The following is of more interest in the case of normed vector spaces, but there is no
harm in stating it in this more general setting. You should verify that the functions
described in the following definition are all continuous.
Definition 6.7.1Let f : X → Y where
(X,d)
and
(Y,ρ)
are metric spaces. Thenf is said to be Lipschitz continuousif for every x,
ˆx
∈ X, ρ
(f (x ),f (ˆx))
≤ rd
(x,ˆx)
.The function is called a contraction map if r < 1.
The big theorem about contraction maps is the following.
Theorem 6.7.2Let f :
(X, d)
→
(X,d)
be a contraction map and let
(X,d)
be a completemetric space. Thus Cauchy sequences converge and also d
(f (x),f (xˆ))
≤ rd
(x,ˆx)
wherer < 1. Then f has a unique fixed point. This is a point x ∈ X such that f
(x )
= x. Also,if x_{0}is any point of X, then
d(x ,f (x ))
d(x,x0) ≤ ---0----0--
1− r
Also, for each n,
n d(x0,f (x0))
d(f (x0),x0) ≤ 1− r ,
and x = lim_{n→∞}f^{n}
(x )
0
.
Proof: Pick x_{0}∈ X and consider the sequence of iterates of the map,
2
x0,f (x0),f (x0),⋅⋅⋅.
We argue that this is a Cauchy sequence. For m < n, it follows from the triangle
inequality,
m n n∑−1 ( k+1 k ) ∑∞ k
d (f (x0),f (x0)) ≤ d f (x0),f (x0) ≤ r d(f (x0),x0)
k=m k=m
which shows that this is indeed a Cauchy sequence. Therefore, there exists x such
that
lim fn(x ) = x
n→ ∞ 0
By continuity,
( )
f (x) = f lim fn(x0) = lim fn+1(x0) = x.
n→∞ n→∞
Also note that this estimate yields
n d(x0,f (x0))
d(x0,f (x0)) ≤ 1− r
Now d
(x0,x )
≤ d
(x0,fn(x0))
+ d
(fn(x0),x)
and so
d (x0,x) − d (fn(x0),x) ≤ d-(x0,f-(x0))
1 − r
Letting n →∞, it follows that
d(x0,f (x0))
d(x0,x) ≤ ---1−-r----
It only remains to verify that there is only one fixed point. Suppose then that x,x^{′} are
two. Then
d (x,x′) = d (f (x),f (x′)) ≤ rd (x′,x)
and so d
′
(x,x )
= 0 because r < 1. ■
The above is the usual formulation of this important theorem, but we actually proved
a better result.
Corollary 6.7.3Let B be a closed subset of the complete metric space
(X,d)
and letf : B → X be a contraction map
d (f (x ),f (ˆx)) ≤ rd(x,ˆx), r < 1.
Also suppose there exists x_{0}∈ B such that the sequence of iterates
{fn(x0)}
_{n=1}^{∞}remains in B. Then f has a unique fixed point in B which is the limit of the sequence ofiterates. This is a point x ∈ B such that f
(x )
= x. In the case that B =B
(x0,δ)
, thesequence of iterates satisfies the inequality
d(fn(x0),x0) ≤ d-(x0,f (x0))
1 − r
and so it will remain in B if
d(x0,f (x0))< δ.
1 − r
Proof: By assumption, the sequence of iterates stays in B. Then, as in the proof
of the preceding theorem, for m < n, it follows from the triangle inequality,
m n n∑−1 ( k+1 k )
d (f (x0),f (x0)) ≤ d f (x0),f (x0)
k=m
∑∞ k -rm--
≤ r d (f (x0),x0) = 1− rd (f (x0) ,x0)
k=m
Hence the sequence of iterates is Cauchy and must converge to a point x in X. However,
B is closed and so it must be the case that x ∈ B. Then as before,
( )
x = lnim→∞ fn(x0) = nl→im∞ fn+1(x0) = f lnim→∞ fn (x0) = f (x)
As to the sequence of iterates remaining in B where B is a ball as described, the
inequality above in the case where m = 0 yields
d (x0,fn (x0)) ≤ -1--d(f (x0),x0)
1− r
and so, if the right side is less than δ, then the iterates remain in B. As to the
fixed point being unique, it is as before. If x,x^{′} are both fixed points in B, then
d
′
(x,x )
= d
′
(f (x) ,f (x))
≤ rd
′
(x,x )
and so x = x^{′}. ■
Sometimes you have the contraction depending on a parameter λ. Then there is a
principle of uniform contractions.
Corollary 6.7.4Suppose f : X × Λ → X where Λ is a metric space and X is acomplete metric space. Suppose f satisfies
d
(f (x,λ),f (y,λ))
≤ rd
(x,y)
for each λ ∈ Λ.
λ → f
(x,λ)
is continuous as a map from Λ to X.
Then if x
(λ)
is the fixed point, it follows that λ → x
(λ)
is continuous.
Proof: Pick x_{0}∈ X and consider the above sequence of iterates,
{fn(x,λ)}
. Let ρ
be the metric on Λ. Then there is a fixed point and if x
(λ)
is this unique fixed
point,
d(f (x ,λ),x )
d (x (λ),x0) ≤-----0-----0-
1 − r
In particular, you could start with x_{0} = x
(μ)
and conclude that
d(f (x(μ),λ),x-(μ-))
d(x (λ),x (μ )) ≤ 1 − r
≤ d(f (x(μ),λ),f (x-(μ-),μ-))-+ d(f (x(μ),μ),x-(μ-))
1− r 1 − r
= d(f (x(μ),λ),f (x-(μ-),μ-))
1− r
Now by continuity of λ → f
(x,λ )
, it follows that if ρ
(λ,μ)
is small enough, the above is
no larger than
ε(1− r)
-1-−-r- = ε
Hence, if ρ
(λ,μ)
is small enough, we have
d(x (λ) ,x (μ)) < ε.■
This is called the uniform contraction principle.
The contraction mapping theorem has an extremely useful generalization. In order
to get a unique fixed point, it suffices to have some power of f a contraction
map.
Theorem 6.7.5Let f :
(X,d)
→
(X, d)
have the property that for some n ∈ ℕ,f^{n}is a contraction map and let
(X, d)
be a complete metric space. Then thereis a unique fixed point for f. As in the earlier theorem the sequence of iterates
{fn(x0)}
_{n=1}^{∞}also converges to the fixed point.
Proof:From Theorem 6.7.2 there is a unique fixed point for f^{n}. Thus
fn(x) = x
Then
fn(f (x )) = fn+1(x) = f (x)
By uniqueness, f
(x)
= x.
Now consider the sequence of iterates. Suppose it fails to converge to x. Then there is
ε > 0 and a subsequence n_{k} such that
nk
d(f (x0),x) ≥ ε
Now n_{k} = p_{k}n + r_{k} where r_{k} is one of the numbers
{0,1,2,⋅⋅⋅,n − 1}
. It follows that
there exists one of these numbers which is repeated infinitely often. Call it r and let the
further subsequence continue to be denoted as n_{k}. Thus
There is a remarkable result concerning compactness and uniform continuity.
Theorem 6.7.8Let f :
(X,d)
→
(Y,ρ)
be a continuous function and let K be acompact subset of X. Then the restriction of f to K is uniformly continuous.
Proof: First of all, K is a metric space and f restricted to K is continuous. Now
suppose it fails to be uniformly continuous. Then there exists ε > 0 and pairs of points
x_{n},
ˆx
_{n} such that d
(xn,xˆn )
< 1∕n but ρ
(f (xn),f (xˆn ))
≥ ε. Since K is compact, it is
sequentially compact and so there exists a subsequence, still denoted as
{xn}
such that
x_{n}→ x ∈ K. Then also
ˆx
_{n}→ x also and so
ρ(f (x),f (x )) = lnim→∞ ρ(f (xn),f (ˆxn)) ≥ ε
which is a contradiction. Note the use of Lemma 6.7.7 in the equal sign. ■
Next is to consider the meaning of convergence of sequences of functions.
There are two main ways of convergence of interest here, pointwise and uniform
convergence.
Definition 6.7.9Let f_{n} : X → Y where
(X,d)
,
(Y,ρ)
are two metric spaces.Then
{fn}
is said to converge poinwise to a function f : X → Y if for everyx ∈ X,
nli→m∞ fn (x) = f (x)
{fn}
is said to converge uniformly if for all ε > 0, there exists N such that if n ≥ N,then
sup ρ(f (x),f (x)) < ε
x∈X n
Here is a well known example illustrating the difference between pointwise and
uniform convergence.
Example 6.7.10Let f_{n}
(x)
= x^{n}on the metric space
[0,1]
. Then this functionconverges pointwise to
{ 0 on [0,1)
f (x) = 1 at 1
but it does not converge uniformly on this interval to f.
Note how the target function f in the above example is not continuous even though
each function in the sequence is. The nice thing about uniform convergence is that it
takes continuity of the functions in the sequence and imparts it to the target function. It
does this for both continuity at a single point and uniform continuity. Thus uniform
convergence is a very superior thing.
Theorem 6.7.11Let f_{n} : X → Y where
(X,d)
,
(Y,ρ)
are two metric spaces andsuppose each f_{n}is continuous at x ∈ X and also that f_{n}convergesuniformly to fon X. Then f is also continuous at x. In addition to this, if each f_{n}is uniformlycontinuous on X, then the same is true for f.