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CHAPTER 4. THEORY 89<br />

So, we have<br />

Kmax<br />

|G2(u)(x, k) − G2(u)(y, k)|dk<br />

0<br />

<br />

e −Cx<br />

x<br />

C Cz<br />

k e k dz +<br />

y k C −Cy<br />

[e k − e<br />

k −Cx<br />

y<br />

k ] e<br />

0<br />

Cz<br />

<br />

k dz dk<br />

= 4τ <br />

Kmax <br />

2KmaxT ∞uX 1 − e<br />

h<br />

0<br />

C(y−x) <br />

k + e −Cy<br />

k − e −Cx<br />

<br />

k e Cy <br />

k − 1 dk<br />

= 4τ <br />

Kmax <br />

2KmaxT ∞uX 1 − e<br />

h<br />

0<br />

C(y−x) <br />

k + 1 − e C(y−x)<br />

k − e −Cy<br />

k + e −Cx<br />

<br />

k dk<br />

= 4τ <br />

Kmax <br />

2KmaxT ∞uX 2 − 2e<br />

h<br />

C(y−x)<br />

k + e −Cx<br />

k − e −Cy<br />

k dk.<br />

≤ 4τ <br />

Kmax<br />

2KmaxT ∞uX<br />

h<br />

0<br />

We will now break the k interval (0,Kmax] into two intervals. Choose ɛk so that<br />

Kmax > ɛk > 0. Then the two intervals are (0,ɛk) and[ɛk,Kmax]. For the first<br />

interval we have:<br />

4τ <br />

ɛk <br />

2KmaxT ∞uX 2 − 2e<br />

h<br />

0<br />

C(y−x)<br />

k + e −Cx<br />

k − e −Cy<br />

k dk<br />

≤ 4τ <br />

ɛk<br />

2KmaxT ∞uX |2 − 2e<br />

h<br />

0<br />

C(y−x)<br />

k + e −Cx<br />

k − e −Cy<br />

k |dk<br />

≤ 4τ <br />

ɛk <br />

2KmaxT ∞uX 2+2e<br />

h<br />

0<br />

C(y−x)<br />

k + e −Cx<br />

k + e −Cy<br />

k dk<br />

≤ 4τ <br />

ɛk<br />

2KmaxT ∞uX [2+2+1+1]dk<br />

h<br />

0<br />

= 24τ <br />

2KmaxT ∞uXɛk.<br />

h<br />

0

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