# Examples of Bivariate Analytic, Globally Continuous Functions, whose Set of Zeros are Boundaries of Polygons

Are there any known examples of analytic, globally continuous functions $f(x,y): (x,y)\in\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ such that
$$f(x,y)\lt 0\Leftrightarrow (x,y)\not\in\mathcal{P}$$ $$f(x,y)=0\Leftrightarrow (x,y)\in\partial\mathcal{P}$$ $$f(x,y)\gt 0\Leftrightarrow (x,y)\in\mathcal{P} \setminus\partial\mathcal{P}$$

where $\mathcal{P}$ is a connected region, whose boundary $\partial\mathcal{P}$ is a finite collection of disjoint simple polygons?

Without the requirement, that $f(x,y)$ has different sign inside and outside of $\mathcal{P}$, examples of such functions can be easily defined with $\sqrt{(x+\frac{l}{2})^2+y^2}+\sqrt{(x-\frac{l}{2})^2+y^2}-l)$ which serve as models for factors that vanish on a line segment of length $l$, but nowhere else

In view of the example that apparently isn't analytic, I have repeated the requirement that $f(x,y)$ be analytic and added the requirement, that it be globally continuous.

Remark:
The functions I am looking for with the relaxed condition of being $C^\infty$, zero on $\partial P$ and, positive in $P\setminus\partial P$ could also be used as testfunctions (cf e.g. http://en.wikipedia.org/wiki/Distribution_%28mathematics%29 ) for generalized functions on polygonal domains.

• $$f(x,y)=\begin{cases}1&\text{if (x,y)\in P\setminus\partial P},\\0&\text{if (x,y)\in\partial P},\\-1&\text{if (x,y)\not\in P}\end{cases}$$ Maybe, you want a function with some special properties? Commented Feb 1, 2014 at 18:14
• @AlexDegtyarev: According to Wikipedia, an analytic function is given locally by convergent power series en.wikipedia.org/wiki/Analytic_function; could you please supply the powerseries for your example? Maybe I should have repeated the term "analytic" in the description of my problem (and I will do so), but I definitely don't understand what the reasons for downvoting are. Commented Feb 1, 2014 at 19:14
• The reason for downvoting is a misstated problem. But I can undo this if you take it so personally. Accidentally, the function in your example is not analytic: it is not even differentiable. The zero locus of a real (?) analytic function is a real analytic variety; I am not quite sure how to prove that right away, but I doubt that it can be a polygon as you wish. Commented Feb 1, 2014 at 19:45
• @AlexDegtyarev: No need to revert the downvoting, but it would help to get reasons for doing so. Concerning my function-example: it was meant for giving an example of a function, that is zero on a line segment; I believe I can provide an example of $f(x,y)$ that is globally differentiable and that is zero on a finite line segment. Commented Feb 1, 2014 at 20:01
• For any closed subset $A\subset\mathbb{R}^n$, there is a $C^\infty$ function $f\colon\mathbb{R}^n\to[0,1]$ such that $A=f^{-1}(0)$. A similar game would give you a $C^\infty$ function as you want. (E.g., take $f+g$, where $f$ and $g$ are as above for $P$ and the closure of $\mathbb{R}^2\setminus P$, respectively.) However, as I said, I strongly doubt that you can find an analytic one. Commented Feb 1, 2014 at 20:09

OK, here is a proof (mimic one-variable complex calculus). Assume that a power series $f(x,y)=\sum a_{ij}x^iy^j$ vanishes on $y=0$, $-\epsilon<x\le0$. Then, as usual, $i!a_{i0}=\partial^if/\partial x^i(0,0)=0$. (As we assume that the partial derivatives exist, we can use negative values to compute them.) But then $f(x,0)=0$ for all $x$, hence $f$ vanishes on the whole line $y=0$. Thus, you cannot have zero locus made out of line segments that are not whole lines.
• So either the class of functions, that are analytic on $\mathbb{R}\times\mathbb{R}$ is too small to contain the functions I am looking for or, $\mathbb{R}\times\mathbb{R}$ is too big and the functions I am looking for can not be analytic in the corners of $\partial P$ or even not on $\partial P$ Commented Feb 2, 2014 at 10:47
• @ManfredWeis: The proof is purely local: a corner cannot be the zero set of an analytic germ. Of course, the reason is the fact that analytic functions are way too rigid; you should downgrade to $C^\infty$. Commented Feb 2, 2014 at 11:56
• as you may already have concluded, I am not an expert in those matters and so your feedback is valuable input for me. I will follow your advice and go to $C^\infty$ functions, that are not defined piecewise, i.e. that do not require 'inspection' of the argument for evaluation. Commented Feb 2, 2014 at 12:09