α
u
 ε /(2
2)
and
is twice continuously differentiable on D and for each
b(x)  β(x)
h(x)  f(x)
in two variables so that
 1  ε /2
2
 ( ε /2)
2
x D
,
and
. For g take
g  (1  ε /2)
1
h
. For each
x D
,
 D
h(D)
and
f(x)  h(x)
2

1
h(x)  (1  ε /2)
 ε /2  (ε /2) (1  ε /2)
h(x)
1
h(x)
2
2
 ε
.
Lemma: If there is a continuous f : D  D with no fixed point, then there is a twice continuously
differentiable h : D   D so that h(x) = x for all x   D .

f(x)  x
Pf: D is compact and
m(x)
m  min {
: x D } 0
f(x)  x
2
continuously differentiable
in D,
m 
f(x)  x
2

h(x)  x  t(x) [x  g(x)]
is continuous on D. m(x) must attain a minimum value and
since f has no fixed points. By the preceding lemma there is a twice
g:D  D
f(x)  g(x)
2
2
with

f  g
g(x)  x
u
2
 m/2
 m/2 
. g is also fixed point free because for each x
g(x)  x
2
. Define
, where t(x) denotes the unique non-negative scalar for which
h(x)
2
1
. By
the quadratic formula
1
2
h(x)
2

2
x
2
 2 t(x)  x, x  g(x)  
2
x  g(x)
2
t(x)
2
has non-negative
solution
  x, x  g(x)  
2
 x, x  g(x) 
t(x) 
x  D
then
 x, g(x)  
x
g(x)
2
2
to point b , meaning
σ(b)
 b
 σ : [a, b]  C
σ(t)  (x(t), y(t))
.
,
y  y(t)
,
x
2
1 ]
C  R
2
,
0   x, x  g(x) 
, t(x) = 0 and h(x) = x.
be a simple, smooth curve oriented from point
which is 1-1, onto and continuously differentiable with
σ(a)
 a
a
and
is called a parameterization of C with t the parameter. Given functions P(x,y)
and Q(x,y) continuous on C, the line integral
x  x(t)
2
2
[
2
 1  x, x 
Computational Review of Line Integrals: Let
2
x  g(x)
2
x  g(x)
If

d x  x (t) d t
,
d y  y (t) d t
L 
C
Pdx  Qdy
is calculated by formally substituting
and then integrating from lower limit t = a to upper limit t =
b. The value of the line integral is independent of the parameterization but not the orientation. Reversing
the orientation changes the sign of the line integral.
Illustrative Example: Let C be the arc of the circle
from
a  (1,0)
(a) Let
to
x  cos(t)
b  (0,1)
and
. Calculate
y  sin (t)
π /2
L 
0
2
cos(t) [sin(t)]
for
L 
2
2
C
 y
2
1
which lies in the first quadrant and runs
using these two parameterizations.
x y dx
0  t  π /2
[  sin(t)]
x
.
d t  (  1/4) sin
4
t  π /2
(t)
  1/4
t0
(b) Use x itself as the parameter and
y 
1 x
2
.
. Integrating from lower limit x = 1 to upper limit x = 0
reflects the given orientation of C.
0
L 
1
x[
1 x
2
]
2
2
4
d x  [ x /2  x /4 ]
t0
  1/4
t 1
Green's Theorem for the Disk: If A(x,y) and B(x,y) are continuously differentiable real valued functions
on D then
 D
A x  B y d x dy 
 D
Ady  Bdx
.
Pf: The integral equality will follow by combining these two separate equalities;
(*)
For the first,
 D
A x d x dy 
 D
Ady
and (**)
 D
B y d x dy  
 D
Bdx
.
 D
1
A x d x dy 

1 y
1 
1 y
2
1

1
A(
1
1
A x (u, y) d u dy 
2
1 y
2
1 y
A(
1
, y) dy 
1
A (
1 y
2
2
, y)  A ( 
1 y
to
( 0 ,  1) .
( 0 ,1 )
, y) dy
, y) dy
On the preceding line the first integral is the line integral up the right hand half of
( 0 ,  1)
2
, and the second is the line integral down the left hand half of
x
2
x
 y
2
 y
2
1
2
1
from
from
( 0 ,1 )
to
This proves (*).
For the second,

 D
1
B y d x dy  

1 
1 x
2
2
1 x
1

1
B (x,
1
B y (x, v) d v dx  
1 x
2
1
1
) dx 
1
B (x, 
1 x
B (x,
1 x
2
2
and the second is the line integral over the bottom half of
x
2
2
1 x
) dx
) dx
On the preceding line the first integral is the line integral over the top half of
(  1, 0 ) ,
)  B (x, 
 y
2
1
x
2
 y
from
2
1
from
(  1, 0 )
(1, 0 )
to
to (1, 0 ) . This
proves (**).
Brouwer Theorem for the Disk: Every continuous
f :D  D
has a fixed point.
Proof: To get a contradiction suppose here is a twice continuously differentiable
identity on
D
. Write
h(x, y) 
partial derivatives. Define
(g y ) x  (g x ) y
 f(x,
y), g(x, y)
h : D  D
 with f and g having continuous first and second order
J(x, y)  ( f g y ) x  ( f g x ) y
. Continuity of the second order partials gives
and
fx
J(x, y)  ( f g y ) x  ( f g x ) y  f x g y  f (g y ) x  f y g x  f (g x ) y  det 
f
 y
Claim (1):
0  (f
2
 g
J(x, y)  0
2
which is the
. Since
(f, g)   D
)y  2 f fy  2 g g y
,
f
2
 g
2
 1 , 0  (f
2
gx 

gy 

2
 g )x  2 f fx  2 g g x
. Writing the last two equalities in matrix form,
fx
2
f
 y
.
and
gx f

g y   g


0
   

 

0
The square matrix has a non-trivial kernel and must have determinate J(x,y) = 0.
Claim (2):
 D
J(x, y) d x dy  π
. Parameterize
D
by
x  cos(t)
and
x  sin (t)
,
0  t  2π
Green's Theorem
 D
J(x, y) d x dy 

On
D
(x, y)  h(x, y) 
 D
2π
0
 f(x,
( f g y ) x  ( f g x ) y d x dy 
f [ g x x   g y y]d t 
y), g(x, y)
,
2π
0
 D
( f g x ) dx  ( f g y ) dy
f(x, y) [ g(x, y) ]  d t
f(x, y)  x  cos(t)
,
g(x, y)  y  sin(t)
, and
. By
.
 D
2π
J(x, y) d x dy 
0
f(x, y) [ g(x, y) ]  d t 
The two claims are clearly contradictory.
2π
0
cos(t) [ sin(t) ]  d t 
2π
0
cos
2
(t) d t  π
.