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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

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