# Can we represent the interior of a polygon as the graph of a continuously differentiable function?

Let $P$ be a polygon in $\mathbb R^3$. Can we find an open set $U\subseteq\mathbb R^2$ and a continuously differentiable function $f:U\to\mathbb R$ such that $$\text{int}(P)=\left\{\left(x,f(x)\right):x\in U\right\}\;?$$ The polygon can be assumed to be "drawable in 2D" and we can assume, that its edges do not cross.

• By a polygon in $\mathbb{R}^3$, do you mean a polygon in $\mathbb{R}^2$ where $\mathbb{R}^2$ is included in $\mathbb{R}^3$ as a hyperplane? And I'm assuming you don't want to include the edges. – Eric Auld Nov 9 '15 at 16:05
• @EricAuld Yes, the polygon should be "drawable in 2D" and there should be no crossings of the edges. In fact, It doesn't matter to me whether or not the edges (the boundary) are (is) included or not, but I didn't expect to find an open $U$ if the boundary is included. – 0xbadf00d Nov 9 '15 at 16:08
• @0xbadf00d: You should add that to the question. – copper.hat Nov 9 '15 at 16:10

If I understand your question correctly, then yes. A hyperplane in $\mathbb{R}^3$ is the graph of a linear function of two variables on some of the coordinate axes. That is, the projection operator that takes one of the variables to zero will be a diffeomorphism. The catch is that you can't always choose the same projection operator for each hyperplane. For instance, the hyperplane $z=2$ can't be diffeomorphically projected onto anything but the $xy$ plane. (That is, project "along the $z$ axis").

To find the right projection operator, consider a vector $v \in \mathbb{R}^3$ orthogonal to the hyperplane, and project along any axis where $v$ doesn't have zero component in that direction.

Now just find the right projection operator $\pi$, and your polygon $P$ will be a graph of the linear function on $\pi(P) \subset \mathbb{R}^2$. Projections are open maps, so if you start with an interior of a polygon, $\pi(P)$ will be open.

Example: If $(0,0,0)$, $(1,1,0)$, $(1,1,1)$ is the polygon (triangle) given to me, I notice that it lies in the simple hyperplane given by $\{(x,y,z) \mid x - y =0 \}$. In other words, an orthogonal vector to this hyperplane is $(1,-1,0)$. That means I can choose to project along either the $x$ or $y$ axes and obtain a diffeomorphism. If I project along the $x$ axis, I find that my polygon is the graph of the triangle $(0,0)$, $(1,0)$, $(1,1)$ in the $yz$ plane.

• Can you provide an example for the case when $P$ is a triangle given by its vertices $a,b,c\in\mathbb R^3$? – 0xbadf00d Nov 9 '15 at 16:10
• Let $U=\{(x_1,x_2)| x_k >0, x_1+x_2 <1 \}$, $f(x) = x_1 a + x_2 b + (1-x_1-x_2) c$. – copper.hat Nov 9 '15 at 16:15
• @copper.hat This is related to Barycentric coordinates, isn't it? Can we extend this to an arbitrary polygon? – 0xbadf00d Nov 9 '15 at 16:20
• Can we extend this to $P$, i.e. find an $f$ such that $P$ (not only its interior) is the graph of $f$? Given @copper.hat's solution, I assume not, since then $U$ would not be open. – 0xbadf00d Nov 9 '15 at 16:23
• @0xbadf00d Why do you want it to be the graph of an open set in particular? – Eric Auld Nov 9 '15 at 17:34