Mathematische Zeitschrift

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1 Math. Z. DOI /s y Mathematische Zeitschrift Pointwise behaviour of M 1,1 Sobolev functions Juha Kinnunen Heli Tuominen Received: 2 January 2006 / Accepted: 12 December 2006 Springer-Verlag 2007 Abstract Our main objective is to study Hajłasz type Sobolev functions with the exponent one on metric measure spaces equipped with a doubling measure. We show that a discrete maximal function is bounded in the Hajłasz space with the exponent one. This implies that every such function has Lebesgue points outside a set of capacity zero. We also show that every Hajłasz function coincides with a Hölder continuous Hajłasz function outside a set of small Hausdorff content. Our proofs are based on Sobolev space estimates for maximal functions. Mathematics Subject Classification (2000) 46E35 1 Introduction If the gradient of a function is locally integrable to a power which is higher than the dimension of the underlying space, then by the Sobolev embedding theorem the function is locally Hölder continuous. It is a more delicate question to study Sobolev functions if the gradient is integrable to a power which is smaller than the dimension, see for example [3,6,9,13,19]. In this work, we employ the definition Sobolev spaces on metric measure spaces introduced by Hajłasz [4]. This work is a continuation of [6,13] and the purpose of this work is to deal with the case when the Hajłasz gradient is integrable to the power one. In this case, we have new challenges and new phenomena. The basic problem is that the Hardy Littlewood maximal function is not bounded on L 1. We overcome this J. Kinnunen () Department of Mathematical Sciences, University of Oulu, P.O. ox 3000, Oulu, Finland juha.kinnunen@oulu.fi H. Tuominen Department of Mathematics and Statistics, University of Jyväskylä, P.O. ox 35 (MaD), Jyväskylä, Finland tuheli@maths.jyu.fi

2 J. Kinnunen, H. Tuominen problem by using appropriate versions of Sobolev Poincaré inequalities, see [5]. In the Euclidean case with the Lebesgue measure for exponents which are strictly greater than one the Hajłasz space coincides with the standard Sobolev space W 1,p (R n ), but for the exponent one we have a strictly smaller class of functions. Indeed, by an unpublished result of Koskela and Saksman this space coincides with the space of integrable functions, whose first order weak partial derivatives belong to the Hardy space H 1. Our results apply only to the Hajłasz type Sobolev space and it is not clear to us how to obtain the corresponding result for the standard Sobolev space. We show that functions which belong to Hajłasz type Sobolev space with the exponent one have Lebesgue points outside a set of capacity zero and that they coincide with Hölder continuous Sobolev functions outside a set of small Hausdorff content. Our proofs are based on maximal function arguments. More precisely, we use a maximal function which is related to discrete convolution approximations of the function. The discrete maximal function is comparable by two-sided pointwise estimates with the Hardy Littlewood maximal function, but it seems to be smoother than the standard maximal function. Thus, it can be used as a test function for the capacity. Indeed, we show that the discrete maximal function is bounded in the Hajłasz space also with the exponent one. This may be somewhat unexpected, since the discrete maximal function is not bounded on L 1. The corresponding result in the Euclidean space for exponents which are strictly bigger than one has been studied in [12,14], see also [8] and references therein. For the metric case we refer to [13]. Recently, Luiro [16] observed that the maximal operator is continuous in W 1,p (R n ) when 1 < p <. 2 Notation and preliminaries 2.1 asic assumptions Throughout the paper, is a metric measure space equipped with a metric d and a orel regular outer measure µ. We assume that µ is doubling, that is, there is a fixed constant C µ > 0, a doubling constant of µ, such that µ ( (x,2r) ) C µ µ ( (x, r) ) for each x, andallr > 0. Here (x, r) ={y : d(y, x) <r} is the open ball of radius r centered at x. If0< t < and = (x, r) is a ball in, thent = (x, tr). We also assume that the measure of every open set is positive, and that the measure of each bounded set is finite. Recall that the doubling condition of µ implies that there exists a constant C 0 > 0 such that whenever 0 = (x 0, r 0 ) and = (x, r) are balls with x 0 and 0 < r r 0,then µ() ( r ) s, µ( 0 ) C 0 (1) r 0 where s = log 2 C µ, (see for example [7, Lemma 14.6]). In this paper, s denotes the smallest exponent for which (1) holds and it is called the doubling dimension of µ. We recall the definitions of two maximal functions. Let 0 α<, 0<β<, R > 0, and u L 1 loc (). The (restricted) fractional maximal function of u is M α,r u(x) = sup r α u dµ. 0<r R (x,r)

3 Pointwise behaviour of M 1,1 Sobolev functions If R =, then there is no restriction for the radii and we denote M α, u = M α u.if α = 0andR =, then we obtain the usual Hardy Littlewood maximal function and write M 0, u = M u. The (restricted) fractional sharp maximal function of u is u # β,r u(x) = sup r β u u (x,r) dµ. 0<r R (x,r) Again, if R =, we denote u # β, = u# β. Above, u = udµ = µ() 1 udµ is the integral average of u over the ball. The local space L 1 loc () consists of functions that are integrable in each ball. The Hausdorff t-content of a set E is the number H t (E) = inf i rt i,wherethe infimum is taken over all countable coverings of E by balls of radius r i. y χ E, we denote the characteristic function of a set E. In general, C will denote a positive constant whose value is not necessarily the same at each occurrence. y writing C = C(τ, λ), we indicate that the constant depends only on τ and λ.ifthere is a positive constant C 1 such that the two-sided estimate C 1 1 u v C 1u holds, we write u v, and say that u and v are comparable. 2.2 Sobolev spaces M 1,p () and capacity A measurable function g 0isa generalized gradient of a measurable function u, g D(u), ifthereisasete with µ(e) = 0 such that u(x) u(y) d(x, y) ( g(x) + g(y) ) (2) for all x, y \ E. The Sobolev space M 1,p (), 1 p <, defined by Hajłasz [4], consists of functions u L p () for which there exists a function g L p () D(u). The space M 1,p (), equipped with the norm u M 1,p () = ( u p L p () + inf g p L p ()) 1/p, (3) where the infimum is taken over all functions g L p () D(u), isaanachspace [5, Theorem 8.3]. The space M 1,p () can be defined for all 0 < p <, but (3) isa norm only when p 1. The Sobolev capacity in M 1,p () for 1 < p < has been studied [15]. The definition of [15] extends for all 1 p < in a natural way. The p-capacity with 1 p < of the set E is defined by setting C p (E) = inf { u p M 1,p () : u A(E)}, (4) where A(E) = { u M 1,p () : u 1 in an open neighborhood of E } is the set of admissible functions (test functions) forc p (E). IfA(E) =,thenweset C p (E) =. Carefully reading the proofs of [15], we note that most properties of p-capacity hold also for p = 1. In particular, the p-capacity is an outer measure [15, Theorem 3.2] and an outer capacity [15, Remark 3.3], that is, C p (E) = inf { C p (U) : E U, U open }.

4 J. Kinnunen, H. Tuominen It is easy to see that µ(e) C p (E), in particular, sets of zero p-capacity are of zero measure, see [15, Lemma 4.1]. The doubling property of µ gives an upper bound for the p-capacity of a ball of radius 0 < r 1, C p () Cr p µ(), (5) where the constant C depends only on the doubling constant of µ and p. Forthe proof, we observe that the 1/r-Lipschitz function with support in 2, used in[15, Theorem 4.6], is a suitable test function also for the 1-capacity. A function u : R is p-quasicontinuous if for every ε>0, there is an open set U such that C p (U) <εand the restriction of u to \ U is continuous. y the definition, functions of M 1,1 () are defined only up to sets of measure zero. However, it was shown in [15, Corollary 3.7] that each Sobolev function u M 1,p () has a p-quasicontinuous representative, that is, there is a p-quasicontinuous function u M 1,p () such that u = u µ-almost everywhere in. The proof remains valid also for p = 1 because continuous functions are dense in M 1,1 () by [5, Theorem 8.4] and M 1,1 () is complete. Since the p-capacity is an outer capacity, and the norm of M 1,p () does not see sets of zero measure, a result of Kilpeläinen [11] implies that the p-quasicontinuous representative u of u M 1,p (),1 p <, is unique. Indeed, if two p-quasicontinuous functions f and g coincide µ-almost everywhere, then the p-capacity of the set where f = g is zero. 3 Lebesgue points One of our main results shows that the 1-quasicontinuous representative of an M 1,1 - function u is obtained as a limit of integral averages of u over small balls. This result implies that almost every point, in the 1-capacity sense, is a Lebesgue point of u. Theorem 1 Let u M 1,1 (). Then there is a set E with C 1 (E) = 0 such that lim udµ = u (x) (6) r 0 (x,r) for all x \ E, where u is the 1-quasicontinuous representative of u. The proof is presented in the end of this section. Remark 1 y using Theorem 1 for functions u q k,whereu M 1,1 () and (q k ) is an enumeration of rational numbers, and basic properties of capacity, we see that if u M 1,1 (), then lim u u (x) dµ = 0 r 0 (x,r) for all x \ F with C 1 (F) = 0. Hence, 1-quasi every point is a Lebesgue point of u, see [9, Remark 2.8(2)]. The following stronger result corresponds the second part of [13, Theorem 4.5].

5 Pointwise behaviour of M 1,1 Sobolev functions Theorem 2 Let u M 1,1 () with the 1-quasicontinuous representative u,andlet 0 < q s/(s 1). Then there is a set F with C 1 (F) = 0 such that lim u u (x) q dµ = 0 (7) for all x \ F. r 0 (x,r) 3.1 The discrete maximal operator To prove Theorem 1, we will use the maximal operator for a discrete convolution approximation, defined in [13]. We begin with a covering of and a corresponding partition of unity, see also [2,10,20]. For r > 0, let { i } beacoveringof by balls i = (x i, r) such that the enlarged balls 6 i have bounded overlap, that is, there is a constant N = N(C µ ) N such that χ 6i (x) N for all x. Let then (ϕ i ) be a partition of unity related to the covering { i } such that i ϕ i(x) = 1forallx, 0 ϕ i 1in, ϕ i C in 3 i, supp ϕ i 6 i, and that each ϕ i is L/r-Lipschitz. Here the constants C > 0andL > 0 depend only on the doubling constant of µ. We define u r, the discrete convolution of u, by setting u r (x) = ϕ i (x)u 3i, x. For the definition of the discrete maximal operator, we numerate the positive rationals and choose for each radius r j a covering consisting of balls j i = (x i, r j ) and a corresponding partition of unity as above. Then we define the discrete maximal function of u L 1 loc () related to coverings {j i} by M u(x) = sup u rj (x), x. (8) j The maximal operator M depends on the covering. However, the estimates for M are independent on the covering. The operator M has the useful property of being comparable with the usual Hardy Littlewood maximal operator, C 1 M u(x) M u(x) C M u(x) (9) for each u L 1 loc () and all x, wherec = C(C µ), see[13, Lemma 3.1]. This means that for almost all practical purposes we may use the discrete maximal function instead of the Hardy Littlewood maximal function. If p > 1, then (9) together with the boundedness of the maximal operator M imply that there is a constant C = C(C µ, p) such that M u L p () C M u L p () C u L p () for all u L p (). In Lemmas 1 and 2 below, we will show that if u M 1,1 (), then both u r and M u are in M 1,1 (), in particular, the discrete operator is bounded in M 1,1 (). Sincethe Hardy Littlewood maximal operator is not bounded in L 1 (),thecasep = 1 requires a different proof than the case p > 1. We will follow the proofs of [13, Lemma 3.3, Theorem 3.6] and use the Sobolev Poincaré inequality to overcome the difficulties caused by the case p = 1.

6 J. Kinnunen, H. Tuominen y the Sobolev Poincaré inequality, we mean the following result of Hajłasz [5, Theorem 8.7]. If = (x, r) is a ball, σ > 1ands/(s + 1) p < s, then each u M 1,p (σ ) with g D(u) L p (σ ) belongs to u L p () and 1/p 1/p u u p dµ Cr g p dµ, (10) where C = C(p, C µ, σ)and p = sp/(s p). Note that if u M 1,1 (),thenu M 1,s/(s+1) (σ ) for each ball.since(s/(s+1)) = 1, inequality (10) and the Hölder inequality imply that u u dµ Cr g s/(s+1) dµ σ σ (s+1)/s Cr g q dµ for all s/(s + 1) q 1 and all balls = (x, r) in. The constants C in Lemmas 1 and 2 depend only on the doubling constant of µ and on the constants of the Sobolev Poincaré inequality (10). Lemma 1 Let u M 1,1 (), g D(u) L 1 (), r> 0, ands/(s + 1) q < 1. Then u r M 1,1 () and there is a constant C such that C(g + (M g q ) 1/q ) belongs to D(u r ) L 1 (). Moreover, there is a constant C such that u r M 1,1 () C u M 1,1 (). σ 1/q (11) Proof Let u M 1,1 () and g D(u) L 1 (). Fix1<σ <2 for the Sobolev Poincaré inequality (10). The proof consists of two steps. First we will find a generalized gradient for u r, and then we will show that both u r and the gradient are in L 1 (). Generalized gradient of u r : For each x we have, since i ϕ i(x) = 1, that u r (x) = ϕ i (x)u 3i = u(x) + ϕ i (x)(u 3i u(x)), where the sum is over finitely many terms only by the bounded overlap of the balls 6 i. Hence, by the definition of generalized gradient, the function g + g i D(u r ), where g i is a generalized gradient of ϕ i (u 3i u). To find suitable gradients g i, we first note that by the Leibniz differentiation rule, [6, Lemma 5.20], and the properties of the functions ϕ i, the function (Lr 1 u u 3i +g)χ 6i belongs to D(ϕ i (u 3i u)). To estimate u u 3i,letx 6 i.then3 i (x,9r) 15 i,and u(x) u 3i u(x) u (x,9r) + u (x,9r) u 3i. (12)

7 Pointwise behaviour of M 1,1 Sobolev functions The first term in the right-hand side of (12) is estimated by a standard telescoping argument. We use the doubling property of µ and (11) and obtain u(x) u (x,9r) u (x,3 2 j r) u (x,3 1 j r) j=0 C Cr j=0 (x,3 2 j r) j=0 3 2 j u u (x,3 2 j r) dµ (x,σ 3 2 j r) g q dµ 1/q Cr ( M g q (x) ) 1/q whenever x is a Lebesgue point of u. For the second term, we use the doubling property of µ and (11), and obtain 1/q u (x,9r) u 3i C u u (x,9r) dµ Cr g q dµ y (12) (14) we have that (x,9r) σ (x,9r) (13) Cr ( M g q (x) ) 1/q. (14) u(x) u 3i Cr ( M g q (x) ) 1/q for all Lebesgue points of u, and hence for almost all x. Hence,wecanselect g i = (C(M g q ) 1/q + g)χ 6i. Using the bounded overlap of the balls 6 i, we conclude that the function g ur = C ( g + (M g q ) 1/q) belongs to D(u r ). Integrability of u r and g ur : y the properties ϕ i = 0in \ 6 i,0 ϕ i 1, the bounded overlap of the balls 6 i, and the doubling property of µ, we have that u r dµ ϕ i u 3i dµ = u 3i dµ C u dµ, (15) 6 i and hence u r L 1 (). For the integrability of g ur, it suffices to show that (M g q ) 1/q L 1 (). Since g L 1 (), we have g q L 1/q (). The boundedness of the maximal operator for 1/q > 1 implies that M g q L 1/q () and (M g q ) 1/q dµ C (g q ) 1/q dµ = C gdµ, (16) and hence g ur L 1 () with g ur L 1 () C g L 1 (). y choosing the generalized gradient g of u so that g L 1 () 2 u M 1,1 () and combining estimates (15)and(16), the claim u r M 1,1 () C u M 1,1 () follows.

8 J. Kinnunen, H. Tuominen Lemma 2 Let u M 1,1 (), g D(u) L 1 () and s/(s + 1) q < 1. ThenM u M 1,1 () and there is a constant C such that the function C(g + (M g q ) 1/q ) belongs to D(M u) L 1 (). Moreover, there is a constant C such that M u M 1,1 () C u M 1,1 (). Proof We begin with the integrability of M u.y(9), it is enough to show that M u L 1 (). Fix1<σ <2 for the Sobolev Poincaré inequality (10). Step 1 M u L 1 () for all balls : Let be a ball of radius r.y(10), u L 1 () and 1/1 u 1 dµ u u 1 dµ σ 1/1 + u 1 dµ 1/1 Cr µ() 1/1 gdµ + µ() 1/1 1 u dµ. (17) Since 1 = s/(s 1) >1, M u L 1 () and (M u)1 dµ C u 1 dµ. Hence, using the Hölder inequality, (17) and the doubling property of µ, we obtain 1/1 M udµ (M u) 1 dµ µ() 1 1/1 and conclude that M u L 1 (). Cr σ gdµ + C u dµ, (18) Step 2 M u L 1 (): We cover by balls of radius 1/5, and use the 5r-covering theorem to obtain balls (y i,1) such that the balls (y i,1) cover, the balls (y i,1/5) are pairwise disjoint, and that there is a constant N = N(C µ ) such that i χ (y i,2)(x) N for all x. Using (18), the bounded overlap of the balls (y i,2), and the assumption 1 <σ <2, we have that M udµ M udµ C (y i,1) C (y i,σ) gdµ + gdµ + (y i,1) u dµ u dµ, (19) which shows that M u is in L 1 (). Now M u is in L 1 () by (19) and(9), and hence it is finite almost everywhere in. Since the function C(g + (M g q ) 1/q ) L 1 () belongs to D(u rj ) D( u rj ) for

9 Pointwise behaviour of M 1,1 Sobolev functions all j N, the claim that M u M 1,1 () with generalized gradient C(g + (M g q ) 1/q ) follows from Lemmas [13, Lemma 2.6] and 1. As in the proof of Lemma 1, we obtain the desired norm estimate by choosing the generalized gradient g of u so that g L 1 () 2 u M 1,1 (), and combining estimates (19), (9), and (16). Remark 2 Note that u r C M u almost everywhere in by (9). Hence, the integrability of u r follows also from (19). Note also that by a similar arguments as in the proof of Lemma 1, we see that u r (x) u(x) as r 0 for almost every x, and that u r u in L 1 (). Namely, u r (x) u(x) ϕ i (x) u(x) u 3i i u(x) u 3i Cr ( M g q (x) ) 1/q, (20) where the last sum is taken over all indices i for which x 6 i. The right-hand side of (20) tends to zero as r 0 for almost every x because (M g q ) 1/q L 1 (). In [9, Theorem 2.11] the first author and Harjulehto gave several equivalent conditions for a differentiation basis, which give the validity of (6)foru M 1,p () p-quasi everywhere. We will use part of this result in the proof of Theorem 1 below. Proof (Proof of Theorem 1)y[9, Theorem 2.11], the claim follows if we manage to show that there is a constant C > 0 such that C 1 x : lim sup u dµ >λ Cλ 1 u M r 0 1,1 () (x,r) for all λ>0and every u M 1,1 () (note that the proof of [9, Theorem 2.11 (ii) (i)] holds also for p = 1). Since lim sup u dµ M u(x) r 0 (x,r) for all x, it suffices to show that C 1 ({x : M u(x) >λ}) Cλ 1 u M 1,1 (). (21) To show that the weak type estimate (21) holds, we proceed as in the proof of [13, Lemma 4.4]. Let u M 1,1 (), λ>0, and let M u be the discrete maximal function of u. y(9), {x : M u(x) >λ} E λ,where E λ ={x : C M u(x) >λ}. The set E λ is open by the lower semicontinuity of M u,andcλ 1 M u is a test function for C 1 (E λ ). Hence, using the definition of the p-capacity and Lemma 2, we have that C 1 (E λ ) Cλ 1 M u M 1,1 () = Cλ 1 M u M 1,1 () Cλ 1 u M 1,1 (), from which (21) follows by the monotonicity of the capacity.

10 J. Kinnunen, H. Tuominen Note that instead of using the sufficient condition for the existence of the differentiation basis that differentiates M 1,1 (), we could have followed the proof of [13, Theorem 4.5]. Namely, the main tools used in the proof; density of continuous functions in M 1,1 (), (21), and the basic properties of the p-capacity, hold also in our case. Proof (Proof of Theorem 2) Letu M 1,1 (), g D(u) L 1 (), andletu be the 1-quasicontinuous representative of u. y Theorem 1, thereisasete with C 1 (E) = 0 such that u (x,r) u (x) as r 0 whenever x \ E. Sinceg 0 is in L 1 (), the proof of [13, Lemma 4.3], which uses the 5r-covering theorem, the subadditivity of capacity, the doubling condition of µ,estimate(5), and the absolutely continuity of the integral, implies that the set { D = x : lim sup r r 0 (x,r) } gdµ>0 has zero 1-capacity. Let = (x, r) beaballin, andlet0< q 1,where1 = s/(s 1). ythe Hölder inequality and the Sobolev Poincaré inequality (10), we have that 1/q 1/1 u u q dµ u u 1 dµ Cr gdµ. Hence lim r 0 (x,r) u u q dµ = 0 whenever x \ D. The subadditivity of the 1-capacity implies that C 1 (E D) = 0. For x \ (E D) we have that 1/q u u (x) q dµ (x,r) (x,r) u u (x,r) q dµ 1/q σ + u (x,r) u (x). (22) y the selection of the sets E and D, the limit as r 0 of the right-hand side of (22) is zero, and hence the claim follows for F = E D. 4 Hölder quasicontinuity In this section, we show that Hölder continuous functions are dense in M 1,1 () both in norm and Lusin sense, see Theorem 4. This is a generalization of the main result of [6] tothecasep = 1.

11 Pointwise behaviour of M 1,1 Sobolev functions The first theorem provides a characterization of M 1,1 () using a Poincaré inequality or fractional sharp maximal function, see [6, Theorem 3.4] for the case p > 1. Theorem 3 Let u L 1 (). The following three conditions are equivalent: 1. u M 1,1 (), 2. there is a function g L 1 (), g 0, and0 < q < 1 such that the pair u, g satisfies a (1, q)-poincaré inequality, that is, there exist constants C > 0 and σ 1 such that 1/q u u dµ Cr g q dµ for each ball = (x, r), 3. u # 1 L1 (). Proof If u M 1,1 (), thena(1, q)-poincaré inequality for s/(s + 1) q < 1 follows from the Sobolev Poincaré inequality (11). Suppose that 2. holds for u and g. Then for = (x, r), 1/q r 1 u u dµ C g q dµ, and hence u # 1 (x) C(M gq (x)) 1/q.Sinceg L 1 (), the function g q belongs to L 1/q (). The assumption 0 < q < 1 together with the boundedness of the Hardy Littlewood maximal operator implies that M g q L 1/q (). Thus (M g q ) 1/q,and hence also u # 1 is in L1 (). Suppose then that u # 1 L1 (). The integrability of u implies that the oscillation of u is controlled by the fractional sharp maximal function; for 0 <β<, thereisa constant C = C(β, C µ )>0such that u(x) u(y) C d(x, y) β( u # β,4 d(x,y) (x) + u# β,4 d(x,y) (y)) (23) for almost all x, y,see[6, Lemma 3.6], [17]. Hence 1. follows from the integrability of u # 1. The next result is a useful tool in the proof of Theorem 4, see[6, Corollary 3.10] for the case p > 1. Corollary 1 Let u M 1,1 (), g L 1 () D(u), 0 α<1, R> 0, σ > 1, and s/(s + 1) q < 1.Then u # 1 α,r (x) C( M αq,σ R g q (x) ) 1/q for all x. Proof Let x, 0< r < R, and = (x, r). y the Sobolev Poincaré inequality (11) we have that r α 1 u u dµ Cr α 1+1( ) 1/q g q dµ σ = C (r αq σ σ σ g q dµ) 1/q C ( Mαq,σ R g q (x) ) 1/q.

12 J. Kinnunen, H. Tuominen The claim follows by taking supremum over r. As in [6, Sect. 5], we define ũ by setting ũ(x) = lim sup r 0 (x,r) udµ (24) for a function u of M 1,1 (). y Theorem 1, the limit of the right-hand side of (24) exists and equals u, the quasicontinuous representative of u, except on a set of 1-capacity zero. In this section, we use the representative ũ for u, and denote it by u. Then, by the proof of (23)([6, Lemma 3.6]), the inequality u(x) u(y) C d(x, y) β( u # β,4 d(x,y) (x) + u# β,4 d(x,y) (y)) (25) holds for every x, y and for all 0 <β 1, see [6] for the discussion on infinite values of u. Hence,u is Hölder continuous with exponent β if u # β <. Theorem 4 Let u M 1,1 () be defined pointwise by (24),andlet0 <β 1. Then for each ε>0, there is a function v and an open set O such that 1. u = vin\ O, 2. v M 1,1 () and it is Hölder continuous with exponent β on every bounded set of, 3. u v M 1,1 () <ε, 4. H s (1 β) (O) <ε. Since the Hardy Littlewood maximal function is not bounded in L 1,thecasep = 1 requires a different proof than the case p > 1. When showing that the approximation of u has a generalized gradient, we use the Sobolev Poincaré inequality. In the proof we first assume that u vanishes outside a ball. The general case follows by using a localization argument as in [6, Theorem 5.3]. Proof Let u M 1,1 () and g L 1 () D(u). Let also s/(s + 1) q < 1and 1 <σ <2, and recall that (11) holds for the pair u, g. This will be an important tool for us. Step 1 Suppose that the support of u is in (x 0,1) for some x 0. Let λ>0, and denote E λ = { x : u # β (x) >λ}. The set E λ is open, and by (25), u is Hölder continuous with exponent β in \ E λ. We will correct the values of u in the bad set E λ using discrete convolution. For that, we use a Whitney type covering of E λ from [2, Theorem III.1.3], [18, Lemma 2.9]. Let ={ i }, i = (x i, r i ), be balls such that r i = d(x i, \ E λ )/10, 1. the balls (x i, r i /5) are pairwise disjoint, 2. E λ = i (x i, r i ), 3. (x i,5r i ) E λ, 4. if x (x i,5r i ),then5r i d(x, \ E λ ) 15r i, there is x i \ E λ such that d(x i, x i )<15r i,and χ (xi,5r i )(x) M for all x E λ.

13 Pointwise behaviour of M 1,1 Sobolev functions Let (ϕ i ) be a partition of unity corresponding to the collection for which supp ϕ i 2 i,0 ϕ i 1, each ϕ i is K/r i -Lipschitz, and ϕ i (x) = χ Eλ (x), see for example [18, Lemma 2.16]. For each x i,letx i be the closest point in \ E λ given by 5. We begin with the properties of the set E λ. y a similar argument as in [6, Theorem 5.3], using the assumption supp u (x 0,1) and the doubling property of µ, itis easy to see that there is λ 0 > 0 such that Claim 1 µ(e λ ) 0asλ. E λ (x 0,2) for all λ>λ 0. (26) Proof Since 0 < q < 1, the Hölder inequality implies that for all R > 0andα 0, 1/q r αq g q dµ r α gdµ M α,r g(x) (x,r) (x,r) whenever x and r R.Hence(M αq,r g q (x)) 1/q M α,r g(x). Moreover, if α 1 and R 1, then M α,r g(x) R M g(x). y (26), Corollary 1, and the estimate above, we have for x E λ, λ>λ 0,that u # β (x) = u# β,1 (x) C( M (1 β)q,σ g q (x) ) 1/q C M 1 β,σ g(x) C M g(x). (27) y (27) and the weak type estimate for the maximal operator M, we have that ( {x } ) µ(e λ ) µ : M g(x) >Cλ Cλ 1 gdµ. (28) The claim follows because g L 1 () and the right-hand side of (28) tends to zero as λ. Now, we define the function v = v λ as a Whitney type extension of u to the set E λ by setting { u(x), if x \ E λ, v(x) = ϕ i (x)u 2i, if x E λ. We will select the open set O to be E λ for sufficiently large λ > λ 0.Hence,the claim 1 of Theorem 4 follows from the definition of v. Sincesupp u (x 0,1) and E λ (x 0,2) for λ>λ 0, the support of v is in (x 0,2). For the proof of 2 the Hölder continuity of v, we refer to the proof of [6, Theorem 5.3]. In the following, we denote x ={ i : x 2 i } for x E λ. y the bounded overlap of the balls 2 i, there is a bounded number of balls in x. Proof of 2 v M 1,1 () We have to show that v L 1 (), and that it has an integrable generalized gradient. Since v = u in \ E λ, to show the integrability of v, it suffices to estimate E λ v dµ.

14 J. Kinnunen, H. Tuominen y the properties of the functions ϕ i, the bounded overlap of the balls 2 i E λ,and the doubling property of µ, we have that v dµ u 2i dµ C u dµ C u dµ, (29) 2 i 2 i E λ E λ and hence v L 1 () with v L 1 () C u L 1 (). Concerning the gradient, we will show that the function g v, g v (x) = C ( g(x) + (M g q (x)) 1/q), x (30) belongs to D(v) L 1 (). We will consider four cases. (i) Since g D(u), v(x) v(y) = u(x) u(y) d(x, y)(g(x) + g(y)) d(x, y)(g v (x) + g v (y)) for almost all x, y \ E λ. (ii) If x, y E λ and d(x, y) δ, whereδ = 1/4 max { d(x, \ E λ ), d(y, \ E λ ) }, then by the properties of functions ϕ i, we have that ( v(x) v(y) = ϕi (x) ϕ i (y) )( ) u(x) u 2i C d(x, y) i u(x) u 2i, (31) x y r 1 where u(x) u 2i u(x) u (x,28ri ) + u (x,28ri ) u 2i (32) and 2 i (x,28r i ) by the properties of the Whitney covering. y (11) and the doubling property of µ, we have ( 1/q. u (x,28ri ) u 2i C u u (x,28ri ) dµ Cr i g dµ) q (x,28r i ) (x,28σ r i ) This together with the telescoping argument for u(x) u (x,28ri ) as in (13) shows that ( u(x) u 2i Cr i M g q (x) ) 1/q (33) for almost all x. Since the cardinality of x y is bounded, the estimates (31) (33) show that v(x) v(y) C d(x, y) ( M g q (x) ) 1/q C d(x, y)(gv (x) + g v (y)) (iii) for almost all x, y E λ with d(x, y) δ. Let x, y E λ with d(x, y) δ. Using the properties of the functions ϕ i,the fact that g D(u), similar estimates for u(x) u 2i and u(y) u 2i as in the previous case, and the properties of the Whitney covering to conclude that

15 Pointwise behaviour of M 1,1 Sobolev functions r i d(x, \ E λ ) for all i x (and similarly for y ), we have that v(x) v(y) x u(x) u 2i + y u(y) u 2i + u(x) u(y), C d(x, \ E λ ) ( M g q (x) ) 1/q + C d(y, \ Eλ ) ( M g q (y) ) 1/q + d(x, y) ( g(x) + g(y) ) c d(x, y) ( g v (x) + g v (y) ). (iv) If y E λ and x \ E λ,then v(x) v(y) = u(x) v(y) = ϕ i (y)(u(x) u 2i ) u(x) u 2i, (34) y where, by the assumption that g D(u), and by a similar calculation as for (32), we have u(x) u 2i u(x) u(y) + u(y) u 2i d(x, y) ( g(x) + g(y) ) ( + Cr i M g q (y) ) 1/q. (35) Since for i y, r i d(y, \ E λ ) and d(y, \ E λ ) d(x, y), (34) and(35) show that v(x) v(y) C d(x, y) ( g(x) + g(y) + ( M g q (y) ) 1/q) C d(x, y) ( g v (x) + g v (y) ) for almost all y E λ and x \ E λ. We conclude that the function g v is a generalized gradient of v. The integrability of g v follows similarly as that of g ur in the proof of Lemma 1; the assumption g L 1 () and the boundedness of the maximal operator for 1/q > 1, imply that (M g q ) 1/q dµ C (g q ) 1/q dµ = C gdµ. (36) Hence, g v L 1 () with g v L 1 () C g L 1 (). y choosing the generalized gradient g of u so that g L 1 () 2 u M 1,1 () and combining estimates (29)and(36), we have that v M 1,1 () C u M 1,1 (). Proof of 3 approximation in norm We will show that v u in M 1,1 () as λ. Using the fact that v = u in \ E λ and (29), we have that u v dµ = E λ u v dµ C E λ u dµ, which tends to 0 as λ because µ(e λ ) 0asλ.Hencev u in L 1 () as λ. Next, we have to find a generalized gradient g λ of u v for which g λ L 1 () 0 as λ. We claim that the function g λ = g v χ Eλ = C ( g + ( M g q) 1/q) χeλ

16 J. Kinnunen, H. Tuominen is in D(u v), that is, inequality (2) holds for u v and g λ almost everywhere in. If x, y \ E λ,thenu v = 0and(2) holds trivially. Inequality (2) for almost all x, y E λ follows because g D(u) and g v D(v). Ifx E λ and y \ E λ,then (u v)(y) = 0andg λ (y) = 0. Similar arguments as for (31) show that u(x) v(x) x u(x) u 2i C x r i ( M g q (x) ) 1/q C d(x, \ E λ ) ( M g q (x) ) 1/q C d(x, y) ( M g q (x) ) 1/q. Hence g λ D(u v). Sinceg v is in L 1 (), soisg λ, too. Moreover, as µ(e λ ) 0as λ, we have that g λ L 1 (), and hence also u v M 1,1 () tends to 0 as λ. Proof of (4) Hausdorff content of E λ Using (26), a similar estimate as in (27), and Corollary 1, we see that for λ>λ 0, E λ { x (x 0,2) : ( M (1 β)q,σ g q (x) ) 1/q } > Cλ { x (x 0,2) : M (1 β),σ g(x) >Cλ }. y a weak type estimate for the Hausdorff content of the fractional maximal function, we conclude that H s (1 β) (E λ ) Cλ 1 gdµ, where C 5 s (1 β) (2 diam((x 0,2))) s µ((x 0,2)) 1, see, for example [1, Lemma 3.2], [6, Lemma 2.6]. Hence H s (1 β) (E λ ) 0asλ. Step 2 General case. Let ε > 0. We cover by balls of radius 1/10, and use the 5r-covering theorem to obtain pairwise disjoint balls (a j,1/10) from this covering such that j=1 (a j,1/2) and that the balls (a j,2) have bounded overlap. Let (ψ j ) be a partition of unity for this covering such that j=1 ψ j (x) = 1forallx, each ψ j is L-Lipschitz, 0 ψ j 1, and supp ψ j (a j,1) for all j N. For u M 1,1 () with g D(u) L 1 (), we have u(x) = u j (x), (37) where u j = uψ j,forallx. y[6, Lemma 5.20], each u j is in M 1,1 () and j=1 g j = (g + L u )χ (aj,1) is a generalized gradient of u j.sincesupp u j (a j,1), the first step of the proof shows there are functions v j M 1,1 () and open sets O j (a j,2) such that (i) v j = u j in \ O j, supp v j (a j,2), (ii) v j is Hölder continuous with exponent β, (iii) u j v j M 1,1 () < 2 j ε, (iv) H s (1 β) (O j )<2 j ε, (v) h j = C(g j + (M g q j )1/q ) is a generalized gradient of v j.

17 Pointwise behaviour of M 1,1 Sobolev functions We define O = j=1 O j, and claim that the function v = j=1 v j together with the open set O satisfy requirements 1 4 of the theorem. For 1, letx \ O. Then, by (i) and (37), we obtain v(x) = v j (x) = u j (x) = u(x). j=1 The Hausdorff content estimate 4 for O follows from (iv) using the subadditivity of H s (1 β). y the Hölder continuity of v j, we have j=1 v j (x) v j (y) Cλ j d(x, y) β for all x, y. Since, by the proof of the Hölder continuity, the constant λ j depends on ε and on j, the Hölder continuity of the functions v j and the fact that supp v j (a j,2) give Hölder continuity of v only in bounded subsets of. To prove the first part of 2 and 3, we have to show that v M 1,1 () and that u v M 1,1 () <ε. y (iii), we have u j v j M 1,1 () < 2 j ε = ε, (38) j=1 that is, the series j=1 (u j v j ) convergences absolutely, and hence converges in the anach space M 1,1 (). Sinceu = j=1 u j is in M 1,1 (), also j=1 v j converges in M 1,1 (). Moreover, by (38) we have that u v M 1,1 () u j v j M 1,1 () <ε. This completes the proof of Theorem 4. j=1 Remark 3 Note that j=1 h j,whereh j is as in (v) above, is a generalized gradient of v but it is not necessarily integrable. If we would like to construct an integrable g v D(v), we need a cut-off function for each j; let j : [0, 1] be a Lipschitzfunction that equals 1 in (a j,2) and vanishes outside (a j,3). Since the support of v j is in (a j,2), v j j = v j in. Moreover, the function j=1 g vj = (h j + C v j )χ (aj,3) is in D(v j ) by [6, Lemma 5.20]. This together with the fact that supp v j (a j,2) and the bounded overlap of the balls (a j,2) shows that the function g v = j=1 is a generalized gradient of w. Integrability of v and g v follow since supp v j (a j,2), supp g vj (a j,3) and the balls (a j,3) have bounded overlap, using integral estimates from the first part of the proof. g vj

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