# Open Mapping, Riemann Mapping, Casorati-Weierstrass

## Spring 2020.6 (Prove the open mapping theorem) #complex/qual/completed

Prove the open mapping theorem for holomorphic functions: If $$f$$ is a non-constant holomorphic function on an open set $$U$$ in $$\mathbb{C}$$, then $$f(U)$$ is also an open set.

Idea: Let $$f: U\to {\mathbf{C}}$$. Pick $$w_0\in W$$ with $$f(z_0) = w_0$$ for some $$z_0\in U$$; we want to show that $$w_0$$ is an interior point of $$f(U)$$, so we’re looking for a disc containing $$w_0$$ and contained in $$f(U)$$.

Write \begin{align*} g_0(z) \coloneqq f(z) - w_0 ,\end{align*} so $$g_0$$ is holomorphic and has a zero at $$z_0$$. Since zeros of holomorphic functions are isolated, there is some $$U' \coloneqq{\mathbb{D}}_r(z_0)$$ where $$g_0$$ is nonvanishing. The claim is that if we choose $${\varepsilon}$$ small enough, we can arrange so that $$W_{\varepsilon}\coloneqq{\mathbb{D}}_{\varepsilon}(w_0) \subseteq f(U)$$. This will follow if for every $$w\in W_{\varepsilon}$$, the equation $$f(z) = w$$ has a solution in $$U$$, i.e.  Define a function that counts the number of zeros: \begin{align*} F(w) &\coloneqq{1\over 2\pi i}\int_{{{\partial}}U' } {f(z) \over f(z) - w_1 }\,dz\\ &= {1\over 2\pi i}\int_{{{\partial}}U' } {{\frac{\partial }{\partial z}\,}\qty{f(z) - w} \over f(z) - w }\,dz\\ &= {\sharp}Z(f(z) - w, U' ) ,\end{align*} which is the number of zeros of $$f(z) - w$$ in $$U'$$ by the argument principle. Now $$F$$ is a $${\mathbf{Z}}{\hbox{-}}$$valued function, and the only obstruction to continuity is if $$f(z) - w = 0$$ in the integrand for some $$z$$. The claim is that $${\varepsilon}$$ can be chosen such \begin{align*} z\in {{\partial}}U' \implies {\left\lvert {f(z) - w} \right\rvert} > 0 \qquad \forall w\in W_{\varepsilon} .\end{align*} The theorem then follows immediately: $$F(w): U' \to W_{\varepsilon}$$ is a continuous and $${\mathbf{Z}}{\hbox{-}}$$valued, thus constant. Then noting that $$F(w_0) = 1$$ since $$z_0\in U'$$ and $$w_0\in W_{\varepsilon}$$, we have $$F\equiv 1 > 0$$ for all $$w$$.

Choose \begin{align*} {\varepsilon}\coloneqq\min_{z\in {{\partial}}U'}{\left\lvert {f(z) - w_0} \right\rvert} .\end{align*} Now if $${\left\lvert {w-w_0} \right\rvert} < {\varepsilon}$$ and $${\left\lvert {z-z_0} \right\rvert} = r$$, we have $${\left\lvert {f(z) - w} \right\rvert} > {\varepsilon}> 0$$.

## Fall 2019.4, Spring 2020 HW 3 SS 3.8.14, Tie’s Extras Fall 2009, Problem Sheet (Entire univalent functions are linear) #complex/qual/completed

Let $$f: \mathbb{C} \rightarrow \mathbb{C}$$ be an injective analytic (also called univalent) function. Show that there exist complex numbers $$a \neq 0$$ and $$b$$ such that $$f(z)=a z+b$$.

Hint: Apply the Casorati-Weierstrass theorem to $$f(1/z)$$.

Note that $$f$$ is non-constant, since a constant function is extremely non-injective. Consider the singularity at $$\infty$$:

• If it is removable, then $$f$$ is bounded outside of a large disc, and bounded inside of it as a continuous function on a compact set, making $$f$$ entire and bounded and thus constant by Liouville.

• If it is essential, then by Casorati-Weierstrass there is a large disc of radius $$R$$ such that $$f(\overline{{\mathbb{D}}_R}^c) \subseteq {\mathbf{C}}$$ is dense. By the open mapping theorem, $$f({\mathbb{D}}_R) \subseteq {\mathbf{C}}$$ is open, so by density it intersects $$f(\overline{{\mathbb{D}}_R}^c)$$, but $${\mathbb{D}}_R \cap\overline{{\mathbb{D}}_R}^c$$ is empty so this contradicts injectivity.

So we can conclude $$\infty$$ is a pole of some order $$N$$, so $$f\qty{1\over z} = \sum_{0\leq k\leq N} c_k z^{-k}$$ and thus $$f(z) = \sum_{0\leq k\leq N} c_k z^k$$ is a polynomial of degree $$N$$. However, a polynomial of degree $$N$$ is generically $$N$$-to-one locally, so injectivity forces $$N=1$$ and $$f(z) = c_0 + c_1 z$$, where $$c_1\neq 0$$ since $$f$$ is nonconstant.

Write $$g(z) \coloneqq f(1/z)$$, which has a singularity at $$z=0$$. The claim is that this is a pole.

If $$z=0$$ is a removable singularity, $$g$$ is bounded on some closed disc $${\left\lvert {z} \right\rvert} \leq {\varepsilon}$$, so $$f$$ is bounded on $${\left\lvert {z} \right\rvert} > {\varepsilon}$$. Moreover $$f$$ is continuous and $${\left\lvert {z} \right\rvert}\leq {\varepsilon}$$, $$f$$ is bounded on this disc. This makes $$f$$ an entire bounded function and thus constant by Liouville, contradicting injectivity.

If $$z=0$$ is essential, then by Casorati-Weierstrass pick a punctured disc $$D = \left\{{{\left\lvert {z} \right\rvert} \leq {\varepsilon}}\right\}$$ where $$g(D)$$ is dense in $${\mathbf{C}}$$. Writing $$D^c \coloneqq\left\{{{\left\lvert {z} \right\rvert} > {\varepsilon}}\right\}$$, this means that $$f(D^c)$$ is dense. But $$U\coloneqq\left\{{{\left\lvert {z} \right\rvert} < {\varepsilon}}\right\}$$ is open and by the open mapping theorem $$f(U)$$ is open, so by density there is a point $$w\in f(D^c) \cap f(U)$$ while $$U \cap D^c = \emptyset$$, again contradicting injectivity.

So $$z=0$$ is a pole of $$g$$, and $$g$$ admits a Laurent expansion \begin{align*} g(z) = \sum_{k\geq -N} c_k z^k .\end{align*} Since $$f$$ is entire, it equals its Laurent expansion at $$z=0$$, so equating the two series yields \begin{align*} f(z) = \sum_{k\geq 0} d_k z^k &\implies g(z) = \sum_{k\geq 0} {d_k \over z^k} = \sum_{1\leq k\leq N} {c_k\over z^k} + \sum_{k\geq 0} c_k z^k \\ &\implies \sum_{k\geq 0} c_k z^k = 0 \\ &\implies f(z) = \sum_{0\leq k \leq N} c_k z^k ,\end{align*} making $$f$$ a polynomial of degree at most $$N$$.

Now $$f$$ can not be degree zero, since constant maps are not injective. Moreover $$f$$ can not be degree $$N\geq 2$$, since any polynomial of degree $$N$$ has $$N$$ roots in $${\mathbf{C}}$$ by the fundamental theorem of algebra, and any two distinct roots will be points where injectivity fails. Finally, ruling out the case of roots with multiplicity, if $$f(z) = c(z-a)^N$$, then $$f$$ has exactly $$N$$ preimages in a neighborhood of $$a$$. Letting $$p$$ be any such point, we can find $$N$$ complex points mapping to it: \begin{align*} p = c(z-a)^N &\implies {p\over c} = (z-a)^N \\ &\implies \qty{p\over c}^{1\over N}\zeta_N^k = z-a \quad k=0,1,\cdots, n-1 \\ &\implies z_k\coloneqq\qty{p\over c}^{1\over N}\zeta_N^k + a \xrightarrow{f} p .\end{align*}

So $$f$$ must be degree exactly 1, i.e. $$f(z) = az+b$$.

## Tie’s Extra Questions: Spring 2015 #complex/exercise/completed

• Let $$f$$ be analytic in $$\Omega: 0<|z-a|<r$$ except at a sequence of poles $$a_n \in \Omega$$ with $$\lim_{n \rightarrow \infty} a_n = a$$. Show that for any $$w \in \mathbb C$$, there exists a sequence $$z_n \in \Omega$$ such that $$\lim_{n \rightarrow \infty} f(z_n) = w$$.

• Explain the similarity and difference between the above assertion and the Weierstrass-Casorati theorem.

DZG: I think it’s also necessary to state that $$z_n \to a$$. As in the proof of Casorati-Weierstrass, fix $$w$$ and suppose toward a contradiction that no sequence sequence exists. Then there is some $${\varepsilon}, R$$ such that \begin{align*} f({\mathbb{D}}_{\varepsilon}(a)) \subseteq {\mathbb{D}}_R(w)^c ,\end{align*} for otherwise one could construct the desired sequence. In particular, $${\left\lvert {f(z) - w} \right\rvert} > R$$ for $${\left\lvert {z-a} \right\rvert} < {\varepsilon}$$, so define \begin{align*} G(z) \coloneqq{1\over f(z) - w} \implies {\left\lvert {G(z)} \right\rvert} \leq R^{-1}< \infty \qquad \text{in }{\mathbb{D}}_{\varepsilon}(a) .\end{align*} Since $$G$$ is bounded in this disc, any singularities here must be removable. Since the $$a_k$$ are poles of $$f$$, they are zeros of $$G$$ – this is because if $${\left\lvert {f(z)} \right\rvert}\to\infty$$ as $$z\to a_k$$ then $${\left\lvert {G(z)} \right\rvert}\to 0$$. So $$G(a_k) = 0$$ for all $$k$$ and $$G$$ extends holomorphically over the removable singularity $$a$$, and by continuity must satisfies $$G(a) = 0$$. But now $$G$$ is zero on a set with a limit point, hence $$G\equiv 0$$ by the identity principle. This is a contradiction since if $$G\equiv 0$$ on an open set, $$f$$ has poles on an open set, contradicting that $$f$$ is holomorphic on $$\Omega$$.

The difference to Casorati-Weierstrass: the singularity at $$a$$ is not essential, since in particular it is not isolated. The conclusion is nearly the same though: this says that every $$w\in {\mathbf{C}}$$ is a limit point for $$f(\Omega)$$, so $$w$$ is in the closure of $$f(\Omega)$$, making the image dense in $${\mathbf{C}}$$.

## Dense images #complex/exercise/work#stuck

Suppose $$f: {\mathbb{H}}\cup{\mathbf{R}}\to {\mathbf{C}}$$ satisfies the following:

• $$f(i) = i$$
• $$f$$ is continuous
• $$f$$ is analytic on $${\mathbb{H}}$$
• $$f(z) \in {\mathbf{R}}\iff z\in {\mathbf{R}}$$.

Show that $$f({\mathbb{H}})$$ is a dense subset of $${\mathbb{H}}$$.

Ideas:

• If an entire function doesn’t have dense image, it’s constant by Liouville using the proof idea of Casorati-Weierstrass.
• Conjugate $$f$$ by $$T:{\mathbb{H}}\to {\mathbb{D}}$$ where $$T(z) = {z-i\over z+i}$$, then $$\tilde f(0) = 0$$
• Use that $$T({\mathbf{R}}) = S^1$$, so $${\left\lvert {\tilde f(z)} \right\rvert} = 1$$ when $${\left\lvert {z} \right\rvert} = 1$$.
• Schwarz reflection applies to $$\tilde f$$ to define an entire function – if $$f$$ isn’t dense, then the extension of $$\tilde f$$ isn’t dense…?
• No clue how to use $$f(i) = i$$, although it implies $$\tilde f(0) = 0$$ and Schwarz applies.

## Tie’s Extra Questions: Spring 2015 #complex/exercise/work

Let $$f(z)$$ be an analytic function on $${\mathbb C} \backslash \{ z_0 \}$$, where $$z_0$$ is a fixed point. Assume that $$f(z)$$ is bijective from $${\mathbb C} \backslash \{ z_0 \}$$ onto its image, and that $$f(z)$$ is bounded outside $$D_r(z_0)$$, where $$r$$ is some fixed positive number. Show that there exist $$a, b, c, d \in \mathbb C$$ with $$ad-bc \neq 0$$, $$c \neq 0$$ such that $$\displaystyle f(z) = \frac{az + b}{cz + d}$$.