When you want to show a function is holomorphic on a set $U$ it suffices to show that it is holomorphic on any disc in $U$.

1) Why is this true?

Also I was doing a problem to show the uniform limit of a sequence of holomorphic functions was also holomorphic in the set. This involved using Morera's Thm.

The next part of the problem was to show the sum from $r=0$ to $r=\infty$ of $\frac{1}{(r-z)^2} $ was a holomorphic. Using the previous part if you show it converges uniformly in ($\mathbb{C}$ set difference $\mathbb{N}$) you are done In order to do this, one of the hints was the following: " The series does not converge (UNIFORMLY) on the whole of ($\mathbb{C}$ set difference $\mathbb{N}$) but for any $z$ in ($\mathbb{C}$ set difference $\mathbb{N}$) convergence is uniform on some neighbourhood of $z$.

2) How can something converge uniformly on every disc in the set but not on the whole thing?

I would appreciate answers to $1)$ and $2)$


1 Answer 1

  1. Because by definition, $f$ is holomorphic on an open set $U$ iff it is complex differentiable in a neighbourhood of every point of $U$.
  2. I hope the hint did not really say "The series does not converge on the whole of ...". If it converges uniformly on some neighbourhood of every point in an open set $U$, it certainly converges on all of $U$. But it might not converge uniformly on $U$. A simple example is the sequence $z/n$, which converges to $0$ uniformly on every disc in $\mathbb C$ but does not converge uniformly on $\mathbb C$.
  • $\begingroup$ yes sorry i forgot the word uniformly. the hint said what you said. ill change it. Thanks $\endgroup$
    – Arcane1729
    Dec 21, 2015 at 23:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.