# Joining boundary points through affine/billiards trajectories

Assume we have a smooth bounded domain $$\Omega \subset \mathbb{R}^d$$, and two points at the boundary $$x, y$$. I want to show that there exists an integer $$n$$ such that one can draw an affine path joining $$x$$ to $$y$$ in $$n$$ steps. By an affine path I mean a path which picks up a direction at every contact with the boundary, follows this direction until it hits the boundary again, and pick a new direction at this rebound, and so on...

So in a strictly convex set, the number $$n$$ is 1 for any couple of points such that $$x \ne y$$, two for the couple $$(x,x)$$ (starting from $$x$$, we can rebound anywhere but at $$x$$ in one step and come back to $$x$$ for the second rebound). In general, starting from $$x$$, we can choose the direction, draw the corresponding segment inside the domain until a point $$z \in \partial \Omega$$, then choose another direction... and so on until we hit $$y$$. In general a result like the density of the set of orbits starting at a point will be a sufficient condition (but will also be way more than what we need, since at every rebound we can go from one orbit to the next one), but I am unsure that it holds without further conditions.

Also note that by connectedness, a path already exist, so it seems quite natural that an affine path should exist as well. I have tried to twist this path to obtain something but could not get what I was looking for.

You can think of pathological geometries such as the domain of $$(x,y)$$ defined by
$$0.9x^2\sin\frac\pi x\le y\le1.1x^2\sin\frac\pi x, x\in[0,1].$$
A path from $$(0,0)$$ to $$(1,0)$$ must make an infinity of bends, though the curve is everywhere differentiable.
• Great example. However it seems to me that this curve is not even $C^2$, and thus that this domain is not smooth (by which I mean it is a $C^{\infty}$-manifold with boundary). Am I missing something ? – Gâteau-Gallois Oct 11 '18 at 8:48