Moments of the weighted Cantor measures

Based on the seminal work of Hutchinson, we investigate properties of {\em $\alpha$-weighted Cantor measures} whose support is a fractal contained in the unit interval. Here, $\alpha$ is a vector of nonnegative weights summing to $1$, and the corresponding weighted Cantor measure $\mu^\alpha$ is the unique Borel probability measure on $[0,1]$ satisfying $ \mu^\alpha(E) = \sum_{ n=0 }^{N-1} \alpha_n\mu^\alpha( \varphi_n^{-1}(E) )$ where $\varphi_n: x\mapsto (x+n)/N$. In Sections 1 and 2 we examine several general properties of the measure $\mu^\alpha$ and the associated Legendre polynomials in $L_{\mu^\alpha}^2[0,1]$. In Section 3, we (1) compute the Laplacian and moment generating function of $\mu^\alpha$, (2) characterize precisely when the moments $I_m = \int_{[0,1]}x^m\,d\mu^\alpha$ exhibit either polynomial or exponential decay, and (3) describe an algorithm which estimates the first $m$ moments within uniform error $\varepsilon$ in $O( (\log\log(1/\varepsilon))\cdot m\log m )$. We also state analogous results in the natural case where $\alpha$ is {\em palindromic} for the measure $\nu^{\alpha}$ attained by shifting $\mu^{\alpha}$ to $[-1/2,1/2]$.


Introduction
In the seminal paper [1], Hutchinson realized a fractal as the invariant compact set, called the attractor, of an iterated function system (IFS), i.e. a family of contraction maps on a complete metric space. Specifically, given an IFS {φn} N−1 n=0 on X, the attractor of the IFS is the unique compact set K ⊂ X satisfying Hutchinson showed the existence and uniqueness of a self-similar Borel probability measure supported on the attractor of an IFS. We denote by △ N the standard simplex in R N and △ * N ⊆ △ N consisting of α = (α 0 , α 1 , ..., α N−1 ) ∈ △ N such that αn < 1 for all n and call elements of △ N weight vectors. We now paraphrase Hutchinson's result.
Theorem 1.1 (Hutchinson,[1]). Suppose {φn} N−1 n=0 is an IFS on a complete metric space X with attractor K, and let α ∈ △ N . There exists a unique Borel regular measure µ α on X supported on K such that for all Borel-measurable E ⊆ X.
Using the terminology of [2], we refer to the measure µ α as the α-equilibrium measure when X = R n or X = C. We will call the measure µ α an α-weighted Cantor measure when the associated IFS {φn} N−1 n=0 on R is given by φn : x ↦ → (x + n)/N. An equilibrium measure is described as having maximal entropy if the associated weights are uniform, i.e. αn is either 0 or 1/k for each n. An equilibrium measure that has attracted a lot of interest in the non-smooth harmonic analysis community is the ternary Cantor measure which arises from the weight vector α = (1/2, 0, 1/2). In [3], Jorgensen and Pedersen addressed the question of when a maximal entropy equilibrium measure µ α is spectral, that is, if there exists some countable set Λ ⊂ R so that the complex exponential functions {e 2πiλx } λ∈Λ form an orthonormal basis for the Hilbert space L 2 µ α [0, 1]. Jorgensen and Pedersen found that, while the quaternary Cantor measure corresponding to α = (1/2, 0, 1/2, 0) is spectral, the ternary Cantor measure is not.
Much effort has been made to remedy this artifact of the ternary Cantor measure. In [4], Dutkay, Picioroaga, and Song constructed an orthonormal basis consisting of piecewise exponentials on the ternary Cantor set. Strichartz in [5] posed the question of the existence of a frame, which is a generalization of an orthonormal basis, on the ternary Cantor set; however, this problem remains open. Polynomial function systems provide a tempting alternative. To this end, we define the Legendre polynomials in L 2 µ α [0, 1] to be the result of applying the Gram-Schmidt algorithm to any sequence of polynomials of degrees 0, 1, 2, . . . , respectively. At each step, it becomes necessary to compute inner products of the form ∫︀ [0,1] x m dµ α (x). These quantities, better known as the moments of the measure µ α , have elicited a lot of attention. Dovgoshey, Martio, Ryazanov, and Vuorinen provide a fairly comprehensive survey of the ternary Cantor function, including moments of the measure for which it is the distribution, in [6]; Jorgensen, Kornelson and Shuman in [2] study the moments of equilibrium measures through an operator theory perspective using infinite matrices.
Our main results are as follows. In Section 2, we make the connection of these measures to a result by Pei, showing that the weighted Cantor measures are singular except in the trivial case of αn = 1/N for all n when the measure is Lebesgue. We then provide more content in the way of characterizing these measures. In Proposition 2.11, we prove a generalization of Bonnet's recursion formula for orthogonal polynomial systems. In Theorem 3.4, we derive an explicit infinite product formula for the Laplacian (and thus the moment generating function) of µ α and estimate in Theorem 3.6 the rapid convergence of the coefficients of the partial product. This leads to Remark 3.8 which outlines a O(log log(1/ε) · m log m) algorithm for estimating the first m moments to uniform error at most ε > 0.

Properties of the weighted Cantor measure
Our first observation motivates the distinction of △ * N from the simplex △ N . It is a direct consequence of the uniqueness of a Borel measure satisfying the invariance relation in Equation (1), and the proof is omitted. Proposition 2.1. Suppose α ∈ △ N with αn = 1 for some n. Then µ α is the Dirac measure centered at n/(N − 1), the fixed point of φ −1 n .
Given a finite Borel measure µ on R, the cumulative distribution function (CDF) Fµ(x) := µ(−∞, x] is the increasing, right-continuous function which uniquely determines the measure. Therefore, to understand the weighted Cantor measure µ α , it is useful to note some basic properties of F µ α .
Recall that the weighted Cantor measure is determined by weighting, scaling and translating under the IFS according to the invariance relation in Equation (1). The next proposition illustrates that this invariant condition applies as well to the weight vector. Precisely, there are α ∈ △ M and β ∈ △ N with M ≠ N so that µ α = µ β .
Proof. It is readily checked that the element of β indexed by n = n 0 + n 1 N + ...
For illustration, we attain the graph of F µ α ,k through F µ β ,1 where β = α ⊗k , as stated above. The benefit of the latter is that it is somewhat simple to take the Kronecker product of vectors up to sufficient resolution in programs such as Mathematica, which was used to produce Figure 1. The next results show that a small variation in α ∈ △ * N leads to a relatively small variation in the corresponding measure. We start with a lemma which is pertinent to those results.
Proof. Let α, β ∈ △ * N . Because F µ α ,k and F µ β ,k are linear interpolations of F µ α and F µ β , respectively, on the set S k , there exists a positive number x ∈ S k such that Then, by Proposition 2.2, we have Repeating this argument sufficiently many times, we attain We conclude the proof by showing the upper bound The inequality is trivial for k = 1. We proceed by induction on k. There are m ℓ ∈ {0, 1, ..., N − 1} such that This concludes the induction. Now, since ‖α‖∞ < 1 and ‖β‖∞ < 1, we may let c(N, k) = kN k .
In particular, we have We note that the transform in Proposition 2.8 is not continuous on the entire simplex △ N since the CDF of the measure associated to α ∈ △ N \△ * N is discontinuous. Now let M be the space of Borel probability measures on [0, 1] with the total variation norm, ‖µ‖ TV = sup E |µ(E)|. We next show that the transform α ↦ → µ α : △ * N → M is continuous.
and, thus, Now let E be a Borel-measurable subset of [0, 1], and let η ′ > 0. From the regularity of the measures, see [8], Since η ′ was arbitrary, we have This concludes that µ β → µ α in the total variation norm.
Conversely, suppose that µ β → µ α in the total variation norm. Then, by Proposition 2.2, we have We conclude this section with a discussion of symmetric weighted Cantor measures. As motivation, note that both CDF's in Figure 1 exhibit rotational symmetry about the point (1/2, 1/2). First, we need a few definitions. Proof. If αn = 1 for some n, then µ α is a Dirac measure centered at n/(N − 1). As such, the measure is symmetric only when N = 2n + 1, when α is palindromic.
So we assume otherwise, that is, αn < 1 for all n. Suppose α is palindromic. For any positive integer k and ⃗ n = (n 0 , n 1 , ..., n k−1 ) ∈ {0, 1, ..., N − 1} k , let I ⃗ n be the open interval By Proposition 2.2, we have As a consequence, any open set satisfies this identity by continuity of the measure. Then, from the regularity of µ α , it follows that the measure is symmetric.
Conversely, suppose µ α is symmetric. By Proposition 2.2, we have Therefore, α is palindromic, completing the proof.
The final observation of this section is a recursive formula for the monic Legendre polynomials associated to any symmetric, finite Borel measure µ on [0, 1], e.g. the ternary Cantor measure. To be clear, we say that a sequence (p 0 , p 1 , . . . ) is a sequence of Legendre polynomials (associated to µ) if each p k is a polynomial of degree k so that for each k ≠ ℓ, p k and p ℓ are orthogonal elements of L 2 µ [0, 1]. Note that for each such measure µ, this definition determines the family of Legendre polynomials uniquely up to scaling each polynomial.
Out of independent interest, we note the following 2-term recursive formula for Legendre polynomials.
Proof. For convenience, we denote q 1 (x) := x − 1/2. By way of the Gram Schmidt algorithm, we generate m n+1 (x) by subtracting off the projections of q 1 (x)mn(x) on each of the monic Legendre polynomials up to degree n. For conciseness, we proceed by induction on n ≥ 0, proving that (A) the parities of mn , m n+1 , and m n+2 match the parities of n, n + 1, and n + 2, respectively, and that (B) Equation (4) holds.
We begin with the base case n = 0. Clearly m 0 (x) = 1 and additionally, m 0 is even. Since µ is symmetric and q 1 (x)m 0 (x) = q 1 (x) is odd, ⟨q 1 m 0 , m 0 ⟩ = 0. It follows that m 1 (x) is a (monic) constant multiple of q 1 (x), so m 1 (x) = x − 1/2 and m 1 is odd. Finally, note that ⟨q 1 m 1 , m 1 ⟩ = 0 since q 1 (x)m 1 (x)m 1 (x) is odd and µ is symmetric. Since m 2 is monic, we need only subtract off the projection of q 1 (x)m 1 (x) in the m 0 direction to find m 2 . So and in particular, m 2 is even, so the base case of the claim holds. Now suppose n ≥ 1 and that the inductive hypothesis holds for n − 1. Since (m 0 , m 1 , m 2 , . . . ) is an orthogonal basis of L 2 µ [0, 1] and q 1 (x)m n+1 (x) is a polynomial of degree n + 2, it follows that we may write for some constants c 0 , c 1 , . . . , c n+2 . Note first that if k ≤ n − 1, then since m n+1 is orthogonal to any polynomial of degree less than n + 1. So c k = 0 and we may write Since q 1 (x)m n+1 (x) and m n+2 (x) are monic polynomials of degree n + 2 and both m n+1 and mn have lower degree, it follows that c n+2 = 1. Finally, since q 1 (x)m n+1 (x) and m n+1 have opposite parity, ⟨q 1 m n+1 , m n+1 ⟩ = 0 and c n+1 = 0. So finally By Equation (5) and the inductive hypothesis, it follows that the parity of m n+2 matches the parity of n + 2, so claim (A) holds. For claim (B), it suffices to show that cn = ‖m n+1 ‖ 2 µ /‖mn‖ 2 µ . By rearranging Equation (5) and considering the projection onto mn, it follows that cn = ⟨q 1 m n+1 , mn⟩ We conclude the calculation by first expanding q 1 (x)mn(x), a polynomial of degree n + 1, in terms of m 0 , m 1 , . . . , m n+1 . Thus q 1 (x)mn(x) = ∑︀ n+1 k=0 d k m k (x) for some constants d 0 , d 1 , ..., d n+1 . By inspecting the leading coefficient, it follows that d n+1 = 1 and by projecting onto the m n+1 direction that ⟨q 1 mn , m n+1 ⟩ = ‖m n+1 ‖ 2 µ . So cn = ‖m n+1 ‖ 2 µ /‖mn‖ 2 µ , completing the induction and the proof.
Up to a translation factor, Proposition 2.11 is a reproduction of Bonnet's recurrence formula when the measure is Lebesgue. The drawback of Theorem 2.11 is that the algorithm is dependent on the norm of the monic polynomials. One method to compute the norm of a polynomial is through the moments of the measure, which is the focus of Section 3. In Figure 2, we provide the graph of the first six normalized Legendre polynomials for the ternary Cantor measure.

Moments of the weighted Cantor measure
As previously observed, if α ∈ △ N is a standard basis vector, then µ α is a Dirac measure, and L 2 µ α [0, 1] is 1-dimensional. Therefore, throughout this section, we focus mainly on α ∈ △ * N , but several results remain most general. In this case, integration with respect to µ α presents a difficult calculation. One method is to interpret the problem as a Riemann-Stieltjes integral: for f continuous on [0, 1], Recall the sample set By considering a uniform mesh size of 1/N k , we obtain the left-endpoint approximation of the above Riemann-Stieltjes integral, For any α ∈ △ N and any nonnegative integer m, we define the m-th moment of µ α to be When the weight vector α is understood, we suppress the superscript on the moment notation.
In the following proposition, we derive an invariance identity analogous to the invariance relation in Equation (1). This identity will be essential for the remainder of the paper.
Proof. Since {φn} N−1 n=0 are affine transformations, we note that the right-hand side of Equation (7) is welldefined. The proof follows by a standard bootstrapping argument. First observe, by Equation (1), that (7) holds for any characteristic function, Then, by linearity of the integral, Equation (7) holds for simple functions. We attain the identity for nonnegative measurable functions by an application of the Simple Approximation and the Monotone Convergence Theorems; hence, the result follows in general by linearity of the integral.
We now derive a recurrence relation for the moments of the weighted Cantor measure. We note that the relation exhibits the approximation in (6). While the relation was shown in [2], the proof of Theorem 3.2 as presented in this paper is original.
In particular, Proof. Let β = α ⊗k as in Proposition 2.4. Recall that we showed that µ α = µ β where the corresponding IFS for the weighted Cantor measure with respect to β is {ψn} N k −1 n=0 given by ψn(x) = (x + n)/N k and βn = ∏︀ k−1 ℓ=0 αn ℓ where n = n 0 + n 1 N + ... + n k−1 N k−1 . Then, applying Equation (7) with respect to µ β , we have Next, we expand the product in the above integrand and rearrange the terms and sums.
When i = m in (10) the summand is Im /N km . We subtract this term from the left-hand side of the equation and solve for Im to attain the desired recurrence relation.
For a reference to a large number of the moments I 0 , I 1 , I 2 , . . . of the ternary Cantor measure, see the Online Encyclopedia of Integer Sequence [9]. Since each I k is rational, the sequence of numerators and denominators appear separately under A308612 and A308613, respectively. Additionally, moments of the shifted ternary Cantor measure appear under A308614 and A308615.
Instead of computing the moments recursively, we can individually approximate them from (6). The next result estimates the error of this approximation.
Proof. The first inequality follows from the observation that (6) is a lower approximation of the Riemann-Stieltjes integral Im. For the upper bound, we manipulate (10). Specifically, we subtract the term corresponding to i = 0 to obtain Using as desired.
We define the Laplace transform of a finite measure µ on [0, 1] as the function on R given by Here, we use the Laplace transform of a weighted Cantor measure to approach the moment problem.
Proof. Using the triangle inequality, the power series for exp(·), and Tonelli's theorem, we have We observe that the last sum converges by the ratio test. From this, it follows that f is well-defined and entire.
Applying Equation (7) to f (x) = e −sx , we find Then, from an argument by induction, we have From the Bounded Convergence Theorem, we have lim k→∞ L µ α (︁ s N k )︁ = 1, and the desired identity follows.
The moment generating function (MGF) Gα(s) is defined analogously, It can be seen that We may derive many interesting identities from Gα(s), such as the following recurrence relation.

Proposition 3.5. Let α ∈ △ N be palindromic, and let m be an odd integer. Then
Proof. From Theorem 3.4 and the assumption that α is palindromic, we find This identity, in terms of the power series expansion of Gα(s) and of e s , is then From the uniquess of the coefficients, we have from which the desired identity follows.
Viewing Gα as a function on C, i.e. Gα(z) = f (−z) for f in Theorem 3.4, we note that Gα is entire. A useful consequence of this viewpoint is in estimating the moments. Specifically, we consider the partial product approximations defined for all nonnegative integers k, For any nonnegative integer m, we note that 0 ≤ I m;k ↗ Im. Indeed, this follows immmediately from the fact that G α;0 (z) = 1 and, for all k ≥ 0, G α;k+1 (z) is the product of G α;k (z) and a power series centered at z = 0 with nonnegative coefficients and constant term 1.
Theorem 3.6. Let α ∈ △ N . For any positive integers m, k with m ≥ 2, Proof. We first verify the following as an identity of formal power series, for all positive integers k. Indeed, so by definition of Gα(z), Equation (12) holds as formal power series. To see that Equation (12) holds analytically, it is sufficient to note that both G α;k (z) and Gα(z) are entire functions. Now let m, k be positive integers with m ≥ 2 and let R > 0. By subtracting G α;k (z) from both sides of Equation (12), we obtain the analytic identity Since the coefficients of z m in Gα(z) and G α;k (z) are Im /m! and I m;k /m!, respectively, it follows from the Cauchy integral formula that Note that G α;k (z) is a power series with nonnegative coefficients, so it follows that the first maximum is attained by setting z = R. Likewise since Gα(z/N k )(z) − 1 is a power series with nonnegative coefficients, it follows also that the second maximum is attained by setting z = R. Continuing the calculation, where f : (0, ∞) → R is defined by f (x) := e x x 1−m and the second inequality follows by noting that Gα(x) ≤ G (0,0,...,1) (x) = e x for x > 0 and similarly for G α;k . We minimize this upper bound (for m, k fixed) using elementary calculus. Note first that f (x) is differentiable on (0, ∞) and f (x) → ∞ as x tends to either endpoint. Since it follows that f is minimized at x = m − 1. Evaluating f (m − 1), we have Applying the Stirling approximation (m − 1)! ≤ e )︀ to degree m, it follows that G α;k (s) and F(s) := f (s)f (s/N) · · · f (s/N k−1 ) have identical coefficients up to degree m. For algorithmic simplicity, we may assume that k is a power of two, but this assumption may be circumvented with some care, or absorbed as a factor in O(log(1/ε)). We provide the following pseudocode.
Since a successful termination performs log 2 (k) products of degree m polynomials, by using a Fast Fourier Transform, the overall complexity is reduced to O(log log(1/ε) · m log m), as desired.
In [10], Grabner and Prodinger investigated measures whose distributions are given by Cantor sets and are somewhat similar to the ternary Cantor measure yet in general do not arise from an IFS. The major result in their paper is the following asymptotic behavior of the corresponding moments, where F(x) is a periodic function of period 1 and known Fourier coefficients. In regards to this paper, the ternary Cantor measure is ascertained by letting θ = 1/3. The final result of this paper is a lower bound approximation for the rate of decay of the moments for a weighted Cantor measure. It is intriguing that the bound that we obtain is precisely of the same order as the result of Grabner and Prodinger. So the first claim holds. Now suppose α N−1 > 0. We assume without loss of generality that m > (N − 1) to establish C(α) > 0 which may be adjusted to compensate for the remaining (finitely many) moments. Now note for all positive integers k, µ α [1 − 1/N k , 1] = (α N−1 ) k , so In order to maximize this lower bound on Im, we appeal to elementary calculus to first optimize the differentiable function f on (0, ∞) and then select the most optimal positive integer k, for a given m. From logarithmic differentiation, we find so f ′ has its unique zero at x 0 = log N (︂ 1 + m )︂ > 0. In fact, by the assumption on m, Note that f ′ > 0 on (0, x 0 ) and f ′ < 0 on (x 0 , ∞), so that f (x) is maximized over (0, ∞) at x 0 . Moreover, by monotonicity of f on (0, x 0 ) and (x 0 , ∞) and the fact that x 0 > 1, it also follows that the optimal integer is either ⌈x 0 ⌉ or ⌊x 0 ⌋. Write k 0 = k 0 (m) for the positive integer which maximizes f . We now show that the ratio of f (x 0 ) and f (k 0 ) is bounded above and below by constants (depending only on α). Let ε = ε(m) := k 0 − x 0 ∈ (−1, 1). Note that (1 − N −ε ) ∈ ( (1 − N), (1 − 1/N)) and for some C 1 > 0 depending only on N and , i.e. α. The last inequality follows from observing that sequence of functions {(1 + x/m) m } are positive and converge uniformly to e x on ( (1 − N), (1 − 1/N)). Further note that Thus, we find the bound