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Let $(M,g)$ be a complete 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 of same volume than $\Omega$.

Is that true?

What if $\Omega$ is large, how can one define a geodesic ball (say if the radius is larger than $inj(M,g)$).

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up vote 1 down vote accepted

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 $|\nabla r|=1$) $$ \int_\Omega f(r) dV= \int_0^\infty f(s) Area(\{r=s\}\cap\Omega) ds $$ 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 $$ \int_{\Omega^*} f(r) dV-\int_\Omega f(r) dV= \int_0^\infty f(s) \varphi(s) ds $$ $$ = \int_0^R f(s) \varphi(s)ds + \int_R^\infty f(s)\varphi(s) ds $$ $$ \geq f(R) \int_0^R\varphi(s)ds + f(R) \int_R^\infty \varphi(s) ds = 0. $$

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This is very clear. Thank you. I have an additional question in fact. I am always bugged by the cut locus. Suppose that $f(r) = r^{-n}$. If the radius $R <$ injrad, then $\int_\Omega f(r) dV = \int_{\Omega^\star} r^{-n} dV $ We can then use Gauss' lemma to pass on to "polar coordinates", to get that $ \int_{\Omega^\star} r^{-n} dV = C \int_0^R r^{-n} (r^{n-1} + O(r^2)) dr $ and do some explicit computations. How can one do that if the radius is larger than injrad? –  Henry Jun 14 '13 at 13:32
@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
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