Singular measures and the little Bloch space

Publicacions Matem`atiques, Vol 42 (1998), 211–222. SINGULAR MEASURES AND THE LITTLE BLOCH SPACE Alicia Cant´ on Abstract Aleksandrov, Anderson and Nicolau have found examples of inner functions that are in the little Bloch space with a specific rate of convergence to zero. As a corollary they obtain positive singular measures defined in the boundary of the unit disc that are simoultaneously symmetric and Kahane. Nevertheless their construction is very indirect. We give an explicit example of such measures by means of a martingale argument. 1. Introduction and some known results A function fholomorphic in the unit disk Dis said to be in the little Bloch space B0,if lim |z|→1|f0(z)|(1 −|z| 2 )=0. Inner functions Iare bounded holomorphic functions on D={|z|<1} such that lim r→1|I(reiθ)|=1,for almost every θ∈[0,2π]. Obviously, finite Blaschke products are in B0. Sarason [S] constructed an infinite Blaschke product in B0, from the singular inner function associated to a measure µwhose indefinite integral is in the little Zygmund class. Bishop [B] gave a characterization of inner functions in the little Bloch space in terms of a certain associated measure, which for infinite Blaschke products turns out to be a characterization in terms of the distribution of its zeroes. Aleksandrov, Anderson and Nicolau [AAN] have found examples of inner functions Iwith specified rates of convergence to 0 of (1−|z|2)|I0(z)|. Precisely, 212 A. Cant´ on Theorem. There exists an inner function Isuch that (∗) lim |z|→1 |I0(z)|(1 −|z| 2 ) 1−|I(z)| 2=0. As a corollary they obtain that for every continuous function a(t), defined in [0,1) with a(0) = 0 and a(t)>0ift>0, there exists an inner function Isuch that lim |z|→1 |I0(z)|(1 −|z| 2 ) a(1 −|I(z)| 2 )=0. A (singular) inner function like Ican be obtained by first constructing a positive singular measure µon ∂Dfor which |µ(J)−µ(J0)|=o(µ(J)),as |J|→0, whenever J, J0are contiguous intervals of the same size and then considering the associated inner function I(z) = exp ½−Zeiθ +z eiθ −zdµ(θ)¾. Their construction of the function Isatisfying (∗) is indirect (see [AAN] for details). Anderson has asked for a direct construction. In this note we produce an explicit example. After this paper was written I learnt about some overlapping work of Wayne Smith. His paper is titled: Inner functions in the hyperbolic little Bloch class. I wish to give special thanks to Professor J. M. Anderson for proposing the problem and for discusions about it. I would also like to thank Jos´e L. Fern´andez for many helpful suggestions and to Jos´e G. Llorente and Paul MacManus for their careful reading of the manuscript. 2. The construction From now on we will use the following notation: JvJ0means that J,J0are contiguous intervals of the same length; i.e., clos(J)∩clos(J0) is only one point and |J|=|J0|, where |J|denotes the Lebesgue measure of J. Singular measures and the little Bloch space 213 Theorem. There exists a positive measure µon Rsuch that: (1) µ(A)>0, for every open set A. (2) For every ε>0there exists δ>0such that if |J|<δthen ¯¯¯¯ µ(J) |J|−µ(J0) |J0|¯¯¯¯ <ε for all JvJ0. (3) For all ε>0there exists δ>0such that if |J|<δthen ¯¯¯¯ µ(J)−µ(J0) µ(J)¯¯¯¯ <ε, for all JvJ0. (4) µis singular with respect to the Lebesgue measure. Proof: For the sake of clearness we are going to divide the proof into four stages labelled from A to D. A. We are going to construct a 4-adic martingale {fn}on [0,1]. That is, the sequence of functions {fn}will be adapted to the standard 4-adic filtration of [0,1]. Precisely, we consider the sequence of partitions of [0,1] into 4-adic intervals: Fn=½· j 4n,j+1 4 n¶:j=0,... ,4 n−1 ¾,n=0,1,2,... . Each fnwill be measurable with respect to the algebra generated by Fn and so each fnwill be constant on each J∈F ntherefore, it makes sense to refer to the value of fnon J∈F nas fn(J). The martingale will have the following properties: (a) fn(J)>0, for each J. (b) For every ε>0 there exists δ>0 such that if |J|<δthen |fn(J)−fn(J0)|<ε, for each pair J,J0∈F nwith JvJ0. (c) For all ε>0 there exists δ>0 such that if |J|<δthen ¯¯¯¯ fn(J)−fn(J0) fn(J)¯¯¯¯ <ε, for each pair J,J0∈F nwith JvJ0, (d) fn(x)→0a.e. x. 214 A. Cant´ on Once {fn}has been constructed then one defines a measure µon [0,1] as follows, for J∈F n , µ(J)=f n (J)|J|; this is a consistent definition of a measure on each σ(Fn) since {fn}is a martingale. Any other interval ˜ J⊂[0,1] can be written as a disjoint union of 4-adic intervals, so µ(˜ J)= X J⊂˜ J,J 4-adic µ(J). We extend µin the usual way to a Borel measure on [0,1]. Notice that for 4-adic intervals properties (a) to (d) of the martingale imply trivially properties (1) to (4) of the measure. B. Now we are going to develop the actual construction of the martingale. Let us consider three sequences of positive real numbers {αk}, {εk},{Mk}which have the following properties, αk&0; εk&0, ε0=1;M k%∞,M 1>2; αk−1 εk <1; αk−1 εk &0; εk<ε k−1−α k−1 ; M k ε k−1&0. Define mk= min{m∈N:0<α k−1−mαk≤αk}, and denote ˜ Mk=Qk j=1 Mj. We also require εkand Mkto be multiples of αkand αkto divide αk−1. These last conditions are only needed for technical reasons. We will construct the martingale recursively: we start with a random walk with step α1and absorbing barriers at ε1and M1=˜ M1.At a certain time, say n1, we change the step to a smaller one α2and consider barriers at ε2,M2ε1and ˜ M2. At time n2, we again change the step and the barriers as indicated above. We continue the construction indefinitely in this way. We will refer to εkas the bottom barrier,Mkεk−1 as the middle barrier and ˜ Mkas the top barrier. We have chosen the εk’s and Mk’s to be possible positions of the random walk of step αk,so it could land exactly at any of the barriers. The main idea is to “trap” “many” walks in a narrow band near zero (between εkand Mkεk−1) by waiting long enough. Then we change the step to a smaller one and make the band narrower and closer to zero. The top barrier plays no crucial role but it is useful for technical reasons. From now on the nk’s denote the time at which we change the step in the random walk. The criteria for choosing the nk’s will be given later. Singular measures and the little Bloch space 215 To begin with, define f0(J)=1forJ=[0,1]. To start with the induction, let us suppose we have defined the martingale for n≤nk−1, that is {fn}n≤nk−1and that we have chosen nj’s for j≤k−1. From now on and in order to avoid endless repetitions, we will consider I,I0,J,J04-adic intervals such that J⊂I,J0⊂I0and JvJ0. Also we will write fn(J) only for J∈F nand Inwill always denote an interval In∈F n . For n>n k−1 , we will construct the martingale considering four different cases that depend on what has been happening up to step n−1. (i) If the random walk at time n−1 is between the bottom barrier and the middle barrier, or if it is between the middle barrier and the top barrier, then we let it run “freely”; i.e., if εk<f n−1 (I)<M k ε k−1 ,orifM k ε k−1<f n−1 (I)<˜ M k , then fn(J)=f n−1 (I)+α kζ n (J), where ζn(J)∈{1,−1},PJ⊂I,J∈Fnζn(J) = 0, and the ζn(J) are chosen so that if fn−1(I)−fn−1(I0)>0 then ζn(J)=−1 and ζn(J0)=1. (ii) If the random walk reaches the bottom barrier or the top barrier then we stop it, i.e., if fn−1(I)=ε kor fn−1(I)= ˜ M kthen, fn(J)=f n−1 (I), for all J⊂I. (iii) If the random walk is on the middle barrier and it is the first time it has reached that barrier and has never been above it, then we stop the random walk (so we don’t let it go too high), i.e., if fn−1(I)=M k ε k−1and if for every Ij⊃I, nk−1≤j<n−1 f j (I j )≤M k ε k−1 , then fn(J)=f n−1 (I), for all J⊂I. (iv) Finally, if the random walk is on the middle barrier and it has already been above it at some time before, we let it run “freely”, (since we still have the chance it will reach the lower level at a later time), i.e., if fn−1(I)=M k ε k−1and if there exists a j, nk−1≤j<n−1, Ij⊃I, such that fj(Ij)>M k ε k−1 , then fn(J)=f n−1 (I)+ α k ζ n (J), where ζn(J) is as in (i). 216 A. Cant´ on Notice that fn≥εk. To choose nkwe need the following lemma, whose proof we postpone to Section 3. Lemma 1. Given α>0, let ε,δ,M, be integer multiples of αso that 0<ε<δ<M and let abe an integer greater than 1. Let F0be a function measurable with respect to the σ-algebra generated by FNfor some N∈N, whose values are integer multiples of αand lie between δand M; i.e., range(F0)∈αNand δ≤F0(x)≤Mfor all x∈[0,1]. For J∈F N+n ,I∈F N+n−1 , and I0∈F Nwith J⊂I⊂I0, we define Fn(J)=F n−1 (I)+αηn(J) where ηnis chosen so that, •if δ≤F0(I0)≤aδ then, {Fn}is a random walk of step αthat starts at F0(I0)and has absorbing barriers at εand aδ, •if aδ ≤F0(I0)≤M, then {Fn}is a random walk with step αthat starts at F0(I0)and has absorbing barriers at εand aM. Under these assumptions there exists an integer ˜n>0such that, |{F˜n=ε}| >1−1 a. Remark. The definition of ηnin Lemma 1 is a shorter way to write items (i)-(iv) in the definition of the martingale. In this case, the botton barrier is ε, the middle barrier is aδ and the top barrier is aM. Take F0=fnk− 1(therefore N=nk−1), α=αk,δ=εk−1,ε=εk, M=˜ Mk−1and a=Mk. Notice that Fn=fn+nk−1. Let ˜nkbe the ˜ngiven by the lemma and set nk=˜n k+n k−1 . It follows that, |{fnk=εk}| >1−1 Mk . Observe also that (∗∗)nk>m k+n k−1 , where mk= min{m∈N:0<α k−1−mαk≤αk}. This requires a little argument: by the way the martingale has been constructed, we have for n>n k−1 , that fn(J)≥fnk−1(Ink−1)−(n−nk−1)αk. Singular measures and the little Bloch space 217 Since εk−1is the lowest position the martingale could have reached at time nk−1, that is, fnk−1(Ink−1)>ε k−1 , we have that, fn(J)≥εk−1−(n−nk−1)αk. Recall that εk’s have been chosen so that εk−1>ε k+α k−1 . Then, fn(J)≥εk+αk−1−(n−nk−1)αk. Therefore if fn(J) has reached the bottom barrier n, must be so that αk−1−(n−nk−1)αk≤0 and then n>m k+n k−1 . Observe that if αk−1−(n−nk−1)αk<0 then either the random walk could not have been at the lowest position at time nk−1, that is, fnk−1(Ink−1)>ε k−1or the random walk is on the barrier εkafter certain time. In particular, at time nk, “many” random walks could have reached the bottom barrier, and so nk>m k+n k−1 . Finally, notice that a similar argument works with the top and middle barriers. That means that the random walk cannot reach a barrier before time mk+nk−1. To construct the martingale for times n≥nkwe restart the process again with new step αk+1 and barriers at εk+1,Mk+1εkand ˜ Mk+1. C. Next we verify that the martingale we have just constructed satisfies the properties (a)-(d). (a) fn>0 for all n∈N. In particular, if n≤nkthen fn≥εk>0. (b) |fn(J)−fn(J0)|→0, |J|→0, JvJ0. We have to consider several situations, depending on whether the intervals are children of the same parent or not. b.1. Same parent: J, J0⊂I, then |fn(J)−fn(J0)|=½0 2αk for nk−1≤n<n k . b.2. Different parents: J⊂I,J0⊂I0. We have to distinguish between two different situations, that depend on whether the martingale has changed its step. (i) With change in the step of the martingale, i.e.,nk−1=n−1< n<n k . 218 A. Cant´ on Assume that 0 ≤fn−1(I)−fn−1(I0)≤2αk−1. We will see in (ii.2) that it is, in fact, the only possibility since the time we change the step nk−1=n−1, is bigger than mk−1+nk−2(see (∗∗) after Lemma 1). So, in this case we have, fn(J)=f n−1 (I)−α k f n (J 0 )=f n−1 (I 0 )+α k, and thus −2αk≤fn(J)−fn(J0)≤2(αk−1−αk). (ii) With no change on the step, i.e.,nk−1<n−1<n≤n k .We will use an induction argument, but again several cases come into consideration. (ii.1) Suppose first that at time n−1, neither fn−1(I) nor fn−1(I0) have reached a barrier, and that 0 ≤fn−1(I)−fn−1(I0)≤ 2αk(that is, for example Iand I0are children of the same parent or that n≥mk+nk−1), then fn(J)=f n−1 (I)−α k f n (J 0 )=f n−1 (I 0 )+α k and therefore, 0 ≤fn(J0)−fn(J)≤2αk. (ii.2) Let us suppose again that at time n−1, fn−1(I) and fn−1(I0) have reached no barrier, and assume also that 2αk≤ fn−1(I)−fn−1(I0)≤2αk−1−2mαkfor m<m k ,(i.e. n<m k+n k−1so it has not been long enough to avoid the effect of change of the step), then fn(J)=f n−1 (I)−α k f n (J 0 )=f n−1 (I 0 )+α k and so, 0 ≤fn(J)−fn(J0)≤2αk−1−2(m+1)α k. Note that when time runs long enough, i.e.,n≥mk+nk−1,we get |fn(J)−fn(J0)|≤2α k , and we are ready again to continue with the induction argument. In particular for nk,|fnk(J)− fnk(J0)|≤2α k , and so, as mentioned, the assumption in case (i) of change of step turns out to be always true. (ii.3) And the remaining case is when fn−1(I) has reached a barrier, and fn−1(I0) has not (when both have reached a barrier, trivially we get fn(J)=f n (J 0 )). Since the random walk has already reached a barrier, by the observation after Lemma 1, n−1>m k+n k−1 , and then, |fn−1(I)−fn−1(I0)|≤2α k ,so f n (J)=f n−1 (I) f n (J 0 )=f n−1 (I 0 )+ζ n (J0 )α k therefore, |fn(J)−fn(J0)|≤α k . Singular measures and the little Bloch space 219 In both cases b.1 and b.2 we get, |fn(J)−fn(J0)|≤2(αk−1−αk) and αk&0, therefore property (b) holds. (c) Now we will prove that |fn(J)−fn(J0)| |fn(J)|→0 for intervals JvJ0,nk−1<n≤n k , then ¯¯¯¯ fn(J)−fn(J0) fn(J)¯¯¯¯ ≤2αk−1 εk , and αkand εkhave been chosen so that αk−1 εk &0. (d) To verify fn(x)→0 a.e. xwe just have to check that lim inf n→∞ fn(x)=0 a.e. x: Because of the Martingale Convergence Theorem and since the martingale is positive, we already know that lim n→∞ fn(x) exists a.e. x.We need to define the following sets. Let Bk={x:εk≤fnk(x)≤εk−1Mk} Ak={x:fnk(x)=ε k }. Clearly, because of the way the martingale was constructed Bk⊃Ak−1. So, by the observation made after Lemma 1, we obtain |Bk|≥|A k−1 |> 1−1 M k−1, which together with εk&0, εk−1Mk&0 and Mk%∞, yields fn(x)→0 for a.e. x. As we have mentioned above, properties (a) to (d) of the martingale imply trivially properties (1) to (4) of the measure for 4-adic intervals. D. For general intervals one should use Kahane’s argument in [K, p. 190], where essentially he compares the µ-length of a general interval ˜ J, with the µ-length of the smallest 4-adic interval Jthat intersects ˜ Jwith |J|>|˜ J|. In our case, extra care needs to be taken because of the change of step in the martingale, but this causes only minor changes.