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or
for all x X and y Y , we have h · x>c>h · y .
n
Here the hyperplane is a set of the form
{
z
∈ R
|
h
·
z
=
c
}
.
Definition 2.18
Let ( X , d ) be a metric space. A function f : X
X is said to be
a contraction mapping if there is a constant ʴ with 0
ʴ< 1 such that
d ( f ( x ), f ( y ))
ʴ
·
d ( x , y )
for all x , y
X .
In the above definition, the constant ʴ is strictly less than 1. It entails the following
property whose proof crucially relies on that constraint on ʴ and is worth writing
down.
Theorem 2.8 (Banach Fixed-Point Theorem) Every contraction on a complete
metric space has a unique fixed point.
Proof Let ( X , d ) be a complete metric space, and f be a contraction mapping on
( X , d ) with constant ʴ . For any x 0
=
X , define the sequence ( x n )by x n + 1 :
f ( x n )
for n
0. Let a :
=
d ( x 0 , x 1 ). It is easy to show that
ʴ n
d ( x n , x n + 1 )
· a
by repeated application of the property d ( f ( x ), f ( y ))
ʴ · d ( x , y ). For any ʵ> 0,
ʴ n
it is possible to choose a natural number N such that
1 ʴ a<ʵ for all n N .Now,
for any m , n
N with m
n ,
d ( x m , x n )
d ( x m , x m + 1 )
+
d ( x m + 1 , x m + 2 )
+···+
d ( x n 1 , x n )
ʴ m
ʴ m + 1
ʴ n 1
·
a
+
·
a
+···+
·
a
ʴ m 1 ʴ n m
1 ʴ
=
a
ʴ m
<
1 ʴ a<ʵ
by repeated application of the triangle inequality. So the sequence ( x n ) is a Cauchy
sequence. Since ( X , d ) is complete, the sequence has a limit in ( X , d ). We define
x to be this limit and show that it is a fixed point of f . Suppose it is not, i.e.
a :
d ( x , f ( x )) > 0. Since ( x n ) converges to x , there exists some N
=
∈ N
such
that d ( x n , x ) < a 2
for all n
N . Then
d ( x , f ( x ))
d ( x , x N + 1 )
d ( x N + 1 , f ( x ))
+
d ( x , x N + 1 )
d ( x N , x )
+
ʴ
·
a 2 +
a 2 = a ,
<
 
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