Kenneth Kuttler

22.7. DIFFERENTIAL EQUATIONS IN BANACH SPACE 599

which is finite. Thus√

T ∗T is self adjoint and in L2 (H,H) and so

∑k

∥∥∥√T ∗T ek

∥∥∥2= ∑

∥∥∥√T ∗T fk

∥∥∥2= ∑

k∥T fk∥2

showing that ∑k ∥Tek∥2 = ∑k ∥T fk∥2. ■There is a whole book on powers of operators. This has just given a short introduction.

See [33].

22.7 Differential Equations in Banach SpaceHere we consider the initial value problem for functions which have values in a Banachspace. Let X be a Banach space.

Definition 22.7.1 Define BC ([a,b] ;X) as bounded continuous functions f whichhave values in the Banach space X. For f ∈ BC ([a,b] ;X) , γ a real number. Then

∥ f∥γ≡ sup

t∈[a,b]

∥∥∥ f (t)eγ(t−a)∥∥∥ (22.25)

Then this is a norm. The usual norm is given by ∥ f∥ ≡ supt∈[a,b] ∥ f (t)∥ .

Lemma 22.7.2 ∥·∥γ

is a norm for BC ([a,b] ;X) and BC ([a,b] ;X) 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. To simplify notation, letT = [a,b]. It is clear that ∥ f∥

γ= 0 if and only if f = 0 and ∥ f∥

γ≥ 0. Also,

∥α f∥γ≡ sup

t∈T

∥∥∥α f (t)eγ(t−a)∥∥∥= |α|sup

t∈T

∥∥∥ f (t)eγ(t−a)∥∥∥= |α|∥ f∥

so it does what is should for scalar multiplication. Next consider the triangle inequality.

∥ f +g∥γ

= supt∈T

∥∥∥( f (t)+g(t))eγ(t−a)∥∥∥≤ sup

t∈T

(∣∣∣ f (t)eγ(t−a)∣∣∣+ ∣∣∣g(t)eγ(t−a)

∣∣∣)≤ sup

t∈T

∣∣∣ f (t)eγ(t−a)∣∣∣+ sup

t∈T

∣∣∣g(t)eγ(t−a)∣∣∣= ∥ f∥

γ+∥g∥

The rest follows from the next inequalities.

∥ f∥ ≡ supt∈T∥ f (t)∥= sup

t∈T

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

≡ e|γ(b−a)| supt∈T

∥∥∥ f (t)eγ(t−a)∥∥∥≤ (e|γ|(b−a)

)2supt∈T∥ f (t)∥=

(e|γ|(b−a)

)2∥ f∥ ■

Now consider the ordinary initial value problem

x′ (t) = F (t,x(t)) , x(t0) = x0, t ∈ [a,b] , t0 ∈ [a,b] (22.26)

where here F : [a,b]×X → X is continuous and satisfies the Lipschitz condition

∥F (t,x)−F (t,y)∥ ≤ K ∥x− y∥ , F : [a,b]×X → X is continuous (22.27)

22.7. DIFFERENTIAL EQUATIONS IN BANACH SPACE 599which is finite. Thus 7*T is self adjoint and in % (H,H) and so2 2Y||vrre| =L\ vera) =Diirsd?k k k. 2 2showing that ¥y ||Tex||" = Le ||7 fell”.There is a whole book on powers of operators. This has just given a short introduction.See [33].22.7 Differential Equations in Banach SpaceHere we consider the initial value problem for functions which have values in a Banachspace. Let X be a Banach space.Definition 22.7.1 Define BC ([a,b];X) as bounded continuous functions f whichhave values in the Banach space X. For f € BC (|a,b];X), ya real number. Thenll,= vp. Ff (re) (22.25)tela,bThen this is a norm. The usual norm is given by || f|| = sup;ejay) ||f (||.Lemma 22.7.2 ||- || is a norm for BC ([a,b] ;X) and BC [a,b] ;X)) is a complete normedlinear space. Also, a sequence is Cauchy in |\-||, if and only if it is Cauchy in ||-\).Proof: First consider the claim about ||-||,, being a norm. To simplify notation, letT = {a,b}. It is clear that || f||, = 0 if and only if f = 0 and || f||, > 0. Also,=|a|sup]] f (reteT= lal llfllylaf, =sup jaf (jetteTso it does what is should for scalar multiplication. Next consider the triangle inequality.f+ ally = sup |(f@) +e yer <sup (ret | + fe er™])teT teT< sup |f(jer™| +sup|e(r)e"| = [iFlly + llallyteT teTThe rest follows from the next inequalities.IF = sup | f (0) || =sup|]f(eM™ Me MOY < el FI,teT teT= el">-4)| sup | f (perateT< (eX) sup|ir(nl) = (2) Ip)teTNow consider the ordinary initial value problemx' (t) =F (t,x(t)), x(to) =xo, t € [a,b] ,t0 € [a, | (22.26)where here F’: [a,b] x X — X is continuous and satisfies the Lipschitz condition||F (t,x) — F (t,y)|| < K ||x—y||, F : [a,b] x X — X is continuous (22.27)