arXiv:1503.08842v1 [math.CA] 30 Mar 2015 3 rpryta h atritraso ie ee btntneces not (but level given a of intervals Cantor the that property iest ntegoer.(e 1 o h ento ftegenera the of definition in the sets for Cantor-like includes (1) generalization (See our example, geometry. For the in diversity o´ nrltdthe Mor´an related etr in vectors xml with example n[o8]Mra nrdcdtento fasmset, sum a of notion the Mor´an introduced 89] [Mo In Introduction 1 Keywords: h e falpsil usm fteseries the of subsums possible all of set the eateto ahmtc,UiesddNcoa eMrdel Mar de Nacional Universidad Mathematics, of Department akn n asoffmaue fCno sets Cantor of measures Hausdorff and Packing 3 2 1 2 nti ae,w eeaieMra’ u e oint emtagrea a permit to notion set Mor´an’s sum generalize we paper, this In M ujc lsicto:2A8 28A80. 28A78, Classification: Subject AMS atal upre yCONICET by supported Partially NSERC by supported 44597 Partially NSERC by supported Partially eateto ahmtc n ttsis cdaUniversi Acadia Statistics, and Mathematics of Department 1 eateto ueMt,Uiest fWtro,Wtlo O Wateloo, Waterloo, of University Math, Pure of Department bu the about ftersubsets. their of esuyagnrlzto fMra’ u es bann in obtaining sets, Mor´an’s sum of generalization a study We R p .Hare K. asoffmaue akn esr,dmnin u set sum , measure, packing measure, Hausdorff ihsmal om.TecasclCno idetidsti one is set middle-third Cantor classical The norms. summable with a i h Hudr and -Hausdorff 3 = h Hudr esr of measure -Hausdorff − i soitdwt series with associated 1 .Asmn utbesprto odto,i M 94] [Mo in condition, separation suitable a Assuming 2. C a = .Mendivil F. ( h uy1,2018 18, July cta Canada Scotia, pcigmaue fteest n certain and sets these of measures -packing X i ∞ =1 Argentina. Abstract ε i a 1 i : C P ε a i 2 0 = otequantities the to a n where , 1 ) .Zuberman L. , a ( = y ofil,Nova Wolfville, ty, lt,MrdlPlata, del Mar Plata, a n R R sasqec of sequence a is ) t,Canada nt., n hc aethe have which aiytegaps) the sarily formation = 3 P lization.) i>n k a ter i k . are all of the same length. Moreover, unlike Mor´an’s sets, our generalized sum sets can have Hausdorff dimension greater than one. We obtain the analogue of Mor´an’s results on h-Hausdorff measures for these generalized sum sets and prove dual results for h-packing measures. We show that for any of these sum sets there is a doubling dimension function h for which the sum set has both finite and positive h-Hausdorff and h-packing measure. We give formulas for the Hausdorff and packing , and show that given any α less than the Hausdorff dimension (or β less than the ) there is a sum subset that has Hausdorff dimension α (or packing dimension β). In fact, there is even a sum subset with both Hausdorff dimension α and packing dimension β provided α/β is dominated by the ratio of the Hausdorff dimension to the packing dimension of the original set. Furthermore, if the Hausdorff and/or packing measure is finite and positive (in the corresponding dimension), then we can choose this sum subset to have finite and positive Hausdorff and/or packing measure.

2 Preliminaries N N Let sn > 0 with n sn < . Fix N and for each n let the nth digit n n n ∞n p ∈ ∈ set = 0= d , d ,...,d R be given. We define Cs,D by D { 1P 2 N }⊂ ∞ i C D = s b : b , (1) s, { i i i ∈ D } i=1 X n the set of all possible sums with choices drawn from and scaled by sn. D i Mor´an’s sum set is the special case when si = ai , N = 2 and = 0,ai/ ai . This generalized sum set is the main object of|| study|| in this paper.D { || ||} For each n define

κ = max dn dn :0 i, j N,i = j n {k i − j k ≤ ≤ 6 } and τ = min dn dn :0 i, j N,i = j . n {k i − j k ≤ ≤ 6 } In Mor´an’s case, κ = τ = 1. We assume that κ := sup κ < , as well as n n n n ∞ τ := inf τn > 0; the intent is that the sequence sn controls the decay rate, not the (possibly varying) geometry of the digit sets n. In addition, we assume the rapid decay condition D κR sup n = M < 1, (2) n τsn where Rn = i>n si. This is the analogue of Mor´an’s separation condition. The quantity Rn is very important for describing the geometry of Cs,D. In certainP situations where we have precise information about the geometry of n, it is possible to assume something weaker than (2) and still have a suitable separationD property to allow for dimensions to be calculated; see Example 8.

2 Example 1. 1. A very simple example is the classical Cantor set with sn = 2 3−n and = 0, 1 . · D { } 2. Consider a finite set Rp, a real number r

′ i i Φ(σ) Φ(σ ) = s (d d ′ ) k − k k i σ(i) − σ (i) k i X n n i i s (d d ′ ) s (d d ′ ) ≥ k n σ(n) − σ (n) k−k i σ(i) − σ (i) k i>n s τ κR > 0. X (3) ≥ n − n This means that Φ is injective and is thus a homeomorphism, so that Cs,D is also totally disconnected and perfect. For a given n N and σ Ξn, we define ∈ ∈ i xσ = sidσi ≤ Xi n and C = x + s b : b i . σ,n σ { i i i ∈ D } i>n X Our condition (2) ensures the non-overlapping of the sets Cσ,n. Using this notation, we see two very important facts. First, Cσ,n = xα n − xσ + Cα,n for any σ, α Ξ . That is, for a fixed n the collection of Cσ,n are all ∈ n translates of each other. Secondly, we can decompose Cs,D into N copies of Cσ,n as n C D = C = x : σ Ξ + C , s, σ,n { σ ∈ } 1,n ∈ n σ[Ξ

3 where by 1 Ξn we mean the element all of whose terms equal to 1. An elementary∈ estimate gives that

C s b s b′ κ s = κR (4) | σ,n|≤k i i − i ik≤ i n i>n i>n i>n X X X where C means the of the set C. | | 3 Hausdorff and packing measures

We first recall some facts about Hausdorff and packing measures (see [Ro 98, Ma 95]). For us, a dimension function is a continuous non-decreasing function h : [0, ) [0, ) with h(0) = 0. It is said to be doubling if there is some constant∞ c>→0 so∞ that h(2x) ch(x) for all x> 0. For two dimension function≤ f,g we say that f g if ≺ lim g(t)/f(t)=0. t→0+

For each δ > 0, a δ-covering of a set E is a countable collection Bi of p { } subsets of R with dominated by δ, that is Bi δ, and for which E B . We define | | ≤ ⊆ ∪i i h(E) = inf h( B ): B is a δ-covering of E Hδ { | i| { i} } i X and the Hausdorff h-measure as

h h (E) = lim δ (E). H δ→0 H

h Notice that in the definition of δ it is sufficient to consider coverings by balls. Now we turn to the h-packingH measure h. A δ-packing of a set E is a P disjoint family of open balls B(xi, ri) with xi E and ri δ. The h-packing pre-measure is given by { } ∈ ≤ h h 0 (E) = lim δ (E) P δ→0 P where h = sup h( B ): B is a δ-packing of E . Pδ { | i| { i} } i X h Unfortunately 0 is not a measure as it is in general not countably additive. Thus we need oneP more step to construct the packing measure h, P h(E) = inf h(E ): E E . P { P0 i ⊂ i} i i X [ The next Theorem gives estimates for the Hausdorff and packing measures of Cs,D. The first two claims about the Hausdorff measure of Cs,D are given in [Mo 94] for the special case of n containing two digits. D

4 Theorem 2. Suppose that h is a doubling dimension function.

n h 1. If lim inf N h(κR )= α then (C D) α. n H s, ≤ n h 2. If lim inf N h(κR )= α> 0 then (C D) > 0. n H s, n h 3. If lim sup N h(κR )= α< then (C D) Nα. n ∞ P s, ≤ n h 4. If lim sup N h(κR )= α> 0 then (C D) > 0. n P s, n h Remark 3. If h is doubling and 0 < lim inf N h(Rn) < , then 0 < (Cs,D) < ∞ n H and so Cs,D is an h-Hausdorff set. Similarly, if lim sup N h(Rn) is positive ∞ n and finite, then Cs,D is an h-packing set. Finally, if lim inf N h(Rn) is pos- n itive and lim sup N h(Rn) is finite, then Cs,D is both an h-Hausdorff and an h-packing set.

n Proof. Item 1) is trivial by considering the covering CI,n for all I 0, 1,...,N , n ∈{ } which consists of N sets all of diameter at most κRn. To prove the rest of the statements, we will use the fact that there is a Borel −n measure µ supported on Cs,D for which µ(Cσ,n)= N for each σ and n. This measure is often called the natural probability measure. n 2): Let β < α so that we have N h(κRn) > β for all large n. Now choose x C D and δ > 0 and let n be such that κR < δ κR − . By a simple ∈ s, n ≤ n 1 modification of Lemma 2 in [Mo 89] there is a q N so that the number of Cσ,n which intersect B(x, δ) is less than q (independent∈ of B and δ). (This is where the condition (2) is used.) But then we have

q qh(δ) µ(B(x, δ)) qµ(C )= qN −n < h(κR ) < . ≤ σ,n β n β

h By the mass distribution principle (see [Fal 86]), we have (C D) α/q. H s, ≥ n 3) Let β > α so that we have N h(κRn) < β for all large n. Now choose x C D and δ > 0 and let n be such that κR < δ κR − . We know ∈ s, n ≤ n 1 that x Cσ,n for some σ and, since Cσ,n κRn < δ, we have that Cσ,n B(x, κR∈ ) B(x, δ). But then | | ≤ ⊆ n ⊆ N −(n−1) h(κR ) h(δ) µ(B(x, δ)) µ(C )= N −n = > n−1 , ≥ σ,n N Nβ ≥ Nβ since h is a nondecreasing function. But then we have that

lim inf µ(B(x, δ))/h(δ) (Nα)−1 ≥ h and so (C D) Nα by Theorem 3.16 in [C 95]. P s, ≤

nj 4) Let 0 <β<α. Then there are nj so that N h(κRnj ) >β for all j. Let x Cs,D be given. For any j we have x Cσj ,nj for some σj . By the same simple∈ modification of Lemma 2 in [Mo 89],∈ there is a q N so that for any ∈

5 δ > 0 and any ball B of radius δ, if m N is the smallest value with κR <δ ∈ m then the number of CI,m which intersect B is less than q (independent of B and

δ). Let δ = κRnj −1, so κRnj <δ = κRnj −1. Then

µ(B(x, κR )) µ(B(x, δ)) qµ(C )= qN −nj 0, let n,m N be such that κRm+1 < x κRm < κRn ∈ τM ≤ ≤ 2x<κR − . Then κR τMs κ R − for all i. Letting θ = τM/κ < 1, n 1 i ≤ i ≤ κ i 1 κR 2x θn−m n < =2 ≤ κRm x and so we have m n ln(2)/ ln(θ). As f(N j )= κR , − ≤− j h(2x) f −1(x) N m+1 = N 2−ln(2)/ ln(θ), h(x) f −1(2x) ≤ N n−1 ≤ and so h is doubling. n Since N h(κRn) = 1 for all n, we have Cs,D is an h-Hausdorff set and an h-packing set for this dimension function h, as desired. The next theorem is a simple consequence of some known results. However, it shows that the set of dimensional subsets of Cs,D is an initial segment in the partially ordered set of all doubling dimension functions. Theorem 6. Let f,h be doubling dimension functions and assume f h. ≺ h 1. If 0 < (Cs,D) < , then for any t > 0 there is a compact and perfect H ∞ f subset E C D so that (E)= t. ⊂ s, H h 2. If 0 < (Cs,D) < , then for any t > 0 there is a compact and perfect P ∞ f subset E C D so that (E)= t. ⊂ s, P

6 f Proof. 1. From Theorem 40 in [Ro 98], we have that (Cs,D)= . Then by ′ H ∞ f ′ Theorem 2 in [La 67] there is some closed subset E Cs,D for which (E )= t. As E′ is a closed subset of a perfect set, it is the union⊂ of a perfectH set E and f f ′ a countable set, so (E)= (E )= t and E is a perfect subset of Cs,D. 2. By the same argumentH H as Theorem 40 in [Ro 98], but adapted to packing f measures, we have that (Cs,D) = . Now, if we obtain a closed subset ′ f P′ ∞ E Cs,D for which (E )= t, then we find a perfect subset in a similar way to the⊂ case 1 before. InP [JP 95], Joyce and Preiss proved that if a set has infinite h-packing measure (for any given h ), then the set contains a compact subset with finite h-packing measure. With∈ D a simple modification of their proof, (in particular their Lemma 6), we obtain a set of finite packing measure greater than t. By Lyapunov’s convexity theorem, there is a subset whose h-packing measure is exactly t [Ru 91, Theorem 5.5].

s We now specialize to the “usual” dimension functions hs(x) = x and let dimH and dimP denote the “usual” Hausdorff and packing dimension. In anal- ogy with the case of a “cut-out” Cantor subset of R (see [BT 54, CMMS 04, GMS 07], we have the following Proposition. Proposition 7. We have that n ln(N) n ln(N) dimH (Cs,D) = lim inf − and dimP (Cs,D) = lim sup − . ln(sn) ln(sn) Proof. First, we note that n ln(N) n ln(N) n ln(N) lim inf − = lim inf − = lim inf − , ln(κRn) ln(Rn) ln(sn) with a similar equality for the limit superior. If β > α := lim inf −n ln(N) , then there is a subsequence (n ) so that ln(κRn) j nj β n β N (κRnj ) < 1. Thus lim inf N (κRn) < 1 and so dimH (Cs,D) α by Theorem 2. ≤ n γ Conversely, if γ < α, then for large n we have N (Rn) > 1 and thus n γ lim inf N (Rn) > 1 and so dimH (Cs,D) α by Theorem 2. The proof for packing dimension is similar.≥

p Example 8. For any α [0,p), it is possible to construct a sum set, Cs,D R , ∈ ⊂n with dim (C D) = α. The simplest way of doing this is to choose = H s, D (ǫ1,ǫ2,...,ǫp): ǫi 0, 1 , the set of all corners of a p-dimensional unit { ∈n { }} −p/α cube, and set sn = λ where λ = 2 . This will generate a self-similar set, Cs,D, that is a product of classical Cantor sets. The problem is that condition (2) requires that λ< 1/(1 + √p), which does not allow the full range of dimensions (and, in fact, gets worse as p increases). However, from the simple geometry of this example, we can see that the sets Cσ,n are non-overlapping provided sn > Rn. Under this (weaker) assumption, Cs,D is a self-similar set satisfying the open set condition and hence its dimensions are as stated in the previous proposition. This separation condition allows for any λ [0, 1/2). ∈

7 In the case of the “usual” dimension functions hs, Theorem 6 has a stronger form in that not only is there a Cantor subset with the correct dimension but this subset corresponds to all the subsums of a subsequence of (sn).

Theorem 9. Suppose that dimH (Cs,D) = A and dimP (Cs,D) = B. Then for any 0 α A and 0 β B, with α/A β/B, there is a subsequence (t ) of ≤ ≤ ≤ ≤ ≤ n (sn) such that dimH (Ct,D)= α and dimP (Ct,D)= β. Proof. We will assume 0 <α

ni If necessary, we take subsequences in order to assure that n1 100, mi 2 , mi ≥ ≥ and ni+1 2 . To obtain the new sequence tk, we remove terms from sn in segments,≥ each in a “uniform” manner with some density ξ (0, 1). To explain, suppose the segment is the set of indices q, q+1,...,ℓ N∈. Then to uniformly remove terms with density ξ from this segment,{ we remove}⊂ all the terms of the form q + i/ξ for i =0,..., ξ(ℓ q) 1 (to make sure we do not remove ℓ). Note that⌊ removing⌋ with density⌊ −ξ is the− ⌋ same as retaining with density 1 ξ. − From the set of indices 1, 2,...,n1 , we remove terms in a “uniform” way with density 1 α . Then from{ the set} of indices n +1,...,m we remove − A { 1 1} terms in a “uniform” way with density 1 β . We continue alternating, removing − B terms with density 1 α from m +1,...,n and with density 1 β from − A { i i+1} − B ni +1,...,mi . Call the resulting sequence tℓ where we have tℓ = sn, with {ℓ = nΘ(n) where} Θ : N [α/A, β/B] is a measure of the “local scaling” of → the index. From the construction we have Θ(nj ) α/A, Θ(mj ) β/B, Θ is increasing on n +1,...,m and decreasing on ≈m +1,...,n ≈. Further, { i i} { i i+1} ℓ ln(N) n ln(N) − = θ(n)− . ln(tℓ) ln(sn)

From here it is straightforward to show that lim inf −ℓ ln(N) = α and also that ln(tℓ) lim sup −ℓ ln(N) = β, as desired. The condition α/A β/B is used to check the ln(tℓ) ≤ new liminf and limsup. Since the original sequence satisfies condition (2), it is easy to see that any subsequence will as well.

Of course, this construction does not guarantee that Ct,D will satisfy 0 < α (Ct,D) < even if it has the proper dimension. Comparing Theorem 6 with HTheorem 9, we∞ trade the ability to specify the t-measure of the subset with the ability to ensure that the subset is of a particularlyH nice form, in Theorem 9 being the full set of subsums of some subsequence. However, if we assume a bit more on Ct,D we can obtain a substantially stronger result. Theorem 10.

8 A 1. Suppose that 0 < (Cs,D) < . Then for any 0 a A there is a H ∞ a ≤ ≤ subsequence (t ) of (s ) such that 0 < (C D) < . n n H t, ∞ B 2. Suppose that 0 < (Cs,D) < . Then for any 0 b B there is a P ∞ b ≤ ≤ subsequence (t ) of (s ) such that 0 < (C D) < . n n P t, ∞ A B 3. Suppose that 0 < (C D) < and 0 < (C D) < . Then for any H s, ∞ P s, ∞ 0 a A and 0 b B with a/A b/B, there is a subsequence (tn) of ≤ ≤ ≤ a≤ ≤ b (s ) such that 0 < (C D) < and 0 < (C D) < . n H t, ∞ P t, ∞ Proof. We prove the third statement as it is the most involved. The other two are similar. As in Theorem 9 we work with sn rather than Rn. N mj B n B nj A Let mi,ni be such that lim N smj = lim sup N sn = S and lim N snj = n A ∈ lim inf N sn = I. In addition, we assume that nj < mj < nj+1 < mj+1, mj /nj , and nj+1/mj . The two cases a/A = b/B and a/A < b/B require→ different ∞ techniques and→ ∞ so we do them separately. Case 1: a/A = b/B If a/A = 1, then there is nothing to prove. We define our subsequence (tn) by defining the indexing function π : N N such that tn = sπ(n). Define πˆ : N N byπ ˆ(i) = (A/a)i . If m ,n →πˆ(N) for all j, then we let π =π ˆ. → ⌊ ⌋ j j ∈ Otherwise, suppose that mj / πˆ(N). Then i := mj a/A 1 we j j ∈ know thatπ ˆ is injective. If we assume that nj mk > 2A/a for all j and k then π is also guaranteed to be injective. Since| [k,− k + A/a| + 1] πˆ(N) is nonempty for any k, we know that 1 π(i) (A/a)i A/a +1 or,∩ more useful for us, − ≤ − ≤ a a a a 1 + π(i) i + π(i) . − − A A ≤ ≤ A A     This means that a/A N ita = N isa N −1−a/AN π(i)(a/A)s(a/A)A = N −1−a/A N π(i)sA i π(i) ≥ π(i) π(i) N −1−a/A(I ǫ)a/A > 0,   ≥ − i a a for large enough i. Thus lim inf N ti > 0 and so (Ct,D) > 0. By construction, there is a sequence q N so that π(q )= n andH so j ∈ j j a/A N qj ta N a/AN π(qj )(a/A)s(a/A)A = N a/A N nj sA N a/A(I+ǫ)a/A < . qj ≤ nj nj ≤ ∞ a  b Thus (C D) < as well. The proof that 0 < (C D) < is similar. H t, ∞ P t, ∞ Case 2: a/A < b/B Let B b 1 γ0 = − A 1 a − and then choose δ > 0 so that γ := γ0 + δ < 1. Define n n′ = j and m′ = γm j γ j ⌊ j⌋  

9 ′ ′ and notice that nj >nj and mj 0 small and all large enough i, we have a/A N ita = N isa N π(i)(a/A)s(a/A)A = N π(i)sA (I ǫ)a/A > 0, i π(i) ≥ π(i) π(i) ≥ − a   and thus (Ct,D) > 0. For i = Qj , we have π(i)= nj and Qj (a/A)nj +1 and so H ≤ a/A N ita = N Qj sa N (a/A)nj sa N = N N nj sA N(I + ǫ)a/A < i nj ≤ nj nj ≤ ∞ a  b and so (C D) < . The argument that 0 < (C D) < is similar. H t, ∞ P t, ∞ Example 11. The simple Example 1 will show that in general we cannot find a subsequence which will give a subset of arbitrary measure. Recall that it was sn = n n 2/3 and = 0, 1 for all n. It is known that dimH Cs,D = d = ln(2)/ ln(3) d D { } and (Cs,D)=1. The key observation is that if tn is a subsequence of sn H d −K constructed by removing only K terms from sn, then (Ct,D)=2 , since K H Cs,D is the union of 2 disjoint copies of Ct,D (these copies correspond to the possible subsums of the removed terms). But this means that it is impossible to d find a subsequence t with (C D)=1/3. n H t,

10 References

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