7.1 Bounded Continuous Functions
As before, F will denote either ℝ or ℂ.
Definition 7.1.1 Let T be a subset of some Fm, possibly all of Fm. Let BC
denote the bounded continuous
functions defined on T.
Then this is a vector space (linear space) with respect to the usual operations of addition and scalar
multiplication of functions. Also, define a norm as follows:
This is a norm because it satisfies the axioms of a norm which are as follows:
is a Cauchy sequence if for every ε >
0 there exists Mε such that if
m,n ≥ Mε, then
Such a normed linear space is called complete if every Cauchy sequence converges. Such a complete normed
linear space is called a Banach space. This norm is often denoted as
I am letting T be a subset of Fn just to keep things in familiar territory. T can be an arbitrary metric
space or even a general topological space.
Now there is another norm which works just as well in the case where T ≡
an interval. This is
described in the following definition.
Definition 7.1.2 For f ∈ BC
, let c ∈
,γ a real number. Then
Then this is a norm. The above Definition 7.1.1 corresponds to γ = 0.
γ is a norm for BC
is a complete normed linear
space. Also, a sequence is Cauchy in
γ if and only if it is Cauchy in
Proof: First consider the claim about
being a norm. To simplify notation, let T
. It is
= 0 if and only if f
so it does what is should for scalar multiplication. Next consider the triangle inequality.
The rest follows from the next inequalities.
Now consider the general case where T is just some set.
Lemma 7.1.4 The collection of functions BC
is a normed linear space (vector space) and
it is also complete which means by definition that every Cauchy sequence converges.
Proof: Showing that this is a normed linear space is entirely similar to the argument in the above for
γ = 0 and T =
be a Cauchy sequence. Then for each
t ∈ T,
is a Cauchy sequence in
completeness of Fn
this converges to some g
. We need to verify that
0 and that g ∈
0 be given. There exists Mε
such that if m,n ≥ Mε,
Let n > Mε
By Lemma 1.13.2
which says you can switch supremums,
so in fact g is bounded. Now by the fact that fn is continuous, there exists δ > 0 such that if
It follows that
Therefore, g is continuous at t. Since t is arbitrary, this shows that g is continuous on T.
Thus g ∈ BC
is large enough so limn→∞
Definition 7.1.5 When limn→∞
, we say that fn converges uniformly to f and speak
of uniform convergence. This norm is also called the uniform norm.
Note that uniform convergence of continuous functions imparts continuity to the limit function. This is
not true of pointwise convergence, that the sequence converges for each t, as can be seen by consideration
for t ∈
The limit function is discontinuous on this interval and is 0 on [0,
1) and 1 at
Now here is a major theorem called the Banach fixed point theorem.This theorem lives on complete
normed linear spaces, more generally on complete metric spaces.
Theorem 7.1.6 Let
be a complete (Cauchy sequences converge.) normed linear space and let
: X → X be a contraction map. That is,
Then F has a unique fixed point, that is a point x ∈ X such that Fx = x. In addition to this, if
and F is only defined on B
then F has a unique fixed point in this ball.
signifies the set of all x such that
≤ R. Also, the sequence
Proof: Pick any x0 ∈ X. Consider the sequence
I will argue that this is a Cauchy sequence.
To see this, suppose n,m ≥ M
with n > m
and consider the following which comes from the triangle
inequality for the norm,
Using this in the above,
since r < 1, this is a Cauchy sequence. Hence it converges to some x. Therefore,
The third equality is a consequence of the following consideration. If zn → z, then
so also Fzn → Fz. In the above, Fnx0 plays the role of zn and its limit plays the role of z.
The fixed point is unique because if you had two of them, x,
and so x =
In the second case, let m = 0 in 7.2 and you get the estimate
It is still the case that the sequence
is a Cauchy sequence and must therefore converge to some
which is a fixed point as before. The fixed point is unique because of the same argument as
Why do we care about complete normed linear spaces? The following is a fundamental existence
theorem for ordinary differential equations. It is one of those things which, incredibly, is not presented in
ordinary differential equations courses. However, this is the mathematically interesting thing. The initial
value problem is to find t → x
is continuous, this is obviously equivalent to the single integral equation
Indeed, if x
is a solution to the initial value problem, then you can integrate and obtain the
above. Conversely, if you find a solution to the above, integral equation, then you can use the
fundamental theorem of calculus to differentiate and find that it is a solution to the initial value
Theorem 7.1.7 Let f satisfy the Lipschitz condition
and the continuity condition
Then there exists a unique solution to the initial value problem,
Proof: It is necessary to find a solution to the integral equation
Let a,b be finite but given and completely arbitrary, c ∈
be the new norm on BC
Then for t ∈
Now consider the case that t ≥ c. The above reduces to
Next consider ∗
when t < c
. In this case, ∗
It follows that for any t ∈
and this shows that F
is a contraction map on BC
Thus there is a unique solution to the above integral equation and hence a unique solution to the initial
value problem 7.5 on
Definition 7.1.8 For the integral equation,
one considers the Picard iterates. These are given as follows. x0
≡ x0 and
Thus letting Fx
ds, the Picard iterates are of the form Fxn
By Theorem 7.1.6, the Picard iterates converge in BC
with respect to
and so they also converge in BC
with respect to the usual norm