Abstract
Abstract
Abstract
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CHAPTER 4. THEORY 87<br />
we have shown that we can choose x and y close enough so that<br />
Kmax<br />
0<br />
|G1(u)(x, k) 2 − G1(u)(y, k) 2 |dk ≤ γ<br />
2 .<br />
Using a very similar argument for the case k0. We want to show that Kmax<br />
−Kmax G2(u)(x, k ′ ) 2 dk ′ is<br />
continuous. So<br />
Kmax<br />
<br />
G2(u)(x, k)<br />
−Kmax<br />
2 Kmax<br />
dk − G2(u)(y, k)<br />
−Kmax<br />
2 Kmax<br />
<br />
dk<br />
= [G2(u)(x, k)<br />
−Kmax<br />
2 − G2(u)(y, k) 2 <br />
<br />
]dk<br />
Kmax<br />
<br />
<br />
<br />
= [G2(u)(x, k) − G2(u)(y, k)][G2(u)(x, k)+G2(u)(y, k)]dk<br />
−Kmax<br />
Kmax<br />
≤ |G2(u)(x, k) − G2(u)(y, k)|[|G2(u)(x, k)| + |G2(u)(y, k)|]dk<br />
−Kmax<br />
≤ 8τ<br />
h<br />
<br />
Kmax<br />
2KmaxuXT ∞ |G2(u)(x, k) − G2(u)(y, k)|dk.<br />
−Kmax
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