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I am trying to show that if $f$ analytic in $0 \lt |z| \lt R$ and there exists $M \gt 0$ such that for every $0 \lt r \lt R$ we have:
$$r\int_{0}^{2\pi}|f(re^{i\theta})|d\theta \lt M$$ then $z=0$ is either a simple pole or removable singularity.

Since I need to show that $z=0$ is either a simple pole or removable singularity, I tried to prove that $\lim_{z\to0}zf(z) = c$

So I first tried to estimate $|f(z)|$ using Cauchy's integral formula for punctured disk, where $C_1 = r_1e^{i\theta}, C_2 = r_2e^{i\theta}, 0 \lt r_1 \lt r_2 \lt R$: $$|f(z)| = |\frac{1}{2\pi i} \int_{C_2}\frac{f(\zeta)}{\zeta-z}d\zeta - \frac{1}{2\pi i} \int_{C_1}\frac{f(\zeta)}{\zeta-z}d\zeta | \le$$ $$\le \frac{1}{2\pi}|\int_{C_2}\frac{f(\zeta)}{\zeta-z}d\zeta| + \frac{1}{2\pi} |\int_{C_1}\frac{f(\zeta)}{\zeta-z}d\zeta | =$$ $$\frac{1}{2\pi}|\int_{0}^{2\pi}\frac{r_2ie^{i\theta}f(r_2e^{i\theta})}{r_2e^{i\theta}-z}d\theta| + \frac{1}{2\pi}|\int_{0}^{2\pi}\frac{r_1ie^{i\theta}f(r_1e^{i\theta})}{r_1e^{i\theta}-z}d\theta| \le$$ $$\le \frac{r_2}{2\pi}\int_{0}^{2\pi}\frac{|f(r_2e^{i\theta})|}{|r_2e^{i\theta}-z|}d\theta + \frac{r_1}{2\pi}\int_{0}^{2\pi}\frac{|f(r_1e^{i\theta})|}{|r_1e^{i\theta}-z|}d\theta$$

if we choose $r_1 \lt r_2 \lt \frac{1}{2}|z|$ then $|\zeta| \le \frac{|z|}{2}$
and then $|\zeta -z| \ge ||\zeta|-|z|| \ge \frac{|z|}{2} \implies \frac{1}{|\zeta-z|} \le \frac{2}{|z|}$

therefore we have: $$ |f(z)| \le \frac{r_2}{\pi|z|}\int_{0}^{2\pi}|f(r_2e^{i\theta})|d\theta + \frac{r_1}{\pi|z|}\int_{0}^{2\pi}|f(r_1e^{i\theta})|d\theta \lt \frac{2M}{\pi |z|} $$

so $|zf(z)| \lt \frac{2M}{\pi}$ and therefore the limit is finite and could be zero in which case $z=0$ is removable singularity or non-zero in which case it's a simple pole.

Can someone confirm that this proof is correct? I the general approach is fine but there's a mistake please point it out. If I am completely way off please give a hint regarding how to approach this, but without a full solution please

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  • $\begingroup$ Can you add more steps to prove that if $r_1 \lt r_2 \lt \frac{1}{2}|z|$ then $|\zeta-z| \ge \frac{1}{2}|z| \implies \frac{1}{|\zeta-z|} \le \frac{2}{|z|}$? It is not obvious to me. $\endgroup$ – mike Feb 14 '15 at 0:00
  • $\begingroup$ i dont see it... you bound the absolute value of zf(z) from ABOVE and hence deduce it equals a non-zero limit? you need to bound it from BELOW $\endgroup$ – Asier Calbet Feb 14 '15 at 0:02
  • $\begingroup$ @mike yes, let me update the question $\endgroup$ – benji Feb 14 '15 at 0:02
  • $\begingroup$ @Assaultous2 I should have been clearer - if the limit is non-zero it's a simple pole, if it is zero it's removable singularity - I will update the question $\endgroup$ – benji Feb 14 '15 at 0:04
  • $\begingroup$ @mike I updated the steps you asked, but now I realize that if $r_1 \lt r_2 \lt \frac{1}{2}|z|$ then $z$ is outside $C_2$ and I don't think this works.. maybe the choice of $r_1$ like that is fine because it lets me bound it and I only need to work with a different $r_2$ $\endgroup$ – benji Feb 14 '15 at 0:30
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Almost correct except for several portions.

A revised version is :
Let $r_2<R$. First we take $z$ with $0<|z|<\frac{r_2}{2}$. Next take $r_1$ with $0<r_1<\frac{|z|}{2}$. We try to estimate $|f(z)|$ using Cauchy's integral formula for punctured disk: \begin{align} |f(z)|&=\left|\frac{1}{2\pi i}\int_{C_2} \frac{f(\zeta)}{\zeta -z} d\zeta- \frac{1}{2\pi i}\int_{C_1}\frac{f(\zeta)}{\zeta-z}d\zeta\right|\\ &\le\frac{1}{2\pi}\int_{C 2} \frac{|f(\zeta)|}{|\zeta -z|} |d\zeta|+\frac{1}{2\pi}\int_{C_1}\frac{|f(\zeta)|}{|\zeta-z|}|d\zeta | \end{align} where $C_1 = r_1e^{i\theta}, C_2 = r_2e^{i\theta}.$ It is easily seen that $$\frac{1}{2\pi}\int_{C 2} \frac{|f(\zeta)|}{|\zeta -z|} |d\zeta|\le \frac{1}{\pi r_2}\int_{C 2} |f(\zeta)|d\zeta|\le \frac{M}{\pi r_2}$$ since $|\zeta-z|\ge \frac{r_2}{2} $ for $\zeta\in C_1$, and $$\frac{1}{2\pi}\int_{C_1}\frac{|f(\zeta)|}{|\zeta-z|}|d\zeta |\le \frac{1}{\pi |z|}\int_{C_1}|f(\zeta)||d\zeta |\le \frac{M}{\pi |z|}$$ since $|\zeta-z|\ge \frac{|z|}{2}$ for $z\in C_1$. Therefore we have $$|f(z)|\le \frac{M}{\pi r_2}+\frac{M}{\pi |z|} \, \left(2r_1<|z|<\frac{r_2}{2}\right).$$ Since we can take $r_1$ arbitrary small, we have $$ |f(z)|\le \frac{M}{\pi r_2}+\frac{M}{\pi |z|} \, \left(0<|z|<\frac{r_2}{2}\right),$$ so $|zf(z)|\le \frac{M}{\pi r_2}|z|+\frac{M}{\pi}$ and therefore the limit is finite and could be zero in which case $z=0 $ is removable singularity or non-zero in which case it's a simple pole.

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  • $\begingroup$ yes, I believe you're right. I had it in front of me in my notes but missed it :( +1. I was able to find a different, shorter approach though :) $\endgroup$ – benji Feb 15 '15 at 16:53
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I was able to find a solution using a different approach:

Using Laurent's theorem we have: $$a_n = \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta^{n+1}}d\zeta$$ where $C=re^{i\theta}$

and then using the given property of $f$: $$ |a_n| = \left| \frac{1}{2\pi i}\int_C \frac{f(\zeta)}{\zeta^{n+1}}d\zeta \right| \le \frac{M}{2\pi r^{n+1}} $$ This expression goes to $0$ as $r \to 0$ for $n<-1$

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