NULL SETS FOR DOUBLING AND DYADIC DOUBLING MEASURES

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1 Annales Academiæ Scientiarum Fennicæ Series A. I. Mathematica Volumen NULL SETS FOR DOUBLING AND DYADIC DOUBLING MEASURES Jang-Mei Wu University of Illinois at Urbana-Champaign Department of Mathematics 73 Altgeld Hall MC West Green Street Urbana IL U.S.A. Abstract. In this note we study sets on the real line which are null with respect to all doubling measures on R or with respect to all dyadic doubling measures on R. We give some sufficient conditions for the former a test for the latter some examples. Our wor is motivated by a characterization of dyadic doubling measures by Fefferman Kenig Pipher [5] by a result of Martio [8] on porous sets sets of total A -harmonic measure zero for certain class of nonlinear A -operators. A measure µ on R is said to have the doubling property with constant λ if whenever I J are two neighboring intervals of same length then µi λµj; denote by Dλ the collection of all doubling measures with constant λ D = λ 1 Dλ. A measure µ on R has the dyadic doubling property with constant λ if µi λµj whenever I J are two dyadic neighboring intervals of same length I J is also a dyadic interval; denote by D d λ D d the corresponding collections of dyadic doubling measures. Given {a n } 0 < α n < 1 a set E R is called {α n }-porous if there exists a sequence of coverings E n = {E nj } of E by intervals with mutually disjoint interiors so that each E nj \ E contains an interval J nj of length α n E nj En+1 E n+1 is contained in En E nj \ J nj sup E nj 0 as n. The Cantor ternary set is { 1 3 }-porous. The porous sets studied by Martio [8] are {α}-porous for some α > 0. 1 αk n Theorem 1. If 0 < α n < 1 = for all K 1 E is {α n }-prorous then E is null for all doubling measures on R. Corollary. There exist sets of Hausdorff dimension one which are null for all doubling measures. The condition given in Theorem 1 cannot be improved: Theorem. If 0 < α n < 1 4 is a decreasing sequence satisfying n αk n < for some K 1 then there exists a perfect set which is {α n }-porous but carries a positive measure for some µ D Mathematics Subject Classification: Primary 8A1. Research partially supported by the National Science Foundation.

2 78 Jang-Mei Wu Let E be a closed set in [0 1] λ > 1. In Theorem 3 we give a deterministic procedure of testing whether E is D d λ-null. In this process an optimal measure µ Eλ among D d λ is selected for E. The precise statement is given in Section. Denote by N the collection of null sets for doubling measures { E : µe = 0 for all µ D } N d its dyadic counterpart { E : µe = 0 for all µ D d }. Clearly N d N N is translation invariant. The assertion that N d N is not suprising however it requires a lot of wor. Theorem 4. There exists a perfect set S [0 1] which is in N \ N d. And corresponding to this S there exists a set T of dimension one so that t+s N d for each t T. It would be interesting to now whether a pair of sets S T can be chosen to satisfy length T > 0 in addition to the properties in Theorem 4. Theorem 5. Let t be any number whose binary expansion has infinitely many zeros infinitely many ones. Then there exists a perfect set S t so that S t N \ N d but t + S t N d. Finally in Section 4 we shall comment on relations betwen sets in N null sets of the harmonic measures with respect to the p-laplacians in the upper half plane. The author would lie to than A. Hinanen for his joint contribution on Theorem 1 R. Kaufman for many conversations. We first state a useful lemma. 1. Proofs of Theorems 1 Lemma 1. Let µ be a dyadic doubling measure on [0 1] I be any subinterval. Then there exists K > 1 depending on the dyadic doubling constant λ only so that 4 I 1/K µ [0 1] µi 1 4 I K µ [0 1]. Proof. Let I 1 I be two adjacent dyadic closed intervals in [0 1] such that I I 1 I I I 1 + I. Then µi µi 1 + µi λ log I 1 λ log I + µ [0 1] 1 + λ 1 + λ I log 1+λ/λ µ [0 1] 4 I log 1+λ/λ µ [0 1]. If I 1 16 log 1+λ/λ 1 A

3 then Null sets for doubling dyadic doubling measures 79 µ [0 1] \ I 8 1 I log 1+λ/λ µ [0 1] 1 µ [0 1]. Hence µi 1 µ [0 1]. If I < A let J be the largest dyadic interval contained in I. Therefore J I /4 µi µj 1 log J log 1+λ µ [0 1] I /4 µ [0 1] 1 + λ µ [0 1] I log 1+λ1 log A 1. Proof of Theorem 1. Assume E [0 1] let E n = {E nj } be the coverings of E {J nj } be the subintervals of E nj \ E in defining {α n }-porosity. Let µ D it follows from Lemma 1 that µj nj 1 4 αk n µe nj for some K 1 depending on µ only. Thus µe nj \ J nj αk n µe nj. Summing over j we obtain µe n αk n j µe nj. Therefore µe n αk n µ [0 1] = 0. Proof of Theorem. Let N n be a rapidly increasing sequence of odd integers with N 1 = 1 N n α 1 n 1 for n. After replacing α n by a number which is at most twice its size we may assume that α n = m n Nn+1 1 for some odd integer m n. The construction of E resembles that of the Cantor set. First we remove the open interval which constitutes the middle α 1 position of [0 1] subdivide the two remaining closed intervals into subintervals of equal length N 1 call this collection of subintervals S 1. This subdivision is possible due to the modification on α n s. On each interval in S 1 remove the open interval which constitutes its middle α portion subdivide the remaining intervals into subintervals of equal length N N 3 1 call this new collection of subintervals S. Continue the process indefinitely let E = n I S n I Clearly E is {α n }-porous. It remains to choose N n so that µe > 0 for some µ D. Our idea comes from Ahlfors Beurling [; Theorem 3]. First we construct a function h which plays the role of 1 + λcosine in []..

4 80 Jang-Mei Wu Lemma. Given 0 < α < 1 4 K > there exists a function h continuous on R of period 1 monotonic in [0 1 ] in [1 1] respectively which satisfies hx dx = hx = { α K 1 on [ 1 1 α α] 1 + α on [0 1 α] [ 1 + α 1] hx dx is in DB K for some absolute constant B >. As an example we may choose on piecewise linear on hx = α K 1 + α K x 1 + α K 1 [ 1 + α 1 + α [ 1 + α + α 4 + α 4 K 1/4K 1 ] K 1/4K ] α with derivatives between 1/4 α 4/ α hx = h1 x for x in [ 1 α 1 1 α] so that the continuity monotonicity 1 hx dx = 1 are 0 satisfied. For this h h dx DB for some absolute constant B >. In the hypothesis α K n < we may assume K >. Corresponding to each pair α n K we fix a function h n which satisfies properties in Lemma with α = α n. Denote by A n = F = = I S I M n = [ α n] [ α n + 1] n N f n x = 1 n h M x. We shall choose N n inductively so that N n+1 N n that f n x is nearly constant on each interval of length M 1 n+1. Recall that N 1 = 1 assume that odd integers N N 3... N n have been chosen so that 1.1 F f x dx 1 α K j 1 n j=1 1

5 Null sets for doubling dyadic doubling measures 81 whenever x x M 1 n < h 1 M 1 x/h 1 M 1 x < < f 1 x/f 1 x < + 1. Note that 1 0 h x dx = 1 [01] A h x dx = 1 α K that f x is uniformly continuous on R for each 1. Let F [0 1] be any measurable set. Then χ F f n is the pointwise a.e. L 1 limit of an increasing sequence of simple functions each of which has the form a j χ Ij where {a j } are constants {I j } are finitely many mutually disjoint open intervals with rational end points. Therefore 1 χ F xf n xχ An+1 Mxh n+1 Mx dx 1 α K n+1 f n x dx 0 as M. Thus a large odd integer N n+1 can be found so that N n+1 > α 1 n hold with = n + 1. Here we have used the fact that F n+1 = {x F n : M n+1 x A n+1 }. Let µ be a wea limit point of f n x dx. Then µe > 0 in view of 1.1 α K n <. To verify that µ is a doubling measure we consider two neighboring intervals I I satisfying In view of 1.3 n 1 M 1 n+1 I = I M 1 n. n f n 1 x n + 1 f n 1 x n whenever x I x I. We note that f m x/f n x has period Mn+1 1 if m n + 1 that n n + 1 < h nm n x h n M n x < n + n + 1 whenever x x M 1 n+1. Writing f m x = f n 1 xh n M n x f mx for m n + 1 f n x we deduce from the fact h n M n x dx DB K that / CB K 1 f m x dx f m x dx CB K I I for every m n + 1. Therefore µ D. F

6 8 Jang-Mei Wu. A test for D d λ-null sets Given a closed set E in [0 1] we shall develop a procedure to test whether E is in D d λ for some λ > 1. Let I nj 1 j n be the dyadic closed intervals in [0 1] of length n Inj r be the closed interval which forms the right half of I nj let Inj l = I nj \ Inj r. Let 1 on Inj r h nj x = 1 on Inj l 0 on R \ I nj. For a fixed integer n define dν n n f n n n j=1 1 + δn jτhnj f n n dx where τ = λ 1/λ + 1 δn j = { 1 E I r nj 1 E Inj r =. Denote by E n = { I n+1j : I n+1j E 1 j n+1}. Let dν n n 1 fn n 1 dνn n δn 1 j = n 1 f n n 1 where j=1 1 + δn 1 jτhn 1j { n 1 ν n In 1 r E n ν n n In 1 l E n 1 otherwise. After defining f n dν n we let dν n 1 fn 1 dνn 1 f n 1 where j=1 1 + δ 1 jτh 1j { n δ 1 j = 1 ν Ir 1j E n ν n Il 1j E n 1 otherwise.

7 Null sets for doubling dyadic doubling measures 83 Continue this process until we arrive at ν n 1. Define µ n ν n 1. Notice that.1 ν n I j = 1 n with the understing that I 0j I 01 [0 1]; that from dν n to dν n 1 total measure in each I 1i is ept unchanged but the total measures of I 1i r I 1i l are redistributed in the most advantageous way. Repeat for each n to obtain a sequence of measures {µ n }. Let µ Eλ be a wea limit point of {µ n } extended to R with period 1. Theorem 3. Among all the measures in D d λ which have mass one on [0 1] µ Eλ has the maximum measure on E. In particular E is D d λ-null if only if µ Eλ E = 0. Proof. It is clear that µ Eλ [0 1] = 1 µeλ D d λ. Let ω be any measure in D d λ with ω [0 1] = 1. We claim in fact that. ωe n µ n E n. Let m be the largest integer in [1 n] if it exists such that there exists at least one interval I mj on which.3 ωimj r ωimj l µ nimj r µ n Imj l. If such m does not exist then ωe n = µ n E n. We shall redistribute the measure ω on these I mj s eep ω unchanged elsewhere. Denote by I m = {I mj :.3 holds on I mj }; let ω m = ω on [0 1] \ Im I mj ωi mj µ n Imj r ωimj r.4 dω m = µ ni mj dω on Ir mj ωi mj µ n Imj l ωimj l µ ni mj dω on Il mj for each I mj I m. Clearly if I mj I m then ω m I r mj ω m I l mj = µ ni r mj µ n I l mj ω m is dyadic on I mj with constant λ. Actually ω m is in D d λ by the following lemma.

8 84 Jang-Mei Wu Lemma 3. Let I be a dyadic subinterval of [0 1] µ ν be dyadic doubling measures on [0 1] I respectively satisfying µi = νi. Then the new measure ω defined by ω ν on I µ on [0 1] \ I is dyadic doubling on [0 1] with constant bounded by the maximum of those of µ ν. First we shall verify that ωe n ω m E n. To show this it is enough to prove.5 ωe n I mj ω m E n I mj for each I mj I m. Fix I mj I m clearly.5 holds when m = n. Thus we assume m < n note from the definition of m that for m + 1 n. Therefore ωii r ωi i = µ nii r µ n I i ωii l ωi i = µ nii l µ n I i Moreover from the construction of µ n.6 ωimj r E n ωimj r = µ nimj r E n µ n Imj r ωi l mj E n ωi l mj = µ ni l mj E n µ n I l mj. µ n I n+1l µ n I i = νn I n+1l ν n I i if 1 n I n+1l I i. Thus by.1 the above identities.7 ωi r mj E n ωi r mj = ν n m+1 Ir mj E n m+1.8 ωi l mj E n ωi l mj = ν n m+1 Il mj E n m+1. Writing I in place of I mj for the rest of this paragraph we obtain [ ωen I r ωi r ωe n I = ωi r ωi + ωe n I l ωi l ] ωi ωi l ωi [ = m+1 ωi ν n m+1 E n I r ωir ωi + νn m+1 E n I l ωil ] ωi m+1 ωi ν n m+1 E n I r A + ν n m+1 E n I l 1 A

9 Null sets for doubling dyadic doubling measures 85 where A = 1 1+τ if νn m+1 E n I r ν n m+1 E n I l A = 1 1 τ otherwise. From the definition of ν m n it follows that [ ωe n I m+1 ωi ν n m+1 E n I r νn m I r ν n m I + νn [ ωen I r µ n I r = ωi ωi r µ n I + ωe n I l ωi l This proves.5 hence ωe n ω m E n. m+1 En I l νn m I l ] ν m n I µ n I l ] = ω m E n I. µ n I We proceed to mae modifications of ω m on each dyadic interval I m 1i of size m+1 on which.3 holds with m j ω replaced by m 1 i ω m respectively according to the rule.4 adapted for m 1 i ω m ; call this new measure ω m 1. Continue to modify ω m 1 on dyadic intervals of size m+ if necessary to obtain ω m.... Finally we arrive at a measure ω 1 obtain ωe n ω m E n ω m 1 E n ω 1 E n ω 1 Imj r ω 1 Imj l = µ nimj r µ n Imj l for all 1 m n 1 j m. Therefore ω 1 E n = µ n E n. is proved. We note that E = m E m. Therefore for any ε > 0 sufficiently large m n with m > mε n > nε m we have µ Eλ E µ Eλ E m ε µ n E m ε µ n E n ε This shows that ωe µ Eλ E. ωe n ε ωe ε. 3. Proofs of Theorems 4 5 Given a ε δ 0 1 εa < δ < ε we choose a sequence of integers {n } satisfying n n n+1 > n + [εlog ]. For + [ 1/δ ] 0 j n 1 denote by [ j L j = [ j I j = n n j n j + 1 ] n + 1 n δ ]

10 86 Jang-Mei Wu [ j + 1 J j = n 1 j + 1 ] n ε n where δ = [δ log ] ε = [εlog ] [ ] is the greatest integer function. Note that intervals L s I s J s are dyadic 3. J j / I j = O δ ε = o1 as 3.3 J j / L +1j = n +1 n [εlog ] >. The construction of a set S N \N d is similar to the Cantor set; collections of nested intervals from {J j } are used. The measure µ in D d to be produced with µs > 0 will satisfy µj j µl j = Jj a L j on infinitely many J j s. Let {K i } be an increasing sequence of integers with K 0 +[ 1/δ ] some other properties to be specified later. Let S 0 = [0 1] C1+K I 0 be the collection of all I 1+K0 j S 0 C1+K J 0 be the collection of all J 1+K0 j S 0. After C I C J have been defined for some 1 + K 0 K 1 1 we let C+1 I = C I { } I +1j S 0 : I +1j is not contained in any interval in C I C J C+1 J = C J { J +1j S 0 : J +1j is not contained in any interval in C I C J } ; let S I 1 = union of all intervals in C I K 1 S 1 = union of all intervals in C J K 1. Next let C1+K I 1 be the collection of all I 1+K1 j S 1 C1+K J 1 be the collection of all J 1+K1 j S 1. And define for each 1 + K 1 K 1 CK+1 I = C I { } I +1j S 1 : I +1j is not contained in any interval in C I C J CK+1 J = C J { J +1j S 1 : J +1j is not contained in any interval in C I C J } S I = union of all intervals in C I K S = union of all intervals in C J K. Clearly S I S 1 S S 1. Continue this procedure to obtain CK I 3 CK J 3 S3 I S 3... so on let S = S m. 1 To construct µ D d with µs > 0 we shall use scale invariant versions of Lemma 3 the following lemma repeatedly.

11 Null sets for doubling dyadic doubling measures 87 Lemma 4. Given a α β 0 1 with α a + β < 1/16 c 1 c 1 there exists a measure µ D d 10 1/a which satisfies µ [0 1] = 1 µ [0 α] = c 1 α a µ [1 β 1] = c β. As an example we may choose µ [0 t] c 1 t a 0 t t 0 1 1/a 8 = 1 8 c c 1 + c t t 0 / 7 8 t 0 t 0 t c t + 1 c 8 t 1. Then extend µ periodically to R with period 1. All measures µ defined below are periodic with period 1. Choose µ 1+K0 D d 10 1/a so that µ 1+K0 L 1+K0 j = L 1+K0 j µ 1+K0 I 1+K0 j = L 1+K0 j 1 + K 0 δ µ 1+K0 J 1+K0 j = L 1+K0 j 1 + K 0 εa for each 0 j 1+K 0 1. After µ is selected for some 1 + K 0 K 1 we choose µ +1 D d 10 1/a so that µ +1 = µ on each interval in C I C J µ +1 is a redistribution of µ on each L +1j which is not contained in any interval in C I C J : 3.4 µ +1 L +1j = µ L +1j 3.5 µ +1 I +1j = 1 + δ µ +1 L +1j 3.6 µ +1 J +1j = 1 + εa µ +1 L +1j. The measure µ K1 so chosen has the properties that because εa < δ. µ K1 S I 1 S 1 = 1 K 1 1+K 0 1 εa δ µ K1 S 1 µ K1 S1 I S εa 1 inf 1+K 0 K 1 εa + δ 1 K 1 1 εa 1 K0 εa δ 1+K 0

12 88 Jang-Mei Wu Next choose µ 1+K1 D d 10 1/a so that µ 1+K1 = µ K1 on S 0 \ S 1 on each L 1+K1 j S 1 it is a redistribution of µ K1 satisfying with = K 1. After µ is constructed for some 1 + K 1 < K build µ +1 from µ following the same steps as in the case 1 + K 0 K 1. The dyadic doubling measure µ K so obtained belongs to D d 10 1/a moreover µ K S I S = 1 µ K S µ K S I S εa inf 1+K 1 K εa + δ µ K S I S 1 K εa δ 1 1 K 1 1 εa 1 1+K 0 K 1 εa δ µ K1 S 1 1+K 1 K 1 εa 1+K 1 1 K0 εa δ 1 K1 εa δ. Whenever µ Km is constructed eep µ 1+Km = µ Km on S 0 \ S m redistribute the mass on each L 1+Km j S m according to with = K m eep the dyadic doubling constant bounded by 10 1/a. Continue this indefinitely. Thus we obtain a sequence of measures µ Km D d 10 1/a with µ Km [0 1] = 1 µ Km S m m 1 i=0 1 K εa i δ1 A i where A i = K 1+i 1+K i 1 εa. Let µ be a wea limit point of {µ Km }. Clearly µ D d 10 1/a. Since εa < 1 it is possible to choose {K i } so that 3.7 i=1 K εa δ i + A i < +. i=1 With respect to this choice of {K i } we have µs > 0 hence µ N d. It remains to show that S N. Let ν D. Recall that J j I j+1 have the common boundary point j + 1/ n ; by the doubling property νj j I j+1 A ε δlog 5 νj j for some A > 1 depending only on the doubling constant of ν. For m intervals in C J K m {I j+1 : J j C J K m } C J K m C I K m may meet in their

13 Null sets for doubling dyadic doubling measures 89 interiors; however because of 3.3 every point in [0 1] is covered by at most three such intervals. Therefore 3ν [0 1] νj j I j+1 A ε δlog Km 1 5 νs m J j C J Km A ε δlog K m 1 5 νs. Hence νs = 0. Therefore S N \ N d. Let { T = t = t n n where t n = 0 or 1 n=1 but t n +[δ log ]+1 = 1 t n +[δ log ]+ = 0 for each integer > K 0 }. In view of 3.1 it has Hausdorff dimension 1. Fix t T ν D d let J j be any interval in C J K m. We note that p + 1 < t + j + 1 n δ n for some integer p because < p n δ q + 1 < tn δ < q for some integer q. Therefore t + J j is contained in the middle half of some dyadic interval [ p M j = p + 1 ]. n δ n δ Recall that the interval I j+1 shares an end point j + 1/ n with J j has length 1/ n δ. Therefore t + Ij+1 M j > n δ The dyadic doubling property of ν 3. Lemma 1 ν t + J j I j+1 M j c ννt + Jj with c ν as. Summing over all J j in CK J m before we obtain 3ν [0 1] ck m 1 ννt + S m ck m 1 ννt + S. reasoning as Letting m we have νt + S = 0. This completes the proof of Theorem 4. It would be interesting to characterize those t s so that t + S is in N d. However this seems difficult.

14 90 Jang-Mei Wu To prove Theorem 5 we note that in the binary expansion of t the event that a digit 1 is followed immediately by a digit 0 occurs infinitely often. Choose ε δ a as in Theorem 4 {n } depending on t so that 3.1 t n +[δ log ]+1 = 1 t n +[δ log ]+ = 0 hold for each > 0. Let S t S in Theorem 4 associated with this sequence {n }. Then S t N \ N d. The proof of t + S t N d is similar to that in Theorem Null sets for p-harmonic measures Consider the p-laplace equation 1 < p < div u p u = 0 in the half plane Ω {x R : x > 0}. For the definition properties of p-harmonic measure the harmonic measure for p-laplacian see [6; Chapter 10]. Let E be a compact set on Ω which has positive p-harmonic measure for some p. Then there exists a nonconstant solution u 0 u 1 of the p- Laplacian in Ω with continuous boundary value 0 on Ω \ E. Following [1] we may apply a linearization technique in [7] or an approximation technique in [4] Theorem 4.5 in [3] to write ux = Kx yfy dωy Ω where K is a limit of ernel functions ω is a wea limit of harmonic measures at a fixed point corresponding to a sequence of uniformly elliptic operators of nondivergence form in Ω with ellipticity constants depending only on p. Moreover ω has the doubling property u has nontangential limit on Ω ω-a.e. Because u has zero boundary value on Ω \ E fy dωy is supported in E. This implies that ωe > 0. Therefore we have Theorem 6. Compact sets in N are null sets for any p-harmonic measure with respect to the half plane {x R : x > 0}. Remar. Martio has defined a version of porosity proved that a porous set on {x = 0} has zero A -harmonic measure with respect to all those nonlinear operators A on {x > 0} considered in [8]; p-laplacians are examples of such operators. We do not now whether Theorem 6 can be extended to all such A -operators. However a compact set E on {x = 0} is A -harmonic measure null for all such A if it satisfies a stronger {a n }-porous condition for some {a n } α K n = for all K > 1 namely in defining {a n }-porosity E E nj is required to lie in the middle 1 α n portion of E nj for each n j. Proof follows by combining the original proof of Martio that of Theorem 1.

15 Null sets for doubling dyadic doubling measures 91 References [1] Avilés P. J. Manfredi: On null sets of p-harmonic measures. - Preprint. [] Beurling A. L. Ahlfors: The boundary correspondence under quasiconformal mappings. - Acta Math [3] Carffarelli L. E. Fabes S. Mortola S. Salsa: Boundary behavior on nongative solutions of elliptic operators in divergence form. - Indiana Univ. Math. J [4] Fabes E. N. Garofalo S. Marín-Malave S. Salsa: Potential theory Fatou theorems for some nonlinear elliptic operators. - Rev. Mat. Iberoamericana [5] Fefferman R.A. C.E. Kenig J. Pipher: The theory of weights the Dirichlet problem for elliptic equations. - Ann. Math [6] Heinonen J. J. Kilpeläinen O. Martio: Nonlinear potential theory of solutions of degenerate elliptic equations. - Oxford University Press to appear. [7] Manfredi J. A. Weitsman: On the Fatou-theorem for p-harmonic functions. - Comm. Partial Differential Equations 13: [8] Martio O.: Harmonic measures for second order nonlinear partial differential equations. - In: Function spaces differential operators nonlinear analysis L. Päivärinta editor Longman Scientific Technical Publishers Received 17 December 1991