Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose $(M,g)$ is a complete Riemannian manifold. $p\in M$ is a fixed point. $d_{p}(X)$ is the distance function defined by $p$ on M (i.e., $d_p(x)$=the distance between $p$ and $x$). Let $\epsilon>0$ be an arbitrary positive number. Is there a smooth function $\tilde{d}_p(x)$ on $M$, such that $$ | d_p(x)-\tilde{d}_p(x) | < \epsilon$$ $$ |\textrm{grad}(\tilde{d}_p)(x)|<2$$ for $\forall x \in M$ ?

Functions satisfying the first condition can be constructed easily by partition of unity and the standard technique of mollifiers. However, I can't see how to control the gradient of the approximate function.

I need this result when I'm trying to follow a proof. The existence of such a function is taken for granted in that proof, and is called the "regularization" of the distance function. This question also seems to be interesting by itself.

I'm not sure if this question is too easy for math overflow. I put it on stack exchange last week but nobody answered it. Could you please help me? Thanks a lot.

share|improve this question

4 Answers 4

up vote 7 down vote accepted

You function $d_p$ is a Lipschitz function (w.r.t. to the Riemannian distance) with the Lipschitz constant $1$ and it is possibly, for every $\varepsilon>0$, to $\varepsilon$-approximate it by a smooth function with the length of the gradient $\le 1 + \varepsilon$.

The proof of this statement, even if we replace 'Riemannian' by 'Finslerian', can be found in the appendix to my paper arXiv:1112.5060; the proof seems to be wellknown for experts so I did not pretend that the proof is new but tried to explain it in all details. In the Riemannian category, the statement was proved in D. Azagra, J. Ferrera, F. Lopez-Mesas, Y. Rangel, Smooth approximation of Lipschitz functions on Riemannian manifolds, J. Math. Anal. Appl. 326(2007) 1370–1378. I should confess that I found it complicated to read this paper but if you need it as a citation and not in order to understand the proof it could be an option

share|improve this answer

If I'm not mistaken, this follows from from Greene-Wu, "$C^\infty$ approximation of convex, subharmonic and plurisubharmonic functions", Annales scientifiques de l'ENS, 1979. Actually you need some kind of mollifying technique that takes into account the geometry. I think this is done in section 2.

I think it might be possible to get that in a cheaper way, at least given some bounds on the curvature and maybe the injectivity radius. Would heat flowing the distance function work ?

edit: I checked Greene-Wu's paper, this is done in section 2. Contrary to the result quoted by Pr Matveev, this only works in finite dimension. However, depending on one's taste, it might be more readable.

share|improve this answer
1  
The results I quote (mine and of Azagra-Ferrera-Lopez-Mesas-Rangel) are also finite-dimensional. Finsler metric is not when we allow infinite dimension but if we replace the scalar product in each tangent space by a norm. And, in my humble oppinion which may be different from oppinion of other mathematicians, my papers are readable :-) –  Vladimir S Matveev Nov 26 '13 at 18:59
    
But isn't an assumption on the curvature needed? –  Deane Yang Nov 26 '13 at 20:53
    
Deane, if it is a question related to my comment/answer, the answer is no –  Vladimir S Matveev Nov 26 '13 at 21:04
1  
@VladimirSMatveev : I had a look at Azagra's paper, the improvement over Greene and Wu is that it handle infinite dimensional manifolds (at the level of the results, not of the proof which use really different methods if I understand). –  Thomas Richard Nov 27 '13 at 2:51
    
@DeaneYang For Greene and Wu, no curvature assumption is required. –  Thomas Richard Nov 27 '13 at 2:52

This is an additional comment about higher derivative bounds. Consider the class of pointed complete noncompact Riemannian manifolds $(M^{n},g,O)$ with $\left\vert \operatorname{sect}\left( g\right) \right\vert \leq K$. For the distance-like function constructed from mollification by Greene and Wu, one also has a uniform upper bound for the Hessian of $u$. To obtain a two-sided bound for the Hessian, one can smooth $u$ by the heat equation. In a paper titled "Construction of an exhaustion function on complete manifolds", Luen-Fai Tam proves that there exists a $C^{\infty}$ function $f:M\rightarrow \mathbb{R}$ with $d\left( \cdot,O\right) +1\leq f\leq d\left( \cdot,O\right) +C$, $\left\vert \nabla f\right\vert \leq C$, and $\left\vert \nabla\nabla f\right\vert \leq C$, where $C$ depends only on $n$ and $K$. Techniques that Tam uses include heat kernel estimates and weighted $L^{2}$ estimates. An exposition of Tam's result is given in Section 4 of Chapter 26 in the book "The Ricci Flow: Techniques and Applications: Part III: Geometric-Analytic Aspects"; see therein for references to previous related work.

share|improve this answer

EDIT: this doesn't work (see Sergei's comment).

If you're worried about a distance function instead of a general Lipschitz function, I think the following construction also works, and even gives you a better gradient bound.

Let $\phi$ be a smooth function with support in $B_\epsilon(p)$ and integral 1, and set $$ \tilde{d}_p(x) = \int_M \mathrm{dist}(x,y)\phi(y) dy $$ The function $\tilde{d}_p(x)$ is smooth because the function $\mathrm{dist}(x,y)$ is smooth for almost every $y$. Moreover, $$ \begin{align} |\tilde{d}_p(x) - d_p(x)| &= \left|\int_M(\mathrm{dist}(x,y) - \mathrm{dist}(x,p))\phi(y) dy \right| \\ &\leq \int_M \mathrm{dist}(p,y) \phi(y) dy \\ &\leq \epsilon \end{align} $$ and $$ \begin{align} |\nabla_V\tilde{d}_p(x)| &= \left|\int_M \nabla_V\mathrm{dist}(x,y)\phi(y)dy\right|\\ &\leq \int_M |V| \phi(y)dy \\ &\leq |V| \end{align} $$ and so we get the stronger statement that $|\mathrm{grad}(\tilde{d}_p(x))| \leq 1$.

share|improve this answer
    
I'm afraid this function may fail to be $C^\infty$. Although $dist(\cdot,y)$ is smooth for a.e. $y$, its higher order derivatives may be large if $y$ approaches a singular point. Then, when computing the respective derivative of $\tilde d_p$, you may get a non-summable integral. –  Sergei Ivanov Nov 29 '13 at 13:19
    
Oops, thanks for setting me straight. –  Peter Smillie Nov 30 '13 at 6:57

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.