I already asked this question at stackexchange with no response - so I'll try here. 


I'm reading a paper on discrete differential geometry:
[Meyer et.al.](http://www.cs.caltech.edu/~mmeyer/Publications/diffGeomOps.pdf)


They define the Laplace-Beltrami operator at a point $P$ by 
$$\vec{K}(p) = 2k\_H(P)\vec{n}(P)$$
where $\vec{n}(p)$ is the normal vector at $p$ and $k\_H(p)$ the mean curvature. Then in Section 3.2 they give an error bound for the discrete Laplace-Beltrami operator obtained in the previous section.


At one step they write $||\vec{K}(x) - \vec{K}(x\_i)||^2 \leq C\_i^2||x-x\_i||^2$ where $C\_i$ is the Lipschitz constant of the Laplace-Beltrami operator. 


Does anyone have a reference that proves that the Laplace-Beltrami operator is Lipschitz (at least for surfaces in $R^3$)? 


Thank you.



Lipschitz constant of Laplace-Beltrami Operator
First, a comment on terminology. What you denote by $\vec K(p)$ seems to me to just be the vector version of the mean curvature (that's what I'm going to call it). This is a vector, not an operator. The "Laplace-Beltrami operator" is a generalization of the Laplacian on a Riemannian manifold (see <http://en.wikipedia.org/wiki/Laplace%E2%80%93Beltrami_operator>); it takes as input a function on the manifold and returns another one (and is thus called an operator). This is at least related to the mean curvature, in that probably applying the Laplace beltrami operator to the embedding $S\to\mathbb R^3$ would give the function $\vec K(p)$, but it's wrong to call $\vec K(p)$ the Laplace-Beltrami operator.


Now to answer your question. Let's call the surface $S\subseteq\mathbb R^3$. A function $K:S\to\mathbb R^3$ (like the mean curvature $\vec K(p)$) is called Lipschitz iff $C:=\sup\_{x,y}\|K(x)-K(y)\|/\|x-y\|$ is finite. So, your equation is basically the definition of the mean curvature vector being Lipschitz.


For any given surface, the mean curvature may or may not be Lipschitz. It amounts to the existence of some higher derivatives of the embedding. In particular, if your surface is parameterized by a thrice-differentiable function (whose derivative is nonzero), then $K$ is Lipschitz. If not, the probably not, and there might not be a $C$ for which your inequality holds for all $x,y\in S$. For instance, on [this](http://mathworld.wolfram.com/Capsule.html) surface, $K$ is not even continuous, let alone Lipschitz, even though the surface doesn't have any "corners".


