Definition 7.1.1Let T be a subset of some F^{m}, possibly all of F^{m}. Let BC
(T;Fn)
denote the bounded continuousfunctionsdefined on T.^{1}Then this is a vector space (linear space) with respect to the usual operations of addition and scalarmultiplication of functions. Also, define a norm as follows:
∥f∥ ≡ sup |f (t)| < ∞.
t∈T
This is a norm because it satisfies the axioms of a norm which are as follows:
∥f + g∥ ≤ ∥f∥+ ∥g∥
∥αf∥ = |α|∥f∥
∥f∥ ≥ 0 and equals 0 if and only if f = 0
A sequence
{fn}
in BC
(T ;Fn )
is a Cauchy sequence if for every ε > 0 there exists M_{ε}such that ifm,n ≥ M_{ε}, then
∥fn − fm∥ < ε
Such a normed linear space is called complete if every Cauchy sequence converges. Such a complete normedlinear space is called a Banach space.This norm is often denoted as
∥⋅∥
_{∞}.
I am letting T be a subset of F^{n} 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 ≡
[a,b]
, an interval. This is
described in the following definition.
Definition 7.1.2For f ∈ BC
([a,b];Fn)
, let c ∈
[a,b]
,γ a real number. Then
|| − |γ(t−c)|||
∥f∥γ ≡ sup |f (t) e |
t∈[a,b]
Then this is a norm. The above Definition 7.1.1corresponds to γ = 0.
Lemma 7.1.3
∥⋅∥
_{γ}is a norm for BC
([a,b];Fn)
and BC
([a,b];Fn)
is a complete normed linearspace. Also, a sequence is Cauchy in
Therefore, g is continuous at t. Since t is arbitrary, this shows that g is continuous on T.
Thus g ∈ BC
(T;Fn)
. By ∗,
∥fn − g∥
< ε when n is large enough so lim_{n→∞}
∥fn − g∥
= 0.
■
Definition 7.1.5When lim_{n→∞}
∥fn − f∥
= 0, we say that f_{n}converges uniformly to f and speakof 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
of f_{n}
(t)
= t^{n} for t ∈
[0,1]
. The limit function is discontinuous on this interval and is 0 on [0,1) and 1 at
1.
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 7.1.6Let
(X,∥⋅∥)
be a complete (Cauchy sequences converge.) normed linear space and letF : X → X be a contraction map. That is,
∥F x− F y∥ ≤ r∥x − y∥, 0 ≤ r < 0
Then F has a unique fixed point, that is a point x ∈ X such that Fx = x. In addition to this, if
∥Fx0 − x0∥
< R
(1 − r)
and F is only defined on B
(x0,R )
then F has a unique fixed point in this ball.Here B
(x0,R )
signifies the set of all x such that
∥x − x0∥
≤ R. Also, the sequence
{F nx0}
converges.
Proof:Pick any x_{0}∈ X. Consider the sequence
{F nx0}
. 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,
n∑−1 n−∑ 1 m
∥∥Fk+1x0 − F kx0∥∥ ≤ rk∥F x0 − x0∥ ≤ r--∥F x0 − x0∥ (7.2)
k=m k=m 1− r
(7.2)
since r < 1, this is a Cauchy sequence. Hence it converges to some x. Therefore,
n n+1 n
x = ln→im∞ F x0 = lni→m∞ F x0 = F nli→m∞ F x0 = F x.
The third equality is a consequence of the following consideration. If z_{n}→ z, then
∥F zn − F z∥ ≤ r∥zn − z∥
so also Fz_{n}→ Fz. In the above, F^{n}x_{0} plays the role of z_{n} and its limit plays the role of z.
The fixed point is unique because if you had two of them, x,
ˆx
, then
∥x− ˆx∥ = ∥Fx − F ˆx∥ ≤ r∥x− ˆx∥
and so x =
ˆx
.
In the second case, let m = 0 in 7.2 and you get the estimate
1
∥F nx0 − x0∥ ≤ 1−-r ∥F x0 − x0∥ < R
It is still the case that the sequence
{Fnx }
0
is a Cauchy sequence and must therefore converge to some x ∈B
(x ,R)
0
which is a fixed point as before. The fixed point is unique because of the same argument as
before. ■
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
(t)
on
[a,b]
such that
x′(t) = f (t,x(t))
x(c) = x
0
Assuming
(t,x)
→ f
(t,x)
is continuous, this is obviously equivalent to the single integral equation
∫
t
x (t) = x0 + c f (s,x (s))ds
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
problem.
Theorem 7.1.7Let f satisfy the Lipschitz condition
|f (t,x)− f (t,y )| ≤ K |x − y| (7.3)
(7.3)
and the continuity condition
(t,x) → f (t,x) is continuous. (7.4)
(7.4)
Then there exists a unique solution to the initial value problem,
∫ t
x (t) = x0 + c f (s,x (s))ds,c ∈ [a,b] (7.5)
(7.5)
on
[a,b]
.
Proof:It is necessary to find a solution to the integral equation
∫ t
x(t) = x0 + f (s,x(s))ds, t ∈ [a,b]
c
Let a,b be finite but given and completely arbitrary, c ∈
[a,b]
. Let Fx
(t)
≡ x_{0} + ∫_{c}^{t}f
(s,x (s))
ds
Thus
F : BC ([a,b],Fn) → BC ([a,b],Fn)
Let
∥⋅∥
_{γ} be the new norm on BC
([a,b],Fn)
.
| |
∥f∥γ ≡ sup ||f (t)e− |γ(t−c)|||
t∈[a,b]
Note that
|x (s) − y (s)| = e|γ(s−c)|e−|γ(s−c)||x(s)− y(s)| ≤ e|γ(s−c)|∥x − y∥γ
Then for t ∈
[a,b]
,
|∫ | |∫ |
|| t || || t ||
|F x(t)− Fy (t)| ≤ | c |f (s,x(s)) − f (s,y (s))|ds| ≤ |c K |x(s)− y(s)|ds|
|∫ | |∫ |
≤ K || te|γ(s−c)|∥x − y∥ ds||= K ∥x − y∥ || te|γ(s− c)|ds|| (*)
| c γ | γ| c |
(*)
Now consider the case that t ≥ c. The above reduces to
∫ t (e|γ|(s−c) )
= K ∥x − y ∥γ e|γ|(s−c)ds = K ∥x− y∥γ -------|tc
(c ) |(γ| )
= K ∥x − y ∥ e|γ|(t−-c) − 1-- = K ∥x − y ∥ e|γ(t−c)|-− 1--
γ |γ| |γ| γ |γ| |γ|
Next consider ∗ when t < c. In this case, ∗ becomes
∫ c ( e|γ|(c−s) )
K ∥x − y∥γ e|γ|(c−s)ds = K ∥x − y∥γ ------|ct
t ( − |γ| )
= K ∥x − y∥ -1--+ e|γ|(c−-t)
γ − |γ| |γ|
( e|γ(t−c)| 1 )
≤ K ∥x − y∥γ --|γ|---− |γ|-
It follows that for any t ∈
[a,b]
,
−|γ(t−c)| −|γ(t−c)|( e|γ(t−c)| 1 )
e |Fx (t)− F y(t)| ≤ K ∥x − y∥γe --|γ|---− |γ|-
1
≤ K ∥x − y∥γ---
|γ|
and so
1
∥F x− F y∥γ ≤ K ∥x − y∥γ |γ|
so let
|γ|
> 2K and this shows that F is a contraction map on BC
([a,b];Fn)
.
Thus there is a unique solution to the above integral equation and hence a unique solution to the initial
value problem 7.5 on
[a,b]
. ■
Definition 7.1.8For the integral equation,
∫ t
x (t) = x0 + f (s,x (s))ds
c
one considers the Picard iterates.These are given as follows. x_{0}
(t)
≡ x_{0}and
∫ t
xn+1 (t) ≡ x0 + f (s,xn (s))ds
c
Thus letting Fx
(t)
≡ x_{0} + ∫_{c}^{t}f
(s,x(s))
ds, the Picard iterates are of the form Fx_{n} = x_{n+1}.
By Theorem 7.1.6, the Picard iterates converge in BC