Abstract
Abstract
Abstract
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CHAPTER 4. THEORY 91<br />
Thisshowsthatifwechoosex and y closetogether,wehavethat<br />
Kmax<br />
|G2(u)(x, k) 2 − G2(u)(y, k) 2 |dk ≤ γ.<br />
−Kmax<br />
So Kmax<br />
−Kmax G2(u)(x, k) 2 is a continuous function. This means that G1(u),G2(u) ∈ X<br />
for u ∈ X, and therefore, Z maps X to X.<br />
Show Z is continuous<br />
Let {un} ⊂X with un → u ∈ X. To show that Z is continuous, we need to show<br />
Z(un) → Z(u), or G1(un) → G1(u) andG2(un) → G2(u). Now for the G1 term we<br />
have for k>0,<br />
|G1(u)(x, k) − G1(un)(x, k)| =<br />
<br />
<br />
C<br />
k<br />
x<br />
0<br />
dze C(z−x)<br />
k h(z,k) ¯<br />
Kmax<br />
−Kmax<br />
x<br />
≤ dz<br />
0<br />
C C(z−x)<br />
e k |<br />
k ¯ Kmax<br />
h(z,k)|<br />
−Kmax<br />
x<br />
≤ dz<br />
0<br />
C C(z−x)<br />
e k <br />
k ¯ <br />
h∞ 2KmaxuX<br />
= ¯ <br />
Cx<br />
−<br />
h∞ 2Kmaxu − unX 1 − e k<br />
≤ ¯ <br />
h∞ 2Kmaxu − unX,<br />
[u(z,k ′ ) − un(z,k ′ )]dk ′<br />
|u(z,k ′ ) − un(z,k ′ )|dk ′
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