Main problem: Let $\mathcal{L}=\left\{z\in\mathbb{C}:Re(z)<0\right\}$ be the left open half-plane of the complex plane and $\mathcal{C}=\left\{z\in\mathbb{C}:|z|<1\right\}$ be the open unit disk of $\mathbb{C}$. Let $p\in\mathbb{P}_n$ be a polynomial of degree $n\in\mathbb{N}$. Can I have a geometric representation of the sets $p(\mathcal{L})$ and $p(\mathcal{C})$? Is that possible for the general class of polynomials?
Additional Questions: Let $K$ be a non-empty convex and compact subset of $\mathcal{C}$ and let $p$ be continuous and have one of the following properties:
- $p$ is linear
- $p$ is convex
- $p$ is a polynomial (!)
In the first case $p(K)$ will be convex and compact and if $A_p$ is the set of extreme points of $K$ then $p(A_p)$ will be the set of extreme points of $p(K)$ and $p(K)=\overline{p(A_p)}$. If $p$ is a convex function (Second case), then $p(K)$ will be a convex set (and compact since $p$ is continuous and $K$ compact). It also holds that $p:\partial K \rightarrow p(\partial K) = \partial p(K)$. So $p(K)=conv(p(\partial K))$. What about the third case? If this is difficult to answer, then is there a class of functions that is more general than convex functions and not as general as the class of polynomials?
Using the boundary: The main question we address here is:
Let $p:\mathbb{C}\rightarrow\mathbb{C}$ be a function and $K\subset\mathbb{C}$ a convex set whose closure is compact. If we know $p(\partial K)$, is it possible to determine $p(K) ?$.
The calculation of the image of a convex set in $\mathbb{C}$ would be much easier if we knew whether (or under which conditions) the following equation holds:
$p(\mathcal{C})=conv \left\{ p(\partial \bar{\mathcal{C}}) \right\} ^o$
where $\bar{\cdot}$ and $\cdot^o$ is the closure and the interior operators and $\partial$ is the boundary of a set. This, attempts to be act as an extension for the Krein-Milman theorem.
It might be true: The above relation might be actually true... If it holds, then for $g(x)=\sqrt{x}$ we have:
$ \partial \mathcal{C} = \left\{ z\in\mathbb{C} : |z|=1 \right\} = \left\{ exp(i\vartheta) : \vartheta\in [0,2\pi) \right\} $
Hence
$ g(\partial \mathcal{C}) = \left\{ \sqrt{exp(i\vartheta)} : \vartheta\in [0,2\pi) \right\} = \left\{ exp(i\varphi) : \varphi\in [0,\pi) \right\} $
which is a semi-circle. And utilizing the above claim, this would give:
$g(\mathcal{C}) = \left\{ \rho exp(i\varphi) : \varphi\in [0,\pi), \rho \in [0,1) \right\}$ - which is correct (can be easily verified without using this claim). Of course this does not prove that the claim holds, but renders its proof challenging. Personally, I don't expect it to hold for arbitrary any (polynomial) function $p$, although it is known that it holds for linear mappings (by the KM theorem).