If $f$ is holomorphic on $U\setminus D$ and $D$ consists of poles isolated singularities, is there any isolated singularity outside $D$?

Question

Let $X$ be a normed vector space.

1.: If $\Omega\subseteq\mathbb C$ is open and $f:\Omega\to X$ is holomorphic, then $z_0\in\mathbb C\setminus\Omega$ is called an isolated singularity of $f$ if there is a neighborhood $N$ of $z_0$ with $N\setminus\{z_0\}\subseteq\Omega$.

2.: If $U\subseteq\mathbb C$ is open, then $g$ is called a $X$-valued holomorphic function on $U$ up to isolated singularity if there is a $D\subseteq U$ with

1. $U\setminus D$ is open;
2. $g:U\setminus D\to X$ is holomorphic;
3. Every point in $D$ is an isolated singularity of $g$.

Let $\Omega:=U\setminus D$ and $$S:=\{z_0\in\mathbb C\setminus\Omega:z_0\text{ is an isolated singularity of }f\}.$$ By assumption, $D\subseteq S$. But does it necessary hold $D=S$?

If $z_0\in S$, then clearly $$z_0\in\mathbb C\setminus\Omega=(\mathbb C\setminus U)\cup D\tag1.$$ On the other hand, since $z_0$ is an isolated singularity of $f$, there is a neighborhood $N$ of $z_0$ with $$N\setminus\{z_0\}\subseteq\Omega=U\setminus D\tag2.$$ However, that doesn't necessary yield $z_0\in D$ ...

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I first thought that a construction of the following kind could yield a counterexample: Let

• $z_1\in\mathbb C$;
• $\varepsilon_1>0$ and $U:=B_{\varepsilon_1}(z_1)$;
• $D:=\{z_1\}$;
• $z_2\in\mathbb C\setminus U$ and $\varepsilon_2>0$ with $$U\cap B_{\varepsilon_2}(z_2)=\emptyset$$ and $$f:U\setminus D\to\mathbb C\;,\;\;\;z\mapsto\frac1{z-z_1}+\frac1{z-z_2}.$$

$f$ does clearly satisfy the definition above. But while I first thought that $z_2\in S\setminus D$, it does not satisfy the definition of being an isolated singularity of $f$ above, since there is no neighborhood $N$ of $z_2$ with $N\setminus\{z_2\}\subseteq\Omega=B_{\varepsilon_1}(z_1)\setminus\{z_1\}$.

Answer 1

There is no normed space, $U$ is a connected open subset of $\Bbb{C}$,

$f$ is complex-valued holomorphic (complex differentiable at every point, thus analytic) on $U-D$ where $D$ is a discrete subset of $U$ (ie. locally $D$ is finitely many points). Nothing complicated here, everything is crystal clear, the isolated singularities of $f$ are at each point of $D$ and $f$ is not defined outside of $U$.

Some points of $D$ can be "removable singularities" (try with $\sin(z)/z$ which is entire, no singularity at $0$), some may be poles (ie. $(z-a)^k f$ is analytic at $a$), the others are essential singularities.

Next we can consider the analytic continuations of $f$, if $f$ extends to an analytic function on some larger connected open then this continuation is unique and is defined uniquely by the restriction of $f$ to any neighborhood.

Sometimes an analytic continuation exists, sometime not, of course the continuation may have some isolated singularities outside of $U$.

Sometimes the set of analytic continuations is complicated, try with $\log z,|z-1|<1$, the continuation to $arg(z)\in (-\pi/2,3\pi/2)$ is not the same as the continuation to $arg(z)\in (-3\pi/2,\pi/2)$ (there is a $2i\pi$ difference at the negative reals).

$\sum_{k\ge 0} z^{2^k},|z|<1$ has a natural boundary on $|z|=1$, no continuation exists.

Hope you realize there is nothing abstract in those things.