SPARSE POTENTIALS WITH FRACTIONAL HAUSDORFF DIMENSION

Transcription

1 SPARSE POTENTIALS WITH FRACTIONAL HAUSDORFF DIMENSION ANDREJ ZLATOŠ Abstract We construct non-random bounded discrete half-line Schrödinger operators which have purely singular continuous spectral measures with fractional Hausdorff dimension in some interval of energies To do this we use suitable sparse potentials Our results also apply to whole line operators, as well as to certain random operators In the latter case we prove and compute an exact dimension of the spectral measures Introduction In the present paper, we consider discrete half-line Schrödinger operators H φ on l Z + = l,, }, given by H φ ux ux + + ux + V xux for x Z +, with the potential V and boundary condition φ π, π ] u0 cosφ + u sinφ = 0 Here defines u0, which then enters in for x = Hence H 0 is the Dirichlet operator with u0 = 0, and H φ = H 0 tanφδ where δ is the delta function at x = H π/ is the Neumann operator with u = 0 All these are rank one perturbations of H 0 A function u on Z + 0} is a generalized eigenfunction of the above operators for energy E and boundary condition φ if ux + + ux + V xux = Eux 3 for x Z + and holds Since such u is uniquely given by its values at x = 0,, the space of generalized eigenfunctions for any energy is -dimensional The unimodular matrix T E x, y which takes uy+ uy to ux+ ux whenever u is a generalized eigenfunction for energy E, is called the transfer matrix for E It is immediate that x x E V j T E x, y = T E j, j = 0 j=y+ j=y+ We denote T E x T E x, 0 We let µ φ be the spectral measures of the above operators The aim of this paper is to construct a non-random bounded potential V such that these measures are purely singular continuous and have fractional not 0 or Hausdorff dimension in some interval of energies We consider the sparse potential with equal barriers V v,γ given by 6 below, with v 0 and γ Here is our main result: Date: 8 October, Mathematics Subject Classification Primary: 47A0; Secondary: 34L40, 47B39 Keywords Schrödinger operators, sparse potentials, singular continuous spectrum, fractional dimension

2 ANDREJ ZLATOŠ Theorem Let H φ be the discrete Schrödinger operator on Z + with potential V v,γ given by 6, and boundary condition φ Let µ φ be its spectral measure For any closed interval of energies J, there is v 0 > 0 and γ 0 N such that if 0 < v < v 0 and γ γ 0 v is an integer, then for any φ, the measure µ φ has fractional Hausdorff dimension in J This is proved in Section 5 as Theorem 5 From the rest of our results we would like to single out Theorems 4, 63 random case and 7 Our motivation is a paper by Jitomirskaya-Last [4], which relates the power growth/decay of eigenfunctions and the Hausdorff dimension of spectral measures We will apply ideas from [5] and use sparse potentials, which allow us to control this growth We mention that [4] also provides an example of potentials with the above properties, but these are unbounded and hence so are the operators First we recall some basic facts about dimension of sets and measures If S R and α [0, ], then the α-dimensional Hausdorff measure of S is [ I n ] α h α S lim δ 0 inf δ-covers Here a δ-cover is a covering of S by a countable set of intervals I n of lengths at most δ Notice that h 0 is the counting measure and h the Lebesgue measure For any S there is a number α S [0, ] such that h α S = 0 if α > α S, and h α S = if α < α S This α S is the dimension of S If µ is a measure on R, we say that µ is α-continuous if it is absolutely continuous with respect to h α, and µ is α-singular if it is singular to h α Hence α-continuous measures do not give weight to sets S with h α S = 0 eg, to sets S such that dims < α, and α- singular measures are supported on sets S with h α S = 0 and so dims α We say that µ has fractional Hausdorff dimension in some interval I if µi is α-continuous and α-singular for some α > 0 Finally, µ has exact local dimension in I if for any E I there is an αe, and for any ε > 0 there is δ > 0 such that µe δ, E + δ is both αe ε-continuous and αe + ε-singular We do not prove an exact dimension for our measures µ φ corresponding to 6, but we do it for the random potential case which we consider in Section 6 In the present paper, we will sometimes say that α-continuous measures have dimension at least α and that α-singular measures have dimension at most α We will mainly use two results from [4] Corollaries 44 and 45 which relate eigenfunction growth and spectral dimension Here, however, these results will be restated in terms of the EFGP transform of eigenfunctions Propositions and 3 below, rather than in terms of the eigenfunctions themselves The EFGP transform R, θ of an eigenfunction u with energy E, is a Prüfer-type transform which makes the growth/decay of u more transparent It is defined as follows We let k 0, π be such that E = cosk and set n= ux coskux = Rx cosθx, sinkux = Rx sinθx These equations define Rx > 0 and θx mod π uniquely, and we write u R, θ 4

3 If we set then 3 becomes see [5] SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 3 θx θx + k and v k x V x sink, cot θx + = cot θx + v k x, Rx + Rx = + v k x sin θx + v k x sin θx 5 Notice that 5 only determines θx + mod π There are two ways to deal with this The first is to examine 4 more closely and conclude as in [6] that sgn sinθx + = sgn sinθx, and if this is 0, then sgn cosθx+ = sgn cosθx This fact and 5 determine θx + mod π uniquely, but we will not need this extra condition here The second way is to realize that sinθ, sin θ and cotθ all have period π, and so this ambiguity in θ does not affect the values of R, which are of main interest to us Notice also, that once u0 and u are fixed and k is varied, ux is a polynomial in E of degree x Therefore ux, and by 4 also Rx and θx viewed as a function on the unit circle, are well-defined C functions of k Moreover, there is completely no ambiguity in θx, k which will be of significant importance in our considerations The relations 5 are of interest to us for two reasons The first is that they are particularly useful when dealing with sparse potentials, which we will consider here Sparse potentials are non-zero only at sites x = x n such that x n x n This is because on the gaps the intervals where the potential is zero the propagation of R and θ is especially transparent Namely, R is constant and θ increments by k when passing from x to x + The second reason is that 5 provides a good control of the growth of R, which is the same as the growth of u in the sense of the following lemma Let us define L u L ux Then we have x= Lemma There are constants c, c > 0 depending only on k 0, π such that if u R, θ is any generalized eigenfunction for energy cosk and L, then c u L R L c u L Remark The proof shows that c and c can be chosen uniformly for k I, with I any closed sub-interval of 0, π Proof From 4 we have with d i = + i cosk Hence Rn = un + un coskunun [d un + un, d un + un ] d u L R L d u L + u0

4 4 ANDREJ ZLATOŠ The result follows from the fact that u0 = [ cosk V u u ] [ cosk V + ] [ u + u ] Let us denote by u φ,k R φ,k, θ φ,k the generalized eigenfunction for energy E = cosk satisfying the boundary condition φ We are now ready to state, in terms of R rather than u, the abovementioned results from [4] These will be our main tools for proving fractional dimension of measures Proposition [4] Let 0 < α and let A be a Borel set of energies If for every E A and every generalized eigenfunction u R, θ for energy E R L lim <, L L α then for any φ the restriction µ φ A is α-continuous This says that if all eigenfunctions for all energies in some support of µ φ have a small power growth, then µ φ cannot be very singular Proposition 3 [4] Let 0 < α and let A be a Borel set of energies If for every E A R φ,k L lim = 0, L L α where k is such that E = cosk, then the restriction µ φ A is α-singular An eigenfunction v for energy E is called a subordinate solution if v L lim = 0 L u L for any other eigenfunction u with the same energy The Gilbert-Pearson subordinacy theory [3] shows that µ φ is supported off the set of energies for which a subordinate solution exists, but does not satisfy the boundary condition φ Hence Proposition 3 says that the existence of a power decaying eigenfunction which is then by standard arguments the subordinate solution for all energies in some support of µ φ implies certain singularity of µ φ Moreover, Lemma below shows that the existence of a power growing eigenfunction u implies for the potential 6 the existence of a power decaying subordinate solution v, and the power of decay of v is the same as the power of growth of u Thus we only need to estimate the power of growth of eigenfunctions We will specifically concentrate on the generalized eigenfunctions with Dirichlet boundary condition φ = 0 Let us denote by u k the eigenfunction for energy E = cosk with k 0, π, such that u k 0 = 0 and u k = Let u k R k, θ k, and let θ k x = θ k x + k Notice that R k = Recall that R k x, θ k x and θ k x are C functions of k As we mentioned before, we will use sparse potentials, which are non-zero only for x x n } n= It turns out that the set of possible candidates is quite small Firstly, for our considerations we will need to have a good control of θ k kx n, in order to estimate the longrun behavior of R k using 5 This turns out to be hard if x n } n= grows sub-geometrically and so we will not consider this case On the other hand we have

5 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 5 Theorem 4 Let x n N be an increasing sequence, a n R and let µ φ be the spectral measure for the discrete Schrödinger operator on Z + with boundary condition φ and potential a n x = x n for some n, V x = 0 otherwise Then if lim x n /x n+ < and a n 0, then the dimension of µ φ is everywhere in, ; if x n /x n+ 0 and sup a n <, then the dimension of µ φ is everywhere in, Remark It is known [5, 6] that in the type of the spectrum depends on n= a n If this is finite, then we have purely ac spectrum, whereas if it is infinite, we have purely sc spectrum Proof Let I 0, π be a closed interval For k I let w k P k, ψ k be the generalized eigenfunction with energy E = cosk such that w k 0 = and w k = 0 There are γ > and n 0 N such that x n /x n > γ and x n > γ n for all n > n 0 Since a n 0, for each ε > 0 there are c, c > 0 such that R k L c ε n+ and P k L c + ε n for L [x n +, x n+ ] and k I This follows from 5 Also, L x n c 0 L with c 0 = γ > 0 if n > n 0 Let β < Then for n > n 0 R k L P k β L L x n c ε n Lc + ε n β cl β α n c γ β α n with α = ε/ + ε β We can choose ε small enough so that γ β α > and we obtain R k L lim L P k β L = Then Theorem from [4] implies β -continuity of µ +β 0 Since this is true for any β < and for any pair of generalized eigenfunctions with energy E = cosk, as well as for any I, the result follows We know that a n / sink M < for any n and k I I as above It is easy to show that then there are 0 < a < b < such that a n sink sinθ + a n sin k sin θ a, b for any n, θ and k I Therefore in the previous argument we have to replace ε and + ε with a and b On the other hand, x n /x n+ 0 implies that for any γ there is n 0 such that x n /x n > γ and x n > γ n for n > n 0 Thus for any β < choose γ large enough so that γ β a/b β > The rest of the proof is identical with This leaves us with non-decaying potentials and x n growing geometrically Therefore we will consider the following natural choice of potential We take v 0 and an integer γ, and define v x = x n γ n for some n, V v,γ x 6 0 otherwise

6 6 ANDREJ ZLATOŠ Notice that the increasing gaps in V v,γ ensure that the interval of energies [, ], which is the essential spectrum of the free operator, is contained in the spectrum of each H φ Indeed, by the discrete version of a theorem of Klaus Theorem 33 in [], the essential spectrum of H φ is [, ] sgnv } 4 + v In what follows, we will show that for suitable v and γ, the spectral measures µ φ are α- continuous and α-singular for some α > 0 in some sub-interval of [, ], and so have fractional Hausdorff dimension there It turns out that we will need small v and large γ More precisely, we will show, that for certain v and γ, we have that for most k in a sense to be specified later within a given closed interval I 0, π the function R k has a suitable power growth with some power β 0, Then Propositions, 3, together with Lemma below, can be used to compute bounds on the dimension of the spectral measures Notice that since by 5 R k is constant on [x n +, x n ], we only need to look at the growth of R k x n + } n= In what follows, we will not only prove fractional Hausdorff dimension, but also give bounds on it, and therefore we will need to enumerate those constants appearing in our argument, which affect the power of growth of R k We will denote these C i In the present paper, we will consider spectral measures wrt k, rather than wrt E = cosk This will not affect the validity of the results because on any closed interval I 0, π of k s and we consider only such the function cosk is C with bounded non-zero derivative Therefore the dimensional properties of µ φ wrt k I and wrt E J cosi are identical Nevertheless, our results will be stated in terms of E Finally, we mention that in 6 γ does not need to be integral If one only requires γ > and sets, for instance, x n = γ n, all our results continue to hold The rest of the paper is organized as follows In Section we present the main ideas of our proofs and results Section 3 contains the abovementioned estimates on k θ kx n In Section 4 we prove fractional Hausdorff dimension of the spectral measures for almost all boundary conditions, along with bounds on this dimension Theorems 4, 4 Section 5 contains the same results for all boundary conditions Theorems 5, 5 We distinguish these two cases because in the case of ae φ the bounds we provide are considerably better than those for all φ In Section 6 we consider certain randomization of the potential V v,γ given by 6 and prove exact fractional Hausdorff dimension for ae realization of the potential and ae boundary condition Theorem 63 Finally, Section 7 contains the corresponding whole-line results Theorem 7 It is a pleasure to thank Barry Simon for many useful discussions and suggestions My thanks also go to David Damanik, Wilhelm Schlag and Boris Solomyak Growth of Eigenfunctions In this section, we will present a short tour of the proof of our main result Technical details are left for later Let J, be a given closed interval of energies We let I 0, π be such that cosi = J We define v k v for k I, w sink min k I v k } > 0, w max k I v k } <, x n γ n for n and x 0 0 Here v 0 and γ are to be determined later

7 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 7 We consider the half-line discrete Schrödinger operator H φ given by, with the potential 6, and we denote its spectral measure µ φ We will prove fractional Hausdorff dimension of µ φ I for suitable v, γ and φ As mentioned earlier, we need to estimate the growth of R k x n + As in [5], using 5 and the Taylor series of ln + x, one obtains for n ln R k x n + ln R k x n + = ln + v k sinθ k x n + vk sin θ k x n = v k sinθ k x n + 4 v k sin θ k x n sin θ k x n + g n k = v k cos4θ k x n cosθ k x n + g n k 8 + v k sinθ kx n + v k 8 where g n k < C 0 v k 3 for some C 0 > 0 Here g n k is the sum of all third and higher order terms in v k, and the last equality comes from sin θ sin θ = cosθ + cos4θ Now take v 0 small enough so that w + w < and C 0 w 3 < w /8, and define d w /8 C 0 w 3 and d w /8 + C 0 w 3 If we let 0 < C < d and d < C <, then we have vk 8 + g nk [d, d ] C, C for any n and k I From now on, v and C 0 and thus also C and C will be fixed Here is our main idea It follows from that the contribution of the terms vk /8+g nk to the size of lnr k x n + is within the interval [d n, d n] If we could show that for large n the contribution of the remaining three oscillating terms in is small compared to this, we would obtain estimates proving positive power of the growth of R k It is reasonable to hope for this because the oscillating terms change sign when n is varied, and so we can expect cancellations This is the central idea of [5] So let us assume for a while that for some A I with A = 0 A being the Lebesgue measure of A N sinθ k x n = on k I\A 3 n= and that the same holds for the other two oscillating terms For any k I\A we have for large n by R k x n + e C n, e C n = x β n, x β n 4 where β i C i / lnγ If we choose γ to be large enough so that 0 < β < β <, we get c L +β R k L c L +β 5 for large L and c i = c i k Then Proposition, along with the theory of rank one perturbations [0], proves for ae boundary condition φ that the dimension of µ φ is at least β see the last paragraph of the proof of Theorem 4 To obtain a good upper bound for the dimension, we need to prove an appropriate decay of the corresponding subordinate solutions We will use the following result, the proof of which we postpone until the end of this section

8 8 ANDREJ ZLATOŠ Lemma Let x < x < be such that T E x n, x n B for some E, and B < Let us assume that u R, θ is a generalized eigenfunction for energy E such that Rx n + = e α n where α n = n Z j + X j with Z j [d, d ] for some 0 < d d < and n X j = on Then there exists a subordinate solution v P, ψ for energy E such that for any 0 < d < d and for all sufficiently large n we have P x n + e dn Remark In principle, the lemma shows that the power of decay of the subordinate solution is the same as the power of growth of all other solutions for the same energy Hence, in a sense, the result is optimal Notice that information on only one growing solution is needed, but this has to satisfy a steady growth condition Compare with Lemma 87 in [5], where two growing solutions are involved The lemma can be applied here because we have and 3 Notice also that all the powers of the free transfer matrix E 0 for energy E, are uniformly norm-bounded Hence, we also have a uniform in n bound on xn x E v T E x n, x n 0 E n 0 So for k I\A we let and Z j = v k 8 + g jk X j = v k sinθ kx j + v k cos4θ k x j cosθ k x j 8 If we now take d such that C < d < d say d = C + ε, we obtain the existence of a vector uk sub R which generates the subordinate solution u sub k with u sub k L L β ε/ lnγ Hence by Proposition 3 and the Gilbert-Pearson subordinacy theory [3], the dimension of µ φ is at most β for ae φ see the proof of Theorem 4 Moreover, by the proof of Lemma, T E x n as n whenever E cosi\a Theorem from [7] then implies absence of ac spectrum in I for all φ The next paragraph proves absence of pp spectrum if w + w < lnγ Thus for γ large enough, we obtain purely sc spectrum in I for all φ But we can do even better It turns out that with a slightly stronger assumption than 3, we can show fractional dimension for all boundary conditions! First notice that ln R k x n + ln R k x n + < w + w

9 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 9 and so for all k I we have R k x n + < x w +w / lnγ n Since R k is constant on [x n +, x n ], Proposition implies that the dimension is at least w + w/ lnγ on all of I Since w + w <, this is strictly positive if γ 3 Now replace 3 with the stronger assumption that for any ε > 0 there exists A ε I, a set with dimension smaller than such that lim N N N sinθ k x n ε k I\Aε, 6 n= and the same is true for the other two oscillating terms Let us consider the Dirichlet measure µ 0 Since it has no ac part, it is supported on the set of k for which u k is the subordinate solution It follows that µ 0 I\A ε = 0 Indeed, if k I\A ε, then for large n we have } w R k x n + > exp C ε + 3w n 8 If ε is small enough so that the exponent is positive, Lemma again shows the existence of a subordinate solution u sub k, and this must be different from u k by which we mean that u sub k is not a multiple of u k This implies that µ 0 is supported on A ε Hence the dimension of µ 0 on I which we know is positive is at most dima ε <, and therefore fractional Considering instead of u k the generalized eigenfunction for energy cosk satisfying boundary condition φ 0, and assuming 6 for the corresponding θ φ,k x n, we can obtain the same result for any φ For details see the proof of Theorem 5 Our considerations have, however, a pitfall Neither 3 nor 6 need hold for such a large set of k s as we want We cannot hope for this because we have only limited control of the argument of the sin term Fortunately, we do not really need 3 and 6 in the presented form This is because the non-oscillatory term vk /8 gives us some space We can sacrifice part of it, just as we did when we joined it with the g n k term, and still keep a power growth of R k More precisely, we will divide the sin term into two, one of which will be small with respect to the non-oscillatory term, whereas the other one will have enough regularity for 3 and 6 to hold The two cos terms can be treated similarly Proof of Lemma Let d 0 = d + d and let v P, ψ be any solution for energy E different from u Let p n P x n +, r n Rx n + and t n T E x n The first ingredient in the proof is Theorem 3 from [5] which states that there are c, c > 0 such that c maxp n, r n } t n c maxp n, r n } Since by hypothesis r n e d 0n /c for large n, it follows that for large enough n On the other hand, if d 3 > maxd, lnb}, then for large n Hence if then d 0 δ d 3 t n e d 0n 7 t n e d 3n lnt n δ lim n n,

10 0 ANDREJ ZLATOŠ From 7 we know that n= T E x n, x n T E x n < This is the assumption of our second ingredient, Theorem 8 from [7], the proof of which namely inequalities 85, 87 yields the following There is a vector v R and c 0 > 0 such that T E x n v c 0 t ns n + t n with s n = m=0 t n+m Our aim is to show that for large enough n this is smaller than e dn The abovementioned Theorem 8 also asserts that v generates the subordinate solution for energy E One expects that this is different from u, generated by u u0, u, which is a growing solution We will now prove this claim Let δ < δ < δ be such that δ < δ From the definition of δ we know that for large enough n s n e δ n Also there are n j } j= such that t nj < e δ n j and so as j Since T E x nj v c 0 e δ n j e 4δ n j + e δ n j 0 T E x nj u = ux nj + + ux nj Rx n j +, the subordinate solution is indeed different from u Let us take it for v, and change P, p n, c and c accordingly Since Theorem 8 from [7] also states that T E x n v T E x n u 0, it then follows that p n < r n for large n Hence c r n < t n < c r n for large n Let a n = α n /n If n is large enough, then a n d 0, d 3 Let b N be such that d 0 b + > d 3 Pick ε > 0 small enough so that ε εb + < d 0 d/ Since a n n = n Z j + X j, for large n we have n Z j a n εn and n X j εn Since n+m lnr n+m = a n+m n + m = Z j + X j a n εn + d m εn + m, for m bn we have Hence for large n s n c bn m=0 e a n ε n d m + c r n+m e a n ε n+d m m=bn e d0n+m c e a n ε n + e d 0b+n

11 for some c > 0 It follows that SPARSE POTENTIALS WITH FRACTIONAL DIMENSION T E x n v cc 0 c e a n 4a n ε n + e a n 4d 0 b+n + c e a nn c e dn + e d 3n + e d 0n 3c e dn The rest is an easy computation From 4 we have P x n + vx n + + vx n = T E x n v 6c e dn for large n Since this holds for any d < d, the result follows 3 Estimates on Growth of θ From now on let us write θx n, k instead of θ k x n As mentioned before, the key to our results are estimates on θx k n, k which we denote θ x n, k Differentiating the first equation of 5 with respect to k, and using sin θ = + cot θ, we get as in [6] θx + = k k sin θx + cosk θx V x sin k sin θx [ ] cosθx V x sinθx sink For our potential V v,γ and for x = x n, the denominator is and so a n,k = + v k sinθx n, k + v k sin θx n, k θ x n+, k = x n+ x n + θ x n +, k = x n+ x n + θ x n, k a n,k + v k cotk sin θx n, k a n,k The denominator a n,k is in the interval [C 3, C 4 ] w + w, + w + w with 3 C i + w + i w + w 4 Notice that C 3 C 4 = The last fraction in 3 is in some interval M, M for any n and any k I From 3 one can show that as n, θ x n, k gets close to x n Indeed, if for some n we have θ x n, k/x n, for all k I and some i R, then θ x n+, k x n+ + γ C 4 M γ n+, + γ C 3 + M γ n+ Iterating this, one proves that for any D < C 5 γ /γ C 3 and D > C 6 γ /γ C 4 there is n 0 such that for n > n 0 and all k I θ x n, k x n D, D 3 This is because γ /γ C i are the fixed points of the map + /C i γ, and C3 = C 4 Thus D, D is an interval containing, which can be made as small as we need by taking γ large enough This will play an important role in our considerations

12 ANDREJ ZLATOŠ For n > n 0 let K n, and K n, be the smallest and largest numbers in I such that θx n, K n,i is an integral multiple of π Notice that by 3 the distance of these numbers from the corresponding edges of I is smaller than πd γ n Let k n, = K n, < k n, < < k n,jn = K n, be all numbers in I such that θx n, k j is an integral multiple of π Let I n,j = [k n,j, k n,j+ ] Then 3 implies that I n,j πd γ n, πd γ n 33 We will slightly alter the oscillating sin term on each I n,j This way we will obtain the previously mentioned more regular term for which we can prove 3 and later 6 This is the content of Sections 4 and 5 But before that, we present an additional argument It turns out that by considering θ x n, k one can improve 3 and 33 on small scale ie, the scale of I n,j This improvement is not essential for our results; it only yields better numerical estimates for large γ Differentiating 3 with respect to k one obtains θ x n+, k = θ x n, k a n,k [ ] θ v k cosθx n, k + vk x n, k sinθx n, k + b a n,k θ x n, k + b n,k n,k where b n,k, b n,k M, M for some M and all n, k Then for n > n 0 since a n,k C 4, θ x n+, k x n+ C 4 θ xn, k + D C4w + w + MD γ x n γ γ n+ Similarly as above, by iterating this we obtain that for large enough n say n > n 0 for a new n 0 and all k I θ x n, k x n D C4w + w γ This is because the fixed point of the mapping C 4 /γ + DC 4w + w/γ is DC 4w + w/γ C 4, and because C 4 < If n > n 0, let D n,j and D n,j be such that for any k I n,j θ x n, k x n [D n,j, D n,j ], 3 and the interval [D n,j, D n,j ] is smallest possible From 33 and the obtained estimate on θ x n, k we have that D n,j D n,j π DC 4w + w D γ n γ γ n γ n = πd C4w + w D γ Notice that D D c/γ, and so as γ, the estimate 3 is better than 3 We also have the obvious improvement of 33, namely Now let I n,j [ πd n,j γ n, πd n,j γ n] 33 πd C 7 min C4w + w, D } D Dγ D

13 Since D D n,j D n,j D, we have for all j and n > n 0 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 3 D n,j D n,j D n,j C 7 D D D 4 Fractional Dimension for AE Boundary Condition Let us now turn to the announced division of the sin term in two Since C 6 C 5 /C 5 = C 4 C 3 /γ C 4, we can pick D, D close to C 5, C 6 so that D D /D is arbitrarily close to C 4 C 3 /γ C 4 Since C 3 and C 4 as well as C and C are independent of γ, and C 7 D D /D, we can make C 7 arbitrarily small by taking γ large enough Let us do this so that C 7 π w + 3w < C 4 8 For k I and n > n 0 we let In,j In,j Q n k sinθx n, κ dκ k I n,j 0 k I\[K n,, K n, ] and define Notice that for any j Also, by 3 X n k sinθx n, k Q n k sinθx n, k X n k = Q n k π I n,j X n k dk = 0 4 D n,j D n,j D n,j + D n,j C 7 π for any k I and n > n 0 This is because Q n is maximal possible on I n,j if θ equals D n,j γ n when sinθ is positive and D n,j γ n when sinθ is negative or vice-versa And in that case we have equality in the first inequality above Therefore we have N n=n 0 + sinθx n, k w v k C 7 π N + v k N n=n 0 + X n k If we do the same with the other two oscillating terms the corresponding I n,j s and X n s will be slightly different, the three terms containing C 7 will add up to C 7 w π + 3w N < C N 8 So if we prove 3 for X n in place of sin and similarly for the two cos terms, we will still keep a power growth of R k We will be able to do this using 4

14 4 ANDREJ ZLATOŠ To this end we need estimates on the covariances of the X n s, so take n > m > n 0 Then for all j we have 4, whereas from 3 it follows that X m takes on I n,j values within an interval of length at most D γ m I n,j πd D γ m n Therefore X m X n dk I n,j πd γ m n I n,j D Notice that this might not be true if I n,j contains some k m,l, points where X m may be discontinuous But this is not a problem because the total length of such I n,j s is of order γ m n Since also I\[K n,, K n, ] γ n and the X n s are bounded, we get for n > m > n 0 X m X n dk cγm n By Cauchy-Schwarz inequality N N X n dk c I n=n 0 + = c [ I n=n 0 + N I n=n 0 + I X n dk X n + I / n 0 <m<n N X m X n dk ] / c N + cn / cn / the value of c changes in each inequality, but is independent of N Thus N 4 n=n 0 + X n N 4 dk c N which means that for ae k N 4 n=n 0 + X n 0, N 4 because N < Since the X n s are bounded and [N + 4 N 4 ]/N 4 0, it follows that for these k N n=n 0 + X n 0 N and so N n= X n 0 N Theorem 4 Let H φ be the discrete Schrödinger operator on Z + with potential V v,γ given by 6, and boundary condition φ Let µ φ be its spectral measure For any closed interval of energies J, there is v 0 > 0 and γ 0 N such that if 0 < v < v 0 and γ γ 0 is an integer, then each µ φ is purely singular continuous in J, and for ae φ the measure µ φ has fractional Hausdorff dimension in J Remark A priori, γ 0 depends on v However, since in 4 we have C = Ov, C 7 = Ovγ and w = Ov as v 0, γ, we can choose γ 0 uniformly for all small v

15 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 5 Proof Let I 0, π be such that cosi = J The above discussion and show that there is A I with A = 0, such that for any k I\A there is n k such that for n > n k where R k x n + e c n, e c n = x β n, x β n, c i C i + i C 7 π w + 3w > 0 i =, 8 and β i c i / lnγ Take γ large enough so that 0 < β < β < It follows from the constancy of R k on [x n +, x n+ ] that 5 holds for k I\A Using Lemma with d i + i C 7 π w + 3w 8 i =, in place of d, d, one gets for these k the existence of a vector uk sub R such that the following holds If u sub k P k, ψ k is the generalized eigenfunction for energy cosk generated by uk sub, then for some small ε P k x n + e c +εn = x β ε/ lnγ n Since P k is constant on [x n +, x n ], we have P k L c L β ε/ lnγ 43 Since u sub k is the subordinate solution for energy cosk, all other solutions for this energy grow in power no faster than R k Also, µ φ restricted to I is supported on the set of those k, for which u sub k satisfies the boundary condition φ this is because µ φ has no ac part; see [3], and so P k = R φ,k Then we have by Propositions, 3 and by 5, 43 that for any φ the restriction µ φ I\A is β -continuous and β -singular By the theory of rank one perturbations Theorem 8 in [0] we know that µ φ A = 0 for ae φ, and so for ae φ we have the same continuity/singularity of µ φ I To get numerical bounds on the dimensions, one needs to evaluate constants C i These depend on w i and therefore it is best to consider for given v and large γ a small interval I around each k, so that w i v k One then obtains bounds on local dimension of µ φ which will be k-dependent We will, however, first present an additional argument which will considerably facilitate this by eliminating constants C 0, C and C as well as improve the obtained bounds Let us push the above ideas a little bit further At the beginning of this argument we introduced the term g n k as the sum of all third and higher order terms in v k One can, however, write down all these terms explicitly, using the Taylor series of ln + x ln R k x n + ln R k x n + = ln + v k sinθx n, k + vk sin θx n, k = a,b 0 a+b a+b+ a + b a + b a v a+b k sin a θx n, k sin b θx n, k If v k + vk <, then the sum of the amplitudes of the terms of this sum with n fixed converges absolutely

16 6 ANDREJ ZLATOŠ Now notice that all terms with odd a are oscillating, whereas terms with even a do not change sign when varying n This will allow us to obtain more precise bounds on the dimension, since, once again, the contribution of oscillating terms will be for ae k negligible in comparison with that of non-oscillating terms This will be proved by the same methods as above We will replace the oscillating term sin a θx n, k sin b θx n, k by a more regular term sin a θx n, k sin b θx n, k Q n,a,b k, with Q n,a,b k small The fact that we now have an infinite number of terms will not cause problems because the sum of their amplitudes converges absolutely First we need to do what we have already done once We will split each non-oscillating term in two, a constant and an oscillating term The latter will be treated as the other oscillating terms We define If a, then this is 0 and F a,b π π 0 sin a θ sin b θ dθ W n,a,b k sin a θx n, k sin b θx n, k is oscillating If a, the corresponding term is non-oscillating and we split it into F a,b and Let W n,a,b k sin a θx n, k sin b θx n, k F a,b F v k a,b 0 a+b a+b+ a + b Gv k a + b a,b 0 a+b a + b a a + b a v a+b k F a,b v k a+b Then F v k is the sum of all the constant terms, and Gv k is strictly larger than the sum of the amplitudes of the oscillating terms for fixed n Notice that and see [8, 6] F v k = π π 0 Gv k = ln v k v k ln + v k sinθ + vk sin θ dθ = ln + v k 4 We take γ large so that C 7 Gv k /π < F v k for all k I Now fix any a, b and consider all the W n,a,b s We can do everything as above Take the intervals I n,j these will be different for a and a, just as they were for the sin and cos terms As before, the integral of W n,a,b over any I n,j is close to 0 it would be exactly 0 if θ x n, k were constant on I n,j Thus from 3 we know that there is Q n,a,b k such that Q n,a,b k C 7 /π, and for X n,a,b k W n,a,b k Q n,a,b k we have I n,j X n,a,b k dk = 0

17 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 7 As before, one can use this to prove that N n= X n,a,b 0 44 N for ae k This holds for any a, b Since the number of these pairs is countable, we have that for ae k 44 holds for any a, b Using the fact that the sum of the amplitudes of the oscillating terms is finite, we obtain that N N n= a,b 0 a+b a+b+ a + b a + b a v a+b k X n,a,b 0 for ae k Then for each such k and for large n > n k we have R k x n + e c n, e c n, where c i = c i k F v k + i C 7 Gv k /π remember that Gv k is strictly larger than the sum of the amplitudes We know from [0] that for ae φ the spectral measure µ φ is supported on the set of these k s because it is a set of full measure Let us now estimate C 7 for v k + vk < Let I be a small interval around k, so that w is arbitrarily close to v k Then C 4 C 3 < v k + vk and C 4 <, and one can pick D, D so that D D /D < v k + vk /γ and D /D < γ/γ We trivially have πw + w/γ < 7 v k + vk /γ for small I and γ 5 So we obtain 8 vk + vk C 7 < min, v k + vk } C γ γ 7 45 Let I be small enough so that for any k I c k, c k F v k C 7 π Gv k, F v k + C 7 π Gv k Repeating the proof of Theorem 4 we obtain Theorem 4 With the notation of Theorem 4 let γ 5, F x = Gx = ln x x For k 0, π let Ek = cosk, v k = v α k F v k + α k F v k 8 v k +v k πγ Gv k lnγ 8 v k +v k πγ Gv k lnγ, sink, ln + x 4 and Then for ae φ we have for all k such that v k + v k < and α k <, that on a small interval around Ek the measure µ φ is α k-continuous and α k-singular Remarks Notice that the hypotheses imply that α k > 0 whenever α k < So under the above conditions we know that for ae φ, the local dimension of the spectral measure is within an interval which is close to, and for large γ its size is small compared to its distance from

18 8 ANDREJ ZLATOŠ 5 Fractional Dimension for all Boundary Conditions Now we turn to the proof for all boundary conditions We return to considering the term g n k and three oscillating terms first We again restrict our considerations to the sin term, the two cos terms are treated similarly It turns out that the X n from Section 4 are not regular enough to prove 6 with sin replaced by X n Therefore we need to make a new breakup of the oscillating term to obtain a more regular X n This is done in the Appendix in the proof of Theorem A with f n k = θx n, k, β =, δ = D and δ = D Using the reasoning from Section we then obtain our main result: Theorem 5 Let H φ be the discrete Schrödinger operator on Z + with potential V v,γ given by 6, and boundary condition φ Let µ φ be its spectral measure For any closed interval of energies J, there is v 0 > 0 and γ 0 N such that if 0 < v < v 0 and γ γ 0 v is an integer, then for any φ, the measure µ φ has fractional Hausdorff dimension in J Remark A priori, γ 0 depends on v However, 5 below gives ε = Ov as v 0 since C = Ov, C 7 = Ovγ and w = Ov So if γ = Oε = Ov, then in 5 αε < because D, D as v 0 Compare with the remark after Theorem 4 Proof Let I 0, π be such that cosi = J Let α w + w/ lnγ > 0 with w + w < and γ 3 We already know that by Proposition µ φ is α -continuous on I By the Appendix, the absolute value of the small term sinθx n, k X n k from the new breakup is at most πδ + δ δ πc 7, and for all ε > ε 0 = ε 0 D, D, γ ln + 0 D γ D / there is A ε with dima ε < such that 6 with X n k in place of sinθx n, k is satisfied Here C 7, ε 0 0 as D, D and γ, which is guaranteed when γ Hence we proceed as follows This time we let γ, D and D be such that n=n 0 + π C 7 w + 3w 8 < C this is possible for all large γ Then for n > n 0 we have N v k sinθx n, k w π C v k 7N + N n=n 0 + X n k We do the same with the other two oscillating terms, and the three terms containing C 7 add up to π C w 7 + 3w N < C N 8 Next, we choose ε with 0 < ε w + 3w < C π 8 C w 7 + 3w 8 5

19 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 9 and we take γ large enough so that ε 0 < ε By the Appendix there is a set A ε such that ε ln + 0D dima ε γd αε 5 lnγ and 6 holds with sin replaced by X n Similarly for the other two oscillating terms, and we let A ε be the union of the three sets From the above discussion it follows that for k I\A ε and n large enough we have where R k x n + > e c n = x β n, 53 π c = C C w 7 + ε + 3w > 0 8 and β = c / lnγ It is known see [4] that for ae k wrt µ 0 we have lim L R k L L / lnl < Given 53, this can be true only if µ 0 I\A ε = 0 Thus µ 0 I is supported on A ε Hence on I the measure µ 0 is dima ε + δ-singular for any δ > 0 Since D, D can be arbitrarily close to C 5, C 6, and since C 6 /C 5 < γ/γ if w + w <, we have ε ln + 0 dima ε < α γ lnγ So µ 0 is α -singular on I If in this argument we replace u k by the generalized eigenfunction for the same energy satisfying boundary condition φ, we obtain the same singularity for µ φ Now take γ large so that α <, and the result follows At this point we can do the same as what we did after proving Theorem 4: consider an infinite sum of terms instead of g n k Now we use a different type of regularization of the oscillating terms, but everything can be done as before, with one adjustment We need to use Theorem A3 in place of Theorem A, which gives us a different bound for the difference of the oscillating term sin a θx n, k sin b θx n, k and its regularization X n,a,b k, namely πa + bc 7 The derivative of the oscillating term enters here, and we use the very crude estimate [sin a θ sin b θ] a + b [sinθ] So this time we need to take F v k and Gv k as before the latter will be the coefficient for ε, and also Gv k = a + b v k a+b = v k + vk a v k v a,b 0 k a+b which will be the coefficient for π C 7 We take γ large so that π C 7 Gv k < F v k, and take ε > 0 such that εgv k < F v k π C 7 Gv k

20 0 ANDREJ ZLATOŠ Then given any pair a, b we construct the set A ε a,b for X n,a,b We show, as above, that its dimension is at most αε Letting the countable union of these sets play the role of A ε in the proof of Theorem 5, and considering 45, we obtain Theorem 5 With the notation of Theorem 5 let γ 5 For k 0, π let Ek = cosk, v k = v, sink ε k ln + v k4 4π v k +vk γ v k vk ln v k vk, α k v k + vk, lnγ α k ε k ln + 0 γ lnγ If k is such that v k + v k <, ε k > 0 and α k <, then on a small interval around Ek each µ φ is α k-continuous and α k-singular Remarks Notice that the hypotheses imply that α k > 0 So for large γ, the dimension of each spectral measure µ φ is within an interval which is close to, and its size is comparable to its distance from The estimate for ae φ in Section 4 is better by a factor of γ To illustrate the obtained results we provide an Example Let us assume that v, γ and I are such that γ 0 4 v, and v k for any k I Estimating the quantities in Theorems 4 and 5, one can obtain ln + v k4 [α k, α k] 0 ln + v k4 lnγ γ lnγ, + 0 lnγ γ lnγ and [ [α k, α k] v k + vk, lnγ ] vk 500 lnγ notice that v k = v/ 4 E For instance, take v =, γ = 0 06 and J = [ 9, 9] Then for ae φ the local dimension of µ φ at any E J is in ln E 6 ln0 0, ln and for all φ it is in [ ] 0 ln0 4 E, 0 5 ln04 E E 6 ln0 + 0 [ 0, ] 0 6

21 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 6 Random Operators In this section we consider potentials of the form 6, with certain randomness in the position of the special sites x n More precisely, x n will be a random variable uniformly distributed over γ n n,, γ n + n} The fact that the size of these sets grows will yield certain averaging of lnr k x n + lnr k x n + and thus a constant growth in the limit n of lnr k x n + for ae realization of the potential As a consequence we will be able to compute the exact dimension of the spectral measures for these random potentials We begin with a standard result Lemma 6 For ae x R there is q 0 such that for any integer q > q 0 we have distqx, Z > q 6 Proof For any n N the measure of the set of x [n, n +, for which 6 fails, is q This is summable in q, and so by the Borel-Cantelli Lemma the measure of those x [n, n + for which 6 fails infinitely often is zero Let X be the set of all such x and let K = πx Notice that K is a set of full measure, not intersecting πq The averaging result we need is also well-known Lemma 6 Let k K and let f C 4 R have period π with π ft dt = 0 Then there 0 is C = Ck, f < such that for every θ R and n N n fθ + lk C l= Proof Let q= a qe iqt be the Fourier series of ft Since f is C 4, we know that a q cq 4 for some c Thanks to this fact all the sums appearing in this argument are pointwise absolutely convergent Also, a 0 = 0 by the hypothesis We have n fθ + lk q l= n a q e iqlk = a q l= q 0 π q 0 e iqnk e iqk where x = k/π X By Lemma 6 this is bounded by c + c q C < where c < depends on x and c q= a q distqk, πz c q 4 distqx, Z Let ω n be a random variable uniformly distributed over n, n +,, n} and let Ω, ν be the product probability space for these ω n s n N For ω = ω, ω, Ω let V v,γ ω v x = x ω n γ n + ω n for some n, x 6 0 otherwise q=

22 ANDREJ ZLATOŠ It is a consequence of the smallness of ω n compared to γ n, that all previous results for V v,γ apply to each V v,γ ω as well In this random case, however, we will prove and compute exact local dimension for the spectral measures for ae ω and ae boundary condition φ Moreover, our results will apply to any v, k and γ, not only to small v k and large γ Let us fix k K and define X n ω ln + v k sin θ k x ω n + vk sin θ k x ω n ln + v k 4 Obviously X n ω < M for some M = Mk < We will prove that for ae ω N n= X nω 0, 63 N which in turn implies see below that the exact dimension of µ ω φ for ae ω and ae φ is ln + v ln + k4 v 4 E = lnγ lnγ We use the same methods as earlier We exploit Lemmas 6 and 6 to show that the expectations of the cross terms X m X n for m < n are small, which implies 63 for ae ω This time, however, we consider expectations wrt ω, rather than k Let fθ ln + v k sinθ + vk sin θ ln + v k 4 Then fθ = fθ + π, π fθ dθ = 0, and f is C 4, so f is as in Lemma 6 Let δ, ω Ω 0 be such that δ j = ω j for j =,, n, and δ n = ω n + l for some l Then the recursive relation for the EFGP transform of generalized eigenfunctions implies that if X n ω = fθ for some θ, then X n δ = fθ + lk This shows why Lemma 6 enters in our argument For m < n we have with E = E ω E } EX m X n max Xm X n ω = c,, ω n = c n c,c,,c n max M E X n ω = c,, ω n = c n } c,c,,c n By Lemma 6 there exists C < such that the last expectation is at most C/n + Hence for fixed k K there is D = Dk < such that for m < n Then E EX m X n < D n N N X n E X n n= n= N 4 As before, using summability of E N 4 63 holds for ae ω / n= X n M N + DN / = cn / and boundedness of X n, one shows that

23 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 3 We have proved 63 for ae k and ae ω By the Fubini Theorem, 63 holds for ae ω and ae k Thus for these ω, k we obtain an exact power of the growth for the solution u k Namely we have that for any c < ln + v k < c 4 there is n 0 such that for n > n 0 R k x ω n + e c n, e c n Also, Lemma gives us the existence of a subordinate solution u sub k cosk, so that for large n P k x ω n + e c n P k, ψ k for energy Since for any such ω, for ae φ the spectral measure µ ω φ is supported on the set of the corresponding k s, one can conclude by methods presented earlier the following Theorem 63 Let Ω, ν be the above probability space For ω Ω let H ω φ be the discrete Schrödinger operator on Z + with potential V v,γ ω, given by 6 with v 0 and γ >, and boundary condition φ Let µ ω φ be its spectral measure Let J 4 v γ, 4 v γ if v < 4γ, and J otherwise Then for ae ω and for ae φ the measure µ ω φ purely singular continuous in J, with local dimension ln + v 4 E, lnγ and it is dense pure point in the rest of the interval [, ] is 7 Whole Line Operators With Symmetric Potentials It is readily seen that our results also apply to certain whole-line operators satisfying for x Z Let us consider the operator H on l Z with potential Ṽv,γ given by 6 for x and by Ṽv,γx = Ṽv,γ x for x 0 This potential is reflected about One easily sees that H = H E H O where E = u l Z u x = ux for all x Z} and O = u l Z u x = ux for all x Z} are, respectively, the even and odd subspaces of l Z On the other hand, if δ is the delta function at x =, then H E = H0 + δ = H 3π, 4 H O = H0 δ = H π 4 7 by simply taking the restriction of u to Z + Hence µ = µ π/4 + µ 3π/4 is a spectral measure for H, and all we have proved about the measures for the half-line operators applies to µ

24 4 ANDREJ ZLATOŠ Notice that variation of boundary condition ie, of V in the half-line case correponds to varying Ṽ = Ṽ 0, as can be seen from 7 We conclude: Theorem 7 Let H be the discrete Schrödinger operator on Z, with potential Ṽ v,γ given by 6 for x and reflected about Let Hφ H tanφδ 0 + δ and let µ φ be the spectral measure of H φ Then the statements of Theorems 4, 4, 5 and 5 hold for H φ If instead of H and Ṽv,γ we consider H ω and Ṽ v,γ ω given by 6 and reflected about, and H ω φ H ω tanφδ 0 + δ, then the statement of Theorem 63 holds for H ω φ Appendix A Dimensional Estimate Theorem A Let γ >, β > 0 and δ, δ [0, Let I be a finite interval and let f n CI be such that for all n N and ae k I Let f nk βγ n [ δ, + δ ] A F k lim N N N sin f n k and A ε k F k ε} Then there is ε 0 = ε 0 δ, δ, γ with ε 0 0 as δ, δ, γ 0 such that for ε > ε 0 the set A ε has Hausdorff dimension less than Remarks In fact, we obtain dima ε whenever ε > π δ + δ δ n= ε π δ +δ δ ln + 0 γ lnγ +δ δ Similar questions have been studied using dynamical systems and Riesz measures see, eg, [, 9] The methods, however, seem to require δ = δ = 0 and γ N 3 This result is only useful if we can take ε smaller than Hence even in the case δ = δ = 0 it applies only when γ is large larger than 0/ e 4 See Section 5 for the application of this result in the present paper The rest of this appendix is devoted to the proof of the above estimate First, we explain the presence of the term π δ +δ δ The reason is the same as in Section 4 We need more regularity than the function sinf n k possesses, and so we will need to break it up in two terms One of them will be small, with absolute value not more than π δ +δ δ, whereas the other one will be regular enough so that we will be able to prove for it the above theorem with ε replaced by just ε These two facts then yield the theorem as stated We note that the regular term, denoted X n, will be different from the X n term from Section 4, which is not regular enough for the purposes of this argument

25 SPARSE POTENTIALS WITH FRACTIONAL DIMENSION 5 Before we perform this breakup, notice that if we define intervals I n,j = [k n,j, k n,j+ ] in the same way as in Section 3, but with θx n, k replaced by f n k ie, so that f n k n,j are multiples of π, then A implies [ ] π I n,j βd γ, π A n βd γ n with D δ and D + δ We now define for k I fn k n,j + π k k n,j I ϕ n k n,j k I n,j f n k k I\[K n,, K n, ] Notice that ϕ n k n,j = f n k n,j for all j, and ϕ n k is linear on each I n,j So if we let X n k sinϕ n k, then X n is a series of exact sin waves on intervals I n,j This is the type of regularity we need and X n will be the regular term in our breakup Now we want to estimate the small term sinf n k X n k To do that, we need an upper bound on f n k ϕ n k This will be maximal if f nk equals βd γ n on some interval k n,j, k n,j + c and βd γ n on k n,j + c, k n,j+ or vice-versa, and the maximum will occur at k n,j + c Since in such case βd γ n c + βd γ n I n,j c = π by the definition of k n,j, we can compute I n,j and ϕ n in terms of c, and then maximize for c We obtain D D f n k ϕ n k π D + D which yields the above claimed estimate sinf n k X n k f n k ϕ n k π D D π 4D Now we only need to treat the term X n We start with a technical j=0 δ + δ δ Lemma A There is a constant c 0 such that for any n and any 0 ε n εn n c 0 n n e εn j we have Remark By the normal approximation to the binomial distribution, the left-hand side is roughly n Φ ε n < n e εn, where Φ is the standard normal distribution function The extra factor n is added for convenience of proof and can be removed Proof The sum is obviously smaller than n n, so we will estimate this By Stirling s n εn formula we have for n n n n n c n < n! < c n e e for some c, c > 0 Let n εn = δ n hence ε δ Then n c n n n [ e δ n [ ] c δn δn [ ] = c c n +δn +δn 0 δ δ +δ] n + δ e e

26 6 ANDREJ ZLATOŠ Since ε δ, it is sufficient to prove that δ δ + δ +δ e δ for 0 δ This can be done by taking the logarithm of both sides of the inequality and observing that the resulting quantities have the same value and first derivative for δ = 0, and the left hand side has a larger second derivative in 0, Let us now study a simple example which will illustrate our strategy Let us take X n k sgnsin n πk for k [0, ] with sgn0 Thus X n takes only values and, alternatively on intervals of lengths n Let us denote N A ε X k lim X } n= n k > ε N N We will show that the dimension of A ε X is smaller than Let S N be the union of those intervals [j N, j + N ] for j = 0,, N, in which N X n= n k εn Their number equals the number of sequences of N symbols from the alphabet, } such that the number of occurences of is at most ε N For any N the set N S N is obviously a N -cover of A ε X, and we have by Lemma A h α A ε X lim αn N N=N lim N = lim N ε N n=0 N n ε N αn c 0 N N e N=N N c 0 N α e ε N=N If α < is such that α e ε <, we get h α A ε X = 0 Since we can do the same for N B ε X n= k lim X n k N N < ε }, it follows, that the set k lim N N X } n= n k N > ε = A ε X B ε X has dimension at most α < We would like to prove now a similar result for our X n s There is, however, a problem The technique used in the above example was applicable to finite-valued functions only Thus we have to discretize X n via another breakup into a small and a nice term Pick