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.

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.

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    $\begingroup$ $$f(x,y)=\begin{cases}1&\text{if $(x,y)\in P\setminus\partial P$},\\0&\text{if $(x,y)\in\partial P$},\\-1&\text{if $(x,y)\not\in P$}\end{cases}$$ Maybe, you want a function with some special properties? $\endgroup$ Commented Feb 1, 2014 at 18:14
  • $\begingroup$ @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. $\endgroup$ Commented Feb 1, 2014 at 19:14
  • $\begingroup$ 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. $\endgroup$ Commented Feb 1, 2014 at 19:45
  • $\begingroup$ @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. $\endgroup$ Commented Feb 1, 2014 at 20:01
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    $\begingroup$ 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. $\endgroup$ Commented Feb 1, 2014 at 20:09

1 Answer 1


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.

  • $\begingroup$ 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 $\partial P$ $\endgroup$ Commented Feb 2, 2014 at 10:47
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    $\begingroup$ @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 $C^\infty$. $\endgroup$ Commented Feb 2, 2014 at 11:56
  • $\begingroup$ 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. $\endgroup$ Commented Feb 2, 2014 at 12:09
  • $\begingroup$ @ManfredWeis: Judging by your last remark, I begin to suspect that you are interested in something quite different than analytic or smooth. It may help if you explain the original problem in more details. It is very unlikely that there is a "formula" as you want (as "formulas" tend to be analytic). $\endgroup$ Commented Feb 2, 2014 at 12:52

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