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:

  1. $p$ is linear
  2. $p$ is convex
  3. $p$ 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:

$$ \partial \mathcal{C} = \left\{ z\in\mathbb{C} : |z|=1 \right\} = \left\{ exp(i\vartheta) : \vartheta\in [0,2\pi) \right\} $$


$$ g(\partial \mathcal{C}) = \left\{ \sqrt{exp(i\vartheta)} : \vartheta\in [0,2\pi) \right\} = \left\{ exp(i\varphi) : \varphi\in [0,\pi) \right\} $$

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).

  • 1
    $\begingroup$ By $p(\mathcal{L})$ do you just mean the image? Also, note that there is no global square-root function on the complex plane! Basically, if you go around the unit circle $e^{i\theta}$ say starting at $\sqrt{e^0}=1$, you'd get $\sqrt{e^{2i\pi}}=-1$. $\endgroup$ – Aaron Mazel-Gee Mar 28 '11 at 9:48
  • $\begingroup$ Yes, $p(\mathcal{L})$ denotes the image. $\endgroup$ – Pantelis Sopasakis Mar 28 '11 at 10:03

For any bounded open set $U$ and any function $f$ analytic on a neighbourhood of $\overline{U}$, $\partial f(U) \subseteq f(\partial U)$. This is a consequence of the Open Mapping Theorem and the compactness of $\overline{U}$. Moreover, if $\partial U$ is a simple positively oriented closed contour $\Gamma$, and $p$ is not on $f(\Gamma)$, the Argument Principle says the winding number of $f(\Gamma)$ around $p$ is the number of zeros of $f(z) - p$ inside $\Gamma$. So the image of the unit disk $\cal C$ under a polynomial $f$ contains everything inside the curve $f(e^{i\theta})$, $0 \le \theta \le 2 \pi$ (in the sense that the winding number of the curve around it is nonzero).
For an unbounded open set $U$ things can be a bit more complicated. However, if $f$ is a polynomial of degree $d$ and $U$ contains a sector of angle $> 2 \pi/d$, $f(U)$ contains a neighbourhood of $\infty$.

  • $\begingroup$ Thank you for your answer - it is really illuminating! Could you please give me a reference (e.g. a book) where I can find these results with proofs (I refer mainly to the last argument in regard to images containing a neighborhood of $\infty$). $\endgroup$ – Pantelis Sopasakis Mar 28 '11 at 23:22

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