ANALYSIS IN METRIC SPACES

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1 ANALYSIS IN METRIC SPACES HELI TUOMINEN Contents 1. About these notes 2 2. Classical extension results Tietze(-Urysohn) extension theorem Extension of Lipschitz functions Whitney covering, partition of unity and extension About Sobolev extension domains in R n ( ) 9 3. Doubling measures, covering theorems and Hardy-Littlewood maximal function asic properties of doubling measures Covering theorems Hardy-Littlewood maximal function and the Lebesgue differentiation theorem Noncentered maximal function Lebesgue differentiation theorem Mapping properties of the maximal function Regularity of the maximal function in the Euclidean setting Hardy-Littlewood maximal function of Lipschitz funtion in metric spaces The discrete convolution and the discrete maximal operator Sobolev spaces in metric measure spaces - M 1,p (X) Sobolev spaces in metric measure spaces - N 1,p (X) Modulus of a curve family Properties of p-weak upper gradients How to find p-weak upper gradients Minimal p-weak upper gradient Convergence properties of p-weak upper gradients Definition and basic properties of N 1,p (X) Nontriviality of spaces N 1,p (X) About Poincaré inequalities i-lipschitz invariance of the Poincaré inequality p-poncaré inequality - improvement and dependence of p Removability for Sobolev spaces and Poincaré inequality Classical results How does the removability property of a fixed set depend on p? 70 Versio: April 7,

2 About the metric setting Extension results for Sobolev spaces in the metric setting Measure density from extension Extension from measure density 79 References About these notes You are reading the lecture notes of the course Analysis in metric spaces given at the University of Jyväskylä in Spring semester We start with the classical extension results of Tietze, McShane and Whitney and hope to end up to recent extension results for Sobolev spaces in the metric setting. The main topics of the course are doubling measures, covering theorems, maximal functions, Lipschitz-functions, Poincaré inequalities, Sobolev spaces and extension domains. Prerequisites for the course are courses topology 1 (metric spaces), measure and integration theory, course real analysis and familiarity with classical Sobolev spaces are recommended. Notation and standard assumptions: We assume that X = (X, d, µ) is a metric measure space equipped with a metric d and a orel regular, doubling outer measure µ for which the measure of every ball is positive and finite. The doubling property means that there exists a fixed constant c D > 0, called the doubling constant, such that µ((x, 2r)) c D µ((x, r)) for every ball (x, r) = {y X : d(y, x) < r}. The distance between sets A X and X is d(a, ) = inf{d(a, b) : a A, b }, and for x X, the distance of x to the set A is The diameter of the set F X is d(x, A) = d({x}, A) = inf{d(x, a) : a A}. diam F = sup{d(y, z) : y, z F }. If u: X R is function and t R, we sometimes use a short notation {u > t} for sets {x X : u(x) > t}. The Lebesgue measure of a mesurable set E R n is denoted by L n (E) or by E. We sometimes say that a measurable function u: X R is p-integrable, if X up dµ <. y χ E, we denote the characteristic function of a set E X. In general, C is a positive constant whose value is not necessarily the same at each occurrence. y writing C = C(c 1, c 2 ), we mean that the constant C depends only on the constants c 1 and c 2. If there is a positive constant C 1 such that C1 1 A C 1 A, we say that A and are comparable. 2

3 2. Classical extension results This section contains the classical extension results of Tietze, McShane and Whitney for continuous and Lipschitz functions. We also discuss Whitney decomposition, a covering of a complement of a closed set in R n by cubes whose diameters are comparable to the distance of the cube to the closed set, and a corresponding partition of unity. We close the section by a brief history about Sobolev extension domains Tietze(-Urysohn) extension theorem. Tietze extension theorem says that we can always find a continuous extension for a continuous, real valued function defined on a closed set. It is equivalent to the Urysohn lemma, which says that whenever E X and F X are disjoint, nonempty closed sets, then there exists a continuous function g such that g E = 0, g F = 1 and 0 < g(x) < 1 for all x X \ (E F ). Theorem 2.1 (Tietze). Let F X be a closed set and let u: F R be a continuous function. Then there is a continuous function U : X R such that U F = u. Moreover, if there is a constant M > 0 such that u(x) M for all x F, then U(x) < M for all x X \ F. There are several different proofs for the Tietze extension theorem. elow, we give two different proofs. In the first one, we use the following lemma. Lemma 2.2. Let F X be a closed set. If u: F R is a continuous function and M > 0 is a constant such that u(x) M for all x F, then there is a continuous function v : X R such that (i) v(x) 1 M for all x F, 3 (ii) v(x) < 1 M for all x X \ F, 3 (iii) u(x) v(x) 2 M for all x F. 3 Proof. Let A = { x F : u(x) 1 3 M} and = { x F : u(x) 1 3 M}. The sets A and are disjoint and, by the continuity of u and the assumption u M, also closed. If A and, then, using the continuity of the distance function and the definition of the sets A and, it is easy to see that the function v : X R, v(x) = M d(x, A) d(x, ) 3 d(x, A) + d(x, ) has the required properties. If one of the sets A and is empty and the other, denoted by C, is not, then ( ) v(x) = M 3 1 min{1, d(x, C)} satisfies (i) (iii). If A = and =, then we can take v = 1 6 M. Proof 1 of Theorem 2.1. Assume first that u(x) M for all x F. We will use Lemma 2.2 and induction to find continuous functions v i such that the function series i=0 v i converges and gives us the extension U. 3

4 Let v 0 : X R, v 0 = 0. Assume that k N and that there exists continuous functions v 0, v 1,..., v k such that (2.1) u(x) k ( 2 ) km v i (x) for all x F. 3 i=0 Lemma 2.2 applied to continuous function u k i=0 v i and constant ( 2 3 )k M gives us a continuous function v k+1 for which (i) v k+1 (x) 1( )k M for all x F, (ii) v k+1 (x) < 1( )k M for all x X \ F, (iii) u(x) k+1 i=0 v i(x) ( 2 3 )k+1 M for all x F. Since the geometric series k=0 ( 2 3 )k converges, the Weierstrass theorem implies that the function series k=0 v k+1, and hence also k=0 v k, converges uniformly on X to a continuous function U. y (2.1), U = u in F. Moreover, by (ii), and the fact that v 0 = 0, we have, for each x X \ F, U(x) = v k = v k+1 v k+1 < k=0 k=0 Hence the theorem follows for continuous and bounded functions. k=0 k=0 1 2 ) km = M. 3( 3 Assume then that u is unbounded. Let h: R ( 1, 1) be a strictly increasing, continuous function that is onto and let f : F ( 1, 1), f = h u. Since f is bounded and continuous,the first part of the proof gives a continuous function g : X ( 1, 1) such that g F = f. Now the function U : X R, U = h 1 g, is the required extension of u. The second proof of the Tietze extension theorem shows also that (2.2) sup x F u(x) = sup U(x) and inf u(x) = inf U(x). x X x F x X Proof 2 of Theorem 2.1. We can assume that there are no isolated points in X \ F. Assume first that u is bounded and that u 0. We will show that the function U : X R, { u(x), if x F, U(x) = 1 2d(x) M d(x) d(x) x (r) dr if x X \ F, where d(x) = d(x, F ) and for r > 0, M x (r) = sup u(y), y F (x,r) is the desired extension of u. Clearly 0 U(x) sup F u for all x X. Note that for each x X, M x is integrable on bounded intervals of [0, ) as a bounded and increasing function, and that d(x) > 0 for each x X \ F because F is closed. 4

5 To show the continuity of U, let x X. If x is an interior point of F, then the continuity follows from the continuity of u. If x F, then by the definition of U, for each y X \ F, we have inf u U(y) sup u. F (x,3d(x,y)) F (x,3d(x,y)) This together with the continuity of u in F shows that U is continuous at x. Let then x X \ F and let y X be such that d(x, y) < 1 d(x, F ). 4 y the choice of y and the definition of the function M, we have that d(x) d(y) d(x, y) and M x (r) M y (r d(x, y)) for all r > d(x) and d(y) > 3d(x, y). Hence, with h = d(x, y), U(y) U(x) = 1 d(y) 1 d(y) = 1 d(y) = 2d(y) d(y) 2d(y) d(y) 2d(y) 3h d(y) 1 d(y) + h h d(y)(d(y) + h) 4h sup F u, d(y) M y (r) dr 1 d(x) 2d(x) 1 M y (r) dr d(y) + h M y (r) dr + 1 d(y) 2d(y) 3h d(y) 2d(y) 3h d(y) M y (s) ds M x (r) dr d(x) 2d(y) 2h d(y)+h 2d(y) 2d(y) 3h M y (r h) dr M y (r) dr M y (r) dr + 1 2d(y) M y (r) dr d(y) 2d(y) 3h which implies that U is continuous at x. To show that (2.2) holds, let m = inf F u and M = sup F u. Now the function v : F R, v(x) = u(x) m, satisfies 0 v(x) M m for all x F. Hence, by the first part of the proof, there is an extension V of v for which 0 V (x) M m. The function U : X R, U(x) = V (x) + m, is the desired extension of u. Remark 1. Note that the Tietze theorem does not hold for open sets: consider X = R \ {0} with the euclidean metric and the function u: X [0, 1], u(x) = 0, if x < 0 and u(x) = 1, if x > Extension of Lipschitz functions. Lipschitz functions are smooth functions of metric spaces. Extension problems for Lipschitz mappings between metric spaces are under active study. The main question is, for which metric spaces X and Y, each Lipschitz mapping u: A Y, A X, can be extended to a Lipschitz mapping U : X Y. These problems require tools from several areas of mathematics 5

6 including geometric measure theory, topology and geometry, see for example books [7], [9], [10] and the references therein. We will recall here the classical results of McShane and Kirszbraun. The former says that every Lipschitz function u: A R defined on a subset of a metric space X can be extended to a Lipschitz function in the whole space. Thus, real-valued Lipschitz functions can be assumed to be defined in X. Definition 2.3. Let (X, d X ) and (Y, d Y ) be metric spaces. A mapping u: X Y is Lipschitz continuous (or L-Lipschitz), if there is a constant L 0 such that (2.3) d Y (u(a), u(b)) Ld X (a, b) for all a, b X. The infimum of constants L satisfying (2.3) is called the Lipschitz constant of u and denoted by LIP(u). Example 2.4. There are many Lipschitz mappings in metric spaces. (1) If x 0 X, then the distance function d x0 : X R, d x0 (x) = d(x, x 0 ), is an 1-Lipschitz function. (2) If X,Y and Z are metric spaces and u: X Y and v : Y Z are Lipschitz mappings, then v u: X Z is Lipschitz and LIP(v u) LIP(u) LIP(v). Theorem 2.5 (McShane). Let A X and let u: A R be an L-Lipschitz function. Then there exists an L-Lipschitz function U : X R such that U A = u. Proof. Define functions U, Ũ : X R, (2.4) U(x) = inf {u(a) + Ld(a, x)} and Ũ(x) = sup{u(a) Ld(a, x)}. a A Using the fact that the distance function is 1-Lipschitz and the definitions of U and Ũ, it is easy to see that U and Ũ are L-Lipschitz and that U A = Ũ A = u. The function U in (2.4) is largest possible extension of u in the sense that if V : X R is an L-Lipschitz function and V A = u, then V U. Similarly, Ũ is the smallest one ============================= Remark 2. y applying Theorem 2.5 to the coordinate functions of an L-Lipschitz function u: A R n, A X, we obtain an L n-lipschitz extension U : X R n. If X = R m, then there is actually an L-Lipschitz extension. This is the Kirszbraun theorem from [51], for a proof, see also [38]. y Valentine [78], an extension with the same Lipschitz constant, exists also if X and Y are Hilbert spaces, A X and u: A Y is Lipschitz. A generalization to metric spaces with curvature bounds was given by Lang and Schroeder in [60]. 6 a A

7 2.3. Whitney covering, partition of unity and extension. In [80], Whitney constructed an extension for C k -functions defined on a closed set of R n (differentiability defined using Taylor polynomials and uniformity condition for the remainder). The main tools used in the proof are a decomposition of an open set Ω R n into disjoint cubes whose diameter is comparable to the distance of the cube to R n \ Ω and a partition of unity. Starting from the work of Jones [42], Whitney type covering has been an essential tool when showing that domains satisfying given conditions have an extension property for example for Sobolev functions. We will return to Whitney type coverings in the metric space setting later in the course. elow, by a cube, we mean a closed cube in R n whose sides are paraller to the coordinate axes. If the center of the cube is x i and the side length r i, then we write Q = Q i = Q(x i, r i ). Then diam Q i = nr i. Two cubes Q i and Q j are said to be disjoint if their interiors are disjoint, that is, int Q i int Q j =. The collection of cubes given by Lemma 2.6 is called Whitney decomposition, Whitney covering or Whitney cubes. Lemma 2.6. [Whitney covering] Let Ω = R n \ F, where F R n is a nonempty closed set. Let 0 < ε < 1/4. There is a collection W = {Q 1, Q 2,... } of disjoint cubes such that (1) Ω = i Q i, (2) diam Q i d(q i, F ) 4 diam Q i for each i, (3) if Q i Q j (boundaries have a common point), then 1 4 r i r j 4r i, (4) for each Q i, there are at most 12 n cubes Q j W such that Q i Q j. (5) i χ Q i (x) 12n for each x Ω, where Q i = Q(x i, (1 + ε)r i ). Proof. We will give an idea of the proof only, for details, see for example [75, Chapter VI], [61, Thm C.26]. We begin by dividing R n into disjoint closed cubes having side length 2 k, k Z. The first generation G 0 consists of cubes whose vertices belong to Z n and whose side length is 1. From G 0, we obtain a family {G k } k Z of collections of cubes: the side lenght of each cube Q G k is 2 k and from each such Q we obtain 2 n cubes of side length 2 k 1 by bisecting the sides of Q. The collection G k+1 consists of those smaller cubes. In addition to cubes of different sizes, we construct layers around the set F. For each k Z, let Ω k = {x Ω : n2 k+1 < d(x, F ) } n2 k+2. Then Ω = k Z Ω k. For the first selection of cubes, we take from each G k cubes that intersect the layer Ω k and define W 0 = { } Q G k : Q Ω k. k Z The collection W 0 satisfies properties (1) and (2) but it contains too many cubes. To obtain a covering that satisfies also (3)-(5), we have to throw away unnecessary cubes. Note first that if Q 1, Q 2 W 0 are two cubes whose interiors intersect, 7

8 then one of them is contained in the other: Assume that Q 1 G k, Q 2 G l, and int Q 1 int Q 2. If k = l, then Q 1 = Q 2. If k > l, then Q 1 Q 2. Let Q W 0 and let Q W 0 be a maximal cube that contains Q. Such a cube Q exists by property (2) and the inclusion property discussed above. Now W, the collection of maximal cubes in W 0, consist of disjoint cubes and satisfies (3)-(5). For (5), note that for two cubes Q 1, Q 2 W, the interiors of Q 2 and Q 1 intersect only if Q 1 Q 2. Hence (5) follows from (4). Partition of unity connected to the Whitney covering. A partition of unity connected to a covering of a set consists of compactly supported, nonnegative smooth (in the metric setting Lipschitz) functions, whose sum in each point is 1. It is used to transfer local information to the global level for example in extension and interpolation problems. To construct a partition of unity connected to the Whitney covering W = {Q i } i N, where Q i = Q i (x i, r i ), of an open set Ω R n, let Q 0 = Q(0, 1) be the unit cube centered at 0 and having side length 1. Let 0 < ε < 1 and let ϕ: R n [0, 1] be a smooth function such that ϕ Q0 = 1 and supp ϕ Q(0, 1 + ε). y defining ϕ i : R n [0, 1], ( x ϕ xi ) i (x) = ϕ, r i we obtain a function having similar properties in Q i as the function ϕ has in Q 0. Using these adjusted functions, we define functions ϕ i : R n [0, 1], for which ϕ i (x) = ϕ i (x) i ϕ i (x), ϕ i (x) = 1 for all x Ω. i The family {ϕ i } i N is called a partition of unity connected (subordinated) to the Whitney covering W. Whitney extension. Let F R n be a closed set. elow, we will construct a simple extension operator using Whitney covering and partition of unity. This operator extends functions with zero or one order of differentiability to functions which are smooth in the complement of F. It is the mother of all extension operators used for Sobolev-type functions (that can be defined via pointwise definition). Let u: F R be a function. Let W be a Whitney covering of Ω = R n \ F given by Lemma 2.6 and let {ϕ i } i N be the corresponding partition of unity. For each Q i W, let y i F such that d(q i, y i )=d(q i, F ). Such a point exists for each i because F is closed. Define a function Eu: R n R, { u(x), if x F, (2.5) Eu(x) = i u(y i)ϕ i (x) if x Ω. Note that, by the bounded overlap of dilated cubes Q i, Lemma 2.6 (5), and the fact that the support of ϕ i is in Q i, the sum in (2.5) is finite for each x Ω and the number of nonzero terms does not depend on x. 8

9 Theorem 2.7. Let F R n be a closed set. Let u: F R be a continuous function. Then Eu: R n R, defined by (2.5), is continuous in R n and C in Ω. For the proof of Theorem 2.7, see [75, Section VI.2.2]. Moreover, this simple extension operator E, u Eu, is linear and it maps Lip(α, F ), α-hölder continuous functions on F continuously to Lip(α, R n ). The norm of the extension operator is independent of F About Sobolev extension domains in R n ( ). Recall that, for a domain (open and connected set) Ω R n the Sobolev space W 1,p (Ω), 1 p <, consists of the functions u L p (Ω) whose all first order weak derivatives D j u belong to L p (Ω). We then write n u W 1,p (Ω) = u L p (Ω) + D j u L p (Ω). We say that Ω is a W 1,p -extension domain if there exists a bounded linear extension operator E : W 1,p (Ω) W 1,p (R n ). y the boundedness, there is a constant c E > 0 such that Eu W 1,p (R n ) c E Eu W 1,p (Ω) for all u W 1,p (Ω). (Extension domains for spaces W k,p and for other function spaces are defined similarly.) elow are some important extension results for Sobolev spaces. For exact definitions of the geometric conditions and proofs, see the corresponding articles or books. (1) (Calderón 1960 [12], Stein 1970 [75]) Every Lipschitz domain ( locally a graph of a Lipschitz function ) is a W k,p -extension domain. In [12], 1 < p <, the extension operator E depends on k, and Eu R n \Ω = 0 whenever supp u Ω is compact. In [75], 1 p and the same operator E works for all p and k. (2) (Jones 1982 [42]) Each (ε, δ)-domain ( locally connected in a quantitative sense ) is a W k,p -extension domain for all 1 p. The extension operator E depends on k. If Ω R 2 is a finitely connected W 1,2 -extension domain, then Ω is an (ε, δ)-domain. Hence extension for p = 2 implies extension for all p. (3) (Koskela [53]) If Ω R n is a W 1,p -extension domain and p > n 1, then there exists constants c, δ > 0 such that j=1 (2.6) Ω (x, r) c (x, r) = cr n for all x Ω, 0 < r < δ. Measure density condition (2.6) together with the Lebesgue differentiation theorem implies that no point of Ω is a density point and hence Ω = 0. Recent results in [31], [32] (see also the papers of Shvartsman) show the regularity of the boundary is not necessary condition for the extension property but a kind measure density condition is. Our goal is, at the end of the course, to study a connection of a metric 9

10 version of (2.6) and the extension property for Sobolev spaces defined in metric measure spaces ============================= 3. Doubling measures, covering theorems and Hardy-Littlewood maximal function To do the first order calculus in the metric setting, we need a measure and some kind of substitute for derivatives. We begin with the measure. We assume throughout the course that X = (X, d, µ) is a metric measure space equipped with a metric d and a orel regular outer measure µ. Recall that a set function µ: P(X) [0, ] is called an outer measure if µ( ) = 0 and µ is countably subadditive, that is, (3.1) µ(a) µ(a i ) whenever A µ(a i ). i=1 We call such a µ a measure. The orel regularity of the measure µ means that all orel sets are µ-measurable and that for every set A X there is a orel set D X such that A D and µ(a) = µ(d). We denote open balls in X with a center x X and a radius 0 < r < by The corresponding closed ball is i=1 (x, r) = {y X : d(y, x) < r}. (x, r) = {y X : d(y, x) r}. Note that a ball in a metric space as a set does not necessarily have a unique center and radius. y writing (x, r), we fix the center and the radius. Note also that (x, r) may be a larger set than (x, r), the topological closure of the open ball (x, r). If = (x, r) is a ball and 0 < t <, then t is the ball with the same center as and radius tr. It is possible to develop the basic theory of Sobolev spaces in metric measure spaces without any other special assumption on the measure. However, to get a richer theory, some further assumptions on the measure on the geometry of the space are needed. The standard assumptions are the doubling property of the measure and the validity of a (weak) Poincaré inequality that implies that the space is quasiconvex. Definition 3.1. Measure µ is doubling, if there is a constant c d > 0 such that (3.2) µ((x, 2r)) c d µ((x, r)) for each x X and all r > 0. We call c d a doubling constant of µ. We assume that µ is doubling and that it is nondegenerate in the sense that there exist a ball X such that 0 < µ() <. The doubling condition guarantees that the measure of every open set is then positive and the measure of each bounded set is finite. Such a metric measure space X can be written as a union 10

11 of countable family of open sets with finite measure, for example, if x 0 X, then X = i N (x 0, i). y iterating doubling condition (3.2), we see that (3.3) µ((x, tr)) c d t s µ((x, r)), where s = log 2 c d, for all balls (x, r) and all t asic properties of doubling measures. The pioneering work in analysis in metric spaces with a doubling measure is a book [14] by Coifman and Weiss. They call the (quasi)metric spaces (where a constant is allowed in the triangle inequality) with a doubling measure spaces of homogenous type. Lemma 3.2. Let (x, R) be a ball. If y (x, R), 0 < r R, and δ log 2 c d, then µ((y, r)) ( r ) δ. (3.4) µ((x, R)) 4 δ R Proof. Let k N be such that 2 k r < R 2 k+1 r. Then (x, R) 2 k+2 (y, r) and (y, r) (x, 2R) and the doubling property of µ gives µ((x, R)) c k+2 d µ((y, r)) c k+3 d µ((x, R)). Since δ log 2 c d, we obtain µ((y, r)) c (k+2) d µ((x, R)) = (2 log 2 c d ) (k+2) µ((x, R)) 4 δ (2 k ) δ µ((x, R)) 4 δ ( r R )δ µ((x, R)). Hence the doubling condition gives an upper bound for the dimension of the metric space. The number s = log 2 c d is sometimes called the doubling dimension of µ. Next we recall some definitions for metric spaces. A metric space is proper if all closed and bounded sets are compact. A set F X is totally bounded, if for any ε > 0, there exists a finite set {a 1, a 2,..., a n } A such that A n i=1(a i, ε). (This is equivalent to the existence of a finite ε-net.) In a doubling metric space, bounded sets are totally bounded (exercise). This together with standard results from metric spaces implies the following important result. Lemma 3.3. A doubling metric measure space is proper if and only if it is complete. Proof. Assume first that X is proper. (The doubling property is not needed to prove this direction.) Let (x i ) i N be a Cauchy sequence in X. As a Cauchy sequence, (x i ) i N is bounded, and hence there is R > 0 such that x i (x 1, R) for all i N. Since X is proper, (x 1, R) is compact. Hence the sequence (x i ) i N has a limit in X. Assume then that X is complete. Let F X be closed and bounded. y the doubling property of µ, F is totally bounded. Since a closed subset of a complete metric space is compact if and only if it is totally bounded, we see that F is compact. Hence X is proper. 11

12 Example 3.4 (Metric spaces with a doubling measure, all examples but (2) below are from [5]). (1) (X, d, µ) = (R n,, L n ), Euclidean space with the Lebesgue measure is a doubling metric measure space with c d = 2 n. (2) Let (X, d, µ) = (Ω,, L n Ω ), where Ω satisfies measure density condition L n (Ω (x, r)) cl n ((x, r)) = cr n for all x Ω, r > 0. Then, for each ball = (x, r) in X, we have µ(2) = L n (Ω 2) L n (2) = 2 n L n () 2n c Ln (Ω ) = 2n c µ(), and hence µ is doubling with c d = 2 n /c. (3) If X = (M, g) is a complete Riemannian manifold of dimension n and with nonnegative Ricci curvature, and µ is the canonical measure associated to the metric tensor g, then µ is doubling with c d = 2 n. (4) Let X = [ 1, 0] [ 1, 1] [0, 1] {0} be equipped with the Euclidean metric and with the measure µ = L 2 X + H 1 [0,1] {0}, where H 1 is the 1-dimensional Hausdorff measure. Then µ is doubling with c d = 4 and doubling dimension 2. (5) Cantor sets constructed as follows are metric measure spaces: Let F be a finite set having k points, k 2, and F = {x = (x i ) i N : x i F }. Let a (0, 1). Then d a : F F [0, ), { 0, if x = y, d a (x, y) = a j, if x i = y i for i < j and x j y j, is a metric in F. Let then ν be a uniformly distributed probability measure on F. Define measure µ on F as the product measure of ν infinitely many times. Then one can show that µ((x, a j )) = k j and that (F, d a, µ) is doubling metric measure space with dimension s given by a s = k 1. If k = 2 and a = 1/3, then F is bi-lipschitz equivalent to the standard 1/3-Cantor set ============================= Sometimes we would like to estimate the measure of (x, tr), 0 < t < 1, by cµ((x, r)), 0 < c < 1, and have an upper bound for the ratio of measures of the balls corresponding to (3.4). This is possible if X is complete, or more generally, if all annuli in X are nonempty. (This condition is sometimes called a RD-condition, RD coming from reverse doubling, see for example papers of Dachun Yang and Yuan Zhou.) 12

13 Lemma 3.5. Let X be connected. If 0 < t < 1, then there is a constant c 1 = c 1 (t, c d ) such that 0 < c 1 < 1 and (3.5) µ((x, tr)) c 1 µ((x, r)) for all balls (x, r) with 0 < r < 1/2 diam X. Moreover, there exist constants c 2, α > 0 such that µ((y, r)) r ) α (3.6) 2( µ((x, R)) c R whenever (x, R) is a ball, y (x, R) and 0 < r R. Proof. We begin with (3.5). Let 0 < t < 1 and let (x, r) be a ball. Let y be a point such that d(y, x) = 1 (1 + t)r < r 2 and let y = (y, 1 (1 t)r). Such a point exists by the connectedness of X. Now 2 y (x, r) \ (x, tr) and (x, r) a y, where a = 4/(1 t), and hence, using (3.3), we have Thus µ((x, r)) µ(a y ) c d a log 2 c d µ( y ). µ((x, tr)) µ((x, r)) µ( y ) (1 (c d a log 2 c d ) 1 )µ((x, r)), from which (3.5) follows because c d a log 2 c d > 1. To prove (3.6), let (x, R) be a ball, y (x, R) and 0 < r R. Then (y, R) (x, 2R) and the doubling condition together with an application of (3.5) gives that µ((y, r)) µ((x, R)) c µ((y, r)) r ) α. d 2( µ((y, R)) c R Hence, by Lemmas 3.2 and 3.5, in a connected space X with diam X =, there are constants c > 0 and 0 < α s <, such that 1 ( r ) s µ((y, r)) ( r ) α c R µ((x, R)) c R for all balls (x, R) and (y, r) with y (x, R) and 0 < r R. Note also that by Lemma 3.5, µ({x}) = 0 for all x X. Exercise 1. A metric space is doubling if there is a constant c D such that for each x X and all r > 0, the ball (x, r) can be covered by at most c D balls of radius r/2. Show that if X is a metric space with a doubling measure µ, then X is doubling as a metric space. (In the other direction, if X is a complete, doubling metric space, then is a doubling measure in X, see [62].) 13

14 3.2. Covering theorems. In many situations, we would like to have an opposite inequality to (3.1), a kind of quasiadditivity, ( µ(a i ) cµ A i ), i=1 where sets A i usually come from a covering of a set A X. This is obtained by standard covering theorems. Using covering theorems, we can extract from a given family of balls, a better (disjoint) subfamily such that enlarged balls of the subfamily still cover our set and, moreover, have a bounded overlap. The most general covering theorem, the so-called 5- or 5R-covering theorem, deals with covers of metric spaces by balls having finite radius and is usually proved using the Zorn lemma. The second one is a version of the first with balls of equal radii. Lemma 3.6. Let be a family of balls with sup{diam : } <. Then there is a countable, disjoint subfamily = { i } i N of such that 5 i. i N The countability of the covering follows from the doubling condition since balls have finite and nonzero measure. The claim is true in any separable metric space. Moreover, it holds in any metric space except for the countability of. See for example [6, Thm 2.2.3] or [37, Thm 1.2], for a proof. Lemma 3.7. Let be a family of balls of radius r > 0. Then there is a countable, pairwise disjoint subfamily = { i } i N of and a constant N = N(c d ) > 0 such that 5 i and i N i=1 i χ 5i (x) N for each x X. Proof. According to Lemma 3.6, it is enough to show the bounded overlap property of the balls 5 i. Let x X, and let I x = {i : i, x 5 i }. If i I x, then (x, r) 6 i. This together with the doubling property of µ implies that µ( i ) c 3 d µ(6 i) c 3 d µ((x, r)) c 6 d µ((x, 6r)). On the other hand, i (x, 6r) for each i I x. As the balls i are disjoint, we have µ((x, 6r)) µ( i ) c 6 d µ((x, 6r)). i I x i I x ecause the measure of each ball is positive and finite, there is only a finite number of terms in the above sum. Hence µ((x, 6r)) c 6 d #I xµ((x, 6r)), and we can take N = c 6 d. 14

15 The proof of Lemma 3.7 shows that if we enlarge the balls in the covering by a fixed constant, then the enlarged balls have overlap bounded with a constant depending only on the doubling constant of µ and the fixed constant Hardy-Littlewood maximal function and the Lebesgue differentiation theorem. Maximal functions are important tools for example in geometric analysis, harmonic analysis, in the theory of singular integrals, and in the PDE theory. Usually, they are used as tools but the behaviour of maximal operators is an interesting question as its own. One can study mapping properties, such as boundedness and continuity, of maximal operators between different functions spaces. y the classical Lebesgue differentiation theorem, almost every point is a Lebesgue point for a locally integrable function, lim r 0 (x,r) u(y) dy = u(x) for L n -almost x R n. One way to prove this result is to use weak type inequality for the Hardy-Littlewood maximal function and density of continuous functions in L 1, see [75, Chapter 1] in the case of the Lebesgue measure in R n. This proof generalizes to the metric setting. y saying that a measurable function u: X [, ] is locally integrable, we mean that is integrable on balls. Similarly, the class of functions that belong to L p (), p > 0, in all balls, is denoted by L p loc (X). The integral average of a locally integrable function u over a ball is u = u dµ = 1 µ() u dµ. Definition 3.8. The Hardy Littlewood maximal function of a locally integrable function u is (3.7) Mu(x) = sup 0<r< u dµ, and its restricted version for R > 0, (3.8) M R u(x) = sup 0<r<R (x,r) (x,r) u dµ. It is easy to see that the corresponding Hardy Littlewood maximal operators M: u Mu and M R : u M R u are sublinear, M(u + v) Mu + Mv and M(λu) λmu for all locally integrable u and v and all λ 0. One of the mostly used properties of the maximal operator is the boundedness result in L p -spaces, originally proved by Hardy and Littlewood [35] and Wiener [81]. The maximal operator M maps L 1 (X) to weak-l 1 (X) and is bounded in L p (X) for p > 1. Theorem 3.9 (Maximal function theorem). 15

16 (1) There is constant C 1 = C 1 (c d ) > 0 such that (3.9) µ ( {x X : Mu(x) > t} ) C 1 t X u dµ for all t > 0 and for all u L 1 (X). (2) If p > 1, then there is a constant C p = C p (p, c d ) > 0 such that (3.10) Mu L p (X) C p u L p (X) for all u L p (X) ============================= Proof. ecause the 5-covering lemma (Lemma 3.7) requires the balls to have uniformly bounded diameter, we prove the claim first for the restricted maximal function (3.8) with constants independent of R and then pass to the limit R. For that, let R > 0. Claim (1): Let u L 1 (X), t > 0 and let E t = {x X : M R u(x) > t}. For each x E t, let (x, r x ) be a ball such that 0 < r x < R and (3.11) u dµ > tµ((x, r x )). (x,r x) y Lemma 3.7, there are disjoint balls i = (x i, r i ), i = 1, 2,..., satisfying (3.11) such that E t i 5 i. Hence, using the doubling property of µ and (3.11), we obtain µ(e t ) µ(5 i ) C µ( i ) C u dµ C u dµ, t i i i t X from which the first claim follows by letting R. Claim (2): Let u L p (X). y the definition of Mu, if p =, then C = 1. Assume that 1 < p <. Let t > 0. We divide u into a small and large part, u = uχ { u t/2} + uχ { u >t/2} = g + h. Then, by the sublinearity of the maximal operator M R, for each x X, and hence M R u(x) M R g(x) + M R h(x) t 2 + M Rh(x) {M R u > t} {M R h > t/2}. Using the assumption u L p (X) with p > 1 and the definition of h, we see that ( t 1 p h dµ = u p u 1 p dµ u 2) p dµ, X and hence h L 1 (X). u >t/2 16 i X

17 Using integration in terms of the distribution function (3.12) three times, the first part of the proof for h, and change of variables, we obtain M R u p dµ = p t p 1 µ ( {M R u > t} ) dt p t p 1 µ ( {M R h > t/2} ) dt X 0 0 C t p 1 t 1 u dµ dt C C C 0 0 X X { u >t/2} t p 2 ( t 2 µ( { u > t/2} ) + u p dµ + C u p dµ. 2s The second claim follows by letting R. 0 0 t/2 µ ( { u > s} ) ) ds dt µ ( { u > s} ) t p 2 dt ds y keeping track on the constants in the Maximal function theorem, we see that in (2), we can take C p = C2p, where C depends only on c p 1 d. Note that C p as p 1. In fact, the maximal function of an L 1 -function is hardly ever integrable. It is easy to show that in the Euclidean case with Lebesgue measure, if u L 1 (R n ) and Mu L 1 (R n ), then u = 0. See also [74]. Remark 3. As an application of the Fubini theorem, we obtain the following useful formula, sometimes called Cavalieri principle, for L p -integrals in terms of the distribution function. If µ is a orel measure, u 0 a measurable function and 0 < p <, then (3.12) u p dµ = p t p 1 µ ( {x X : u(x) > t} ) dt. X 0 To see that, we write E t = {u > t} for t 0, and define f : [0, ) X R, { 1, if u(x) > t, f(t, x) = χ Et (x) = 0, if u(x) t. Then p 0 t p 1 µ(e t ) dt = p = = 0 t p 1 X 0 u(x) X 0 X χ Et (x) dµ(x) dt pt p 1 f(t, x) dt dµ(x) pt p 1 dt dµ(x) = X u p dµ. Remark 4. Theorem 3.9 implies that for a function u L p (X), 1 p, Mu is finite almost everywhere. Note also that Maximal function sees things that happen far away... - a local version of the L p -boundedness Mu L p (Ω) C p u L p (Ω) 17

18 does not necessarily hold for all u L p (Ω). It holds if the restriction of µ to Ω, µ Ω, is doubling or if supp u Ω Noncentered maximal function. The Hardy Littlewood maximal function has a variant, where the supremum of integral averages in (3.7) is taken over all balls containing the point x. In some situations, this version is smoother and more useful than Mu. A noncentered maximal function of a locally integrable function u is (3.13) M u(x) = sup x u dµ. It is easy to see that for each locally integrable function u and all x X; (3.14) c 2 d M u(x) Mu(x) M u(x); the first inequality follows from the doubling property of µ and the second from the definitions. Definition (3.13) implies that sets {x X : M u(x) > t} are open for each t > 0 and hence M u is lower semicontinuous. A similar proof as for the maximal function Mu in Theorem 3.9 (1) shows that µ ( {M u > t} ) C u dµ t {M u>t} for all t > 0 and for all u L 1 (X). This together with (3.14) and (3.9) implies that for all u L 1 (X). lim tµ( {M u > t} ) = lim tµ ( {Mu > t} ) = 0 t t 3.5. Lebesgue differentiation theorem. Definition A point x X is a Lebesgue point of a locally integrable function u: X R, if (3.15) lim u dµ = u(x). r 0 (x,r) Theorem 3.11 (Lebesgue differentiation theorem). Let u: X R be a locally integrable function. Then µ-almost every point x X is a Lebesgue point of u. Moreover, (3.16) lim u(y) u(x) dµ(y) = 0 r 0 (x,r) for µ-almost every point x X. Remark 5. In the literature, the existence of limits (3.15) and (3.16) almost everywhere are both used as a definition of a Lebesgue point. Clearly, (3.16) implies (3.15). 18

19 About the proof of Theorem The argument in [75, Chapter 1] for the first claim generalizes to the metric setting. It uses weak type inequality (3.9) for the maximal function and the density of continuous functions in L 1 (X) (which is true in every metric space with a orel measure), see [6, Thm 5.2.6] for the proof in the metric setting. Another way to prove the theorem is to use Vitali covering theorem, see [37, Thm 1.8], [41, ]. The second claim follows by using the first claim for locally integrable functions u i, u i (y) = u(y) q i, with Q = (q i ) i N. 3.2 ============================= 4. Mapping properties of the maximal function In this section, we discuss the mapping properties of the maximal function. We are interested in the following question: Does the maximal operator preserve the regularity properties of functions? In definition (3.7), there are two competing things: the integral average is smoothing whereas supremum usually reduces smoothness. We begin with the boundedness results in the classical Euclidean setting Regularity of the maximal function in the Euclidean setting. In the Euclidean space, many boundedness properties of the maximal operator follow from the fact that it commutes with translations: if u: R n R is locally integrable, then (4.1) (Mu) h (x) = Mu h (x) for all x, h R n. Here, for a function v : R n R, v h (x) = v(x + h) for all x, h R n. In the classical case, not only the noncentered maximal function M u of each locally integrable u but also Mu is lower semicontinuous (Exercise). Moreover, the maximal function of a continuous function u: R n R is continuous. If u: R n R is L-Lipschitz, then (4.1) together with the sublinearity of M implies that for all x, h R n, Mu(x + h) Mu(x) = (Mu) h (x) Mu(x) = Mu h (x) Mu(x) M(u h u)(x) = sup u(y + h) u(y) dy L h, r>0 (x,r) and hence M is also L-Lipschitz (if not identically infinite). Moreover, by the Rademacher theorem, Mu is differentiable almost everywhere. The same continuity argument holds also for Hölder continuous functions. How about the differentiability properties? Since u is not differentiable and the supremum of differentiable functions is not differentiable in general, the maximal function of a differentiable function is not differentiable in general. The study of the properties of the maximal function of a Sobolev function started by Kinnunen 19

20 in [46], see also [34] and [48] for the local case. A simple argument using the characterization of Sobolev spaces W 1,p (R n ), 1 < p <, by integrated difference quotients shows that the maximal operator is bounded in Sobolev spaces. Moreover, there is a pointwise estimate for the weak derivatives of the maximal function in terms of the maximal functions of the weak derivatives of the function itself. Recall that u L p (R n ) belongs to W 1,p (R n ) if and only if lim sup h 0 u h u L p (R n ) h for a proof, see for example [23, Chapter 7.11] <, Theorem 4.1. Let u W 1,p (R n ), 1 < p <. Then Mu W 1,p (R n ) and for almost every x R n. D i Mu(x) MD i u(x) for all i = 1, 2,..., n Proof. We will prove the boundedness of M in W 1,p (R n ). y the fact that M commutes with translations, the sublinearity of M and the boundedness of M in L p (R n ) for p > 1, we obtain (Mu) h Mu L p (R n ) = Mu h Mu L p (R n ) C M(u h u) L p (R n ) C u h u L p (R n ). The claim follows from the characterization of W 1,p (R n ) mentioned above. Theorem can be used to prove that for u W 1,p (R n ), 1 < p <, Mu is p-quasicontinuous (in the Sobolev-capacity sense) and to study the pointwise properties u, see [46]. Remark 6. Since the Hardy-Littlewood maximal operator is not bounded in L 1 (R n ), we cannot expect it to be bounded in W 1,1 (R n ) (consider a smooth bump function u that belongs to W 1,1 (R n ) but Mu is not integrable). Hence the more interesting question is: Is the operator u DMu from W 1,1 (R n ) to L 1 (R n ) bounded? In R, the answer is yes, proven for the noncentered maximal operator by Tanaka in [76] and for M in the recent preprint [58] by Kurka. The case n > 1 is open and the techniques used in the one-dimensional case does not generalize to higher dimensions. For partial results in R n, also for the local case, see [34] and [33] Hardy-Littlewood maximal function of Lipschitz funtion in metric spaces. The proofs using the fact that M commutes with translations cannot be generalized to the metric setting. It turns out that the doubling property of the measure does not guarantee that the maximal function of a Lipschitz function is Lipschitz. The next example of uckley in [11] shows that it can be discontinuous. Example 4.2. Let X be the subset of the complex plane consisting of the real line and the points x on the unit circle whose argument θ lies in the interval [0, π ]. Equip 2 X with the Euclidean metric and the 1-dimensional Hausdorff measure. Let u: X [0, 1] be a Lipschitz function such that u(x) = 0, if x R or Arg(x) π/5, and u(x) = 1, if Arg(x) π/4. 20

21 We will show that Mu has a jump discontinuity at the origin. Since Mu(0) = lim u dµ = 1 r 1+ (0,r) 2 + π u dµ, 2 (0,1) we have that Mu(0) π 2 π π 2 = 3π π. If x < 0, then (x, r(x)), where r(x) = d(x, e iπ/4 ), includes points on the arc if and only if their argument exceeds π. It follows that 4 π 4 lim Mu(x) lim u dµ = x 0 x 0 (x,r(x)) 2 + π = π 8 + π > Mu(0). 4 Annular decay and maximal operator. If the measures of annuli in X behave nicely, then the maximal operator M maps Lipschitz functions to Hölder continuous functions. The space of bounded β-hölder continuous functions is equipped with the norm u(x) u(y) u C 0,β (X) = u L + sup, x y d(x, y) β and β = 1 gives bounded Lipschitz functions on X. Definition 4.3. Let 0 < δ 1. The metric measure space X satisfies the δ-annular decay property, if there exists a constant C > 0 such that for all x X, R > 0, and 0 < h < R, we have (4.2) µ ( (x, R) \ (x, R h) ) ( h ) δµ((x, C R)). R Geodesic spaces, and more generally, length spaces (where the distance between any pair of points equals the infimum of the lengths of the rectifiable paths joining them), satisfy the annular decay for some δ > 0, see [11]. The strong annular decay condition, (4.2) with δ = 1, holds only for few spaces, for example for R n and the Heisenberg group H n. Recall that X is geodesic, if every two points x, y X can be joined by a curve γ for which l(γ) = d(x, y). Theorem 4.4. Let 0 < δ 1 and 0 < β 1. If X satisfies the δ-annular decay property, then M: C 0,β (X) C 0,α (X), where α = min{β, δ}, is bounded. In particular, if X satisfies the strong annular decay property, then M maps Lipschitz functions boundedly to Lipschitz functions. Proof. Let u C 0,β (X) with u C 0,β (X) = 1. Since Mu L (X) = u L (X), we have to estimate only the second part of the norm of Mu C 0,β (X). Let x, y X. It suffices to show that there is a constant C > 0, independent of u, x, and y, such that (4.3) Mu(x) Mu(y) Cd(x, y) α. 21

22 We may also assume that d(x, y) 1. Let r > 0 be such that u dµ Mu(x) d(x, y) α. (x,r) Assume first that r d(x, y). (This is usually the easy case in this type of estimates). Then, by the β-hölder continuity of u, for all z, w (x, r) (y, r). Hence u dµ u dµ = (x,r) (y,r) = u(z) u(w) 3 β d(x, y) β, (x,r) (x,r) (x,r) u u (y,r) dµ u(z) u(w) dµ(w) dµ(z) (y,r) u(z) u(w) dµ(w) dµ(z) (y,r) 3 β d(x, y) β 3 β d(x, y) α. This together with the selection of r > 0 implies that Mu(y) u dµ u dµ 3 β d(x, y) α (y,r) (x,r) Mu(x) (3 β + 1)d(x, y) α, from which (4.3) follows. Assume then that r > d(x, y). A calculation using δ-annular decay (4.2) and the assumption u C 0,β (X), shows that u dµ u dµ Cd(x, y) α, (x,r) (y,r+d(x,y)) from which the claim follows. For details, see [11, Thm 1.1]. 4.2 ============================= 4.3. The discrete convolution and the discrete maximal operator. y Example 4.2, the standard Hardy Littlewood maximal operator does not preserve the smoothness of the functions in a doubling metric measure space without an additional assumption on the space. In this section, we will construct a maximal operator which is based on a discrete convolution. It turns out that this operator has better regularity properties in doubling metric measure spaces than the standard Hardy Littlewood maximal operator. The discrete convolution itself has been an important tool in harmonic analysis in doubling metric spaces since the works [14] and [63]. Nowadays, it is a standard tool in geometric analysis in metric spaces. Since Lipschitz functions are the smooth functions in metric spaces, partition of unity used in the discrete convolution consists of Lipschitz functions. 22

23 The idea behind the discrete maximal function is the fact that in the Euclidean setting, the Hardy Littlewood maximal function is a supremum of convolutions. Namely, for each locally integrable u and each ball (x, r) R n, (x,r) u(y) dy = 1 (x, r) (x,r) where is convolution and χ r = χ (0,r) (0,r). Hence Mu(x) = sup( u χ r )(x). r>0 u(y) dy = u χ r (x), The discrete maximal function was introduced and used to study pointwise properties of Sobolev functions in the metric setting by Kinnunen and Latvala in [47]. Its properties are also studied, for example, in [1], [2], [50] and [36]. We begin the construction of the discrete maximal function with a covering of the space and a partition of unity subordinate to this covering. Covering. Let r > 0. y the 5-covering lemma, Lemma 3.7, there are balls i = (x i, r), i = 1, 2,..., and a constant N = N(c d ), such that X = i and χ 6i N. i i Partition of unity. As in Section 2.3, connected to the covering { i } i, there is a partition of unity {ϕ i } i, consisting of functions ϕ i : X [0, 1], i = 1, 2... with the following properties: There are positive constants ν and L depending only on the doubling constant of µ, such that for each i, ϕ i = 0 in X \ 6 i, ϕ i ν in 3 i, ϕ i is Lipschitz with constant L/r, and ϕ i (x) = 1 for all x X. i To construct such functions, we can first define 1/(3r)-Lipschitz functions ϕ i : X [0, 1], i = 1, 2..., 1, if x 3 i, ϕ i (x) = 2 d(x,x i), if x 6 3 i \ 3 i, 0, if x X \ 6 i. Using these cut-off functions, we define functions ϕ i : X [0, 1], i = 1, 2..., for the partition of unity, ϕ i (x) = ϕ i (x) i ϕ i (x) for all x X. It is easy to see that functions ϕ i satisfy the desired properties (exercise). 23

24 Discrete convolution. The discrete convolution of a locally integrable function u at the scale r is u r : X R, (4.4) u r (x) = i ϕ i (x)u 3i for every x X. Note that by the properties of the covering and the partition of unity, the sum in (4.4) is finite at each point and upper bound for number of the nonzero terms depends only on the doubling constant of µ. elow, when we prove results for the discrete convolution u r, covering of X by balls i and the partition of unity {ϕ} i subordinated to the covering { i } i are as above. Lemma 4.5. Let u: X R be a a locally integrable function and let u r : X R be a discrete convolution of u as above. Then u r is continuous. Moreover, it is Lipschitz on each compact set K X. Proof. Exercise. Discrete maximal function. Let Q + = (r j ) j N, be an enumeration of positive rationals. For each j N, we take a covering of X by balls (x i, r j ), i = 1, 2,..., as above. The discrete maximal function of u in X is M u: X R, M u(x) = sup u rj (x) j for every x X. The construction of the discrete maximal function depends on the choice of the coverings, but estimates proved using it that are independent of the chosen coverings. Lemma 4.6. Let u: X R be a locally integrable function. Then the discrete maximal function M u is lower semicontinuous and the discrete maximal operator M is sublinear. Moreover, the discrete maximal function is comparable to the Hardy-Littlewood maximal function: there is a constant C = C(c d ) 1 such that (4.5) C 1 Mu(x) M u(x) CMu(x) for all x X. Proof. As a supremum of continuous functions, the discrete maximal function is lower semicontinuous and hence measurable. The sublinearity, M (u + v) M (u) + M (v), for all u, v L 1 loc (X), follows directly from the definitions of the discrete convolution and the discrete maximal function. To prove the comparability of Mu and M u, let x X. We begin with the second inequality of (4.5). Let r j Q + and let I x = {i : x (x i, 6r j )}. 24

25 Then (x i, 3r j ) (x, 9r j ) (x i, 15r j ) whenever i I x. This together with the doubling property of µ, the fact that ϕ i X\(xi,6r j ) = 0 for all i and the bounded overlap of the balls (x i, 6r j ) imply that u rj (x) = ϕ i (x) u (xi,3r j ) i ϕ i (x) µ((x i, 3r j )) u dµ µ((x, 9r j )) i I (x,9r j ) x CMu(x) i I x ϕ i (x) CMu(x). The second inequality of (4.5) follows by taking supremum over r j. To prove the first inequality of (4.5), note that by Lemma 3.2, by taking the supremum over radii in Q + instead of all over all r > 0 in the Hardy-Littlewood maximal function, we obtain a comparable maximal function. Thus it suffices to show that sup u dµ CM u(x). r j Q + (x,r j ) Let r j Q +. Since balls (x i, r j ), i = 1, 2..., form a covering of X, there is i N such that x (x i, r j ). Then (x, r j ) (x i, 2r j ) and, by the doubling property of µ and the fact that ϕ i (xi,3r j ) ν, we have u dµ C u dµ Cϕ(x) u dµ CM u(x). (x,r j ) (x i,3r j ) (x i,3r j ) The first inequality of (4.5) follows now by taking supremum over r j. Remark 7. Lemma 4.6 together with Theorem 3.9 shows that the maximal operator M maps L 1 (X) to weak-l 1 (X) and is bounded in L p (X) for p > 1. Using the Lebesgue differentiation theorem, it is easy to show that the discrete convolution approximates locally integrable functions pointwise almost everywhere, u r (x) u(x) as r 0 for almost all x X. When p > 1, this together with the facts that u r M u and M u L p (X) and the dominated convergence theorem, imply that in u r u in L p (X) as r 0, (exercise or see [2, Lemma 4.5]). Using the density of continuous functions in L p (X), p 1, we can show that the convergence is true also in L 1 (X). Recall the Jensen inequality for sums: if k N, a 1,..., a k R, n 1,..., n k 0, and Ψ: R R is a convex function, then ( ) n1 a 1 + n k a k Ψ n 1 + n k n 1Ψ(a 1 ) + n k Ψ(a k ) n 1 + n k. elow (and later), we also use the following easy application of the Hölder inequality to estimate integral averages: for p 1 ( ) 1/p u dµ u p dµ for all locally integrable functions u and all balls. 25