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I was reading this proof here: holomorphic function with bounded real part on punctured neighborhood $\dot{D}_{\epsilon}(z_0)$, which was proving that

Let $f$ be holomorphic on a punctured disk $D_\epsilon(z_0)/\{z_0\}$ and $Re(f(z)) < M$ for all $z \in D_\epsilon(z_0)/\{z_0\}$. Then it implies that $f$ has a removable singularity at $z_0$.

In the answer to the original post, they supposed that $z_0$ is a pole for the sake of contradiction, but why $g = \frac{1}{f}$ holomorphic in a "possible smaller" punctured neighbourhood $D_\delta(z_0)/\{z_0\}$? Also, why is $g(D_\delta(z_0)/\{z_0\})$ an open neighbourhood of $0$? The open mapping theorem might have been used here but why is it a neighbourhood of $0$? Last, why is $\frac{1}{D_r(0)\setminus \{0\}} = \mathbb{C}\setminus \overline{D_{\frac{1}{r}}(0)}$?

Thank you .

beigecamellia
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3 Answers3

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The proof in the link given does not say $g(D(z_0)\setminus\{z_0\})$ is an open neighborhood of $0.$ That doesn't even make sense: $0\notin g(D(z_0)\setminus\{z_0\})!$ What is stated is that $g$ has a removable singularity at $z_0.$ Still using the notation $g$ for the extension, we have $g(z_0)=0.$ It is the extended $g$ that maps $D(z_0)$ onto a neighborhood of $0.$

zhw.
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First question: if $f$ has a pole at $z_0$ then $|f(z)| \to \infty$ as $z \to z_0$ so $f$ does not vanish in some deleted neighborhood of $z_0$. Hence $\frac 1 f$ is analytic there.

Second question: it is already stated in the link that $g$ is defined to be $0$ at $z_0$ (to get rid of removable singularity). Now open mapping theorem shows that $g(D_{\delta} (z_0))\setminus \{z_0\}$ is an open set containing $0$.

Third question: $0<|z| <r$ iff $z\neq 0$ and $|\frac 1 z| >\frac 1 r$.

Kavi Rama Murthy
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  • thank you, for the first question i was asking why is the punctured disk "possible smaller"? – beigecamellia Oct 13 '19 at 00:22
  • If $f$ has a zero in the original disk then $g$ is not defined there. So you have to take a smaller disk to make sure that $f$ has no zeros there and this is possible because $|f(z)| \to \infty$. @beigecamellia – Kavi Rama Murthy Oct 13 '19 at 00:29
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1) Because $f$ can have a zero in $D_\varepsilon(z_0) \setminus \{z_0\}$ so $1/f$ can be non holomorphic on $D_\varepsilon(z_0) \setminus \{z_0\}$. One have to consider a smaller disk: if $f$ has a pole at $z = z_0$, then by continuity, $f$ does not vanish around $z = z_0$ that is to say in $D_\delta(z_0) \setminus \{z_0\}$ for $\delta \leq \varepsilon$, so $g=1/f$ is holomorphic on $D_\delta(z_0) \setminus \{z_0\}$.

2) By the open mapping theorem, $g(D_\delta(z_0) \setminus \{z_0\})$ is open because $D_\delta(z_0) \setminus \{z_0\}$ is open. It is a neighbourhood of 0 because if one takes a sequence $(z_n)_n$ in $D_\delta(z_0) \setminus \{z_0\}$ such that $z_n \to z_0$, then $g(z_n) = 1/f(z_n) \to 0$, so $0 \in \overline{g(D_\delta(z_0) \setminus \{z_0\})}$.

3) $z \in 1/\dot{D}_r(0)$ iff ($\vert 1/z \vert < r$ and $z \neq 0$) iff $ \vert z \vert > r$ iff $z \in \mathbb C \setminus \overline{D_{1/r}(0)}$.

Héhéhé
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  • Your proof of 2) has some problems. $g(D(z_0)\setminus{z_0})$ is not an open neighborhood of $0$ because $0$ is not in that set. And I'm not sure what you're doing with $z_n\to z_0$ etc. – zhw. Oct 13 '19 at 16:24