The classical Calderon-Zygmund decomposition says that if $f\geq 0$ is $L^1$ on a cubes $B$, with average value $\alpha$, then there is a sequence of disjoint cubes $B_j$, such that the average of $f$ on each $B_j$ is in between $\alpha$ and $2^n \alpha$, and $f\leq \alpha$ a.e. away from $\bigcup_j B_j$.

I am wondering if there is a similar result on (compact) manifolds, for example with a volume doubling condition. If one uses a Vitali type argument, it seems there is no control on the radius (they could be too small), so no upper bound in the average value of $f$ on (small) balls obtained by Vitali - in contrast to the $2^n\alpha$ upper bound of Calderon-Zygmund.

  • $\begingroup$ Do you endow the manifold with a Radon measure? It is unclear to me whether these volume conditions would depend on the choice of measure. $\endgroup$
    – Z. M
    Nov 22, 2021 at 9:24
  • 1
    $\begingroup$ On a compact manifold where the measure is induced by the Riemannian metric, could you not just apply the classic result in local coordinates? You get some additional multiplicative constants from this, but since you only need finitely many coordinate charts these are bounded and the result would probably still be enough for most estimates that need the decomposition. $\endgroup$
    – mlk
    Nov 22, 2021 at 9:31
  • $\begingroup$ @mlk It will be ideal if one only assumes a volume doubling condition, i.e. $\nu(B_{2R}(x))\leq 2^\kappa \nu(B_R(x))$, and perhaps the Poincare inequality. Assuming a lower bound in injectivity radius would be too strong in many situations... $\endgroup$
    – Yuval
    Nov 22, 2021 at 11:10
  • $\begingroup$ My impression is that this is done on page 628 of Coifman and Weiss "Extensions of Hardy spaces and their use in analysis," but I'm not familiar with any details $\endgroup$ Nov 22, 2021 at 16:18
  • $\begingroup$ @Yuval I think you might even be able to get away with less. You'll only at some point have to relax your definition of $B_j$ a bit to include slightly less nice sets. E.g. if you glue a tiny handle to the plane and concentrate all of $f$ on there, then the only way out is to use one or more large sets that include parts or all of the handle as well as a good chunk of the plane to draw out the average. $\endgroup$
    – mlk
    Nov 22, 2021 at 17:19

1 Answer 1


Calderon-Zygmund theory generalises without much difficulty to doubling metric measure spaces (or more generally to "spaces of homogeneous type"). See for instance Chapter 1 of

Stein, Elias M., Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals. With the assistance of Timothy S. Murphy, Princeton Mathematical Series. 43. Princeton, NJ: Princeton University Press. xiii, 695 pp. (1993). ZBL0821.42001.

The generalisation of the Calderon-Zygmund decomposition in this setting is given in Section 1.4.

In more recent years, most of Calderon-Zygmund theory has also been extended to the non-doubling case, though one has to make some natural modifications to the statements in order to avoid trivial counterexamples. One reference is Chapter 2 of

Tolsa, Xavier, Analytic capacity, the Cauchy transform, and non-homogeneous Calderón-Zygmund theory, Progress in Mathematics 307. Cham: Birkhäuser/Springer (ISBN 978-3-319-00595-9/hbk; 978-3-319-00596-6/ebook). xiii, 396 p. (2014). ZBL1290.42002.

The Calderon-Zygmund decomposition is given as Lemma 2.14 of that book, though the statement may look slightly different from the classical one.

One common trick in this subject is to replace balls by roughly equivalent "dyadic cubes" that have good geometric properties, such as nesting. See for instance

Hytönen, Tuomas; Kairema, Anna, What is a cube?, Ann. Acad. Sci. Fenn., Math. 38, No. 2, 405-412 (2013). ZBL1288.30066.

for a brief introduction to this topic.


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