Complex Analysis/Open mapping theorem

Remark 1 - Real valued functions - Calculus

This theorem is not true in real values ${\displaystyle \mathbb {R} }$. For example ${\displaystyle f\colon U\to \mathbb {R} }$ with ${\displaystyle f(x)=x^{2}}$ and ${\displaystyle U=\mathbb {R} }$ the function ${\displaystyle f}$ is differentiable and ${\displaystyle U=\mathbb {R} }$ is open and connected. The connectness is true for ${\displaystyle f(U)=\left[0,+\infty \right)}$, but ${\displaystyle f(U)}$ is not an open set.

Remark 2 - Open Mapping Theorem

The Open Mapping Theorem just addresses the openness of ${\displaystyle f(U)}$. The connectedness is an additional property that is true for all continuous functions ${\displaystyle f:U\to \mathbb {C} }$. In the Open Mapping Theorem ${\displaystyle f}$ is holomorphic and therefor also continuous.

Proof 1: Connectedness

We show that if ${\displaystyle f}$ is continuous and ${\displaystyle U}$ is connected, then ${\displaystyle f(U)}$ is also connected.

Proof 2: Connectedness

Let ${\displaystyle w_{1},w_{2}\in f(U)}$ be arbitrarily chosen. Then, there exist ${\displaystyle z_{1},z_{2}\in U}$ such that ${\displaystyle f(z_{1})=w_{1}}$ and ${\displaystyle f(z_{2})=w_{2}}$. Since ${\displaystyle U}$ is connected, there exists a path ${\displaystyle \gamma$ :[a,b]\to U} such that ${\displaystyle \gamma (a)=z_{1}}$ and ${\displaystyle \gamma (b)=z_{2}}$.

Proof 3: Connectedness

Because ${\displaystyle f}$ is continuous and ${\displaystyle \gamma$ :[a,b]\to U} is a continuous path in ${\displaystyle U}$, the composition ${\displaystyle \gamma _{f}:=f\circ \gamma }$ is a continuous path in ${\displaystyle f(U)}$, for which:

${\displaystyle \gamma _{f}(a)=f(\gamma (a))=f(z_{1})=w_{1}}$ and ${\displaystyle \gamma _{f}(b)=f(\gamma (b))=f(z_{2})=w_{2}}$.

Proof 4: Openness

It remains to show that ${\displaystyle f(U)}$ is open. Let ${\displaystyle w_{0}\in f(U)}$ and ${\displaystyle z_{0}\in U}$ such that ${\displaystyle f(z_{0})=w_{0}}$. Now, consider the set of ${\displaystyle w_{0}}$-preimages:

${\displaystyle S(f,w_{0}):={z\in U\ |\ f(z)=w_{0}}}$

Proof 5: Openness - Identity Theorem

According to the Identity Theorem, the set ${\displaystyle S(f,w_{0}):={z\in U\ |\ f(z)=w_{0}}}$ cannot have accumulation points in ${\displaystyle f(U)}$. If ${\displaystyle S(f,w_{0})\subseteq f(U)}$ had accumulation points in ${\displaystyle f(U)}$, the holomorphic function ${\displaystyle f\colon U\to \mathbb {C} }$ would be constant with ${\displaystyle f(z)=w_{0}}$ for all ${\displaystyle z\in U}$.

Proof 6: Openness - Neighborhoods

If the set ${\displaystyle S(f,w_{0})}$ of ${\displaystyle w_{0}}$-preimages of ${\displaystyle f}$ has no accumulation points, one can choose a neighborhood ${\displaystyle V\subseteq U}$ of ${\displaystyle z_{0}}$ where ${\displaystyle z_{0}}$ is the only ${\displaystyle w_{0}}$-preimage. Let ${\displaystyle r>0}$ be such that ${\displaystyle {\bar {D}}_{r}(z_{0})\subseteq V}$.

Proof 7: Openness

We then define the smallest lower bound for the distance of ${\displaystyle f(z)}$ to ${\displaystyle w_{0}}$, where ${\displaystyle z}$ lies on the boundary of the disk ${\displaystyle D_{r}(z_{0})}$:

${\displaystyle \varepsilon$ :=\inf _{z\in \partial D_{r}(z_{0})}|f(z)-w_{0}|>0}

Here, ${\displaystyle \varepsilon >0}$, because ${\displaystyle f}$ is continuous and attains a minimum on the compact set ${\displaystyle \partial D_{r}(z_{0})}$. Since ${\displaystyle {\bar {D}}_{r}(z_{0})\subseteq V}$, no ${\displaystyle w_{0}}$-preimages can lie on the boundary.

Proof 8: Openness - Maximum Principle

We show that ${\displaystyle D_{\frac {\varepsilon }{3}}(w_{0})\subseteq f(U)}$. Let ${\displaystyle |w-w_{0}|<{\frac {\varepsilon }{3}}}$. We prove by contradiction that this arbitrary ${\displaystyle w\in D_{\frac {\varepsilon }{3}}(w_{0})}$ is in the image of ${\displaystyle f}$.

Proof 9: Openness - Maximum Principle

Assume ${\displaystyle f(z)\neq w}$ for all ${\displaystyle z\in {\bar {D}}_{r}(z_{0})}$. Then, ${\displaystyle |g(z)|}$ with ${\displaystyle g(z):=f(z)-w}$ attains a nonzero minimum on ${\displaystyle {\overline {D_{r}(z_{0})}}}$. Since ${\displaystyle f}$ is not constant, this minimum must lie on ${\displaystyle \partial D_{r}(z_{0})}$ (otherwise ${\displaystyle h(z):={\frac {1}{f(z)-w}}}$ would be constant by the Maximum Principle. If ${\displaystyle h}$ were constant, ${\displaystyle f}$ would also have to be constant—a contradiction to the assumption).

Proof 9: Openness

Since ${\displaystyle w_{0}\in f(U)}$ was chosen arbitrarily, and for every ${\displaystyle w_{0}\in f(U)}$, there exists a ${\displaystyle {\frac {\epsilon }{3}}}$-neighborhood ${\displaystyle D_{\frac {\epsilon }{3}}(w_{0})\subseteq f(U)}$, we obtain ${\displaystyle f(U)=\bigcup _{w_{0}\in f(U)}D_{\frac {\epsilon }{3}}(w_{0})}$ as an Norms, metrics, topology, and thus ${\displaystyle f(U)}$ is open.

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• Date: 12/26/2024