6.7 Some Simple Fixed Point Theorems
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 6.7.1 Let f : X → Y where
are metric spaces. Then
f is said to be Lipschitz continuous if for every x,
∈ X, ρ
The function is called a contraction map if r <
The big theorem about contraction maps is the following.
Theorem 6.7.2 Let f :
be a contraction map and let
be a complete
metric space. Thus Cauchy sequences converge and also d
1. Then f has a unique fixed point. This is a point x ∈ X such that f
if x0 is any point of X, then
Also, for each n,
and x = limn→∞fn
Proof: Pick x0 ∈ X and consider the sequence of iterates of the map,
We argue that this is a Cauchy sequence. For m < n, it follows from the triangle
The reason for this last is as follows.
and so forth. Therefore,
which shows that this is indeed a Cauchy sequence. Therefore, there exists x such
Also note that this estimate yields
Letting n →∞, it follows that
It only remains to verify that there is only one fixed point. Suppose then that x,x′ are
and so d
= 0 because
The above is the usual formulation of this important theorem, but we actually proved
a better result.
Corollary 6.7.3 Let B be a closed subset of the complete metric space
: B → X be a contraction map
Also suppose there exists x0 ∈ B such that the sequence of iterates
remains in B. Then f has a unique fixed point in B which is the limit of the sequence of
iterates. This is a point x ∈ B such that f
x. In the case that B
sequence of iterates satisfies the inequality
and so it will remain in B if
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,
Hence the sequence of iterates is Cauchy and must converge to a point x
is closed and so it must be the case that x ∈ B
. Then as before,
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
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
Sometimes you have the contraction depending on a parameter λ. Then there is a
principle of uniform contractions.
Corollary 6.7.4 Suppose f : X × Λ → X where Λ is a metric space and X is a
complete metric space. Suppose f satisfies
for each λ ∈ Λ.
- λ → f
is continuous as a map from Λ to X.
Then if x
is the fixed point, it follows that λ → x
Proof: Pick x0 ∈ X and consider the above sequence of iterates,
be the metric on Λ. Then there is a fixed point and if x
is this unique fixed
In particular, you could start with x0 = x
and conclude that
Now by continuity of λ → f
it follows that if ρ
is small enough, the above is
no larger than
Hence, if ρ
is small enough, we have
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
Theorem 6.7.5 Let f :
have the property that for some n ∈ ℕ,
fn is a contraction map and let
be a complete metric space. Then there
is a unique fixed point for f. As in the earlier theorem the sequence of iterates
n=1∞ also converges to the fixed point.
Proof: From Theorem 6.7.2 there is a unique fixed point for fn. Thus
By uniqueness, f
Now consider the sequence of iterates. Suppose it fails to converge to x. Then there is
ε > 0 and a subsequence nk such that
Now nk = pkn + rk where rk is one of the numbers
. It follows that
there exists one of these numbers which is repeated infinitely often. Call it
and let the
further subsequence continue to be denoted as nk.
In other words,
However, from Theorem 6.7.2, as k →∞,fpkn
which contradicts the above
inequality. Hence the sequence of iterates converges to x
, as it did for f
Definition 6.7.6 Let f :
be a function. Then it is said to be
uniformly continuous on X if for every ε >
0 there exists a δ >
0 such that whenever
are two points of X with d
< δ, it follows that ρ
Note the difference between this and continuity. With continuity, the δ could depend
on x but here it works for any pair of points in X.
Lemma 6.7.7 Suppose xn → x and yn → y. Then d
Proof: Consider the following.
and the right side converges to 0 as n →∞. ■
There is a remarkable result concerning compactness and uniform continuity.
Theorem 6.7.8 Let f :
be a continuous function and let K be a
compact 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
such that d
. Since K
is compact, it is
sequentially compact and so there exists a subsequence, still denoted as
xn → x ∈ K.
n → x
also and so
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
Definition 6.7.9 Let fn : X → Y where
are two metric spaces.
is said to converge poinwise to a function f
: X → Y if for every
x ∈ X,
is said to converge uniformly if for all ε >
0, there exists N such that if n ≥ N,
Here is a well known example illustrating the difference between pointwise and
Example 6.7.10 Let fn
xn on the metric space
. Then this function
converges pointwise to
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 6.7.11 Let fn : X → Y where
are two metric spaces and
suppose each fn is continuous at x ∈ X and also that fn converges uniformly to f
on X. Then f is also continuous at x. In addition to this, if each fn is uniformly
continuous on X, then the same is true for f.
Proof: Let ε > 0 be given. Then
By uniform convergence, there exists N such that both ρ
are less than
3 provided n ≥ N.
Thus picking such an n,
Now from the continuity of fn, there exists δ > 0 such that if d
Hence, if d
Hence, f is continuous at x.
Next consider uniform continuity. It follows from the uniform convergence that if x,
are any two points of
then if n ≥ N,
then, picking such an n,
By uniform continuity of fn there exists δ such that if d
then the term on the
right in the above is less than ε∕
Hence if d
and so f
is uniformly continuous as claimed. ■