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

Let $(M,g)$ be a complete Riemannian, connected, compact manifold (with or without boundary). Let $f(r)$ be a decreasing function of $r =$ geodesic distance. If $\Omega \subset M$, then $$ \int_{\Omega} f(r)\, dV \leq \int_{\Omega^\star} f(r)\, dV \, ,$$ where $\Omega^\star$ is a geodesic ball with the same volume as $\Omega$.

Is that true? What if $\Omega$ is large? How can one define a geodesic ball, say if the radius is larger than $\mathrm{inj}(M,g)$?

Addentum:

After reading the argument given below, there is still something that seems unclear to me. In order to use the co-area formula, one needs to have certain hypotheses, namely smoothness of functions (at least, the statement for manifold I know is for smooth functions : http://en.wikipedia.org/wiki/Smooth_coarea_formula). In particular, when claiming that $|∇r|=1$ in the argument below, it seems true only for $r<injrad(M,g)$. I believe (though I do not have a reference) that the distance function on a complete manifold is Lipschitz, thus differentiable almost everywhere by Rademacher's Theorem. But even with that, I am not sure this is enough to apply the coarea formula on a MANIFOLD.

share|improve this question

2 Answers 2

up vote 4 down vote accepted

This is sometimes called the ``bathtub principle'', and can be proved elementarily in a great generality (metric measure spaces).

Consider $\mu$ the measure on $[0,+\infty)$ obtained by pushing forward the volume measure restricted to $\Omega$ by the function $r$, and define similarly $\mu^*$. Then since $|\Omega|=|\Omega^*|$, $\mu$ and $\mu^*$ have the same total mass and the inequality to be proven reduces to $$\int f(x) d\mu(x) \le \int f(x) d\mu^*(x)$$ and this one-dimensional question is easily handled.

For example, one can proceed as follows (maybe slightly too complicated, I am not sure). By construction of $\mu$ and $\mu^*$ we have $\mu(I)\le \mu^*(I)$ for all initial interval $I$. Therefore, the monotone rearrangement from $\mu$ to $\mu^*$ (which is a measure $\pi$ on $[0,+\infty)^2$ with marginals $\mu$ and $\mu^*$) must be supported on the set of couples $(x,y)$ such that $x\ge y$. Then it comes $$\int f(x) \,d\mu(x) = \int f(x) \,d\pi(x,y) \le \int f(y) \,d\pi(x,y) = \int f(y) \,d\mu^*(y).$$

share|improve this answer
    
I am not familiar with the monotone rearrangement from μ to μ ∗ and I could not find an explicit definition. Would you mind adding such details, please? –  Henry Jul 23 at 17:09
    
You will find definition of the monotone rearrangement in many books on optimal transportation (e.g. one of Villani's books). In words, it is the measure on $[0,+\infty)^2$ that has marginals $\mu$ and $\mu^*$, and which ``maps'' the leftmost part of the mass of $\mu$ to the leftmost part of the mass of $\mu^*$. –  Benoît Kloeckner Jul 23 at 18:09

UPDATE: I notice that the question leaves the possibility of boundary open. The answer below only works as written if there is no boundary. I'm not sure what happens if there is boundary, but one should be quite careful in this case, and I would guess that the claim is false.

--

The answer is "yes" if you define $\Omega^*=\exp_p(B_R(0))$ for $R$ chosen so that $|\Omega| = |\Omega^*|$. Of course, $R$ could be larger than the injectivity radius, but no worries.

To see that you can choose such an $R$, note that $R\mapsto\exp_p(B_R(0))$ forms an exhaustion of your manifold by the complete, connected hypothesis (in fact you could probably drop compact in exchange for just requiring complete and $|\Omega| < \infty$).

Now, to prove your identity we may use the co-area formula to write (using that $|\nabla r|=1$ almost everywhere) $$ \int_\Omega f(r) dV= \int_0^\infty f(s) Area(\{r=s\}\cap\Omega) ds $$ and $$ \int_{\Omega^*} f(r) dV= \int_0^\infty f(s) Area(\{r=s\}\cap\Omega^*) ds $$ Define $\varphi(s):=Area(\{r=s\}\cap\Omega^*)-Area(\{r=s\}\cap\Omega)$. The co-area formula applied again gives that $\int_0^\infty \varphi(s) ds = 0$. Also, by construction $\varphi(s) \geq 0$ for $s\leq R$ and $\varphi(s) \leq 0$ for $s \geq R$.

Thus, subtracting the above identities gives $$ \int_{\Omega^*} f(r) dV-\int_\Omega f(r) dV= \int_0^\infty f(s) \varphi(s) ds $$ $$ = \int_0^R f(s) \varphi(s)ds + \int_R^\infty f(s)\varphi(s) ds $$ $$ \geq f(R) \int_0^R\varphi(s)ds + f(R) \int_R^\infty \varphi(s) ds = 0. $$


There seems to be some confusion about the application of the co-area formula. The co-area formula holds for Lipschitz functions on a manifold. This was first proved by Federer, see Theorem 3.1 here. This reference may be difficult to follow. A more readable discussion can be found in Krantz and Parks's book "Geometric Integration Theory," see Theorem 5.4.8 and the preceding discussion.

share|improve this answer
    
@Henry, I'm not sure I quite understand your question -- $r$ is defined on all of $M$ because its just the distance from $p$. On the other hand, I agree that "polar coordinates" only really make sense before the cut locus.. –  Otis Chodosh Jun 14 '13 at 17:50

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.