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In a course about elliptic regularity probably one sooner or later stubles into the reverse Holder inequalities, and has to introduce the Gehring lemma, which in one of its many versions improves a bit the regularity of a function, given the knowledge that such function already satisfies local bounds via another more integrable function (for example one bounds the integral of $f^2$ on any ball via a power of $f^{2-\epsilon}$ on the ball of double radius).

Even though I saw a proof of this (by Giaquinta-Modica) and I "believe" the result intuitively, I have the feeling that I didn't catch yet the "juice" of it.. for example I don't have the intuition of the following 1) how far it can be extended, or 2) if the more extended/general versions are meaningful to teach (e.g. because one can find a clearer proof, or a more natural one, or because they can show connections with other subjects..).

So I would like to ask you if you know how to put this result in a larger context, or about its connections with topics different than elliptic regularity. Also references to nice expositions of it are quite welcome!

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Not sure what you're looking for but I would just say that this kind of lemma, specifically a refinement of it for tensor-valued functions with additional symmetries (in the tensor and its covariant derivative), played a critical work in fundamental work done by Uhlenbeck, Taubes, Schoen, Yau, and others in Yang-MIlls and Riemannian geometry. –  Deane Yang Jul 20 '11 at 14:10
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I'm not sure about "the larger context", but let me try to dissect the proof a bit so that there will be no mystery left there.

It runs upon 3 main ideas:

1) The "lack of concentration implies better summability" principle. In the nutshell, it is the following. Assume that we have some positive integrable function $f$ with $\int_X f=1$ on some probability measure space and know that for every subset $E$ of measure $\delta$, we have $\int_E f\le c(\delta)$ where $c(\delta)\to 0$ as $\delta\to 0$. Then we can design an increasing function $\Psi$ that grows to infinity and depends on $c$ only such that $\int_X f\Psi(f)<+\infty$. Indeed, assume that $\int_X f=1$. Put $E_k=\{f>2^k\}$. Then $\mu(E_k)\le 2^{-k}$, so, assuming that $\Psi$ is doubling, $\int_X f\Psi(f)$ is comparable to is enough to ensure that $\sum_{k\ge 0} \Psi(2^k)2^k\mu(E_k)\approx \sum_{k\ge 0}\psi(2^k) \int_{E_k}f\le \sum_{k\ge 0}\Psi(2^k)c(2^{-k})$, so we can choose any $\Psi$ for which this series converges. If $c(\delta)=\delta^q$, we can take $\Psi(x)=x^{q-\varepsilon}$ bringing $f$ to some $L^p$ with $p>1$..

This principle (or something like it) is useful in many settings. In general it often happens that we can substantially improve the integral bounds if we know a priori that the functions aren't concentrated on small sets or there are no strong correlations between their values. I'm not ready to give impressive particular examples, but once you digest the idea, you can easily both recognize and use it in other contests.

2) The "multiscale trick". You want to show that $\int_E f$ is small compared to $\int_X f$ when $\mu(E)$ is small. Instead of going from $E$ to $X$ directly, you show first that there is a set $E_1$ of measure just slightly bigger than $E$ such that $\int_{E_1}f$ is noticeably bigger than $\int_E f$. Then you apply this to $E_1$ and so on until you reach $X$. If the path is long, the small improvement at each stage will result in a huge one in the end (another general principle that I wish our students could digest). So, if, say, we can always find $E_1$ such that $\mu(E_1)\le C\mu(E)$ and $\int_{E_1}f\ge a\int_E f$ with $a>1$ (provided that $\mu(E)$ is small enough, of course), we can get the power bound for the integrals over sets of small measure.

This multiscale reasoning is the core of many results in modern analysis. It can be almost trivial like in our case or extremely sophisticated but the idea is always the same.

3) The "geometric structure theorems". They are often presented as covering lemmas and density and maximal function theorems but what they really say is that arbitrary measurable sets in nice spaces can be viewed as finite unions of simple almost disjoint pieces (balls in $\mathbb R^n$, say). For the Gehring lemma, we just take a set $E$ of small measure and create the union of "low but noticeable density balls" surrounding it (for almost each point $x\in E$ we can find a ball $B$ such that the $\mu(B\cap E)\approx \gamma \mu(B)$ where $\gamma$ is some number between $0$ and $1$ we can choose to be whatever we want (provided that $\mu(E)<\gamma$). If $B$ is one such ball, we can estimate the square of the average of $\sqrt f$ over $B$ by $\gamma\int_{B\cap E} f+\int_{B\setminus E}f$. Since the first part is small compared to $\int_B f$ when $\gamma$ is tiny, the condition that these averages are comparable implies that second part should be comparable to the entire integral, so if we extend $E\cap B$ to the entire $B$ (multiplying the measure by $\gamma^{-1}$), we increase the integral some fixed number of times.

That would be the end if we could make our balls truly disjoint and covering the entire $E$ (we would just take $E_1$ to be the union of our balls). Since it is not exactly the case, we have to work a bit more. We can use some standard covering lemma to pass to a sequence of "morally disjoint" balls. If, say, we have a doubling measure in a metric space, we can use Vitaly and say that we can find a set of disjoint balls $B$ in our family such that three times larger balls cover $E$.

This still doesn't seem good enough because if we use $B$'s, we are in danger that they capture only a small part of the integral of $f$ over $E$ and if we use $3B$, they will overlap a lot, so we can count the same piece in the extended integral many times but by now we are pretty convinced that the things should work, so we can go over the steps and do the needed tune-up for this particular theorem.

This tune-up can be done in infinitely many ways. I will do it in the way that allows to illustrate one more idea. Let us use the triple balls $B'=3B$. Let $I=\int_E f$, $\mu=\mu(E)$. Let $E_k$ be the set of points outside $E$ covered by exactly $k$ balls $B'$ and let $\mu_k=\mu(E_k)$, $I_k=\int_{E_k} f$. The above consideration show that $\sum_{k\ge 1}I_k\ge cI$ and the doubling amounts to $\sum_{k\ge 1}\mu_k\le C\mu$, whence we can find non-zero $\mu_k$ with $\frac{I_k}{\mu_k}\ge \varkappa \frac I\mu$ with $\varkappa=\frac cC$. But then adding $E_k$ to $E$, we shall increase the ratio $I/\mu^{\kappa}$. Now just notice that if $E$ is a fixed set of positive measure, then this ratio attains its maximum over all supersets of $E$ and we just showed that this maximum cannot be attained on a set of small measure but for sets of large measure, the ratio is trivially bounded.

This "choose the extremal object" idea can also be found almost everywhere.

Of course, the proof of the Gehring's lemma obtained this way is neither the shortest, nor the slickest. My only excuse for bringing it up is that you seem to be dissatisfied with the proof you know, so I took the liberty to build the argument from the a few standard blocks in modern analysis in the most straightforward way. I also avoided a few technicalities related to localization, but those are totally routine and only obscure the main ideas. I also took the liberty to "misinterpret" your request and talk about the "larger context" for the ideas in the proof rather than for the result itself. In a sense, I tried to show that the Gehring lemma is just a combination of 3-4 well-known and widely used principles and it is given a name just because this particular combination is used as a single block often enough to justify creating a formal reference instead of reproducing the proof every single time. Of course, once you realize what the building blocks are, you can freely play with them and get many other statements in the same style, which, IMHO, is what "catching the juice of a theorem" means. I leave it to others to tell where outside PDE this building block comes handy as a whole but I hope I convinced you that the basic steps of which it consists are ubiquitous.

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Thank you! I think this is a good answer, which points in the direction that "the result is not important per se". Gehring is also equivalent to the self-improving proprety of Muckenhoupt weights, so I would also say L^p-self-improvement could have been the general principle answering of the initial question. But that's not the "juice" as you say..it's more the "summary" –  Mircea Aug 25 '11 at 12:30
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