Advances in Mathematics 401 (2022) 108318 Contents lists available at ScienceDirect Advances in Mathematics www.elsevier.com/locate/aim A Central Limit Theorem for inner functions Artur Nicolau a,∗,1 , Odí Soler i Gibert b,1 a Universitat Autònoma de Barcelona, Departament de Matemàtiques, and Centre de Recerca Matemàtica, 08193 Barcelona, Spain b Julius-Maximilians-Universität Würzburg, Institut für Mathematik, Emil-Fischer-Strasse 40, 97074 Würzburg, Germany a r t i c l e i n f o a b s t r a c t Article history: A Central Limit Theorem for linear combinations of iterates Received 30 November 2020 of an inner function is proved. The main technical tool is Received in revised form 17 January Aleksandrov Desintegration Theorem for Aleksandrov-Clark 2022 measures. Accepted 17 February 2022 Available online 14 March 2022 © 2022 The Authors. Published by Elsevier Inc. This is an Communicated by C. Fefferman open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). MSC: primary 30D05, 30J05 secondary 37A05, 37F10 Keywords: Inner functions Aleksandrov-Clark measures Central Limit Theorem * Corresponding author. E-mail addresses: artur@mat.uab.cat (A. Nicolau), odi.solerigibert@mathematik.uni-wuerzburg.de (O. Soler i Gibert). 1 Both authors have been supported in part by the Generalitat de Catalunya (grant 2017 SGR 395), the Spanish Ministerio de Ciencia e Innovación (project MTM2017-85666-P) and the Spanish Research Agency (María de Maeztu Program CEX2020-001084-M). Second author has also been supported by the ERC project CHRiSHarMa no. DLV-682402. https://doi.org/10.1016/j.aim.2022.108318 0001-8708/© 2022 The Authors. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). 2 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 1. Introduction and main results Inner functions are analytic mappings from the unit disc D into itself whose radial limits are of modulus one at almost every point of the unit circle ∂D. Inner functions were introduced by R. Nevanlinna and after the pioneering work of brothers Riesz, Frostmann and Beurling, they have become a central notion in Analysis. See for instance [17]. Any inner function f induces a mapping from the unit circle into itself defined at almost every point z ∈ ∂D by f (z) = limr→1 f (rz). This boundary mapping will be also called f . It is well known that if f (0) = 0, normalized Lebesgue measure m in the unit circle is invariant under this mapping, that is, m(f −1 (E)) = m(E) for any measurable set E ⊂ ∂D. Several authors have also studied the distortion of Hausdorff measures by this mapping. See [15] and [18]. Dynamical properties of the mapping f : ∂D → ∂D, as recurrence, ergodicity, mixing, entropy and others have been studied by Aaronson [1], Crazier [10], Doering and Mañe [11], Fernández, Melián and Pestana [14], [13], Neurwirth [19], Pommerenke [21], and others. Dynamical properties of inner functions have been recently used in several problems on the dynamics of meromorphic functions in simply connected Fatou components. See [5], [6] and [12]. It is well known that in many senses lacunary series behave as sums of independent random variables. Salem and Zygmund ([23] and [24]) proved a version of the Central Limit Theorem for lacunary series and, a few years later, Weiss proved a version of the Law of the Iterated Logarithm in this context ([26]). Our main result is a Central Limit Theorem for linear combinations of iterates of an inner function fixing the origin. It is worth mentioning that in our result no lacunarity assumption is needed. Recall that a sequence of measurable functions {fN } defined at almost every point in the unit circle converges in distribution to a (circularly symmetric) standard complex normal variable if and only if for any Borel set K ⊂ C such that its boundary ∂K has zero area one has  1 e−|w| 2 lim m ({z ∈ ∂D : fN (z) ∈ K}) = /2 dA(w). N →∞ 2π K As it is usual we denote by f n the n-th iterate of the function f . Theorem 1. Let f be an inner function with f (0) = 0 which is not a rotation. Let {an } be a sequence of complex numbers. Consider  N  N  N −k 2 σN = |an |2 + 2 Re f  (0)k an an+k , N = 1, 2, . . . (1.1) n=1 k=1 n=1 Assume there exists a constant η > 0 such that   sup |an |2 : n ≤ N lim  (1−η)/2 = 0. (1.2) N →∞ N n=1 |a n |2 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 3 Then 1  N √ an f n 2−1 σN n=1 converges in distribution to a standard complex normal variable. Let f be an inner function fixing the origin. Then it is well known that Lebesgue measure m is ergodic. Hence the classical Ergodic Theorem gives that 1  n N lim f (z) = 0 N →∞ N n=1 at almost every point z ∈ ∂D. This can be understood as a version of the Law of Large Numbers. Our result provides the corresponding version of the Central Limit Theorem. Actually taking an = 1, n = 1, 2 . . ., in Theorem 1, one can easily show that 2 limN →∞ σN /N σ 2 = 1, where 1 + f  (0) σ 2 = Re (1.3) 1 − f  (0) and we deduce the following result. Corollary 2. Let f be an inner function with f (0) = 0 which is not a rotation. Then 1  N √ fn 2−1 N n=1 converges in distribution to a complex normal variable with mean 0 and variance σ 2 given by (1.3), that is, for any Borel set K ⊂ C such that its boundary ∂K has zero area, we have  N  1 z ∈ ∂D : (2−1 N )−1/2 e−|w| 2 /2σ 2 lim m f n (z) ∈ K = dA(w). N →∞ n=1 2πσ 2 K Observe that when f  (0) is close to 1 and hence f is close to be the identity map, the variance σ 2 is large. However if f  (0) is close to a unimodular constant different from 1, the variance is small. On the opposite side, if f  (0) = 0, σ = 1. Let H 2 be the Hardy space of analytic functions in D whose Taylor coefficients are square summable. Let {an } be a sequence of complex numbers. It is easy to show (see   Theorem 9) that n an f n converges in H 2 if and only if n |an |2 < ∞. A repetition of the proof of our main result gives the following version of the Central Limit Theorem for the tails. 4 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 Theorem 3. Let f be an inner function with f (0) = 0 which is not a rotation. Let {an } be a square summable sequence of complex numbers. Consider    σ 2 (N ) = |an |2 + 2 Re f  (0)k an an+k , N = 1, 2, . . . (1.4) n≥N k≥1 n≥N Assume there exists a constant η > 0 such that   sup |an |2 : n ≥ N lim  (1−η)/2 = 0. (1.5) N →∞  n≥N |a n |2 Then ∞  1 √ an f n 2−1 σ(N ) n=N converges in distribution to a standard complex normal variable. N Let SN2 = n=1 |an |2 . It is easy to show (see Theorem 9) that there exists a constant κ = κ(f ) > 0 such that κ−1 SN 2 ≤ σN 2 ≤ κSN 2 , N = 1, 2, . . .. When f  (0) = 0 we have σN = SN but in general, both quantities do not coincide. However if the following uniform quasiorthogonality condition holds N −k supk≤N n=1 an an+k lim 2 = 0, (1.6) N →∞ SN then limN →∞ SN /σN = 1 and Theorem 1 gives that 1 N √ an f n 2−1 SN n=1 converges in distribution to a standard complex normal variable. We now make some remarks on the assumption and proof of Theorem 1. Condition  n |an | = ∞, but one can not expect this last condition to be 2 (1.2) implies that sufficient in Theorem 1. However note that if {an } is bounded, both conditions are equivalent. The proof of Theorem 1 uses two relevant properties of the iterates of an inner function fixing the origin. The first one is that the square of the modulus of the partial sums are uncorrelated. More concretely, given a set A of positive integers, consider the corresponding partial sum  ξ(A) = an f n . n∈A If max(A) < min(B), we will show in Theorem 6 that A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 5 ⎛ ⎞⎛ ⎞    |ξ(A)|2 |ξ(B)|2 dm = ⎝ |ξ(A)|2 dm⎠ ⎝ |ξ(B)|2 dm⎠ . (1.7) ∂D ∂D ∂D The second property provides an exponential decay of the higher order correlations of the iterates. More concretely, let εi = 1 or εi = −1 for i = 1, 2, . . . , k and n1 < . . . < nk be positive integers satisfying nj − nj−1 ≥ q ≥ 1, j = 2, . . . , k. Denote ε = (ε1 , . . . , εk ) and n = (n1 , . . . , nk ). For a positive integer n, denote by f −n the function defined by f −n (z) = f n (z), z ∈ ∂D. We will prove in Theorem 13 that there exists a constant C > 0, independent of the indices, such that   k f εj nj dm ≤ C k k!|f  (0)|Φ(ε,n) , k = 1, 2, . . . , (1.8) j=1 ∂D if q is sufficiently large and where Φ is a certain function depending on the choice of indices that satisfies Φ(ε, n) ≥ kq/4. The main technical tool in the proof of both prop- erties (1.7) and (1.8) is the theory of Aleksandrov-Clark measures and more concretely, the Aleksandrov Desintegration Theorem. The paper is organized as follows. In Section 2 we introduce Aleksandrov-Clark mea- sures and use them to prove property (1.7). In Section 3 we estimate the L2 and the L4 norm of ξ(A). In Section 4 we prove estimate (1.8). The proof of Theorem 1 is given in Section 5. 2. Aleksandrov-Clark measures and property (1.7) We start with an elementary auxiliary result which is just a restatement of the invari- ance of Lebesgue measure. Lemma 4. Let f be an inner function with f (0) = 0. (a) Let G be an integrable function on ∂D. Then   G(f (z)) dm(z) = G(z) dm(z) ∂D ∂D (b) Let k < j be positive integers. Then  f k f j dm = f  (0)j−k ∂D Proof of Lemma 4. We can assume that G is the characteristic function of a measurable set E ⊂ ∂D. Since m(f −1 (E)) = m(E), the identity (a) follows. Using (a) and Cauchy formula, we have 6 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318   f k f j dm = zf j−k (z) dm(z) = f  (0)j−k .  ∂D ∂D Given an analytic mapping from the unit disc into itself and a point α ∈ ∂D, the function (α + f )/(α − f ) has positive real part and hence there exists a positive measure μα = μα (f ) in the unit circle and a constant Cα ∈ R such that  α + f (w) z+w = dμα (z) + iCα , w ∈ D. (2.1) α − f (w) z−w ∂D The measures {μα : α ∈ ∂D} are called the Aleksandrov-Clark measures of the function f . Clark introduced them in his paper [9] and many of their deepest properties were found by Aleksandrov in [2], [3] and [4]. The two surveys [20] and [22] as well as [8, Chapter IX] contain their main properties and a wide range of applications. Observe that if f (0) = 0 then μα are probability measures. Moreover, f is inner if and only if μα is a singular measure for some (all) α ∈ ∂D. From the definition it is clear that, in the case that f is an inner function, the mass of μα is carried by the set f −1 ({α}) ⊂ ∂D. Assume f (0) = 0. Computing the first two derivatives in formula (2.1) and evaluating at the origin, we obtain  z dμα (z) = f  (0)α, α ∈ ∂D, (2.2) ∂D and  f  (0) 2 z 2 dμα (z) = α + f  (0) α2 , α ∈ ∂D. (2.3) 2 ∂D Our main technical tool is Aleksandrov Desintegration Theorem which asserts that  m= μα dm(α) (2.4) ∂D holds true in the sense that    G dm = G(z) dμα (z) dm(α), ∂D ∂D ∂D for any integrable function G on the unit circle. Aleksandrov Desintegration Theorem will be used in our next auxiliary result. Lemma 5. Let f be an inner function with f (0) = 0. For k = 1, 2, . . . , p, let nk , jk , be positive integers such that A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 7 max{nk , jk } < min{nk+1 , jk+1 }, k = 1, . . . , p − 1. (2.5) Then   p p   f nk f jk dm = f nk f jk dm. (2.6) ∂D k=1 k=1∂D Proof of Lemma 5. We argue by induction on p. Assume (2.6) holds for p − 1 products. We can assume n1 < j1 . By part (a) of Lemma 4 we have   p   p f nk f jk dm = zf j1 −n1 (z) f nk −n1 (z)f jk −n1 (z) dm(z). ∂D k=1 ∂D k=2 Let {μα : α ∈ ∂D} be the Aleksandrov-Clark measures of the inner function f j1 −n1 . The Aleksandrov Desintegration Theorem (2.4) gives that last integral can be written as    p zα f nk −j1 (α)f jk −j1 (α) dμα (z) dm(α). ∂D ∂D k=2 By (2.2) and part (b) of Lemma 4, we have   j1 −n1 z dμα (z) = f  (0) α =α f n1 f j1 dm. ∂D ∂D Hence ⎛ ⎞   p    p f nk f jk dm = ⎝ f n1 f j1 dm⎠ f nk −j1 f jk −j1 dm ∂D k=1 ∂D ∂D k=2 and we can apply the inductive assumption. The invariance property of part (a) of Lemma 4 finishes the proof.  Our next result is the first important tool in the proof of Theorem 1. Theorem 6. Let f be an inner function with f (0) = 0. Let Ak , k = 1, 2, . . . , p, be finite collections of positive integers such that max{n : n ∈ Ak } < min{n : n ∈ Ak+1 }, k = 1, . . . , p − 1. (2.7) Consider  ξk = an f n . n∈Ak 8 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 Then   p p   |ξk | dm = 2 |ξk |2 dm. ∂D k=1 k=1∂D Proof of Theorem 6. Al almost every point of the unit circle we have   |ξk |2 = |an |2 + (an aj f n f j + aj an f j f n ), n∈Ak  where the last sum is taken over all indices n, j ∈ Ak with j > n. Hence |ξk |2 can be written as a linear combination of terms of the form  f n k f jk , where nk , jk ∈ Ak . Observe that (2.7) gives the assumption (2.5) in Lemma 5. Now Lemma 5 finishes the proof.  3. Norms of partial sums In this Section we will use Aleksandrov-Clark measures to estimate the L2 and L4 norms of linear combinations of iterates of an inner function fixing the origin. The main result of this Section is Theorem 9. It is worth mentioning that the asymptotic behavior of the Aleksandrov-Clark measures of iterates of an inner function has been studied in [16], but we will not use their results. As before, if n is a positive integer, we will use the notation f −n to denote the function defined by f −n (z) = f n (z), for almost every z ∈ ∂D. We start with a technical auxiliary result which will be used later. Lemma 7. Let f be an inner function with f (0) = 0 which is not a rotation. Let εk = 1 or εk = −1, k = 1, 2, 3, 4. (a) Let nk , k = 1, 2, 3, 4, be positive integers with max{n1 , n2 } < min{n3 , n4 }. Then  I = I(ε1 n1 , −ε1 n2 , n3 , n4 ) = f ε1 n1 f −ε1 n2 f n3 f n4 dm = 0. ∂D (b) Let n1 < n2 < n3 be positive integers and  II = II(ε1 n1 , ε2 n2 , ε3 n3 ) = f ε1 n1 (f ε2 n2 )2 f ε3 n3 dm. ∂D Then there exists a constant C = C(f ) > 0 independent of the indices n1 , n2 , n3 , such that |II| ≤ C|f  (0)|n3 −n1 . A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 9 (c) Let n1 < n2 < n3 be positive integers and  III = III(ε1 n1 , ε2 n2 , ε3 n3 ) = (f ε1 n1 )2 f ε2 n2 f ε3 n3 dm. ∂D Then there exists a constant C = C(f ) > 0 independent of the indices n1 , n2 , n3 , such that |III| ≤ 1 if n2 = n1 + 1 and n3 ≤ n2 + 2, and |III| ≤ C|f  (0)|n3 −n1 otherwise. (d) Let n1 < n2 < n3 < n4 be positive integers and  IV = IV (ε1 n1 , ε2 n2 , ε3 n3 , ε4 n4 ) = f ε1 n1 f ε2 n2 f ε3 n3 f ε4 n4 dm. ∂D Then there exists a constant C = C(f ) > 0 independent of the indices n1 , n2 , n3 , n4 , such that |IV | ≤ C|f  (0)|n2 −n1 +n4 −n3 if n4 − n3 > 2, and |IV | ≤ C|f  (0)|n3 −n1 if n4 − n3 ≤ 2. Moreover |IV | = |f  (0)|n2 −n1 +n4 −n3 if ε1 ε2 = ε3 ε4 = −1. Proof of Lemma 7. Let C denote a positive constant which may depend on the function f but not on the indices {ni }, whose value may change from line to line. (a) We can assume that n1 < n2 . Part (a) of Lemma 4 gives that  I= z ε1 f −ε1 (n2 −n1 ) (z)f n3 −n1 (z)f n4 −n1 (z) dm(z). ∂D Let {μα : α ∈ ∂D} be the Aleksandrov-Clark measures of f n2 −n1 . The Aleksandrov Desintegration Theorem (2.4) gives   I= z ε1 α−ε1 f n3 −n2 (α)f n4 −n2 (α) dμα (z) dm(α). ∂D ∂D By (2.2)  z ε1 dμα (z) = aαε1 , α ∈ ∂D, ∂D where |a| = |f  (0)|n2 −n1 . Since f (0) = 0, we deduce   n2 −n1 |I| = |f (0)| f n3 −n2 (α)f n4 −n2 (α) dm(α) = 0. ∂D 10 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 (b) We can assume ε1 = 1. Part (a) of Lemma 4 gives that  II = z(f ε2 (n2 −n1 ) (z))2 f ε3 (n3 −n1 ) (z) dm(z). ∂D Let {μα : α ∈ ∂D} be the Aleksandrov-Clark measures of f n2 −n1 . The Aleksandrov Desintegration Theorem (2.4) gives   II = zα2ε2 f ε3 (n3 −n2 ) (α) dμα (z) dm(α). ∂D ∂D By (2.2)  n2 −n1 z dμα (z) = f  (0) α, α ∈ ∂D. ∂D Hence  n2 −n1 II = f  (0) α1+2ε2 f ε3 (n3 −n2 ) (α) dm(α) ∂D Since 1 + 2ε2 ≤ 3, the modulus of last integral is bounded by C|f  (0)|n3 −n2 if n3 − n2 > 2 and by 1 otherwise. This proves (b). (c) We can assume ε1 = 1. Applying part (a) of Lemma 4 and Aleksandrov Desinte- gration Theorem as before, we have   III = z 2 αε2 f ε3 (n3 −n2 ) (α) dμα (z) dm(α), ∂D ∂D where {μα : α ∈ ∂D} are the Aleksandrov-Clark measures of g = f n2 −n1 . Applying (2.3), we obtain   g  (0) III = α1+ε2 f ε3 (n3 −n2 ) (α) dm(α) + g  (0)2 α2+ε2 f ε3 (n3 −n2 ) (α) dm(α). 2 ∂D ∂D Since 2 + ε2 ≤ 3, both integrals are bounded by C|f  (0)|n3 −n2 if n3 − n2 > 2, and by 1 if n3 − n2 ≤ 2. If n2 − n1 > 1, we have that |g  (0)|/2 + |g  (0)2 | ≤ C|f  (0)|n2 −n1 . If n2 − n1 = 1, we have that |g  (0)|/2 + |g  (0)2 | ≤ 2. This proves (c). (d) We can assume ε1 = 1. Arguing as before we have  n2 −n1 IV = f  (0) α1+ε2 f ε3 (n3 −n2 ) (α)f ε4 (n4 −n2 ) (α) dm(α) ∂D A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 11 If ε2 = −1, we repeat the argument and prove that |IV | ≤ |f  (0)|n2 −n1 +n4 −n3 . Moreover if ε2 = −1 and if ε3 ε4 = −1, we have |IV | = |f  (0)|n2 −n1 +n4 −n3 , as stated in the last part of (d). If ε2 = 1, let {μα : α ∈ ∂D} be the Aleksandrov-Clark measures of g = f n3 −n2 . The Aleksandrov Desintegration Theorem (2.4) gives that last integral can be written as   z 2 αε3 f ε4 (n4 −n3 ) (α) dμα (z) dm(α). (3.1) ∂D ∂D By (2.3)  g  (0) z 2 dμα (z) = α + g  (0)2 α2 , α ∈ ∂D. 2 ∂D Hence the double integral in (3.1) can be written as   g  (0) α1+ε3 f ε4 (n4 −n3 ) (α) dm(α) + g  (0)2 α2+ε3 f ε4 (n4 −n3 ) (α) dm(α). 2 ∂D ∂D Since 2 + ε3 ≤ 3, both integrals are bounded by C|f  (0)|n4 −n3 if n4 − n3 > 2, and by 1 if n4 − n3 ≤ 2. If n3 − n2 > 1, we have that |g  (0)|/2 + |g  (0)2 | ≤ C|f  (0)|n3 −n2 . If n3 − n2 = 1, we just use the trivial estimate |g  (0)|/2 + |g  (0)2 | ≤ 2. This proves (d).  We will now prove an elementary auxiliary result which will be used several times. Lemma 8. Let A be a collection of positive integers and let {an } be a sequence of complex numbers. Fix λ ∈ C with |λ| < 1. Then  |λ|  an ak λk−n ≤ |an |2 . 1 − |λ| n,k∈A,k>n n∈A Proof of Lemma 8. Writing j = k − n we have that    an ak λk−n = λj an an+j , n,k∈A,k>n j>0 n,n+j∈A where the last sum is taken over all indices n ∈ A such that n + j ∈ A. It is also understood that this sum vanishes if there is no n ∈ A such that n + j ∈ A. By Cauchy- Schwarz’s inequality,   an an+j ≤ |an |2 . n,n+j∈A n∈A 12 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 This finishes the proof.   Let H 2 be the Hardy space of analytic functions in the unit disc g(w) = n≥0 an w n , w ∈ D, such that  ∞  g 2 2 = sup |g(rz)|2 dm(z) = |an |2 < ∞. r<1 n=0 ∂D Any function g ∈ H 2 has a finite radial limit g(z) = limr→1 g(rz) at almost every z ∈ ∂D and  2 g 2= |g(z)|2 dm(z). ∂D See [17]. For 0 < p < ∞ let g p denote the Lp norm on the unit circle of the function g. Next result provides estimates of the L2 and L4 norms of linear combinations of iterates of an inner function. It will be applied to finite linear combinations. For t, z ∈ C, let t, z = Re(tz) be the standard scalar product in the plane. Theorem 9. Let f be an inner function with f (0) = 0 which is not a rotation and let  {an } be a sequence of complex numbers with n |an |2 < ∞. Consider ∞  ξ= an f n n=1 and ∞  ∞  ∞  σ2 = |an |2 + 2 Re f  (0)k an an+k . n=1 k=1 n=1 2 (a) We have ξ 2 = σ 2 and ∞  ∞  κ−1 |an |2 ≤ σ 2 ≤ κ |an |2 , n=1 n=1 where κ = (1 + |f  (0)|)(1 − |f  (0)|)−1 . (b) For any t ∈ C we have  1 2 2 t, ξ 2 dm = |t| σ . 2 ∂D (c) There exists a constant C = C(f ) > 0 independent of the sequence {an }, such that ξ 4 ≤ C ξ 2. A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 13 Proof of Theorem 9. At almost every point of the unit circle we have ∞  |ξ| =2 |an |2 + 2 Re h, (3.2) n=1 where  h= an ak f k f n . (3.3) n,k≥1,k>n Part (b) of Lemma 4 gives ∞   ξ 2 2 = |an |2 + 2 Re an ak f  (0)k−n , (3.4) n=1 n,k≥1,k>n which is the identity in (a). Next we prove the estimate in (a). To that end, let us first denote  bk,n = f k f n dm ∂D and recall that, by part (b) of Lemma 4, it holds that bn,k = f  (0)n−k if n ≥ k and k−n bn,k = f  (0) if n < k. Now, rewrite identity (3.4) as  2     n an f  = an ak bn,k  n 2 n,k and consider the Toeplitz matrix T whose entries are bn,k = bn−k,0 , n, k = 1, 2, . . . and its symbol ∞  s(z) = bn,0 z n , z ∈ ∂D. n=−∞ It is well known that T diagonalizes and its eigenvalues are contained in the interval in the real line whose endpoints are the essential infimum and the essential supremum of s. See [7]. Since 1 − |f  (0)|2 s(z) = , z ∈ ∂D, |1 − f  (0)z|2 the eigenvalues of T are between κ−1 and κ. This finishes the proof of part (a). Since f (0) = 0, the mean value property gives that 14 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318  ξ 2 dm = 0 ∂D and (b) follows. We now prove (c). Let C(f ) denote a positive constant only depending on f whose value may change from line to line. The identity (3.2) gives that at almost every point of the unit circle, we have ∞ 2 ∞   |ξ| = 4 |an | 2 + 4 Re h |an |2 + 4(Re h)2 , n=1 n=1 where h is defined in (3.3). Observe that   h dm = an ak f  (0)k−n . ∂D n,k≥1,k>n Hence Lemma 8 gives that  ∞ |f  (0)|  h dm ≤ |an |2 . (3.5) 1 − |f  (0)| n=1 ∂D Next we will prove that there exists a constant C = C(f ) > 0 such that  ∞  2 |h|2 dm ≤ C |an |2 . (3.6) n=1 ∂D Observe that (3.5) and (3.6) give the estimate in (c). Write  cn = an ak f k f n . k>n Using the elementary identity ∞ 2 ∞ ∞     cn = |cn |2 + 2 Re cn cj , n=1 n=1 n=1 j>n we can write  |h|2 dm = A + 2 Re B, ∂D where A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 15 ∞    2 A= |an | 2 ak f k dm n=1 k>n ∂D and ∞      B= an ak aj al f k f n f j f l dm. (3.7) n=1 k>n j>n l>j ∂D By part (a) we have   2  ak f k dm ≤ C(f ) |ak |2 ∂D k>n k>n  2 and we deduce that A ≤ C(f ) n |an |2 . We now estimate B. If n < k and n < j < l, we have  f n f k f j f l dm = |f  (0)|r−n+|l−s| , ∂D where r = min{k, j} and s = max{k, j}. This estimate follows from last statement in part (d) of Lemma 7. Part (b) of Lemma 7 gives that  f n (f k )2 f l dm ≤ C(f )|f  (0)|l−n , ∂D if n < k < l. The sum over j > n in (3.7) will be splitted in three terms corresponding to j > k, j = k and j < k. Then |B| ≤ C(f )(B1 + B2 + B3 ) where     B1 = |an | |ak | |aj | |al ||f  (0)|k−n+l−j , n≥1 k>n j>k l>j    B2 = |an | |ak |2 |al ||f  (0)|l−n , n≥1 k>n l>k     B3 = |an | |ak | |aj | |al ||f  (0)|j−n+|l−k| . n≥1 k>n nj Observe that     B1 = |an | |ak ||f  (0)|k−n |aj | |al ||f  (0)|l−j . n≥1 k>n j>k l>j  2 Applying Lemma 8 we deduce that B1 ≤ C(f ) n≥1 |an | 2 . Similarly 16 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 ⎛ ⎞    B2 ≤ ⎝ |ak |2 ⎠ |an | |al ||f  (0)|l−n , k≥1 n≥1 l>n  2 which again by Lemma 8 is bounded by C(f ) k≥1 |ak |2 . Finally ⎛ ⎞      B3 = |an | |ak | |aj ||f  (0)|j−n ⎝ |al ||f  (0)|l−k + |al ||f  (0)|k−l ⎠ . n≥1 k>n nk jn we deduce that B3 ≤ B4 + B5 where     B4 = |an | |aj ||f  (0)|j−n |ak | |al ||f  (0)|l−k n≥1 j>n k>n l>k and     B5 = |an | |aj ||f  (0)|j−n |ak | |al ||f  (0)|k−l . n≥1 j>n k>n nn l 0 such that |f n (w)| < |f  (0)|n (1 − |w|)−d , w ∈ D, for every n ≥ 1. A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 17 Proof of Lemma 10. This is a minor modification of [21, Lemma 2]. We include the argument for completeness. Let us denote a = |f  (0)| and consider the function a+w ψ(w) = w , w ∈ D, 1 + aw denote its n-th iterate by ψ n and observe that, by Schwarz’s Lemma and induction, we have |f n (w)| ≤ ψ n (|w|), w ∈ D, (4.1) for every n ≥ 1. Next we use the construction of the Königs function of ψ (see [25, pp. 89–93]). Define for each n ≥ 1 the function 1 n gn (w) = ψ (w), w ∈ D. an It is known that {gn } converges uniformly on compact subsets of D to g(w) = w + . . . for w ∈ D, satisfying g(ψ(w)) = ag(w). Moreover, for 0 ≤ x < 1 we have that ψ(ψ n (x)) 1 + an−1 gn (x) gn+1 (x) = = gn (x) ≥ gn (x), an+1 1 + an+1 gn (x) so that gn (x) ≤ g(x) for every n ≥ 1. Next, since a > 0, there exists δ = δ(f ) > 0 such that ψ is univalent on {|w| < δ} and, thus, ψ n and gn are also univalent in this region by Schwarz’s Lemma. By Koebe Distortion Theorem, there exists ε = ε(f ) > 0 such that |g(w)| < 1 if |w| < ε. Now take x0 = ε and, for n ≥ 1, let xn+1 = ψ −1 (xn ). Observe that Schwarz’s Lemma implies that xn+1 > xn for every n ≥ 0. Let d be a positive integer that will be determined later on. We want to show that g(x) < (1 − x)−d , 0 ≤ x ≤ xn , (4.2) for every n ≥ 0. By the choice of x0 , it is clear that (4.2) holds for n = 0. Assume that (4.2) holds for n and let x0 ≤ x ≤ xn+1 . By construction, we have that 0 < ψ(x) ≤ xn . Therefore, we get  d 1 1 1 1 + ax g(x) = g(ψ(x)) < (1 − ψ(x))−d = (1 − x)−d . a a a 1+x Since x ≥ x0 = ε, we get the bound  d 1 1 + aε g(x) < (1 − x)−d . a 1+ε 18 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 Hence, using that a = |f  (0)| < 1, we can choose d = d(f ) large enough and independent of n so that (4.2) holds. Note that, since xn → 1, one has in fact that (4.2) is valid for 0 ≤ x < 1. Taking (4.1) and applying (4.2), we get |f n (w)| ≤ an gn (|w|) ≤ an g(|w|) < an (1 − |w|)−d as we wanted to see.  Lemma 11. Let f be an analytic mapping from the unit disc into itself with f (0) = 0 and a = |f  (0)| < 1. Let k, l, n be positive integers with l ≤ n and consider g(w) = (f n (w))k for w ∈ D. Then there exists n0 = n0 (f ) > 0 such that for n ≥ n0 we have |g l) (0)| ≤ akn/2 . l! Proof of Lemma 11. Observe first that if a = 0, the result holds trivially. Indeed, if f has a zero at the origin of multiplicity m ≥ 1, then g has a zero of multiplicity kmn at the origin. Thus, if m ≥ 2 and l ≤ n, we have that g l) (0) = 0. Assume now that a > 0. In this case, Lemma 10 asserts that there is a positive integer d = d(f ) for which |f n (w)| ≤ an (1 − |w|)−d , w ∈ D, for n = 1, 2, . . .. Hence, Cauchy’s estimate gives |g l) (0)| max{|g(w)| : |w| = r} akn ≤ ≤ l , 0 < r < 1. l! r l r (1 − r)kd Since l ≤ n we obtain |g l) (0)| akn ≤ n , 0 < r < 1. (4.3) l! r (1 − r)kd Fix r such that a1/4 < r < 1. Then there exists n0 = n0 (f, r) such that an/2 an/4 ≤ < 1, rn (1 − r)d (1 − r)d if n ≥ n0 . Since k ≥ 1 we deduce that akn/2 an/2 ≤ n < 1. rn (1 − r)kd r (1 − r)d Hence, estimate (4.3) gives |g l) (0)| ≤ akn/2 .  l! A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 19 Let f be an inner function with f (0) = 0 and let {μα : α ∈ ∂D} be its Aleksandrov- Clark measures. Recall that (2.1) gives that for any α ∈ ∂D, there exists a constant Cα ∈ R such that  α + f (w) z+w = dμα (z) + iCα , w ∈ D. α − f (w) z−w ∂D Expanding both terms in power series, for any positive integer l we have   l  z l dμα (z) = αk f (z)k z l dm(z), α ∈ ∂D. ∂D k=1 ∂D Hence for any integer l, the l-th moment of μα is a trigonometric polynomial in the variable α of degree less or equal than |l|. We will need to estimate the coefficients of this trigonometric polynomial. Lemma 12. Let f be an inner function with f (0) = 0 and a = |f  (0)| < 1. Let l, n be integers with 1 ≤ |l| ≤ n and let {μα : α ∈ ∂D} be the Aleksandrov-Clark measures of f n . Then there exists a constant n0 = n0 (f ) > 0 such that if n ≥ n0 , the coefficients of the trigonometric polynomial  z l dμα (z) ∂D are bounded by an/2 for any α ∈ ∂D. Proof of Lemma 12. We can assume l > 0. Then   l l) gk (0) z l dμα (z) = αk , α ∈ ∂D, l! ∂D k=1 where gk (w) = (f n (w))k , w ∈ D. Lemma 11 gives that l) |gk (0)| ≤ akn/2 l! if n is sufficiently large. Since k ≥ 1, the proof is completed.  We are now ready to prove the main result of this Section. As before, if n is a positive integer, we will use the notation f −n to denote the function defined by f −n (z) = f n (z), for almost every z ∈ ∂D. 20 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 Theorem 13. Let f be an inner function with f (0) = 0 and a = |f  (0)| < 1. Let 1 ≤ k ≤ q be positive integers. Let ε = {εj }kj=1 where εj = 1 or εj = −1, and let n = {nj }kj=1 where n1 < n2 < . . . < nk are positive integers with nj+1 − nj > q for any j = 1, 2, . . . , k − 1. Consider   k I(ε, n) = f εj nj dm j=1 ∂D Then there exist constants C = C(f ) > 0, q0 = q0 (f ) > 0 independent of ε and of n, such that if q ≥ q0 we have I(ε, n) ≤ C k k!aΦ(ε,n) , k = 1, 2, . . . , k−1 where Φ(ε, n) = j=1 δj (nj+1 − nj ), with δj ∈ {0, 1/2, 1} for any j = 1, . . . , k − 1, and with δ1 = 1 and δk−1 ≥ 1/2. In addition, for j = 2, . . . , k − 1 the coefficient δj = 1 if and only if δj−1 = 0. Furthermore, if δj−1 > 0, the coefficient δj depends on εj+1 , . . . , εk and nj , . . . , nk for j = 2, . . . , k − 1. Proof of Theorem 13. We first prove the following estimate Claim 14. We have I(ε, n) ≤ |f  (0)|n2 −n1 max I({ε3 , . . . , εk }, {n3 − n2 , . . . , nk − n2 }),   k   k εi (ni −n2 ) z 2 f (z) dm(z) , z −2 f εi (ni −n2 ) (z) dm(z) i=3 i=3 ∂D ∂D To prove Claim 14 we can assume ε1 = 1. By Lemma 4 and Aleksandrov Desintegra- tion Theorem we have    k I(ε, n) = zαε2 f εi (ni −n2 ) (α) dμα (z) dm(α) , i=3 ∂D ∂D where {μα : α ∈ ∂D} are the Aleksandrov-Clark measures of f n2 −n1 . By (2.2) we have  n2 −n1 z dμα (z) = f  (0) α, α ∈ ∂D. ∂D Hence if ε2 = −1, we obtain I(ε, n) = an2 −n1 I({ε3 , . . . , εk }, {n3 − n2 , . . . , nk − n2 }) A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 21 and if ε2 = 1, we obtain   k I(ε, n) = an2 −n1 z2 f εi (ni −n2 ) (z) dm(z) . i=3 ∂D This proves Claim 14. We now prove Claim 15. For any integers k, l, j with 0 < |l| < j and 0 < j < k, we have   k z l f εi (ni −nj−1 ) (z) dm(z) ≤ i=j ∂D ⎧ ⎫ ⎨  k ⎬ ≤ ja(nj −nj−1 )/2 max|n|≤|l|+1 zn f εi (ni −nj ) (z) dm(z) ⎩ ⎭ i=j+1 ∂D By Aleksandrov Desintegration Theorem (2.4) we have   k    k zl f εi (ni −nj−1 ) (z) dm(z) = z l αεj f εi (ni −nj ) (α) dμα (z) dm(α), i=j i=j+1 ∂D ∂D ∂D where {μα : α ∈ ∂D} are the Aleksandrov-Clark measures of f nj −nj−1 . Since l = 0, according to Lemma 12, the moment  z l dμα (z), α ∈ ∂D, ∂D is a polynomial in the variable α of degree at most |l| < j whose coefficients are bounded by a(nj −nj−1 )/2 . This proves Claim 15. The proof of Theorem 13 proceeds as follows. We first estimate I(ε, n) by the modulus of one of the three integrals in the right hand side of Claim 14 and the factor an2 −n1 , that corresponds to choosing δ1 = 1. Note that any of these three integrals involve k − 2 products of iterates of f . In addition, the integral yielding the maximum in Claim 14 depends only on ε3 , . . . , εk and on n2 , . . . , nk . Now if the integral giving the maximum is the first one, we apply Claim 14 again, obtaining the factor an4 −n3 and this gives δ2 = 0 and δ3 = 1. Otherwise we apply Claim 15, obtaining a factor 2a(n3 −n2 )/2 , which corre- sponds to choosing δ2 = 1/2. Assume that we have applied this procedure to determine the values of δ1 , . . . , δj−1 . We continue applying Claim 14 or 15 depending on which integral is yielding the maximum in the previous step, which depends on εj+1 , . . . , εk and nj , . . . , nk . Observe that when Claim 14 is applied, the number of factors of iterates of f is reduced by two units and we obtain the factor anj+2−nj+1 , which corresponds to fixing δj = 0 and δj+1 = 1. When Claim 15 is applied, we obtain the factor ja(nj+1 −nj )/2 , 22 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 corresponding to taking δj = 1/2, and the number of factors of iterates of f is reduced by one unit. We continue applying this process at least k/2 times and at most k − 2 times, until we reach integrals of the form  z l f εk (nk −nk−1 ) (z) dm(z), |l| < k − 1 ∂D or  f εk−1 (nk−1 −nk−2 ) f εk (nk −nk−2 ) dm. ∂D Let g = f nk −nk−1 . The modulus of the first integral is |g l) (0)|/l!. Since |l| < q < nk − nk−1 , if q is sufficiently large, Lemma 11 gives that last expression is bounded by a(nk −nk−1 )/2 . The modulus of the second integral is bounded by ank −nk−1 . This shows that δk−1 ≥ 1/2 and concludes the proof.  In the proof of Theorem 1 we will split the partial sum into finitely many terms such that the sum of the variances of these terms is asymptotically equivalent to the variance of the initial partial sum. Next auxiliary result provides sufficient conditions for this splitting. Lemma 16. Let {an } be a sequence of complex numbers and λ ∈ C with |λ| < 1. Consider the sequence  N  N  N −k 2 σN = |an |2 + 2 Re λk an an+k , N = 1, 2 . . . n=1 k=1 n=1 For N > 1, let Aj = Aj (N ), j = 1, . . . , M = M (N ), be pairwise disjoint sets of consecutive positive integers smaller than N . Consider    σ 2 (Aj ) = |an |2 + 2 Re λk an an+k , j = 1, 2 . . . , M. n∈Aj k≥1 n∈Aj : n+k∈Aj Let A = ∪Aj . Assume  |an |2 lim n∈A N =1 (4.4) N →∞ n=1 |an |2 and max{|an |2 : n ∈ Aj } lim  = 0. (4.5) n∈Aj |an | j→∞ 2 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 23 Then M j=1 σ 2 (Aj ) lim 2 = 1. N →∞ σN Proof of Lemma 16. Let B be the set of positive integers smaller or equal to N which are not in the collection A. Then  M 2 σN − σ 2 (Aj ) = A + 2 Re B + 2 Re C, j=1 where  A= |an |2 , n∈B  N  B= λk an an+k , k=1 B(k) where B(k) = {n ∈ B, n ≤ N − k} and  M   C= λk an an+k , j=1 k≥1 A(j,k) where A(j, k) = {n ∈ Aj : n ≤ N − k, n + k ∈ / Aj }. According to part (a) of Theorem 9 we have 1 − |λ|  N 2 σN ≥ |an |2 . (4.6) 1 + |λ| n=1 Then, A 1 + |λ| A 2 ≤ 1 − |λ| N n=1 |an | σN 2 which by assumption (4.4), tends to 0 as N → ∞. Similarly N  |B| 1 + |λ| k=1 |λ|k B(k) |an ||an+k | ≤ N 2 σN 1 − |λ| n=1 |an | 2 By Cauchy-Schwarz’s inequality 1/2 1/2    N |an ||an+k | ≤ |an | 2 |an | 2 B(k) n∈B n=1 24 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 and we deduce   2 1/2 B 1 + |λ| n∈B |an | 2 ≤ |λ| (1 − |λ|)2  σN 1/2 N n=1 |an | 2 which according to (4.4), tends to 0 as N → ∞. We now estimate C. For any k ≥ 1, Cauchy-Schwarz’s inequality gives  M    N |an ||an+k | ≤ |an ||an+k | ≤ |an |2 . j=1 A(j,k) n∈A : n≤N −k n=1 Hence, applying (4.6), for any positive integer k0 we have  M  k≥k0 |λ|k j=1 n∈A(j,k) |an ||an+k | 1 + |λ| ≤ |λ|k0 (4.7) σN 2 (1 − |λ|)2 Fix ε > 0 and use assumption (4.5) to pick j0 = j0 (ε) > 0 large enough so that  sup{|an |2 : n ∈ Aj } ≤ ε |an |2 (4.8) n∈Aj if j > j0 . Pick also k0 such that |λ|k0 < ε. Fix k ≤ k0 and note that there are at most k indices n ∈ Aj such that n + k ∈ / Aj . Hence  |an ||an+k | ≤ k|anj ||anj +k |, A(j,k) where nj = nj (k) ∈ Aj is the index in Aj with nj + k ≤ N , where the product |an ||an+k | is maximum. Hence ⎛ ⎞1/2 ⎛ ⎞1/2  M   M  M M |an ||an+k | ≤ k |anj ||anj+k | ≤ k ⎝ |anj |2 ⎠ ⎝ |anj +k |2 ⎠ . j≥j0 A(j,k) j≥j0 j≥j0 j=1 Note that (4.8) gives that     N |anj |2 ≤ ε |an |2 ≤ ε |an |2 . j≥j0 j≥j0 n∈Aj n=1 Since there are at most k indices n ∈ Aj such that n + k ∈ / Aj , we also have  M  N |anj +k |2 ≤ k |an |2 . j=1 n=1 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 25 Applying (4.6) again, we deduce  M  k≤k0 |λ|k j≥j0 A(j,k) |an ||an+k | 1 + |λ| ≤ ε1/2 C1 , (4.9) σN 2 1 − |λ|  where C1 = k≥1 |λ|k k3/2 . The estimates (4.7) and (4.9) give that    |C| k≤k0 |λ|k j 0 such that the following conditions hold (a) supn fn 2 ≤ C and  lim fn dm = 1 n→∞ ∂D (b) gn (z) > −C for almost every z ∈ ∂D and limn→∞ gn 2 = 0. Then  lim fn e−gn dm = 1. n→∞ ∂D Proof of Lemma 17. Cauchy-Schwarz’s inequality gives  fn (e−gn − 1) dm ≤ fn 2 e−gn − 1 2 . ∂D Note that there exists a constant M = M (C) > 0 such that |e−x − 1| ≤ M |x| if x ≥ −C. Hence e−gn − 1 2 ≤ M gn 2 , n = 1, 2, . . .. This finishes the proof.  26 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 5. Proof of Theorem 1 Proof of Theorem 1. Let  N 2 SN = |an |2 , N = 1, 2, . . . n=1 Recall that by part (a) of Theorem 9, we have κ−1 σN2 ≤ SN 2 ≤ κσN2 , N = 1, 2, . . ., where   −1 κ = (1 + |f (0)|)(1 − |f (0)|) . Pick 0 < ε < η, pN = SN and qN = SN 1+ε 1−ε . Let C(f ) denote a positive constant only depending on f whose value may change from line to line. The proof is organized in several steps. 1. Splitting the Sum. In this first step, for N large, we will recursively find indices 0 ≤ Mk < Nk < Mk+1 ≤ N , 1 ≤ k ≤ QN , such that if  Nk  Mk+1 n ξk = an f , ηk = an f n , n=Mk +1 n=Nk +1 we have QN lim = 1, (5.1) N →∞ qN  2 N  QN     an f − n (ξk + ηk ) ≤ 2C(f )pN , (5.2)   n=1 k=1 2  Nk  Mk+1 pN ≤ |an | ≤ 2pN , 2 qN ≤ |an |2 ≤ 2qN , k = 1, 2, . . . , QN , (5.3) n=Mk +1 n=Nk +1   N  QN  1   lim  an f n − ξk  = 0, (5.4) N →∞ σN   n=1 k=1 2 β Mk+1 − Nk ≥ qN , Nk − Mk ≥ pγN , k = 1, 2, . . . , QN − 1, (5.5) where β = (η − ε)(1 − ε)−1 and γ = (η + ε)(1 + ε)−1 . Pick M1 = 0 and let N1 be the smallest positive integer such that  N1 |an |2 ≥ pN . n=1 The minimality of N1 gives that  N1 |an |2 ≤ pN + |aN1 |2 . n=1 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 27 Now let M2 be the smallest positive integer such that  M2 |an |2 ≥ qN . n=N1 +1 As before, the minimality of M2 gives that  M2 |an |2 ≤ qN + |aM2 |2 . n=N1 +1 We repeat this process until we arrive at an index Nk or Mk bigger than N . Let QN be the number of times this process is repeated, that is, k = 1, 2, . . . , QN . Then  N RN = |an |2 ≤ 2pN . (5.6) MQN +1 Since  N  QN  N an f − n (ξk + ηk ) = an f n , n=1 k=1 n=MQN +1 the estimate (5.2) follows from part (a) of Theorem 9. By construction we have  Nk pN ≤ |an |2 ≤ pN + |aNk |2 (5.7) n=Mk +1  Mk+1 qN ≤ |an |2 ≤ qN + |aMk+1 |2 , (5.8) n=Nk +1 for k = 1, 2, . . . , QN . Fix δ > 0. Observe that the assumption (1.2) gives that |aNk |2 + |aMk+1 |2 < δqN if N is sufficiently large. Taking δ < 1 one deduces that (5.3) holds if N is sufficiently large. Moreover the estimates (5.7) give that (pN + qN )QN ≤ SN 2 − RN ≤ (1 + δ)(pN + qN )QN , (5.9) 2 if N is large enough. Since pN qN = SN and because of the estimate (5.6), (5.1) follows from (5.9) tending δ to 0. Observe that  Mk+1 |an |2 ≥ qN = SN 1−ε . (5.10) n=Nk +1 28 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 By (1.2), if N is sufficiently large, we have that |an |2 ≤ SN 1−η for any n ≤ N . We deduce β from (5.10) that SN 1−η (Mk+1 − Nk ) ≥ qN and Mk+1 − Nk ≥ qN . A similar argument shows that Nk − MK ≥ pγN . This proves (5.5). We are now going to prove (5.4). Observe that at almost every point of the unit circle we have 2  QN  QN N −1  Q QN ηk = |ηk | + 2 Re 2 ηk ηj . k=1 k=1 k=1 j>k Since ηk 2 2 ≤ 2C(f )qN , we have QN   |ηk |2 dm ≤ 2C(f )qN QN ≤ 3C(f )qN 2 , (5.11) k=1∂D if N is sufficiently large. On the other hand, if j > k we have   ηk ηj dm ≤ |ar ||at ||f  (0)|t−r , ∂D where the sum is taken over all indices r, t with Nk < r ≤ Mk+1 and Nj < t ≤ Mj+1 . Observe that by (5.5), we have t − r ≥ pγN . Writing l = t − r and applying Cauchy- Schwarz’s inequality, we obtain    γ ηk ηj dm ≤ |f  (0)|l |ar ||al+r | ≤ C(f )|f  (0)|pN SN 2 . ∂D l≥pγ N Hence N −1  Q QN  γ ηk ηj dm ≤ C(f )qN SN |f  (0)|pN . 2 2 (5.12) k=1 j>k ∂D Using (5.11) and (5.12) we obtain that Q 2 N     ηk  ≤ 4C(f )qN 2 ,   k=1 2 2 2 if N is sufficiently large. Since σN > C(f )SN = C(f )pN qN , we deduce that    QN 2  k=1 ηk  2 lim 2 = 0. (5.13) N →∞ σN A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 29 Now (5.2) and (5.13) give (5.4). The main idea in the rest of the proof is that {ηk } are irrelevant while due to (5.5), {ξk } act as independent random variables. 2. Arranging the Fourier Transform. Applying (5.4), the proof of Theorem 1 reduces to show that 1  QN TN = √ ξk , N = 1, 2, . . . 2−1 σN k=1 converge in distribution to a standard complex normal variable. By the Levi Continuity Theorem, it is sufficient to show that for any complex number t we have  eit,TN dm → e−|t| 2 ϕN (t) = /2 , as N →∞ (5.14) ∂D Here t, w = Re(tw) is the standard scalar product in the plane. In this second step of the proof we will show that QN     2 i t, ξk t, ξk lim ϕN (t) − 1+ √ exp − dm = 0 (5.15) N →∞ k=1 2−1 σN σN 2 ∂D Fixed δ > 0, consider the sets Ek = {z ∈ ∂D : |ξk (z)| > δSN }, k = 1, 2, . . . , QN . By part (c) of Theorem 9 we have ξk 44 ≤ C(f )p2N . Chebyshev inequality and (5.1) give  QN C(f )p2N QN 2C(f ) m(Ek ) ≤ 4 ≤ 4 δ 4 SN δ qN k=1 if N is sufficiently large. For μ > 1, consider the set  QN 2 E0 = z ∈ ∂D : t, ξk (z) 2 > μSN . k=1 By part (a) of Theorem 9 we have ξk 2 ≤ C(f )pN . Chebyshev inequality and (5.1) give C(f )|t|2 QN 2C(f )|t|2 m(E0 ) ≤ ≤ , μqN μ if N is sufficiently large. Hence the set  QN E= Ek k=0 satisfies 30 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318   1 |t|2 m(E) ≤ 2C(f ) + . δ 4 qN μ Using the elementary identity   z2 exp (z) = (1 + z) exp + o(|z |) , 2 2 where o(|z|2 )/|z|2 → 0 as z → 0, we deduce QN    2  QN 2 i t, ξk t, ξk t, ξk exp (i t, TN ) = 1+ √ exp − 2 exp o 2 k=1 2−1 σN σN k=1 σN Fix ε > 0. Taking δ > 0 sufficiently small we have  QN t, ξk (z) 2 o 2 ≤ C(f )εμ, z ∈ ∂D \ E. σN k=1 Hence   QN    2 i t, ξk t, ξk exp (i t, TN ) dm − 1+ √ exp − 2 dm ≤ 2−1 σN σN ∂D\E ∂D\E k=1     QN t, ξk 2 1/2 t, ξk 2 ≤ eC(f )εμ − 1 1+ exp − dm ≤ eC(f )εμ − 1. 2−1 σN 2 σN 2 k=1 ∂D\E Last inequality follows from the elementary estimate (1 + x)1/2 e−x/2 ≤ 1 if x ≥ 0. Hence QN     2 i t, ξk t, ξk ϕN (t) − 1+ √ exp − 2 dm ≤ 2m(E) + eC(f )εμ − 1, 2−1 σN σN ∂D k=1 which proves (5.15). Therefore to prove (5.14) it is sufficient to show that for any t ∈ C one has QN     2 i t, ξk t, ξk lim 1+ √ exp − 2 dm = exp (−|t|2 /2). N →∞ −1 2 σN σN k=1 ∂D This will follow from Lemma 17 applied to the functions QN    i t, ξk fN = 1+ √ , k=1 2−1 σN A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 31 1  QN 2 |t|2 gN = 2 t, ξk − . σN 2 k=1 According to Lemma 17 it is sufficient to show sup fN 2 < ∞, (5.16) N lim gN 2 = 0, (5.17) N →∞  lim fN dm = 1. (5.18) N →∞ ∂D 3. Estimating fN 2. Observe that  QN t, ξk 2  QN 1  2 2 1 + −1 2 =1+ t, ξj1 . . . t, ξjk , k=1 2 σN k=1 2−k σN 2k where the last sum is taken over all collections of indices 1 ≤ j1 < . . . < jk ≤ QN . Since 2 t, ξn ≤ |t|2 |ξn |2 , Theorem 6 and part (a) of Theorem 9 give that  2 2 t, ξj1 . . . t, ξjk dm ≤ C(f )k |t|2k pkN . ∂D Since the total number of distinct collections of indices j1 , . . . , jk verifying 1 ≤ j1 <   . . . < jk ≤ QN is QkN , we deduce   QN 2 QN    t, ξk QN C(f )k |t|2k pk 1 + −1 2 dm ≤ 1 + N . 2 σN k 2−k σN 2k ∂D k=1 k=1 2 Since σN ≥ C(f )−1 SN 2 = C(f )−1 pN qN , we deduce   QN 2 QN     QN t, ξk QN C(f )2k |t|2k C(f )2 |t2 | 1+ dm ≤ 1 + = 1+ . 2−1 σN 2 k 2−k qN k 2−1 qN ∂D k=1 k=1 Hence (5.1) gives that fN 22 ≤ exp (3C(f )2 |t|2 ) if N is sufficiently large. This gives (5.16). 4. Estimating gN 2 . Consider the set of indices Ak = {n ∈ N : Mk < n ≤ Nk }, k = 1, . . . , QN . Then  ξk = an f n , k = 1, . . . , QN . (5.19) n∈Ak QN Let A = k=1 Ak . Observe that (5.4) gives 32 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318  n∈A |an | 2 lim 2 = 1. N →∞ SN This is assumption (4.4) of Lemma 16. Assumption (4.5) follows from (1.2). Thus, Lemma 16 gives QN 2 k=1 ξk 2 lim 2 = 1. (5.20) N →∞ σN Denote λ = t/|t|. We have QN   |t|2  σN 2 gN = 2 2|ξk | + λ ξk + λ ξk − 2 2 2 2 2 2 . 4σN QN k=1 Applying (5.20), the proof of (5.17) reduces to show    1 QN    lim  2 ψk  = 0, N →∞  σN  k=1 2 where ψk = 2(|ξk |2 − ξk 2 2) + λ2 ξk2 + λ2 ξk2 . Now Q 2 Q   N  N N −1  Q QN   2  ψk  = ψk + 2 Re ψk ψj dm. (5.21)   2 k=1 2 k=1 k=1 j>k∂D Since |ψk | ≤ 4|ξk |2 + 2 ξk 2 2, parts (a) and (c) of Theorem 9 give that ψk 2 2 ≤ C(f )p2N . Hence  QN ψk 2 2 ≤ C(f )p2N QN k=1 and we deduce 1  QN 2 lim 4 ψk 2 = 0. N →∞ σN k=1 The second term in (5.21) is splitted as N −1  Q QN  ψk ψj dm = A + B + C + D, k=1 j>k∂D where A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 33 N −1  Q QN     A=4 |ξk |2 − ξk 2 2 |ξj |2 − ξj 2 2 dm, k=1 j>k∂D  N −1  Q QN  2   B=2 |ξk | − ξk 2 2 λ2 ξj2 + λ2 ξj2 dm, k=1 j>k∂D N −1  Q QN     C=2 λ2 ξk2 + λ2 ξk2 |ξj |2 − ξj 2 2 dm, k=1 j>k∂D N −1  Q QN     D= λ2 ξk2 + λ2 ξk2 λ2 ξj2 + λ2 ξj2 dm. k=1 j>k∂D By Theorem 6, ξk ξj 2 = ξk 2 ξj 2 if k = j and we deduce A = 0. Since the mean of ξj2 over the unit circle vanishes and at almost every point in the unit circle one has    |ξk |2 = |an |2 + 2 Re an aj f n f j , (5.22) n∈Ak n∈Ak j∈Ak ,j>n the integrals in B can be written as a linear combination of    f n1 f j1 λ2 ξj2 + λ2 ξj2 dm, ∂D where n1 , j1 ∈ Ak and hence max{n1 , j1 } < min{n : n ∈ Aj }. According to part (a) of Lemma 7,  f n1 f j1 ξj2 dm = 0 ∂D and we deduce B = 0. Since the mean of ξk2 over the unit circle vanishes, we have N −1  Q QN  C = 4 Re λ2 ξk2 |ξj |2 dm. k=1 j>k∂D For the same reason, using the formula (5.22), we have   ξk2 |ξj |2 dm = ξk2 Re hj dm, ∂D ∂D where  hj = 2 ar al f r f l . r,l∈Aj ,l>r 34 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 Using formula (5.19) to expand ξk2 , we obtain  ξk2 |ξj |2 dm = E + F, ∂D where      E= a2n (f n )2 ar al f r f l + ar al f r f l dm, n∈Ak r,l∈Aj ,l>r ∂D      F =2 an as f n f s ar al f r f l + ar al f r f l dm. n,s∈Ak : s>n r,l∈Aj ,l>r∂D By part (c) of Lemma 7 we have   (f n )2 f r f l dm + (f n )2 f l f r dm ≤ C(f )|f  (0)|l−n , if n < r < l. ∂D ∂D We deduce that   |E| ≤ C(f ) |an |2 |ar ||al ||f  (0)|l−n . n∈Ak r,l∈Aj ,l>r β According to (5.5), we have r − n ≥ qN for any r ∈ Aj and any n ∈ Ak , j > k. Now  β   |ar ||al ||f  (0)|l−n ≤ |f  (0)|qN |f  (0)|t |ar ||ar+t |. r,l∈Aj ,l>r t≥1 r∈Aj : r+t∈Aj  By Cauchy-Schwarz’s inequality, last sum is bounded by r∈Aj |ar |2 ≤ 2pN . Hence β |E| ≤ C(f )|f  (0)|qN p2N (5.23) Similarly, part (d) of Lemma 7 gives that   f nf sf r f l dm + f n f s f r f l dm ≤ C(f )|f  (0)|l−n , n < s < r < l − 2, ∂D ∂D and   f nf sf r f l dm + f n f s f r f l dm ≤ C(f )|f  (0)|r−n , n < s < r < l, r ≥ l − 2. ∂D ∂D A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 35 Using the trivial estimate |ak | ≤ SN for any k ≤ N , we deduce that   |F | ≤ C(f )SN 4 |f  (0)|l−n . n,s∈Ak : s>n r,l∈Aj ,l>r β As before, l − n ≥ qN for any r ∈ Aj and any n ∈ Ak , j > k. We deduce β |F | ≤ C(f )SN 4 |f  (0)|qN /2 . Now, the exponential decay in (5.23) and (5.16) give that C lim 2 = 0. (5.24) N →∞ σN The corresponding estimate for D follows from the estimate  β ξk 2 ξj 2 dm ≤ C(f )SN 4 |f  (0)|qN , k < j. ∂D As before this last estimate follows from (5.5) and from  f n f s f l f t dm ≤ C(f )|f  (0)|t−n , n < s < l < t − 2, ∂D which follows from part (d) of Lemma 7. This finishes the proof of (5.17). 5. Integrating fN . In this last step we will prove (5.18). Observe that at almost every point in the unit circle we have  QN ik  fN = 1 + t, ξi1 . . . t, ξik , k=1 2−k/2 σN k where the second sum is taken over all collections of indices 1 ≤ i1 < . . . ik ≤ QN . Fix 1 ≤ i1 < . . . ik ≤ QN . The integral    k   t, ξi1 . . . t, ξik dm = 2−k tξin + tξin dm n=1 ∂D ∂D is a multiple of a sum of 2k integrals of the form  r l t t ξiε11 . . . ξiεkk dm, ∂D 36 A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 where r + l = k, εi = 1 or εi = −1 for i = 1, . . . , k and we denote ξi −1 (z) = ξi (z), z ∈ ∂D. Now, each ξi is a linear combination of iterates of f ,  ξj = an f n . n∈A(j) Hence   k  ξiε11 . . . ξiεkk dm = aεnjj f n1 ε1 . . . f nk εk dm, ∂D n∈C j=1 ∂D  where n∈C means the sum over all possible k-tuples n = {nj }kj=1 of indices such that nj ∈ A(ij ) for j = 1, . . . , k. Since |an | ≤ SN , n ≤ N , we have    ξiε11 . . . ξiεkk dm ≤ SN k f n1 ε1 . . . f nk εk dm . ∂D n∈C ∂D k−1 Let ε = {εj }k−1 j=1 be fixed and consider Φ(n) = Φ(ε, n) = j=1 δj (nj+1 − nj ) where δj ∈ {0, 1/2, 1} for j = 1, . . . , k − 1, with δ1 = 1 and δk−1 ≥ 1/2, and with δj = 1 if and only if δj−1 = 0 for j = 2, . . . , k − 1, as defined in Theorem 13. Let a = |f  (0)|. Theorem 13 gives   ξiε11 . . . ξiεkk dm ≤ k!SN k C(f )k aΦ(n) . ∂D n∈C We split the sum over n ∈ C as follows. Let D denote the set of (k−1)-tuples δ = {δj }k−1 j=1 of coefficients that can appear in Φ(n). That is, those tuples with δj ∈ {0, 1/2, 1} for j = 1, . . . , k − 1, with δ1 = 1 and δk−1 ≥ 1/2, and with δj = 1 if and only if δj−1 = 0, for j = 2, . . . , k − 1. Observe that there are less than 2k such tuples. Given a k-tuple n ∈ C, let us denote by δ(n) the (k − 1)-tuple δ of coefficients appearing in Φ(n). Then we have that    aΦ(n) = aΦ(n) . n∈C δ∈D {n∈C : δ(n)=δ} k−1 Given δ = {δj }k−1 j=1 ∈ D, we define Φδ (n) = j=1 δj (nj+1 − nj ) for every n ∈ C. We clearly have that   aΦ(n) ≤ aΦδ (n) . (5.25) n∈C δ∈D n∈C Consider now a fixed δ = (δ1 , . . . , δk−1 ), and recall that δ1 = 1. Let l(1) be the minimum integer such that δl(1)+1 = 0 (we set l(1) = k − 1 if δj = 0 for all 1 ≤ j ≤ k − 1). A. Nicolau, O. Soler i Gibert / Advances in Mathematics 401 (2022) 108318 37 In particular, observe that if l(1) > 1, we have that δj = 1/2 for 2 ≤ j ≤ l(1) by Theorem 13. Thus, to find a bound for the right-hand side of (5.25), we need to estimate sums of the form  l   l  (n2 −n1 )+(nl −n2 )/2 a = a(n2 −n1 )/2+(nl −n1 )/2 (5.26) j=1 nj ∈A(ij ) j=1 nj ∈A(ij ) for some 1 < l ≤ k − 1. Denote here n1 = max A(i1 ), n2 = min A(i2 ) and nl = min A(il ), β and observe that n2 − n1 ≥ qN because of (5.5). Assume l > 2. Summing over n1 and n2 we get that (5.26) is bounded by β  l  CaqN /2 a(nl −n1 )/2 . j=3 nj ∈A(ij ) Next, summing over nj for j up to l − 1 yields the factor |A(i3 )| + . . . + |A(il−1 )|, while summing over nl we get the factor a(nl −n1 )/2 . Here, |A(ij )| denotes the number of indices in the set A(ij ). Using (5.5), we have that |A(ij )| ≥ pγN > qN β for any j = 1, . . . , k and, β β thus, we get that nl − n1 ≥ qN + |A(i2 )| + . . . + |A(il−1 )| > lqN . Hence, we find that  l  β a(n2 −n1 )+(nl −n2 )/2 ≤ CalqN /4 . (5.27) j=1 nj ∈A(ij ) Note that if l = 1 or l = 2, then (5.27) is obvious. Assume now that we have determined l(m − 1). If l(m − 1) < k − 1, then let l(m) be the minimum integer such that l(m − 1) < l(m) ≤ k−1 and such that δl(m)+1 = 0. We iterate this process until we set l(r) = k−1 for some integer 1 ≤ r ≤ k. Observe that, by Theorem 13, we have that l(m) ≥ l(m − 1) + 2. Taking l(0) = 0, the full sum over n ∈ C in the right-hand side of (5.25) becomes a product of sums of the form (5.26) with j ranging from l(m − 1) + 1 to l(m), for m = 1, . . . , r. Thus, applying the bound (5.27) we get that   r β β aΦδ (n) ≤ Ca(l(m)−l(m−1))qN /4 ≤ C k akqN /4 . n∈C m=1 Now, summing over δ ∈ D and using the fact that there are at most 2k such tuples, we get that  β aΦ(n) ≤ C k akqN /4 . n∈C Thus  β ξiε11 . . . ξiεkk dm ≤ k!SN k C(f )k akqN /4 . ∂D 38 A. Nicolau, O. 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