I am having trouble solving the following exercise:
Let $U$ be a bounded domain, and let $f(z)$ be a continuous function on $U \cup \partial U$ that is holomorphic on $U$. Show that $\partial\left( f(U)\right)\subseteq f(\partial U)$, that is, the boundary of the open set $f(U)$ is contained in the image under $f(z)$ of the boundary of $U$.
One of the main issues I have with this exercise is that my teacher has defined the open mapping theorem as:
Open mapping theorem: if $\Omega \subseteq \mathbb{C}$ is a connected open set and $f$ is holomorphic and not constant on $\Omega$, then $f(\Omega)$ is an open set in $\mathbb{C}$.
The additional hypothesis that the domain be connected means that we cannot apply the theorem directly. This is my approach:
Suppose $f$ is not constant (if it is the result is clear). Since $\stackrel{\circ}{U}$ is open, for each $z\in\stackrel{\circ}{U}$ there exists $r_z>0$ such that the open disk of center $z$ and radius $r_z$, $D(z, r_z)$, is contained in $\stackrel{\circ}{U}$. Then we can write $\stackrel{\circ}{U}$ as:
$\stackrel{\circ}{U} = \bigcup_{z\in\stackrel{\circ}{U}} D(z, r_z)$
Hence:
$f(\stackrel{\circ}{U}) = f\left(\bigcup_{z\in\stackrel{\circ}{U}} D(z, r_z)\right) = \bigcup_{z\in\stackrel{\circ}{U}} f\left(D(z, r_z)\right)$
Since $f$ is holomorphic in $\stackrel{\circ}{U}$, disks are open connected sets and the arbitrary union of open sets is open, we can apply the Open mapping theorem to conclude that $f(\stackrel{\circ}{U})$ is an open set. This means that $f$ maps interior points of $U$ to interior points of $f(U)$.
Let $w$ be a point of $\partial f(U)$, then there exists a sequence $w_n \subseteq f(U)$ that converges to $w$, for $\partial f(U)$ is a compact set (1). By definition of $f(U)$, for all $k\in\mathbb{N}$ there exists $z_k\in\stackrel{\circ}{U}$ such that $f(z_k) = w_k$. By hypothesis $U$ is bounded, and therefore $z_k\rightarrow z\in\partial U$ and $f(z_k) \rightarrow f(z) = w\in\partial f(U)$, which implies:
$\forall w \in\partial f(U) : \exists z\in\partial U \mid f(z) = f(w) \implies \partial f(U) \subseteq f(\partial U)$.
One of my concerns is that $U$ is an infinite union of open sets, and I am not sure that $f(\bigcup_{k=0}^{\infty} U_k) = \bigcup_{k=0}^{\infty} f(U_k)$ still holds in such conditions. My other issue is that I am not sure that $(1)$ is true.
Thank you.
Best Answer
Your proof is absolutely correct.
As Yanko said in his comment, each function $\phi : X \to Y$ between sets $X,Y$ has the property $\phi(\bigcup_{\alpha \in A} M_\alpha) = \bigcup_{\alpha \in A} \phi(M_\alpha)$, where $(M_\alpha)_{\alpha \in A}$ is an arbitrary family of subsets of $X$.
However, to show that a non-constant holomorphic $f : U \to \mathbb{C}$ defined on arbitrary open $U \subset \mathbb{C}$ has an open image $f(U)$, it is unnecessary to consider unions. Let $w \in f(U)$. Choose $z \in U$ such that $f(z) = w$. Since $U$ is open, there exists an open disk $D(z,r) \subset U$. Now $V = f(D(z,r))$ is open and we have $w \in V \subset f(U)$.