Suppose U is a punctured disk,if f is holomorphic on U and $\int_{U}$$|f|^2$dxdy<$\infty$,prove that z=0 is a removable singularity
it is obvious to prove f is bounded. however. I do not know where to do first
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Holomorphic functions have Mean Value Property. – Kavi Rama Murthy Feb 05 '21 at 05:00
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@Kavi Rama Murthy,yes and is bounded on any compact subset. this still can not prove the f is bounded near the neighbor of 0 – ymm Feb 05 '21 at 05:14
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Integral of $|f|^{2}$ over any disk conatined in $U$ is bounded by the integral over the whole of $U$. – Kavi Rama Murthy Feb 05 '21 at 05:18
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@Kavi Rama Murthy.i am sorry I still don't know. cause we can not use the Cauchy formula(0 is not know if it is a removable singularity)... – ymm Feb 05 '21 at 07:11
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See https://math.stackexchange.com/a/63472. – Martin R Feb 05 '21 at 07:48
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@KaviRamaMurthy: You can apply the (area) mean value property to a disk of radius $|z|$, but that gives you only $|f(z)| \le C/|z|$. Of course I may have overlooked something. – The Laurent series approach from the above link seems to work. – Martin R Feb 05 '21 at 08:24
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@MartinR Your inequality is good enough. Becasue $zf(z)$ has removable singualiry so $ f(z)$ behaves like $\frac c z$ near $0$ but the the hypotheis fails unless $c=0$. – Kavi Rama Murthy Feb 05 '21 at 08:29
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@KaviRamaMurthy: Yes, you are right, the possibility of a simple pole can be excluded. – Martin R Feb 05 '21 at 08:33
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@KaviRamaMurthy: I wonder if the question should be answered here or closed as a duplicate. – Martin R Feb 05 '21 at 08:35
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1I think it should be closed. I will now vote for closure. @MartinR – Kavi Rama Murthy Feb 05 '21 at 08:37