# Does the R-sphere condition imply that a surface is locally a graph of function on a ball of radius R?

Let $S$ be a $C^2$-regular hypersurface with $S=\partial V$ for some open set $V \subset R^{N+1}$, and let $\nu(P)$ be the exterior unit normal of $S$ with respect to $V$.

Assume that $S$ satisfies the R-sphere condition, that is for every $P\in S$ the tangent balls $B^\pm(P,R):= \{Q \in R^{N+1}: \ |P\pm R \nu(P) - Q|<R\}$ do not contain any points of $S$.

Let the origin $O\in S$ and assume that $\nu(O)=e_{N+1}$. By the implicit function theorem we know that $S$ can be written locally as a graph of function. In particular, there exists $r>0$ and a real valued function $u$ such that $$\{ (x,u(x)): |x|<r \} \subset S .$$

**Can we conclude that $r=R$ **?

I proved that this is true for $N=1$ as sketched in the following. By writing the touching balls at $O$ we obtain that $$|u(x)| \leq \rho - \sqrt{\rho^2-x^2},\quad |x|<\rho.$$ If we assume that $\nu((x_0,u(x_0))) \cdot e_2 = 0$ for some $|x_0|<\rho$, then the centers $C^\pm$ of the two touching balls at $(x_0,u(x_0))$ must satisfy $|C^\pm|\geq \rho$ (otherwise one of the two touching balls will contain the origin). By writing down the inequality and using the above estimate on $u$ we find a contradiction.

Unfortunately, it seems to me that the same argument does not work for $N\geq 2$.

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It is true and not too hard. WLOG, $R=1$

Let $p,q$ be two points on $S$. Let $a$,$b$ be the outer unit normals to $S$ at $p$ and $q$ respectively. Let $v=\overline {pq}$. Since $B(p-a,1)\cap B(q+b,1)=\varnothing$, we have $|a+b+v|^2\ge 4$. Since $B(p+a,1)\cap B(q-b,1)=\varnothing$, we get $|a+b-v|^2\ge 4$. Adding those up, opening the parentheses, dividing by $4$, and expressing $(a,b)=\cos\theta$ where $\theta$ is the angle between $a,b$, we get $\cos\theta+\frac{|v|^2}2\ge 1$, which, for infinitesimally small $v$, means $\theta\le|v|$, i.e., if you move along any curve on $S$, the angle between the normals at the endpoints of the curve is at most the curve length.

The rest is trivial: to define the function at $x\in\mathbb R^n$ with $|x|<1$ just start moving along $S$ at speed $1$ so that the projection of your velocity vector to $\mathbb R^n$ points at $x$. You have no trouble as long as the scalar product of the normal with $e_{n+1}$ stays positive, i.e., the length $\frac\pi 2$ is always yours. But, since the normal to the surface rotates not faster than you go, your projected speed pointed toward $x$ is at least the cosine of the distance your travelled, so you are doomed to reach $x$ (because $\int_0^{\frac\pi 2}\cos \ell\,d\ell=1>|x|$). The continuity and smoothness of the resulting function is just the usual mumbo-jumbo about the dependence of a solution to a parametric ODE on the parameters.

This will make an excellent problem for a qualifier exam in differential geometry. Thanks for the idea! :-)

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I think that once we have that $\cos\theta+\frac{|v|^2}2\ge 1$, then can conclude easily by using that $$|v|^2 \leq |x|^2 + (1-\sqrt{1-|x|^2})^2,$$ for $|x|\leq1$, which follows by writing the touching spheres at $O$. Indeed we find that $\cos\theta \geq \sqrt{1-|x|^2}$ and we conclude. Thank you again for your answer! – pedro May 17 '14 at 13:20