222 CHAPTER 9. BASIC FUNCTION SPACES

Lemma 9.1.6 ∥·∥γ

is a norm for BC ([a,b] ;Fn) and BC ([a,b] ;Fn) is a complete normedlinear space. Also, a sequence is Cauchy in ∥·∥

γif and only if it is Cauchy in ∥·∥.

Proof: First consider the claim about ∥·∥γ

being a norm. That ∥·∥γ

is a norm followsdirectly from the definition. The claim that BC ([a,b] ;Fn) is complete with respect to thisnorm follows from the observation that the two norms ∥·∥ ,∥·∥

γare equivalent so they have

the same Cauchy sequences. This follows from:

∥f∥ ≡ supt∈T|f(t)|= sup

t∈T

∣∣∣f(t)e−|γ(t−c)|e|γ(t−c)|∣∣∣≤ e|γ(b−a)| ∥f∥

γ

≡ e|γ(b−a)| supt∈T

∣∣∣f(t)e−|γ(t−c)|∣∣∣≤ e|γ||b−a| sup

t∈T|f(t)|= e|γ||b−a| ∥f∥ ■

Why do we care about complete normed linear spaces? The following is a fundamen-tal 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 themathematically interesting thing. The initial value problem is to find t→ x(t) on [a,b] suchthat

x′ (t) = f(t,x(t)) , x(c) = x0

Assuming (t,x)→ f(t,x) is continuous, this is obviously equivalent to the single integralequation

x(t) = x0 +∫ t

cf(s,x(s))ds

Indeed, if x(·) is a solution to the initial value problem, then you can integrate and obtainthe above. Conversely, if you find a solution to the above, integral equation, then you canuse the fundamental theorem of calculus to differentiate and find that it is a solution to theinitial value problem.

Theorem 9.1.7 Let f satisfy the Lipschitz condition

|f(t,x)− f(t,y)| ≤ K |x−y| (9.2)

and the continuity condition

(t,x)→ f(t,x) is continuous. (9.3)

Then there exists a unique solution to the initial value problem,

x(t) = x0 +∫ t

cf(s,x(s))ds, c ∈ [a,b] (9.4)

on [a,b] .

Proof: It is necessary to find a solution to the integral equation

x(t) = x0 +∫ t

cf(s,x(s))ds, t ∈ [a,b]

Let a,b be finite but given and completely arbitrary, c ∈ [a,b]. Let

Fx(t)≡ x0 +∫ t

cf(s,x(s))ds