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Let $X$ be a compact manifold with boundary $\partial X = X_0 \cup X_1$, and let $\omega$ be a volume form on $X$. Suppose $f:X \rightarrow [0,\infty)$ is a smooth non-negative function.

Is it always true that there exists a path $c:[0,1] \rightarrow X$ with $c(0) \in X_0$ and $c(1) \in X_1$ and $c(t) \in X-\partial X$ for all $0<t<1$ with the property that

$$\int_X f \omega \ge \int_X \omega \cdot \int_0^1 f(c(t))dt?$$

This seems very plausible, yet I don't know how to try proving it. Could somebody help me?

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by setting $f=1$ I get $\int_X\omega\geq1$. Should we suppose that it's satisfied? –  user8268 Jun 8 '11 at 13:57
    
Yes, sorry I should have said that. Alternatively, I've edited it slightly. –  Dave Jun 8 '11 at 14:07

1 Answer 1

up vote 0 down vote accepted

It can be proved in this way. Let $x\in M$ be a point at which $f$ attains its maximum. For every $\epsilon>0$ there is a smooth path $c$ from $X_0$ to $X_1$ such that $c(t)=x$ if $\epsilon/2<t<1-\epsilon/2$. By choosing $\epsilon$ small enough you get your inequality.

To construct $c$: choose a path from $X_0$ to $x$, another path from $x$ to $X_1$, reparametrize them so that all derivatives vanish when you are at $x$, and put in the middle a constant path waiting at $x$.

If $M$ is the maximum of $f$ and $m$ its minimum then $\int_0^1 f(c(t))dt\geq M-\epsilon (M-m)$. It is easy to see that you get your inequality for $\epsilon$ small enough (e.g. by considering the cases $M>m$ and $M=m$ separately).

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Thanks! That's very helpful. –  Dave Jun 9 '11 at 5:52

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