ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...
ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...
ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...
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δ<br />
0<br />
Then<br />
<br />
f(ξ)<br />
δf(w) = δ<br />
dξ −<br />
|ξ|=1/2 ξ − w<br />
and thus<br />
<br />
<br />
<br />
f(w)<br />
−<br />
<br />
|ξ|=1/2<br />
<strong>ANALYSIS</strong> QUALS 31<br />
f(ξ)<br />
ξ − w dξ<br />
<br />
<br />
<br />
<br />
<br />
≤ 1<br />
δ<br />
<br />
<br />
<br />
<br />
<br />
δ<br />
0<br />
δ<br />
0<br />
<br />
<br />
|ξ|=ε<br />
|ξ|=ε<br />
f(ξ)<br />
ξ − w dξdε<br />
f(ξ)<br />
ξ − w dξdε<br />
<br />
<br />
<br />
<br />
.<br />
We claim that this tends to 0 as δ → 0, which would prove that f(w) =<br />
f(ξ)<br />
1<br />
|ξ|=1/2 ξ−w dξ for |w| < 2 , and this is holomorphic, so any singularity at 0<br />
would be removable.<br />
But using Fubini, Cauchy-Schwarz, and basic estimations, we have that<br />
<br />
f(ξ)<br />
ξ − w dξ<br />
<br />
<br />
<br />
<br />
=<br />
<br />
<br />
δ 2π<br />
f(εe<br />
<br />
<br />
iθ )<br />
εeiθ − w iεeiθ <br />
<br />
<br />
dθdε<br />
<br />
|ξ|=ε<br />
≤<br />
0 0<br />
δ 2π<br />
0<br />
≤ 2<br />
|w|<br />
≤ 2<br />
|w|<br />
≤ 2<br />
|w|<br />
0<br />
2π δ<br />
0<br />
2π<br />
<br />
0<br />
|f(εeiθ )|<br />
|εeiθ − w| εdθdε<br />
0<br />
δ<br />
0<br />
|f(εe iθ )|εdεdθ<br />
|f|<br />
{0