Interpolation of Sobolev spaces, Littlewood-Paley inequalities and Riesz transforms on graphs

Transcription

1 Interpoation of Soboev spaces, Littewood-Paey inequaities and Riesz transforms on graphs Nadine Badr Emmanue Russ Université aude Bernard Université Pau ézanne Apri 6, 009 Abstract. Let Γ be a graph endowed with a reversibe Markov kerne p, and P the associated operator, defined by P f(x) = y p(x, y)f(y). Denote by the discrete gradient. We give necessary and/or sufficient conditions on Γ in order to compare f p and (I P ) / f p uniformy in f for < p < +. These conditions are different for p < and p >. The proofs rey on recent techniques deveoped to hande operators beyond the cass of aderón-zygmund operators. For our purpose, we aso prove Littewood-Paey inequaities and interpoation resuts for Soboev spaces in this context, which are of independent interest. AMS numbers 000: Primary: 60J0. Secondary: 4B0, 4B5. Keywords: Graphs, discrete Lapacian, Riesz transforms, Littewood-Paey inequaities, Soboev spaces, interpoation. ontents Introduction and resuts. Presentation of the discrete framework Statement of the probem The L p -boundedness of the Riesz transform The case when p < The case when p > Riesz transforms and harmonic functions The reverse inequaity An overview of the method Université aude Bernard, Lyon I et NRS UMR 508, Bâtiment Doyen Jean Braconnier, 43 bouevard du novembre 98, 696 VILLEURBANNE edex, FRANE Emai: badr@math.univ-yon.fr Université Pau ézanne, Facuté des Sciences et Techniques de Saint-Jérôme, Avenue Escadrie Normandie-Niémen, 3397 MARSEILLE edex 0, FRANE, and LATP, NRS, UMR 663. E-mai: emmanue.russ@univ-cezanne.fr

2 Kerne bounds 4 3 Littewood-Paey inequaities 4 4 Riesz transforms for p > 0 5 The aderón-zygmund decomposition for functions in Soboev spaces 5 6 An interpoation resut for Soboev spaces 9 7 The proof of (RR p ) for p < 30 8 Riesz transforms and harmonic functions The discrete differentia and its adjoint The proof of Theorem Appendix 40 Appendix 4 Introduction and resuts It is we-known that, if n, f L p (R n ) and ( ) / f L are comparabe uniformy p (R n ) in f for a < p < +. This fact means that the cassica Soboev space W,p (R n ) defined by means of the gradient coincides with the Soboev space defined through the Lapace operator. This is interesting in particuar because is a oca operator, whie ( ) / is not. Generaizations of this resut to geometric contexts can be given. On a Riemannian manifod M, it was asked by Strichartz in [50] whether, if < p < +, there exists p > 0 such that, for a function f 0 (M), / f p df p p / f p, (.) p where stands for the Lapace-Betrami operator on M and d for the exterior differentia. Under suitabe assumptions on M, which can be formuated, for instance, in terms of the voume growth of bas in M, uniform L Poincaré inequaities on bas of M, estimates on the heat semigroup (i.e. the semigroup generated by ) or the Ricci curvature, each of the two inequaities contained in (.) hods for a range of p s (which is, in genera, different for the two inequaities). The second inequaity in (.) means that the Riesz transform d / is L p -bounded. We refer to ([3, 5,, 5]) and the references therein. In the present paper, we consider a graph equipped with a discrete gradient and a discrete Lapacian and investigate the corresponding counterpart of (.). To that purpose, we prove, among other things, an interpoation resut for Soboev spaces defined via the differentia, simiar to those aready considered in [45], as we as L p bounds for Littewood-Paey functionas.

3 . Presentation of the discrete framework Let us give precise definitions of our framework. The foowing presentation is party borrowed from [30]. Let Γ be an infinite set and µ xy = µ yx 0 a symmetric weight on Γ Γ. We ca (Γ, µ) a weighted graph. In the seque, we write most of the time Γ instead of (Γ, µ), somewhat abusivey. If x, y Γ, say that x y if and ony if µ xy > 0. Denote by E the set of edges in Γ, i.e. E = {(x, y) Γ Γ; x y}, and notice that, due to the symmetry of µ, (x, y) E if and ony if (y, x) E. For x, y Γ, a path joining x to y is a finite sequence of edges x 0 = x,..., x N = y such that, for a 0 i N, x i x i+. By definition, the ength of such a path is N. Assume that Γ is connected, which means that, for a x, y Γ, there exists a path joining x to y. For a x, y Γ, the distance between x and y, denoted by d(x, y), is the shortest ength of a path joining x and y. For a x Γ and a r 0, et B(x, r) = {y Γ, d(y, x) r}. In the seque, we aways assume that Γ is ocay uniformy finite, which means that there exists N N such that, for a x Γ, B(x, ) N(here and after, A denotes the cardina of any subset A of Γ). If B = B(x, r) is a ba, set αb = B(x, αr) for a α > 0, and write (B) = 4B and j (B) = j+ B \ j B for a integer j. For any subset A Γ, set A = {x A; y x, y / A}. For a x Γ, set m(x) = y x µ xy. We aways assume in the seque that m(x) > 0 for a x Γ. If A Γ, define m(a) = x A m(x). For a x Γ and r > 0, write V (x, r) instead of m(b(x, r)) and, if B is a ba, m(b) wi be denoted by V (B). For a p < +, say that a function f on Γ beongs to L p (Γ, m) (or L p (Γ)) if ( /p f p := f(x) m(x)) p < +. Say that f L (Γ, m) (or L (Γ)) if x Γ f := sup f(x) < +. x Γ Define p(x, y) = µ xy /m(x) for a x, y Γ. Observe that p(x, y) = 0 if d(x, y). Set aso and, for a k N and a x, y Γ, p 0 (x, y) = δ(x, y) p k+ (x, y) = z Γ p(x, z)p k (z, y). The p k s are caed the iterates of p. Notice that, for a x Γ, there are at most N non-zero terms in this sum. Observe aso that, for a x Γ, p(x, y) = (.) y Γ 3

4 and, for a x, y Γ, p(x, y)m(x) = p(y, x)m(y). (.3) For a function f on Γ and a x Γ, define P f(x) = p(x, y)f(y) y Γ (again, this sum has at most N non-zero terms). Since p(x, y) 0 for a x, y Γ and (.) hods, one has, for a p [, + ] and a f L p (Γ), P f L p (Γ) f L p (Γ). (.4) We make use of the operator P to define a Lapacian on Γ. onsider a function f L (Γ). By (.4), (I P )f L (Γ) and (I P )f, f L (Γ) = p(x, y)(f(x) f(y))f(x)m(x) x,y = p(x, y) f(x) f(y) m(x), x,y (.5) where we use (.) in the first equaity and (.3) in the second one. If we define now the operator ength of the gradient by f(x) = ( ) / p(x, y) f(y) f(x) y Γ for a function f on Γ and a x Γ (this definition is taken from [6]), (.5) shows that (I P )f, f L (Γ) = f L (Γ). (.6) Because of (.3), the operator P is sef-adjoint on L (Γ) and I P, which, by (.6), can be considered as a discrete Lapace operator, is non-negative and sef-adjoint on L (Γ). By means of spectra theory, one defines its square root (I P ) /. The equaity (.6) exacty means that (I P ) / f = f L(Γ) L(Γ). (.7) This equaity has an interpretation in terms of Soboev spaces defined through. Let p +. Say that a scaar-vaued function f on Γ beongs to the (inhomogeneous) Soboev space W,p (Γ) (see aso [45], [37]) if and ony if f W,p (Γ) := f L p (Γ) + f L p (Γ) < +. If B is any ba in Γ and p +, denote by W,p 0 (B) the subspace of W,p (Γ) made of functions supported in B. 4

5 We wi aso consider the homogeneous versions of Soboev spaces. For p +, define Ė,p (Γ) as the space of a scaar-vaued functions f on Γ such that f L p (Γ), equipped with the semi-norm f Ė,p (Γ) := f L p (Γ). Then Ẇ,p (Γ) is the quotient space Ė,p (Γ)/R, equipped with the corresponding norm. It is then routine to check that both inhomogeneous and homogeneous Soboev spaces on Γ are Banach spaces. The equaity (.7) means that (I P ) / f = f L (Γ) Ė, (Γ). In other words, for p =, the Soboev spaces defined by and by the Lapacian coincide. In the seque, we address the anaogous question for p.. Statement of the probem To that purpose, we consider separatey two inequaities, the vaidity of which wi be discussed in the seque. Let < p < +. The first inequaity we ook at says that there exists p > 0 such that, for a function f on Γ such that (I P ) / f L p (Γ), f p p (I P ) / f p. (R p ) This inequaity means that the operator (I P ) /, which is nothing but the Riesz transform associated with (I P ), is L p (Γ)-bounded. Here and after, say that a (sub)inear operator T is L p -bounded, or is of strong type (p, p), if there exists > 0 such that T f p f p for a f L p (Γ). Say that it is of weak type (p, p) if there exists > 0 such that m ({x Γ, T f(x) > λ}) f p λ p p for a f Lp (Γ) and a λ > 0. Notice that he functions f wi be defined on Γ, whereas T f may be defined on Γ or on E. The second inequaity under consideration says that there exists p > 0 such that, for a function f Ė,p (Γ), (I P ) / f p p f p. (RR p ) (The notations (R p ) and (RR p ) are borrowed from [3].) We have just seen, by (.7), that (R ) and (RR ) aways hod. A we-known fact (see [46] for a proof in this context) is that, if (R p ) hods for some < p < +, then (RR p ) hods with p such that /p + /p =, whie the converse is uncear in this discrete situation (it is fase in the case of Riemannian manifods, see [3]). As we wi see, we have to consider four distinct issues: (R p ) for p <, (R p ) for p >, (RR p ) for p <, (RR p ) for p >..3 The L p -boundedness of the Riesz transform.3. The case when p < Let us first consider (R p ) when p <. This probem was deat with in [46], and we just reca the resut proved therein, which invoves some further assumptions on Γ. The first one is of geometric nature. Say that (Γ, µ) satisfies the doubing property if there exists > 0 such that, for a x Γ and a r > 0, V (x, r) V (x, r). 5 (D)

6 Note that this assumption impies that there exist, D > 0 such that, for a x Γ, a r > 0 and a θ >, V (x, θr) θ D V (x, r). (.8) Remark. Observe aso that, since (Γ, µ) is infinite, it is aso unbounded (since it is ocay uniformy finite) so that, if (D) hods, then m(γ) = + (see [43]). The second assumption on (Γ, µ) is a uniform ower bound for p(x, y) when x y, i.e. when p(x, y) > 0. For α > 0, say that (Γ, µ) satisfies the condition (α) if, for a x, y Γ, (x y µ xy αm(x)) and x x. ( (α)) The next two assumptions on (Γ, µ) are pointwise upper bounds for the iterates of p. Say that (Γ, µ) satisfies (DUE) (a on-diagona upper estimate for the iterates of p) if there exists > 0 such that, for a x Γ and a k N, p k (x, x) m(x) V (x, k). (DUE) Say that (Γ, µ) satisfies (UE) (an upper estimate for the iterates of p) if there exist, c > 0 such that, for a x, y Γ and a k N, p k (x, y) m(x) V (x, d (x,y) k) e c k. (UE) Notice that, when (D) hods, the estimate (UE) is aso equivaent to p k (x, y) m(x) V (y, d (x,y) k) e c k, (.9) which wi be of frequent use in the seque. Reca that, under assumption (D), estimates (DU E) and (U E) are equivaent (and the conjunction of (D) and (DU E) is aso equivaent to a Faber-Krahn inequaity, [6], Theorem.). The foowing resut hods: Theorem. ([46]) Under assumptions (D), ( (α)) and (DUE), (R p ) hods for a < p. Moreover, the Riesz transform is of weak (, ) type, which means that there exists > 0 such that, for a λ > 0 and a function f L (Γ), m ({ x Γ; (I P ) / f(x) > λ }) λ f. As a consequence, under the same assumptions, (RR p ) hods for a p < +. Notice that, according to [40], the assumptions of Theorem. hod, for instance, when Γ is the ayey graph of a group with poynomia voume growth and p(x, y) = h(y x), where h is a symmetric bounded probabiity density supported in a ba and bounded from beow by a positive constant on an open generating neighborhood of e, the identity eement of G, and actuay Theorem. had aready been proved in [40]. 6

7 .3. The case when p > When p >, assumptions (D), (UE) and ( (α)) are not sufficient to ensure the vaidity of (R p ), as the exampe of two copies of Z inked between with an edge shows (see [46], Section 4). More precisey, in this exampe, as expained in Section 4 of [46], the vaidity of (R p ) for p > woud impy an L Poincaré inequaity on bas. Say that (Γ, µ) satisfies a scaed L Poincaré inequaity on bas (this inequaity wi be denoted by (P ) in the seque) if there exists > 0 such that, for any x Γ, any r > 0 and any function f ocay square integrabe on Γ such that f is ocay square integrabe on E, f(y) f B m(y) r f(y) m(y), (P ) where y B(x,r) f B = V (B) y B(x,r) f(x)m(x) x B is the mean vaue of f on B. Under assumptions (D), (P ) and ( (α)), not ony does (UE) hod, but the iterates of p aso satisfy a pointwise Gaussian ower bound. Namey, there exist c,, c, > 0 such that, for a n and a x, y Γ with d(x, y) n, c m(x) V (x, d (x,y) n) e n p n (x, y) m(x) V (x, d (x,y) n) e c n. (LUE) Actuay, (LUE) is equivaent to the conjunction of (D), (P ) and ( (α)), and aso to a discrete paraboic Harnack inequaity, see [30] (see aso [4] for another approach of (LU E)). Let p > and assume that (R p ) hods. Then, if f L p (Γ) and n, Indeed, (R p ) impies that P n f p p n f p. (G p ) P n f p p (I P ) / P n f p, and, due to the anayticity of P on L p (Γ), one aso has (I P ) / P n f p p n f p. More precisey, as was expained in [46], assumption (α) impies that does not beong to the spectrum of P on L (Γ). As a consequence, P is anaytic on L (Γ) (see [8], Proposition 3), and since P is submarkovian, P is aso anaytic on L p (Γ) (see [8], p. 46). Proposition in [8] therefore yieds the second inequaity in (G p ). Thus, condition (G p ) is necessary for (R p ) to hod. Our first resut is that, under assumptions (D), (P ) and ( (α)), for a q >, condition (G q ) is aso sufficient for (R p ) to hod for a < p < q: Theorem.3 Let p 0 (, + ]. Assume that (Γ, µ) satisfies (D), (P ), ( (α)) and (G p0 ). Then, for a p < p 0, (R p ) hods. As a consequence, if p 0 is such that /p 0 + /p 0 =, (RR p ) hods for a p 0 < p. 7

8 An immediate consequence of Theorem.3 and the previous discussion is the foowing resut: Theorem.4 Assume that (Γ, µ) satisfies (D), (P ) and ( (α)). Let p 0 (, + ]. Then, the foowing two assertions are equivaent: (i) for a p (, p 0 ), (G p ) hods, (ii) for a p (, p 0 ), (R p ) hods. Remark.5 In the recent work [3], property (G p ) is shown to be true for a p (, ] under the soe assumption that Γ satisfies a oca doubing property for the voume of bas. Remark.6 On Riemannian manifods, the L Poincaré inequaity on bas is neither necessary, nor sufficient to ensure that the Riesz transform is L p -bounded for a p (, ), see [3] and the references therein. We do not know if the corresponding assertion hods in the context of graphs..3.3 Riesz transforms and harmonic functions We aso obtain another characterization of the vaidity of (R p ) for p > in terms of reverse Höder inequaities for the gradient of harmonic functions, in the spirit of [48] (in the Eucidean context for second order eiptic operators in divergence form) and [3] (on Riemannian manifods). If B is any ba in Γ and u a function on B, say that u is harmonic on B if, for a x B \ B, (I P )u(x) = 0. (.0) We wi prove the foowing resut: Theorem.7 Assume that (D), ( (α)) and (P ) hod. Then, there exists p 0 (, + ] such that, for a q (, p 0 ), the foowing two conditions are equivaent: (a) (R p ) hods for a p (, q), (b) for a p (, q), there exists p > 0 such that, for a ba B Γ, a function u harmonic in 3B, ( ) ( ) p u(x) p m(x) p u(x) m(x). (RH p ) V (B) V (6B) x B x 6B Assertion (b) says that the gradient of any harmonic function in 3B satisfies a reverse Höder inequaity. Remember that such an inequaity aways hods for soutions of div(a u) = 0 on any ba B R n, if u is assumed to be in H (B) and A is bounded and uniformy eiptic (see [44]). In the present context, a simiar sef-improvement resut can be shown: Proposition.8 Assume that (D), ( (α)) and (P ) hod. Then there exists p 0 > such that (RH p ) hods for any p (, p 0 ). As a consequence, (R p ) hods for any p (, p 0 ). As a coroary of Theorem. and Proposition.8, we get: oroary.9 Assume that (D), ( (α)) and (P ) hod. Then, there exists ε > 0 such that, for a ε < p < + ε, f p (I P ) / f p. 8

9 .4 The reverse inequaity Let us now focus on (RR p ). As aready seen, (RR p ) hods for a p > under (D), ( (α)) and (DUE), and for a p 0 < p < under (D), (P ), ( (α)) and (G p0 ) if p 0 > and /p 0 + /p 0 =. However, we can aso give a sufficient condition for (RR p ) to hod for a p (q 0, ) (for some q 0 < ) which does not invove any assumption such that (G p0 ). For p < +, say that (Γ, µ) satisfies a scaed L p Poincaré inequaity on bas (this inequaity wi be denoted by (P p ) in the seque) if there exists > 0 such that, for any x Γ, any r > 0 and any function f on Γ such that f p and f p are ocay integrabe on Γ, f(y) f B p m(y) r p f(y) p m(y). (P p ) y B(x,r) y B(x,r) If p < q < +, then (P p ) impies (P q ) (this is a very genera statement on spaces of homogeneous type, i.e. on metric measured spaces where (D) hods, see [39]). The converse impication does not hod but an L p Poincaré inequaity sti has a sef-improvement in the foowing sense: Proposition.0 Let (Γ, µ) satisfy (D). Then, for a p (, + ), if (P p ) hods, there exists ε > 0 such that (P p ε ) hods. This deep resut actuay hods in the genera context of spaces of homogeneous type, i.e. when (D) hods, see [4]. Assuming that (P q ) hods for some q <, we estabish (RR p ) for q < p < : Theorem. Let q <. Assume that (D), ( (α)) and (P q ) hod. Then, for a q < p <, (RR p ) hods. Moreover, there exists > 0 such that, for a λ > 0, m ({ x Γ; (I P ) / f(x) > λ }) λ q f q q. (.) As a coroary of Theorem., Proposition.0 and Theorem., we get the foowing consequence: oroary. Assume that (D), ( (α)) and (P p ) hod for some p (, ). Then, there exists ε > 0 such that, for a p ε < q < +, (RR q ) hods. In particuar, (RR p ) hods..5 An overview of the method Let us briefy describe the proofs of our resuts. Let us first consider Theorem.3. The operator T = (I P ) / can formay be written as ( + ) T = a k P k, (.) where the a k s are defined by the expansion k=0 ( x) / = + k=0 a k x k (.3) 9

10 for < x <. The precise meaning of (.) is the foowing statement, which wi be proved in Appendix 8.: Lemma.3 Define E := { f L (Γ); f = (I P ) / g for some g L (Γ) }. Then, E is dense in L (Γ) and, for a f E, ( n ) a k P k f (I P ) / f in L (Γ), (.4) k=0 The kerne of T is therefore given by ( + ) x a k p k (x, y). k=0 It was proved in [47] that, under (D) and (P ), this kerne satisfies the Hörmander integra condition, which impies the H (Γ) L (Γ) boundedness of T and therefore its L p (Γ)- boundedness for a < p <, where H (Γ) denotes the Hardy space on Γ defined in the sense of oifman and Weiss ([]). However, the Hörmander integra condition does not yied any information on the L p -boundedness of T for p >. The proof of Theorem.3 actuay reies on a theorem due to Auscher and Marte (see [6]), which, given some p 0 (, + ], provides sufficient conditions for an L -bounded subinear operator to be L p -bounded for < p < p 0. Let us reca this theorem here in the form to be used in the seque for the sake of competeness (see [6], Theorem 3.7, and aso [5], Theorem., [], Theorem.): Theorem.4 Let p 0 (, + ]. Assume that Γ satisfies the doubing property (D) and et T be a subinear operator acting from a dense subset of L (Γ) into L (Γ). For any ba B, et A B be a inear operator acting on L (Γ), and assume that there exists > 0 such that, for a f L (Γ), a x Γ and a ba B x, and V / (B) T (I A B)f L (B) ( M( f ) ) / (x) (.5) V /p 0 (B) T A B f L p 0 (B) ( M( T f ) ) / (x). (.6) If < p < p 0, then there exists p > 0 such that, for a f L (Γ) L p (Γ), T f L p (Γ) p f L p (Γ). Notice that, to simpify the notations in our foregoing proofs, the formuation of Theorem.4 is sighty different from the one given in [] and in [5], since the famiy of operators (A r ) r>0 used in these papers is repaced by a famiy (A B ) indexed by the bas B Γ, see Remark 5 after Theorem. in []. Observe aso that this theorem extends to vector-vaued 0

11 functions (this wi be used in Section 3). Finay, here and after, M denotes the Hardy- Littewood maxima function: for any ocay integrabe function f on Γ and any x Γ, Mf(x) = sup f(y) m(y), B x V (B) where the supremum is taken over a bas B containing x. Reca that, by the Hardy- Littewood maxima theorem, since (D) hods, M is of weak type (, ) and of strong type (p, p) for a < p +. Foowing the proof of Theorem. in [5], we wi obtain Theorem.3 by appying Theorem.4 with A B = I (I P k ) n where k is the radius of B and n is an integer ony depending from the constant D in (.8). As far as Theorem. is concerned, note first that (RR p ) cannot be derived from (R p ) in this situation (where /p+/p = ), since we do not know whether (R p ) hods or not under these assumptions. Foowing [3], we first prove (.). The proof reies on a aderón- Zygmund decomposition for Soboev functions, which is the adaptation to our context of Proposition. in [3] (see aso [] in the Eucidean case and [6], Proposition 9., for the extension to a weighted Lebesgue measure): Proposition.5 Assume that (D) and (P q ) hod for some q [, ) and et p [q, + ). Let f Ė,p (Γ) and λ > 0. Then one can find a coection of bas (B i ) i I, functions (b i ) i I Ė,q (Γ) and a function g Ė, such that the foowing properties hod: y B f = g + i I b i, (.7) g λ, (.8) supp b i B i, b i q (x)m(x) λ q V (B i ), (.9) x B i V (B i ) λ p f p (x)m(x), (.0) i I x Γ χ Bi N, (.) where and N ony depend on q, p and on the constants in (D) and (P q ). i I As in [3], we rey on this aderón-zygmund decomposition to estabish (.). The argument aso uses the L p (Γ)-boundedness, for a < p < +, of a discrete version of the Littewood- Paey-Stein g-function (see [49]), which does not seem to have been stated before in this context and is interesting in itsef. For a function f on Γ and a x Γ, define ( g(f)(x) = (I P )P f(x) ) /.

12 Observe that this is indeed a discrete anaogue of the g-function introduced by Stein in [49], since (I P )P = P P + can be seen as a discrete time derivative of P and P is a Markovian operator. It is easy to check that the subinear operator g is bounded in L (Γ). Indeed, as aready said, the assumption ( (α)) impies that the spectrum of P is contained in [a, ] for some a >. As a consequence, P can be written as so that, for a integer, and, for a f L (Γ), It foows that, for a f L (Γ), P = (I P )P = a a λde(λ), ( λ)λ de(λ) (I P )P f = ( λ) λ de f,f (λ). g(f) = a (I P )P f = ( λ) λ de f,f (λ) a ( ) λ = de f,f(λ) a + λ f. It turns out that, as in the Littewood-Paey-Stein semigroup theory, g is aso L p -bounded for < p < + : Theorem.6 Assume that (D), (DUE) and ( (α)) hod. Let < p < +. There exists p > 0 such that, for a f L p (Γ), g(f) p p f p. Actuay, this inequaity wi ony be used for p > in the seque, but the resut, which is interesting in itsef, does hod and wi be proved for a < p < +. Before going further, et us mention that, in [3], N. Dungey estabishes, under a oca doubing property for the voume of bas, the L p -boundedness for a p (, ] of another version of the Littewood-Paey-Stein functiona, invoving the gradient instead of the time derivative and the (continuous time) semigroup generated by I P. Athough we do not use Dungey s resut here, it may prove usefu to study the boundedness of Riesz transforms on graphs. The proof of Theorem.6 for p > reies on the vector-vaued version of Theorem.4, whie, for p <, we use the vector-vaued version of the foowing resut (see [], Theorem. and [6] for an earier version):

13 Theorem.7 Let p 0 [, ). Assume that Γ satisfies the doubing property (D) and et T be a subinear operator of strong type (, ). For any ba B, et A B be a inear operator acting on L (Γ). Assume that, for a j, there exists g(j) > 0 such that, for a ba B Γ and a function f supported in B, for a j and V / ( j+ B) T (I A B)f L ( j (B)) g(j) f V /p 0 (B) L p 0 (.) V / ( j+ B) A Bf L ( j (B)) g(j) V /p 0 (B) f L p 0 (.3) for a j. If g(j) Dj < + where D is given by (.8), then T is of weak type (p 0, p 0 ), j and is therefore of strong type (p, p) for a p 0 < p <. Going back to Theorem., once (.) is estabished, we concude by appying rea interpoation theorems for Soboev spaces, which are aso new in this context. More precisey, we prove: Theorem.8 Let q [, + ) and assume ) that (D), (P q ) and ( (α)) hod. Then, for a q < p < +, Ẇ,p (Γ) = (Ẇ,q (Γ), Ẇ, (Γ) As an immediate coroary, we obtain: q p,p. oroary.9 (The reiteration theorem) Assume that Γ satisfies (D), (P q ) for some q < + and ( (α)). Define q 0 = inf {q [, ) : (P q ) hods}. For q 0 < p < p < p +, if p = θ + θ, then p p Ẇ,p,p (Γ) = (Ẇ (Γ), Ẇ,p (Γ)). θ,p oroary.9, in conjunction with (.), concude the proof of Theorem.. Notice that, since we know that Soboev spaces interpoate by the rea method, we do not need any argument as the one in Section.3 of [3]. For the proof of Theorem.7, we introduce a discrete differentia and go through a property anaogous to (Π p ) in [3], see Section 8 for detaied definitions. As far as Proposition.8 is concerned, its proof is entirey simiar to the one of Proposition. in [3] (and wi therefore be skipped in the present paper). Let us just mention that it reies on an eiptic accioppoi inequaity (anaogous to the Eucidean version, see [36]), Proposition.0 and Gehring s sef-improvement of reverse Höder inequaities ([35]). The pan of the paper is as foows. After recaing some we-known estimates for the iterates of p and deriving some consequences (Section ), we first prove Theorem.6, which is of independent interest, in Section 3. In Section 4, we prove Theorem.3 using Theorem.4. Section 5 is devoted to the proof of Proposition.5. Theorem.8 is estabished in Section 6 by methods simiar to [0] and, in Section 7, we prove Theorem.. Finay, Section 8 contains the proof of Theorem.7 and of Proposition.8. 3

14 Kerne bounds In this section, we gather some estimates for the iterates of p and some straightforward consequences of frequent use in the seque. We aways assume that (D), (P ) and ( (α)) hod. First, as aready said, (LU E) hods. Moreover, we aso have the foowing pointwise estimate for the discrete time derivative of p : there exist, c > 0 such that, for a x, y Γ and a N, p (x, y) p + (x, y) m(y) V (x, d (x,y) ) e c. (.) This time reguarity estimate, which is a consequence of the L anayticity of P, was first proved by hrist ([9]) by a quite difficut argument. Simper proofs have been given by Bunck ([5]) and, more recenty, by Dungey ([3]). Thus, if B is a ba in Γ with radius k, f any function supported in B and i, one has, for a x i (B) and a, P f(x) + (I P )P f(x) 4 i k V (B) e c f L. (.) This off-diagona estimate foows from (UE) and (.) and the fact that, for a y B, by (D), ( ( ) ) D V (y, k) k V (y, k) V (B) and V (y, ) sup,. Simiary, if B is a ba in Γ with radius k, i and f any function supported in i (B), one has, for a x B and a, P f(x) + (I P )P f(x) 4 i k V ( i B) e c f L. (.3) Finay, for a ba B with radius k, a i, a function f supported in i (B) and a, P f L e c 4i k f (B) L ( i (B)). (.4) See Lemma in [46]. If one furthermore assumes that (G p0 ) hods for some p 0 >, then, by interpoation between (.4) and (G p0 ), one obtains, for a p (, p 0 ), a f supported in i (B) and a, P f p L e c 4i k p f (B) L p ( i (B)). (.5) Inequaities (.4) and (.5) may be regarded as Gaffney type inequaities, in the spirit of [34]. 3 Littewood-Paey inequaities In this section, we estabish Theorem.6. The proofs rey on the foowing estimates: 4

15 Lemma 3. Let p 0 (, + ). For a positive integer n, a ba B = B(x 0, k) Γ, a f L p 0 (Γ) supported in B and for a integer j :.. x j (B) ( j +)k x j (B) >( j +)k (I P )P (I P k ) n f(x) (I P )P (I P k ) n f(x) m(x) e V cj (j B) f V L p 0 (B) p 0, (3.) m(x) j( D n) V ( j B) p 0 V p 0 (B) f L p 0. (3.) Lemma 3. For a positive integer n, a ba B = B(x 0, k) Γ, a j and a f L (Γ) supported in j (B):.. x B ( j +)k x B >( j +)k (I P )P (I P k ) n f(x) (I P )P (I P k ) n f(x) m(x) e cj V (B) V ( j B) f L, (3.3) m(x) j( 3D n) V (B) V ( j B) f L. (3.4) Proof of Lemma 3.: Let us first prove (3.). Let 0 q j be an integer and consider such that qk < (q + )k. We use the expansion (I P )P (I P k ) n f(x) = Fix 0 m n. For a x j (B), one has (I P )P +mk f(x) n ( ) m n m (I P )P +mk f(x). m=0 ( + mk )V (x, + mk ) e c 4 j k +mk y B ( + mk )V p 0 (x, f L + mk p 0 ) e c 4 j k +mk ( + mk )V p 0 (x 0, f L + mk p 0 ) e c 4 j k +mk ( + mk )V p 0 (B) 5 f L p 0, e c d (x,y) +mk f(y) m(y)

16 where the first inequaity foows from (.), the second one from the Höder inequaity and Lemma 3 in [46], the third and the fourth one are due to (.8). More precisey, the fourth inequaity is trivia when m since V (B) V (x 0, + mk ), and when m = 0, (.8) shows that ( ) D V (x 0, k) k V (x 0, ) e α k, for any α > 0, which is enough to concude. As a consequence, qk <(q+)k x j (B) (I P )P +mk f L ( j (B)) V / ( j B) V p 0 (B) Summing up on (qk, (q + )k ], one obtains (I P )P +mk f(x) m(x) Noticing that ( j +)k x j (B) qk <(q+)k e c 4 j k +mk + mk f L p 0. qk <(q+)k c e 4j k +mk ( + mk ) c e 4j k +mk (q+)k + c te 4j k t+mk ( + mk ) (t + mk ) dt qk q+m+ e c 4j v q+m we sum up on q [0, j ] in (3.5), which yieds ( (I P )P P +mk f(x) j +n+ 0 dv v, e c 4j v e V cj (j B) f V L p 0 (B) p 0. V (j B) f V L p 0 (B) p 0. (3.5) ) dv V ( j B) f v V L p 0 (B) p 0 Summing up on m [0, n] yieds the desired concusion. Let us now turn to (3.). Assume that qk < (q + )k for some integer q > j. onsider first the case when is even and write = m. For any function g L p 0 (Γ) and a x j (B), one has (I P )P m g(x) ( ) mv (x, e cq 0 d q 0 (x,y) m m(y) g m) L p0 y Γ mv p 0 (x, g L p 0 m) mv p 0 (x 0, jd p 0 g L p 0 m) 6

17 where p 0 + q 0 =, the first inequaity foows from (.) and Höder again, the second one from Lemma 3 in [46] and the ast one is due to the fact that V (x 0, m) V (x, m + d(x, x 0 )) V (x, m + j+ k) V (x, ( ) D m) + j+ k m V (x, m) jd, by (.8) and the fact that qk < (q + )k. As a consequence, (I P )P m g L ( j (B)) V ( j B) m V p 0 (x 0, jd p 0 g L m) p 0. (3.6) Moreover, sincep k is a Markov operator and is anaytic on L (Γ) (more precisey, P sk P (s+)k for a integer s, with a constant > 0 independent from k), s it is aso anaytic on L p 0 (Γ) since p 0 (, + ) (see [8], p. 46). This means that, if q denotes the greatest integer such that q q, (I P k ) n P m f (I P k ) n P q k f L p 0 q n f L p ( ) 0 k n f Lp0. L p 0 (3.7) ombining (3.6) and (3.7), one obtains (I P )P (I P k ) n f L ( j (B)) = (I P )P m (I P k ) n P m f V ( j B) V p 0 (x 0, ) ( k L ( j (B)) ) n jd p 0 f L p 0. We argue simiary when is odd, writing = m + m +, and obtain the same estimate. Summing up on > ( j + )k yieds x j (B) >( j +)k (I P )P (I P k ) n f(x) which is the desired concusion. m(x) jd p 0 V ( j B) f L p 0 >( j +)k V j D p 0 n ( ) k n p 0 (x 0, ) V ( j B) f V L p 0 (B) p 0, (3.8) 7

18 The proof of Lemma 3. is identica with the obvious modifications. The main difference is that one has to repace (3.8) by x B) >( j +)k (I P )P (I P k ) n f(x) m(x) jd V (B) f L >( j +)k In the ast inequaity, we use the fact that { V (x 0, j k) if j V (x 0, k, ) ( ) D j k if < j k, ( ) k n V (x 0, ) j( 3D n) V (B) V ( j B) f L. ( ) D and since > ( j + )k, j jd k. Proof of Theorem.6 when < p < : We appy Theorem.7 with T = g and p 0 (, ) and, for a ba B with radius k, A B defined by A B = I (I P k ) n, where n is a positive integer, to be chosen in the proof. Let us first check (.). Let p 0 (, ), B := B(x 0, k) and f supported in B. hoose n > D. If j, Lemma 3. shows that V ( j B) x j (B) (I P )P (I P k ) n f(x) m(x) j D n p 0 V p 0 (B) f L p 0, which means that (.) hods with g(j) := j( D p 0 n), which satisfies j g(j)dj < +. Let us now check (.3). Since A B = n n( ) p p P pk, p= it is enough to prove that, for a j and a p n, P pk f g(j) f V / ( j+ B) L ( j (B)) V L p 0 (B). (3.9) p 0 (B) For a x j (B), (.) yieds P pk f(x) e c 4 j p V p 0 (B) 8 f L p 0 (B)

19 if j, and P pk f(x) for j =, just by (UE). As a consequence, V p 0 (B) f L p 0 (B) P pk f L ( j (B)) 4 j e c p V / ( j+ B) f V L p 0 (B), p 0 (B) so that (3.9) hods. This ends the proof of Theorem.6 when < p <. Proof of Theorem.6 when < p < + : we appy Theorem.4 with the same choice of A B and p 0 = +. Let us first check (.5), which reads in this situation as V / (B) T (I A B)f L (B) ( M ( f )) / (y) for a f L (Γ), a ba B Γ and a y B. Fix such an f, such a ba B and y B. Write f = fχ j (B) := f j. j j The L -boundedness of g and A B and the doubing property (D) yied V / (B) T (I A B)f L (B) Let j. It foows from Lemma 3. that V (B) T (I A B )f j (x) m(x) = x B V / (B) f L (4B) ( M ( f )) / (y). V (B) (I P )P (I P k ) n f j (x) m(x) x B e cj V ( j B) f j L + j( 3D n) V ( j B) f j L j( 3D n) M( f )(y). Summing up on j therefore yieds (.5) provided that n > 3D. 4 To prove (.6), it suffices to estabish that, for a j n, a ba B Γ and a y B, T P jk f ( M ( T f ) (y) ) /. L (B) Let x B. By auchy-schwarz and the fact that p jk (x, z) = z Γ for a x Γ, one has, for any function h L (Γ), P jk h(x) ( P jk h (x)) /. 9

20 It foows that, for a, P jk ( (I P )P f)(x) ( P jk (I P )P f ) (x), so that P jk ( ( (I P )P f)(x) P jk (I P )P f ) (x) ( = P jk T f ) (x) M ( T f ) (y), which is the desired estimate (note that the ast inequaity foows easiy from (U E)). Thus, (.6) hods and the proof of Theorem.6 is therefore compete. 4 Riesz transforms for p > In the present section, we estabish Theorem.3, appying Theorem.4 with A B = I (I P k ) n. One has A B, =. In view of Theorem.4, it suffices to show that and T (I P k ) n f ( M( f ) ) / (x) (4.) V / (B) L (B) ( ) T I (I P k ) n f ( M( T f ) ) / V /p 0 (B) L p (x) (4.) 0 (B) for a f L (Γ), a x Γ and a ba B Γ containing x. Fix such data f, x and B. Proof of (4.): Set f i = fχ i (B) for a i. The L -boundedness of T (I P k ) n yieds V / (B) T (I P k ) n f L (B) V / (B) f L (Γ) ( M( f ) ) / (x). (4.3) Fix now i. In order to estimate the eft-hand side of (4.) with f repaced by f i, we use Lemma.3 (observe that f i E), which yieds ( ) + ) ((I P ) / (I P k ) n f i = a P (I P k ) n f i ( =0 + ) n = a n( ) j j P +jk f i ( =0 j=0 + ) = d P f i, =0 where d = ( ) j na j jk 0jn, jk 0

21 (reca that, for a 0, a > 0). It foows that T (I P k ) n f i (x) + = d P f i (x) for a x B. Indeed, if x B and = 0, P f i (x) = f i (x) = 0 because f i is supported in i (B). Thus, one has According to (.4), one has T (I P k ) n L + f i d P f i L. (B) (B) = T (I P k ) n L + f i d e c 4 i k f L (B) ( i+ B\ i B). (4.4) We caim that the foowing estimates hod for the d s: = Lemma 4. There exists > 0 ony depending on n with the foowing properties: for a integer, (i) if there exists an integer 0 m n such that mk < < (m + )k, d (ii) if there exists an integer 0 m n such that = (m + )k, d, (iii) if > (n + )k, d k n n. mk, We postpone the proof of this emma to the Appendix and end the proof of (4.). According to (4.4), one has T (I P k ) n f i L (B) For S, Lemma 4. yieds + n d e c 4 i k f L mk <<(m+)k ( i+ B\ i B) n d(m+)k e c 4i k m+ k m + f L ( i+ B\ i B) m=0 m=0 + d e c 4 i k >(n+)k f L ( i+ B\ i B) := S + S + S 3. (4.5) S n m=0 mk <<(m+)k e c 4 i k f mk L ( i+ B\ i B).

22 But, for each m n, mk <<(m+)k e c 4 i k mk (m+)k mk 0 e c4i, e c 4 i k t t mk t dt e c 4 i n(+w) w(w + ) dw where, c > 0 ony depend on n. For m = 0, 0<<k e c 4 i k 0 4i c du e u u e c4i. Therefore, As for S, Lemma 4. gives at once S e c4i f L ( i+ B\ i B). (4.6) S e c4i f L ( i+ B\ i B), (4.7) where, c > 0 ony depend on n once more. Finay, for S 3, Lemma 4. provides But one ceary has so that, since k, >(n+)k n S 3 k n e c 4i k n >(n+)k + e c 4i k f L ( i+ B\ i B). (n+)k t n e c 4i k t t + = (4 i k ) n u n e c du u n+ u 4 i + k n 4 in u n e c du u u 4 in, 0 S 3 4 in f L ( i+ B\ i B). (4.8) Summing up the upper estimates (4.6), (4.7) and (4.8) and using (4.5), one obtains T (I P k ) n L f i 4 in f L (B) ( i+ B\ i B). (4.9) The definition of the maxima function and property (.8) yied f L ( i+ B\ i B) V / ( i+ B) ( M( f )(x) ) / (i+)d/ V (B) / ( M( f )(x) ) /. dt

23 hoosing now n > D 4 and summing up over i, one concudes from (4.3) and (4.9) that ( + ) T (I P k ) n f i( D n) V (B) ( / M( f )(x) ) /, L (B) i=0 which ends the proof of (4.). Proof of (4.): We use the foowing emma: Lemma 4. For a p (, p 0 ), there exists, α > 0 such that, for a ba B Γ with radius k, a integer i and a function f L (Γ) supported in i (B), and for a j {,..., n} (where n is chosen as above), one has ( V (B) /p ) P jk f e α4i L p (B) k V ( i+ B) / f L (Γ). Proof of Lemma 4.: This proof is very simiar to the one of Lemma 3. in [5], and we wi therefore ony indicate the main steps. First, (.5) yieds P jk f L p (B) P jk f k L p (Γ). (4.0) Using (UE), and noticing that, by (D), for y B, V (y, k j) V (B), one has, for a x Γ and a y B, p jk (x, y) ( ) V (B) exp c d (x, y) m(y). jk As a consequence, for a x Γ, P jk f(x) V / (B) f L (4B). (4.) The L contractivity of P shows that P jk f f L L (Γ) (4B), (4.) so that, gathering (4.) and (4.), P jk f V (B) p f L L p (Γ) (Γ). (4.3) Finay, (4.3) and (4.0) yied the concusion of Lemma 4. when i =. onsider now the case when i. Let χ the characteristic function of (B) for a. One has, for a x Γ, P jk f(x) P jk χ P jk f(x) =: g (x). 3

24 By (.5) and (.8), ( ) V ( + /p V /p (B) g B) e c4 L p (B) P jk f V (B) k V /p ( + B) e c4 (+)D/p P jk f k V /p ( + B) L p ( + B\ B) Using (UE) and arguing as in the proof of Lemma 3. in [5], one obtains and, for a x + B \ B, where P jk f V ( + B) L ( ) L p ( + B\ B) K i V ( i+ B) f L ( i ) (4.4) P jk f(x) K i (i+)d V / ( i+ B) f L ( i+ B\ i B), (4.5) e c4i if i, K i = if i i +, e c4 if i +. Interpoating between (4.4) and (4.5) therefore yieds P jk f V /p ( + B) L p ( ) K i (i+)d( p) V / ( i+ B) f L ( i ). Summing up in, one ends the proof of Lemma 4. as in [5]. To prove (4.), it is enough to show that, if p (, p 0 ), there exists p > 0 such that, for a j {,..., n}, a function f L oc (Γ) with f L oc (Γ), a ba B Γ with radius k and any point x B, But, since for a 0, P =, one has P jk f ( M( f ) ) / (x). V /p (B) L p (B) P f = P (f f 4B ),. so that P jk f = P jk (χ (f f 4B )). One concudes the proof of (4.) as in [5], using the Poincaré inequaity and Lemma 4.. 4

25 5 The aderón-zygmund decomposition for functions in Soboev spaces The present section is devoted to the proof of Proposition.5, for which we adapt the proof of Proposition. in [3] to the discrete setting. Let f Ė,p (Γ), λ > 0. onsider Ω = {x Γ : M( f q )(x) > λ q }. If Ω =, then set g = f, b i = 0 for a i I so that (.8) is satisfied thanks to the Lebesgue differentiation theorem and the other properties in Proposition.5 obviousy hod. Otherwise the Hardy-Littewood maxima theorem gives m(ω) λ p ( f) q q p q ( ) = λ p f p (x)m(x) < +. x p (5.) In particuar, Ω is a proper open subset of Γ, as m(γ) = + (see Remark.). Let (B i ) i I be a Whitney decomposition of Ω ([]). That is, Ω is the union of the B i s, the B i s being pairwise disjoint open bas, and there exist two constants > >, depending ony on the metric, such that, if F = Γ \ Ω,. the bas B i = B i are contained in Ω and have the bounded overap property;. for each i I, r i = r(b i ) = d(x i, F ) where x i is the center of B i ; 3. for each i I, if B i = B i, B i F ( = 4 works). For x Ω, denote I x = {i I; x B i }. By the bounded overap property of the bas B i, there exists an integer N such that I x N for a x Ω. Fixing j I x and using the properties of the B i s, we easiy see that r 3 i r j 3r i for a i I x. In particuar, B i 7B j for a i I x. ondition (.) is nothing but the bounded overap property of the B i s and (.0) foows from (.) and (5.). The doubing property and the fact that B i F yied: f q (x)m(x) f q (x)m(x) λ q V (B i ) λ q V (B i ). (5.) x B i x B i Let us now define the functions b i s. Let (χ i ) i I be a partition of unity of Ω subordinated to the covering (B i ) i I, which means that, for a i I, χ i is a Lipschitz function supported in B i with χ i and χ i (x) = for a x Γ (it is enough to choose χ i (x) = r i i I ( ) ( d(x i, x) ( ) ) d(x k, x) ψ ψ, where ψ D(R), ψ = on [0, ], ψ = 0 on r i k r k 5

26 [ +, + ) and 0 ψ ). Note that χ i is actuay supported in + B i, so that χ i is supported in 3 B i Ω, where 3 = + + <. We set b i = (f f Bi )χ i. It is cear that supp b i B i. Let us estimate x B i b i q (x)m(x). Since b i (x) = ((f f Bi )χ i )(x) max y x χ i(y) f(x) + f(x) f Bi χ i (x) and since χ i (y) for a y Γ, we get by (P q ) and (5.) that ( ) b i q m(x) f q (x)m(x) + f f Bi q (x) χ i q (x)m(x) x B i x B i x B i λ q V (B i ) + q r q r q i i λ q V (B i ). x B i f q (x)m(x) Thus (.9) is proved. Set g = f b i. Since the sum is ocay finite on Ω, g is defined everywhere on Γ and i I g = f on F. It remains to prove (.8). Since i I χ i (x) = for a x Ω, one has g = fχ F + i I f Bi χ i where χ F denotes the characteristic function of F. We wi need the foowing emma: Lemma 5. There exists > 0 such that, for a j I, a u F 4B j and a v B j, g(u) g(v) λd(u, v). Proof: Since i I χ i = on Γ, one has g(u) g(v) = f(u) f Bi χ i (v) i I = (f(u) f Bi ) χ i (v). i I (5.3) For a i I such that v B i, f(u) f Bi + k=0 fb(u, k r i ) f B(u, k r i ) + fb(u,ri ) f Bi. 6

27 For a k 0, (P q ) yieds fb(u, k r i ) f B(u, k r i ) = ( ) V (u, k r i ) f(z) fb(u, k r i ) m(z) z B(u, k r i ) f(z) fb(u, V (u, k r i ) k r i ) m(z) z B(u, k r i ) q f(z) f V (u, k r i ) B(u, q k r i ) m(z) z B(u, k r i ) q k r i f(z) q m(z) V (u, k r i ) z B(u, k r i ) k r i (M ( f) q ) q (u) k λr i k λr j, (5.4) where the penutimate inequaity reies on the fact that u F and the ast one from the fact that B i B j and r i r j. Moreover, since u 4B j, B(u, r i ) B(x j, r i + d(u, x j )) B(x j, r i + 4r j ) 7B j. Since one aso has B i 7B j, one obtains, arguing as before, fb(u,ri ) f Bi fb(u,ri ) f 7Bj + f7bj f Bi V (7B j ) λr j. z 7B j f(z) f7bj m(z) (5.5) It foows from (5.4) and (5.5) that f(u) f Bi λr j λd(u, v), since r j = d(x j, F ) d(x j, u) d(x j, v) + d(v, u) r j + d(v, u). This ends the proof of Lemma 5. because of (5.3). To prove (.8), it is ceary enough to check that g(x) g(y) λ for a x y Γ. Let us now prove this fact, distinguishing between three cases:. Assume that x, y Ω. Then, χ F (x) = χ F (y) = 0. It foows that g(y) g(x) = i I ( fbi f Bj ) (χi (y) χ i (x)), 7

28 so that g(y) g(x) i I fbi f Bj χi (x) := h(x). We caim that h(x) λ. To see this, note that, for a i I such that χ i (x) 0, we have f Bi f Bj r j λ. Indeed, d(x, B i ), which easiy impies that r i 3r j + 4r j, hence B i 0B j. As a consequence, we have, arguing as before again, f Bi f 0Bj f(y) f 0Bj m(y) V (B i ) y B i f(y) f 0Bj m(y) V (B j ) y 0B j r j q f q (y)m(y) V (0B j ) y 0B j r j λ (5.6) where we used Höder inequaity, (D), (P q ) and the fact that ( f q ) 0Bj M( f ) q (z) for some z F B j. Anaogousy f 0Bj f Bj r j λ. Hence h(x) = (f Bi f Bj ) χ i (x) i I; x B i f Bi f Bj r i i I; x B i Nλ.. Assume now that x F \ F (reca that F was defined in Section.) and y Γ, so that y F. In this case g(y) g(x) = f(y) f(x) λ by the definition of F. 3. Assume finay that x F. i. If y F, we have g(y) g(x) = f(x) f(y) f(x) λ. ii. onsider now the case when y Ω. There exists j I such that y B j. Since x y, one has x 4B j, Lemma 5. therefore yieds g(x) g(y) λd(x, y) λ. Note that the case when x Ω and y F is contained in ase 3.ii) by symmetry, since y F. Thus the proof of Proposition.5 is compete. Remark 5. It is easy to get the foowing estimate for the b i s: for a i I, V (B i ) b i V (B i ) /q b i q λr i. Indeed, the first inequaity foows from Höder and the fact that b i is supported in B i. Moreover, by (P q ) and (5.), V (B i ) /q b i q = V (B i ) f f /q B i L q (B i ) r i V (B i ) f /q L q (B i ) λr i. 8

29 6 An interpoation resut for Soboev spaces To prove Theorem.8, we wi characterize the K functiona of interpoation for homogeneous Soboev spaces in the foowing theorem (see for instance [4] for a genera reference on the K functiona). Theorem 6. Under the same hypotheses as Theorem.8 we have that. there exists such that for every f Ẇ,q (Γ) + Ẇ, (Γ) and a t > 0 K(f, t q, Ẇ,q, Ẇ, ) t q ( f q ) q (t);. for q p <, there exists such that for every f Ẇ,p (Γ) and every t > 0 K(f, t q, Ẇ,q, Ẇ, ) t q ( f q ) q (t). Proof: have We first prove item. Assume that f = h + g with h Ẇ,q, g Ẇ,, we then h Ẇ,q + t q g Ẇ, h q + t q g K( f, t q, L q, L ) t q ( f q ) q (t). Hence we concude that K(f, t q, Ẇ,q, Ẇ, ) t q ( f q ) q (t). We prove now item. Let f Ẇ,p, q p <. Let t > 0, we consider the aderón- ) Zygmund decomposition of f given by Proposition.5 with λ = λ(t) = (M( f ) q q (t). Thus we have f = i I b i + g = b + g where (b i ) i I, g satisfy the properties of the proposition. We have the estimate b q q ( q b i ) (x)m(x) x Γ i I N b i q (x)m(x) i I x B i λ q (t) V (B i ) i I λ q (t)m(ω), where the B i s are given by Proposition.5 and Ω is defined as in the proof of Proposition.5. The ast inequaity foows from the fact that χ Bi N and Ω = B i. Hence i I i b q λ(t)m(ω) q. Moreover, since (Mf) f (see [3], hapter 3, Theorem 3.8), we obtain λ(t) = (M( f ) q ) q (t) ( f q ) q (t). 9

30 Hence, aso noting that m(ω) t (see [3], hapter, Proposition.7), we get K(f, t q, Ẇ,q, Ẇ, ) t q f q q (t) for a t > 0 and obtain the desired inequaity. Proof of Theorem.8: The proof foows directy from Theorem 6.. Indeed, f q p,p, Ẇ,p (Ẇ,q, Ẇ, ) q p,p and f q p,p f Ẇ,p. Hence item. of Theorem 6. gives us that (Ẇ,q, Ẇ, ) q p,p Ẇ,p and f Ẇ,p whie item. gives us that Ẇ,p = (Ẇ,q, Ẇ, ) q p,p with equivaent norms. 7 The proof of (RR p ) for p < In view of Theorem.8 and since (RR ) hods, it is enough, for the proof of Theorem., to estabish (.). Proof of (.): We foow the proof of (.9) in [3]. onsider such an f and fix λ > 0. Perform the aderon-zygmund decomposition of f given by Proposition.5. We aso use the foowing expansion of (I P ) / : (I P ) / = + k=0 a k (I P )P k (7.) where the (a k ) s were aready considered in Section 4. For each i I, pick the integer k Z such that k r(b i ) < k+ and define r i = k. We spit the expansion (7.) into two parts: (I P ) / = ri a k (I P )P k + k=0 + k=r i + a k (I P )P k := T i + U i. We first caim that Indeed, one has m ({ x Γ; (I P ) / g(x) > λ }) λ q f q q. (7.) m ({ x Γ; (I P ) / g(x) > λ }) λ (I P ) / g and since g λ on Γ and g q f q, we obtain = λ g, g λ q g q q λ q f q q, which ends the proof of (7.) (we have used the fact that g q f q, which foows from the fact that b q f q, which foows itsef from (.9)). We now caim that, for some constant > 0, ({ }) m x Γ; T i b i (x) > λ λ q f q q. (7.3) i I 30

31 To prove (7.3), write ({ }) ( ({ m x Γ; T i b i (x) > λ m 4B i )+m x / i i i I Observe first that, by (D) and Proposition.5, ( ) m 4B i V (4B i ) λ q f q q. i i I 4B i ; }) T i b i (x) > λ. i I (7.4) As far as the second term in the right-hand side of (7.4) is concerned, we foow ideas from [7], and estimate it by ({ m x / }) 4B i ; T i b i (x) > λ T i b i (x) m(x) λ i i I i I x/ 4B i + T i b i λ L ( j+ B i \ j B i ). If i, j are fixed, since (I P )b i is supported in B i, i I j= T i b i L ( j+ B i \ j B i ) = ri a k (I P )P k b L i ( j+ B i \ j B i ) k=0 ri a k (I P )P k b L i ( j+ B i \ j B i ) k= Given k r i, one has, for a x j+ B i \ j B i, using (.), (I P )P k b i (x) y B i p k (x, y) p k+ (x, y) b i (y) y B i kv (y, d (x,y) k) e c k b i (y) m(y). Using (.8) and arguing as in [3] (reying, in particuar, on Remark 5.), we obtain (I P )P k b L i r ( ) D i ri ( j+ B i \ j B i e c 4 ) k k j r i k V ( j+ B i )λ. Since a k kπ (see Appendix), it foows that T i b i L ( j+ B i \ j B i ) λe c4j V ( j+ B i ) λe c4j jd V (B i ). 3

32 One concudes, using (.0), that A i I e c4j jd V (B i ) λ q f q q, j which shows that (7.3) hods. What remains to be proved is that ({ }) m x Γ; U i b i (x) > λ λ q f q q. (7.5) Define, for a j Z, so that, for a j Z, One has For a k > 0, define i I β j = b i, r i i I; r i = j i I; r i = j b i = j β j. U i b i = a k (I P )P k b i i I i I k>ri = a k (I P )P k b i k>0 i I; ri <k = a k (I P )P k k>0 i I; ri =j <k = a k (I P )P k j β j. k>0 f k = j; 4 j <k j k β j. j; 4 j <k It foows from the previous computation and Theorem.6 that ( + ) / U i b i k f k. q q i I To see this, we estimate the eft-hand side of this inequaity by duaity, as in [3] and use the fact that a k k for a k. Since, by auchy-schwarz, f k j; 4 j <k k= j k β j, b i 3

33 one obtains ( + k= ) / k f k q By the bounded overap property, ( ) / β k k Z so that, using Remark 5., one obtains x Γ i I q q x Γ ( ) / β k. q k Z b i (x) q m(x), i I b i (x) q r q m(x) λ q V (B i ). i i I As a concusion, ({ }) m x Γ; U i b i (x) > λ V (B i ) λ q f q q, i I i I which is exacty (7.5). The proof of (.) is therefore compete. 8 Riesz transforms and harmonic functions Let us now prove Theorem.7. The proof goes through a property anaogous to (Π p ) in [3], the statement of which requires a notion of discrete differentia. 8. The discrete differentia and its adjoint To begin with, for any γ = (x, y), γ = (x, y ) E (reca that E denotes the set of edges in Γ), set d(γ, γ ) = max(d(x, x ), d(y, y )). It is straightforward to check that d is a distance on E. We aso define a measure on subsets on E. For any A E, set µ(a) = µ xy. (x,y) A We caim that E, equipped with the metric d and the measure µ, is a space of homogeneous type. Indeed, et γ = (a, b) E and r > 0. Assume first that r 5. Then, by (D), µ(b(γ, r)) = µ xy ( r ) µ xy = V (a, r) V a,. 00 d(x,a)<r, d(y,b)<r d(x,a)<r y Γ r q i But V ( r ) a, = 00 d(x,a)< r 00 d(y,x) µ xy 33 d(x,a)< r d(y,b)< r ( ( µ xy = µ B γ, r )),

34 since, when d(x, a) < r r and d(y, x), then d(y, b) < + r Assume now that r < 5. One has, using (D) again, ( µ(b(γ, r)) V (a, r) V (a, 0) V a, ) = m(a) µ ab µ(b(γ, r)), since, whenever x y, one has αm(x) µ xy by ( (α)). The caim is therefore proved. We can then define L p spaces on E in the foowing way. For p < +, say that a function F on E beongs to L p (E) if and ony if F is antisymmetric (which means that F (x, y) = F (y, x) for a (x, y) E) and F p L p (E) := F (x, y) p µ xy < +. (x,y) E Observe that the L (E)-norm derives from the scaar product F, G L (E) = F (x, y)g(x, y)µ xy. x,y Γ Finay, say that F L (E) if and ony if F is antisymmetric and F L (E) := sup F (x, y) < +. (x,y) E Our notion of discrete differentia is the foowing one: for any function f on Γ and any γ = (x, y) E, define df(γ) = f(y) f(x). The function df is ceary antisymmetric on E and is reated to the ength of the gradient of f. More precisey, it is not hard to check that, if ( (α)) hods, then for a p [, + ] and a function f on Γ, df L p (E) f L p (Γ). Indeed, if p < +, for a function f and a x Γ, ( f(x) p p(x, y) f(y) f(x) y x y x y x p p (x, y) f(y) f(x) p p(x, y) f(y) f(x) p ) p where the ast ine is due to ( (α)). As a consequence, f p L p (Γ) p(x, y) f(y) f(x) p m(x) x Γ y x df(x, y) p µ xy x,y Γ = df p L p (E). 34