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Let $(M,g)$ be a non-compact Riemannian manifold, with finite volume (or compactly exhausted, or any nice condition you would like, except for compactness). Suppose I have a tensor $T$ on $M$ of which I have "good" estimate locally, and I would like to have a global estimate of $T$. However, since $M$ is non-compact and I'm not allowed to take a finite cover, I have to avoid the following scenario: the "good" local estimate is only obtained because I can arbitrarily shrink the neighborhood. Hence some scaling-invariance is needed for the local estimate. A model would be the following

Question: Let $(M,g)$ be a non-compact Riemannian manifold of dimension $n$, with finite volume. If $T$ is a $q$-form on $M$, and there is a fixed number $H$ such that for every $x\in M$, $$ \sigma^{q-n}\int_{B_{\sigma}(x)}|T|\mathrm{vol}_g<H $$ for sufficiently small $\sigma$, where $B_{\sigma}(x)$ is the geodesic ball of radius $\sigma$ centered at $x$. Can one deduce $$ \int_M|T|\mathrm{vol}_g<C_{M,H} $$ from the local estimate? (If yes, how? If no, what should be the correct statement?)

Better: Can the scaling-invariant local estimate be phrased in a way that only involves "neighborhoods," instead of "radius of geodesic balls?" (And still yielding a global estimate, of course.)

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  • $\begingroup$ Should it be $|T|^2$? $\endgroup$
    – Deane Yang
    Commented Dec 27, 2013 at 23:02
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    $\begingroup$ Have you tried testing your conjecture for functions and vectors (in particular, constant ones) on $R^n$? $\endgroup$
    – Deane Yang
    Commented Dec 27, 2013 at 23:18
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    $\begingroup$ Isn't the second integral unbounded if the function is constant? $\endgroup$
    – Deane Yang
    Commented Dec 27, 2013 at 23:59
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    $\begingroup$ Why? $M$ is supposed to have finite volume, and the constant $C_{M,H}$ depends on both $M$ and $H$. $\endgroup$
    – A curious cat
    Commented Dec 28, 2013 at 0:03
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    $\begingroup$ If all you have is a scale invariant estimate away from the points, then there's no way to get a global bound. You should be able to construct a counterexample (use a function) that blows up at one of the points and does not have a finite $L^p$ bound. Just use whatever blow-up rate (as a function of distance to the point) that gives the scale-invariant integral bound you stated. $\endgroup$
    – Deane Yang
    Commented Dec 28, 2013 at 0:58

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