SINGULAR MEASURES AND THE LITTLE BLOCH SPACE

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.


Introduction and some known results
A function f holomorphic in the unit disk D is said to be in the little Bloch space B 0 , if lim Obviously, finite Blaschke products are in B 0 .Sarason [S] constructed an infinite Blaschke product in B 0 , from the singular inner function associated to a measure µ whose 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 I with specified rates of convergence to 0 of (1−|z| 2 )|I (z)|.Precisely,

A. Cantón
Theorem.There exists an inner function I such that As a corollary they obtain that for every continuous function a(t), defined in [0, 1) with a(0) = 0 and a(t) > 0 if t > 0, there exists an inner function I such that A (singular) inner function like I can be obtained by first constructing a positive singular measure µ on ∂D for which whenever J, J are contiguous intervals of the same size and then considering the associated inner function Their construction of the function I satisfying ( * ) 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 José L. Fernández for many helpful suggestions and to José G. Llorente and Paul MacManus for their careful reading of the manuscript.

The construction
From now on we will use the following notation: J J means that J, J are contiguous intervals of the same length; i.e., clos(J) ∩ clos(J ) is only one point and |J| = |J |, where |J| denotes the Lebesgue measure of J.
Theorem.There exists a positive measure µ on R such that: (1) µ(A) > 0, for every open set A.
(2) For every ε > 0 there exists δ > 0 such that if for all J J .(4) µ is 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 {f n } on [0, 1].That is, the sequence of functions {f n } 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: Each f n will be measurable with respect to the algebra generated by F n and so each f n will be constant on each J ∈ F n therefore, it makes sense to refer to the value of f n on J ∈ F n as f n (J).The martingale will have the following properties: (a) f n (J) > 0, for each J.
(b) For every ε > 0 there exists δ > 0 such that if |J| < δ then for each pair J, J ∈ F n with J J .(c) For all ε > 0 there exists δ > 0 such that if |J| < δ then

A. Cantón
Once {f n } has been constructed then one defines a measure µ on [0, 1] as follows, for this is a consistent definition of a measure on each σ(F n ) since {f n } is a martingale.Any other interval J ⊂ [0, 1] can be written as a disjoint union of 4-adic intervals, so µ( J) = J⊂ J, J 4-adic µ(J).We extend µ in 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 {α k }, {ε k }, {M k } which have the following properties, We also require ε k and M k to be multiples of α k and α k to divide α k−1 .These last conditions are only needed for technical reasons.
We will construct the martingale recursively: we start with a random walk with step α 1 and absorbing barriers at ε 1 and M 1 = M1 .At a certain time, say n 1 , we change the step to a smaller one α 2 and consider barriers at ε 2 , M 2 ε 1 and M2 .At time n 2 , we again change the step and the barriers as indicated above.We continue the construction indefinitely in this way.We will refer to ε k as the bottom barrier, M k ε k−1 as the middle barrier and Mk as the top barrier.We have chosen the ε k 's and M k 's to be possible positions of the random walk of step α k , 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 ε k and M k ε k−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 n k 's denote the time at which we change the step in the random walk.The criteria for choosing the n k 's will be given later.
To begin with, define f 0 (J) = 1 for J = [0, 1].To start with the induction, let us suppose we have defined the martingale for n ≤ n k−1 , that is {f n } n≤n k−1 and that we have chosen n j 's for j ≤ k − 1.
From now on and in order to avoid endless repetitions, we will consider I, I , J, J 4-adic intervals such that J ⊂ I, J ⊂ I and J J .Also we will write f n (J) only for J ∈ F n and I n will always denote an interval 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., (ii) If the random walk reaches the bottom barrier or the top barrier then we stop it, i.e., (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., (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.,

A. Cantón
Notice that f n ≥ ε k .To choose n k we need the following lemma, whose proof we postpone to Section 3. Lemma 1.Given α > 0, let ε, δ, M , be integer multiples of α so that 0 < ε < δ < M and let a be an integer greater than 1.
Let F 0 be a function measurable with respect to the σ-algebra generated by F N for some N ∈ N, whose values are integer multiples of α and lie between δ and M ; i.e., range(F 0 ) ∈ αN and δ ≤ F 0 (x) ≤ M for all x ∈ [0, 1].
For J ∈ F N+n , I ∈ F N+n−1 , and I 0 ∈ F N with J ⊂ I ⊂ I 0 , we define where η n is chosen so that, • if δ ≤ F 0 (I 0 ) ≤ aδ then, {F n } is a random walk of step α that starts at F 0 (I 0 ) and has absorbing barriers at ε and aδ, • if aδ ≤ F 0 (I 0 ) ≤ M , then {F n } is a random walk with step α that starts at F 0 (I 0 ) and has absorbing barriers at ε and aM .Under these assumptions there exists an integer ñ > 0 such that, Remark.The definition of η n in 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 Notice that F n = f n+n k−1 .Let ñk be the ñ given by the lemma and set n k = ñk + n k−1 .It follows that, Observe also that where This requires a little argument: by the way the martingale has been constructed, we have for Since ε k−1 is the lowest position the martingale could have reached at time Recall that ε k 's have been chosen so that Therefore if f n (J) has reached the bottom barrier n, must be so that )α k < 0 then either the random walk could not have been at the lowest position at time or the random walk is on the barrier ε k after certain time.
In particular, at time n k , "many" random walks could have reached the bottom barrier, and so 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 m k + n k−1 .
To construct the martingale for times n ≥ n k we restart the process again with new step α k+1 and barriers at ε k+1 , M k+1 ε k and Mk+1 .
C. Next we verify that the martingale we have just constructed satisfies the properties (a)-(d). (a) We have to consider several situations, depending on whether the intervals are children of the same parent or not.b.1.Same parent: J, J ⊂ I, then b.2.Different parents: J ⊂ I, J ⊂ I .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., We will see in (ii.2) that it is, in fact, the only possibility since the time we change the step n k−1 = n−1, is bigger than m k−1 +n k−2 (see ( * * ) after Lemma 1).So, in this case we have, (ii) With no change on the step, i.e., n k−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 f n−1 (I) nor f n−1 (I ) have reached a barrier, and that 0 (ii.2) Let us suppose again that at time n−1, f n−1 (I) and f n−1 (I ) have reached no barrier, and assume also that 2α n < m k + n k−1 so it has not been long enough to avoid the effect of change of the step), then and so, 0 ≤ f n (J) − f n (J ) ≤ 2α k−1 − 2(m + 1)α k .Note that when time runs long enough, i.e., n ≥ m k + n k−1 , we get |f n (J) − f n (J )| ≤ 2α k , and we are ready again to continue with the induction argument.In particular for n k , |f n k (J) − f n k (J )| ≤ 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 f n−1 (I) has reached a barrier, and f n−1 (I ) has not (when both have reached a barrier, trivially we get f n (J) = f n (J )).Since the random walk has already reached a barrier, by the observation after Lemma 1, In both cases b.1 and b.2 we get, |f n (J) − f n (J )| ≤ 2(α k−1 − α k ) and α k 0, therefore property (b) holds.
(c) Now we will prove that and α k and ε k have been chosen so that (d) To verify f n (x) → 0 a.e.x we just have to check that lim inf n→∞ f n (x)=0 a.e.x: Because of the Martingale Convergence Theorem and since the martingale is positive, we already know that lim n→∞ f n (x) exists a.e.x.We need to define the following sets.Let Clearly, because of the way the martingale was constructed B k ⊃ A k−1 .So, by the observation made after Lemma 1, we obtain 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 J that intersects J with |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.

Proof of Lemma 1
As a matter of fact we will prove a little more, that is, we will show that there exists a ñ > 0 such that We will need a lemma about random walks, see [F, p. 344-345] for references.
Lemma 2. Given j, M, k ∈ N, M > k > j, let S n be the random walk with step α > 0 and absorbing barriers at αj and αM .Then Here Therefore, there exists ñ > 0 such that, Because ε is an absorbing barrier, once the random walk reaches ε it never escapes.Consequently, which proves Lemma 1.