# Convex sets with analytic boundary, using angles to parametrize boundary

Suppose that $$D$$ is a bounded open convex subset of $$\mathbb{R}^2$$ with analytic boundary. You can parametrize the boundary of $$D$$ using the angles of the support lines at each point, but it isn't obvious that the parametrization you get is a real analytic function of the angles. Is there a quick reference to show that such a parametrization will in fact be analytic?

Such a parametrization will not be in general real analytic, because the angle of the support line may be varying too slowly at some points. E.g., let the boundary of the convex set $$D$$ be $$C:=\{(x,y)\in\mathbb R^2\colon x^4/2+(y-1)^2=1\}.$$ Then the points on $$C$$ in a neighborhood of the point $$(0,0)\in C$$ will be on the graph of the function $$f$$ given by $$f(x):=1-\sqrt{1-x^4/2}$$ for real $$x$$ with $$|x|<2^{1/4}$$. For $$x\to0$$, we have $$f'(x)\sim x^3$$, whence $$x\sim f'(x)^{1/3}=\tan^{1/3}\theta\sim\theta^{1/3}$$, where $$\theta$$ is the angle between the horizontal axis and the support line to $$D$$ at the point $$(x,f(x))\in C$$. So, $$(x,f(x))$$ is not a real analytic function of the angle $$\theta$$.

Added in response to the additional question in a comment by the OP: The boundary (say) $$C$$ of any bounded open convex subset $$D$$ of $$\mathbb R^2$$ can be real-analytically parametrized by the (say central) angle between rays from any point in $$D$$. Indeed, without loss of generality this point in $$D$$ is $$(0,0)$$. Take any point $$(x_0,y_0)\in C$$. Let $$(-1,1)\ni t\mapsto(x_t,y_t)$$ be a real-analytic parametrization of $$C$$ in a neighborhood of the point $$(x_0,y_0)\in C$$ such that the tangent vector $$(x'_0,y'_0)$$ to $$C$$ at the point $$(x_0,y_0)\in C$$ is nonzero. Then for all real $$t$$ close enough to $$0$$ and for the angle $$a_t$$ between the vectors $$(x_t,y_t)$$ and $$(x_0,y_0)$$ we have $$\cos a_t=\frac{x_tx_0+y_ty_0}{\sqrt{x_t^2+y_t^2}\sqrt{x_0^2+y_0^2}}$$ and hence $$\frac{da_t}{dt}\Big|_{t=0}=\pm\frac{x_0 y'_0-y_0 x'_0}{x_0^2+y_0^2}\ne0, \tag{1}$$ since the tangent vector $$(x'_0,y'_0)$$, parallel to the support line to the convex domain $$D$$ at the boundary point $$(x_0,y_0)$$ of $$D$$ cannot be collinear with the radial vector going from the interior point $$(0,0)$$ of $$D$$ to the point $$(x_0,y_0)$$ (the choice of sign in (1) depends on whether $$(x_t,y_t)$$ moves counterclockwise ($$+$$) about the origin or clockwise ($$-$$)). Thus, the function $$t\mapsto a_t$$ is invertible in a neighborhood of $$0$$, and its inverse (say) $$a^{-1}$$, is real analytic. So, the function $$\alpha\mapsto(x_{a^{-1}(\alpha)},y_{a^{-1}(\alpha)})$$ for real $$\alpha$$ close enough to $$0$$ provides a real-analytic parametrization of $$C$$ by the central angle $$\alpha$$ in a neighborhood of any given point $$(x_0,y_0)\in C$$.

• Thank you for your answer. This makes me wonder whether the boundary can be analytically parametrized by the angle that the points on the boundary make with some fixed point in the interior, instead of the angle of the support line. There are a lot of things I don't know about analytic curves so I would be particularly interested in a good reference! Dec 19, 2019 at 14:08
• Yes, I thought about something similar to your updated comment, but what if the original analytic parametrization, $(x_t,y_t)$ in your notation, doubles back on itself (is not 1-to-1)? Then you can't assert that $\frac{da_t}{dt} \ne 0$. But maybe it is obvious that you can define an analytic parametrization in a neighborhood of any point that is locally 1-to-1. Dec 19, 2019 at 15:27
• @BrianLins : The original real-analytic parametrization needs to be only local, and we may choose any such parametrization. If one such parametrization exists, then infinitely many other ones exist as well. Dec 19, 2019 at 15:41