Problem related to continuous complex mapping.

Question

We are given with a map $g:\bar D\to \Bbb C $, which is continuous on $\bar D$ and analytic on $D$. Where $D$ is a bounded domain and $\bar D=D\cup\partial D$.

1) I want to show that: $\partial(g(D))\subseteq g(\partial D).$

And further, I need two examples:

a) First, to show that the above inclusion can be strict, that is: $\partial(g(D))\not= g(\partial D).$

b) Second example, I need to show that conclusion in (1) is not true if $D$ is not bounded.

So basically we have to show that the boundary of the open set $g(D)$ is contained in image of boundary of $D$ (and sometimes strictly contained). I think that we will use open mapping theorem. But how this theorem will help us here that is not clear.

Answer 1

You know by the open mapping theorem that $g(D)$ is open. (I am assuming $g$ is non-constant.) But you also know $\bar D$ is compact (closed and bounded), so $g(\bar D)$ is also compact, and hence closed. Hence the closure of $g(D)$ is contained in $g(\bar D)$. You should be able to argue at this point that $\partial(g(D)) = g(\bar D) \setminus g(D)$.

Answer 2

@Stephen, As we discussed yesterday, If we take $g(z)= z^2$ and $D$ =\begin{cases}z, & \text{where 1<|z|<2} \\\end{cases} Now, we want to prove that $g(\partial D)\not\subset \partial(g(D)) $. Therefore, we need to show that $\exists $ some $z\in g(\partial D)$ but $z \not\in \partial(g(D))$.

First, I want to talk about domain $D$ and its image by map $g$. Please check it:

??

Answer 3

It's better to assume the function $g$ to be bounded. If the image curve $f(\partial D)$ forms infinitely many loops everywhere, then $\partial f(D)\subsetneqq f(\partial D)$.