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I've seen here that a holomorphic function with bounded imaginary part on a punctured neighborhood of $0$ has a removable singularity at $0$.
I just wanted to know if this result could be also extended to get this:

Let $\epsilon >0$, $z_0 \in \mathbb{C}$ and $f$ be holomorphic on a punctured neighborhood $\dot{D_{\epsilon}}(z_0)$. Futhermore it holds for all $z \in \dot{D_{\epsilon}}(z_0)$ that $Re(f(z))< K \in \mathbb{R}$
This implies that $z_0$ is a removable singularity of $f$ $(|f(z)|$ is bounded ?$)$.

If the answer is yes I'm searching for a proof
Thanks for help !

Community
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Thomas Produit
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1 Answers1

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Yes. $z_0$ is an isolated singularity of $f$, so there are three possibilities,

  1. it is a removable singularity,
  2. it is a pole,
  3. it is an essential singularity.

If $z_0$ is a pole of $f$, then $g = \dfrac{1}{f}$ is holomorphic in a possibly smaller punctured neighbourhood $\dot{D}_\delta(z_0)$ of $z_0$, and has a removable singularity in $z_0$, which becomes a zero of $g$ after removing it. Since $g$ is not identically $0$, $g(D_\delta(z_0))$ is an open neighbourhood of $0$, hence contains a disk $D_r(0)$ for some $r > 0$. Thus

$$f(\dot{D}_\delta(z_0)) = \frac{1}{g(\dot{D}_\delta(z_0))} \supset \frac{1}{\dot{D}_r(0)} = \mathbb{C}\setminus \overline{D_{1/r}(0)}.$$

That means that neither the real nor the imaginary part of $f$ is bounded on $\dot{D}_\delta(z_0)$.

If $z_0$ is an essential singularity of $f$, the Casorati-Weierstraß theorem asserts that $f(\dot{D}_\delta(z_0))$ is dense in $\mathbb{C}$ for all $0 < \delta \leqslant \varepsilon$, hence neither the real nor the imaginary part of $f$ is bounded there.

The only remaining possibility is that $z_0$ is a removable singularity.

Daniel Fischer
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  • Shouldn't it be "[...] Casorati-Weierstraß theorem asserts that $f(\dot{D}_{\delta}(z_0))$ [...]" ? – Thomas Produit Feb 04 '14 at 14:04
  • Yes, typo. Thanks. – Daniel Fischer Feb 04 '14 at 14:04
  • Great answer though thanks a lot ! – Thomas Produit Feb 04 '14 at 14:05
  • @DanielFicher By rereading your answer I presume that (by the same arguments) it would also hold if $Im(f)$ is bounded, wouldn't it ? – Thomas Produit Feb 05 '14 at 14:34
  • Yes, a bound on $\operatorname{Im} f$ is completely analogous. (Superficially) More generally, a bound on $\operatorname{Re} (e^{-i\alpha}f)$ for any $\alpha\in\mathbb{R}$, which ensures the image of $f$ is contained in some half-plane. – Daniel Fischer Feb 05 '14 at 14:38
  • Thanks for this rapid but enlightening comment ! – Thomas Produit Feb 05 '14 at 14:42
  • @DanielFischer Did you mean $f(\dot{D}\delta(z_0)) = \frac{1}{g(\dot{D}\delta(z_0))}$? Don't we need to assume that $g$ has no zeros in $\dot{D}_\delta(z_0)$? – Twnk Sep 14 '20 at 22:51
  • @Twink Yes, that should have been a $z_0$. Thanks for spotting the typo. That $g$ has no zeros in $\dot{D}{\delta}(z_0)$ need not be explicitly assumed, it follows from $g = 1/f$. Thus $g$ can only have zeros at poles of $f$, and our hypotheses tell us that $f$ has no poles in $\dot{D}{\varepsilon}(z_0)$ (and $\delta$ is chosen such that $f$ has no zeros in $\dot{D}_{\delta}(z_0)$, hence $g$ is holomorphic and nonzero there). – Daniel Fischer Sep 17 '20 at 18:57