# Casorati-Weierstrass

If $$f$$ is holomorphic on $$\Omega\setminus\left\{{z_0}\right\}$$ where $$z_0$$ is an essential singularity, then for every $$V\subset \Omega\setminus\left\{{z_0}\right\}$$, $$f(V)$$ is dense in $${\mathbf{C}}$$.

Equivalently, suppose $$z_{0}$$ is an essential isolated singularity of $$f(z)$$. Then for every complex number $$w_{0}$$, there is a sequence $$z_{n} \rightarrow z_{0}$$ such that $$f\left(z_{n}\right) \rightarrow w_{0}$$. The image of a punctured disc at an essential singularity is dense in $${\mathbf{C}}$$.

Let $$f$$ have an essential singularity. Suppose toward a contradiction that there exists a $$z_0\in {\mathbf{C}}$$, a punctured neighborhood $$\Omega$$ of $$z_0$$, and $${\varepsilon}> 0$$ with $$f(\Omega) \cap{\mathbb{D}}_{\varepsilon}(z_0)$$ empty. So $${\left\lvert {f(z) - z_0} \right\rvert}>{\varepsilon}$$ for every $$z\in \Omega$$. Write $$\tilde \Omega \coloneqq\Omega\cup\left\{{z_0}\right\}$$.

Define $$g(z) \coloneqq{1\over f(z) - z_0}$$, noting that $$z_0$$ is the only singularity of $$g$$ in $$\tilde\Omega$$, making $$g$$ holomorphic on $$\Omega$$. We have $${\left\lvert {g(z)} \right\rvert} < {\varepsilon}^{-1}< \infty$$ for all $$z\in \Omega$$, so $$g$$ is bounded in $$\Omega$$ and $$z_0$$ must be a removable singularity. By Riemann’s removable singularity theorem, $$g$$ extends to a holomorphic function on all of $$\tilde\Omega$$.

We can now write $$f(z) = {1\over g(z)} + z_0$$. If $$g(z_0) = 0$$, then it is a zero of some finite order for $$g$$, and thus a pole of finite order for $$f$$, contradicting that $$z_0$$ is essential for $$f$$. Otherwise, if $$g(z_0) = w_0\neq 0$$, then \begin{align*} {\left\lvert { f(z_0)} \right\rvert} = {\left\lvert { {1\over w_0} + z_0} \right\rvert} \leq {\varepsilon}+ {\left\lvert {z_0} \right\rvert} < \infty ,\end{align*} making $$z_0$$ removable and again yielding a contradiction.

Pick $$w\in {\mathbf{C}}$$ and suppose toward a contradiction that $$D_R(w) \cap f(V)$$ is empty. Consider \begin{align*} g(z) \coloneqq{1\over f(z) - w} ,\end{align*} and use that it’s bounded to conclude that $$z_0$$ is either removable or a pole for $$f$$.

In detail, from Gamelin:  # Exercises

Show that if $$f$$ is entire and there exists a disc $${\mathbb{D}}_r(a)$$ not intersecting $$f({\mathbf{C}})$$, then $$f$$ must be constant.

Write $$a\not\in f({\mathbf{C}})$$, and replace $$f$$ with $$f(z) - a$$ so that $$f(a) = 0$$ and there is some $${\mathbb{D}}_R(0)$$ not intersecting $$f({\mathbf{C}})$$. Then $$f(z) \in {\mathbb{D}}_R^c$$ for all $$z\in {\mathbf{C}}$$, so $${\left\lvert {f(z)} \right\rvert} > R$$ for all $$z$$. Defining $$G(z) \coloneqq{1\over f(z)}$$ yields $${\left\lvert {G(z)} \right\rvert} < R^{-1}$$ for all $$z$$, making $$G$$ bounded. The singularity of $$G$$ at $$z=0$$ is thus removable, so $$G$$ extends to an entire function by Riemann’s removable singularity theorem. By Liouville, $$G$$ is constant, so $$f$$ is constant.

Find all entire functions $$f$$ that satisfy \begin{align*} {\left\lvert {f(z)} \right\rvert} \geq e^{{\left\lvert {z} \right\rvert}} && \forall z\in {\mathbf{C}} .\end{align*}

Claim: there are no such functions. Consider $$g(z) \coloneqq f(z)/e^z$$, which is entire since $$e^z$$ is nonvanishing. Now $$g$$ is entire and $${\left\lvert {g} \right\rvert} \geq 1$$ everywhere, so $$\operatorname{im}(g) \cap{\mathbb{D}}$$ is empty. This contradicts Casorati-Weierstrass, which requires that $$\operatorname{im}g$$ be dense in $${\mathbf{C}}$$.

Alternatively, note $$f\neq 0$$ by the inequality, so $$1/f$$ is bounded and entire and thus constant. However, $$f(z) = c$$ contradicts the inequality, since $$e^{{\left\lvert {z} \right\rvert}}\to \infty$$ as $${\left\lvert {z} \right\rvert}\to \infty$$.

Alternatively, note that $$e^{{\left\lvert {z} \right\rvert}} \geq 1$$ for all $$z$$, so $$\operatorname{im}(f) \cap{\mathbb{D}}$$ is empty, again contradicting Casorati-Weierstrass.

Let $$f$$ be entire and define a function \begin{align*} m: [0, \infty) &\to {\mathbf{R}}\\ r &\mapsto \min_{{\left\lvert {z} \right\rvert} = r} {\left\lvert {f(z)} \right\rvert} .\end{align*} Suppose $$\tilde m \coloneqq\lim_{r\to \infty}m(r)$$ exists and is a finite positive real number, and show that $$f$$ is constant.

Hint: consider $$g(z) \coloneqq f(1/z)$$.

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