CENTRAL LIMIT THEOREM ON CHEBYSHEV POLYNOMIALS

Abstract

Let $T_l$ be a transformation on the interval [-1, 1] defined by Chebyshev polynomial of degree $l(l{\geq}2)$, i.e., $T_l(cos{\theta})=cos(l{\theta})$. In this paper, we consider $T_l$ as a measure preserving transformation on [-1, 1] with an invariant measure $\frac{1}{\sqrt[\pi]{1-x^2}}dx$. We show that If f(x) is a nonconstant step function with finite k-discontinuity points with k < l-1, then it satisfies the Central Limit Theorem. We also give an explicit method how to check whether it satisfies the Central Limit Theorem or not in the cases of general step functions with finite discontinuity points.

1. INTRODUCTION

Let (X, μ) be a probability measure space. A measurable transformation T : X → X is said to be measure preserving if μ(T−1E) = μ(E) for every measurable subset E. A measure preserving transformation T on X is called ergodic if f(Tx) = f(x) holds only for constant functions and it is called weakly mixing if the constant function is the only eigenfunction with respect to T [3, 5].

Let 1E be the characteristic function of a set E and consider the behavior of the sequence which equals the number of times that the points Tkx visit E. The Birkhoff Ergodic Theorem applied to the ergodic transformation Bernoulli shift on gives the Laws of the Large Numbers.

Let T be a transformation which is piecewise expanding on the unit interval X = [0, 1) and be a function of bounded variation, where T'(x) is the appropriate one-sided derivative at the discontinuities. Then it is well-known that there exists an absolutely continuous invariant measure with respect to the Lebesgue measure. Furthermore if T is weakly mixing with respect to the T-invariant absolutely continuous measure, f(x) is a bounded variation function and the functional equation

f = g ○ T − g + c

does not have any solution g(x) for any constant c ∈ ℝ, then we can apply the Central Limit Theorem to the function f(x) [2].

For each natural number l(l ≥ 2), let Tl be the transformation on the interval [−1, 1] defined by Chebyshev polynomial of degree l. In this paper, we consider Tl as a measure preserving transformation on [−1, 1] with an invariant measure We show that if f(x) is a step function with finite k-discontinuity points(k < l − 1) then it satisfies the Central Limit Theorem. We also give a explicit method how to check whether it satisfies the Central Limit Theorem or not in the cases of general step functions with finite discontinuity points. It is known that the entropy of is log l for each l ≥ 2[1].

 

2. PROPERTIES OF CHEBYSHEV POLYNOMIALS

Let Tl be the Chebyshev polynomial of degree l (l ≥ 2). Recall that Tl is defined by

Tl(cos x) = cos(lx)

on [−1, 1]. Chebyshev polynomials are orthogonal in the Hilbert space

H = L2([−1, 1], ρ(x) dx)

where

Let T : (X, μ) → (X, μ) and Λ : (Y, ν) → (Y, ν) be measure preserving. Two measure preserving transformations are said to be measure theoretically isomorphic if there exists an isomorphism ψ : (X, μ) → (Y, ν) such that ψ ○ T = Λ ○ ψ, in other words, the following diagram commutes:

From now on, let ν be the Lebesgue measure on [0, 1] and μ be an absolutely continuous measure on [−1, 1] with the density function ρ(x). i.e., the measure μ is defined by

Definition 1. For each l ∈ ℕ , let Λl be a map on [0, 1] defined by

for k = 0, 1, ⋯ , l − 1.

It is well-known that Λl preserves the Lebesgue measure ν and it is weakly mixing.

Lemma 1. Let Tl be the l-th Chebyshev polynomial of order l ≥ 2. Then Tl preserves the measure μ on ([−1, 1]) and is measure theoretically isomorphic to the transformation Λl on ([0, 1], ν) by a topological homeomorphism ψ(x) = arccos(x).

Proof. Let ϕ(y) be the inverse function of ψ(x), i.e., ϕ(y) = cos(πy) from [0, 1] to [−1, 1]. It is obvious that ϕ○Λl = Tl ○ϕ holds. Hence ψ○Tl = Λl ○ψ. So it is enough to show that ϕ is a measure theoretical isomorphism. Note that the inverse image of [ϕ(y), 1] under ϕ is [0, y], which has Lebesgue measure equal to y. For ϕ to be an measure theoretical isomorphism, it must satisfy

μ([ϕ(y), 1]) = y

for all 0 ≤ y ≤ 1. Thus μ([x, 1]) = ψ(x) and μ([0, x]) = 1−ψ(x) for all −1 ≤ x ≤ 1, because μ is a probability measure on [−1, 1]. Since

ϕ is a isomorphism and the following diagram commutes.

Hence Tl is a measure preserving transformation on ([−1, 1], μ) and weakly mixing.

 

3. THE CENTRAL LIMIT THEOREM

The following lemma gives a suffcient condition for a special class of transformations on which the Central Limit Theorem holds [2]. In Lemma 2, μ is an arbitrary absolutely continuous measure.

Lemma 2. Let T be a piecewise continuously differentiable and expanding transformation on an interval [a, b], i.e., there exists a partition

a = a0 < a1 < ⋯ < ak−1 < ak = b

such that T is continuously differentiable on each [ai−1, ai] (1 ≤ i ≤ k) and | T'(x)| > B for some constant B > 1(At the endpoints of an interval we consider directional derivatives). Assume that is a function of bounded variation. Suppose that T is weakly mixing with respect to an invariant probability measure μ. Let f(x) be a function of bounded variation such that the equation

f(x) = g(Tx) − g(x) + c,

where c is constant, has no solution g(x) of bounded variation. Then

and, for every α,

where

and

Since Tl is measure theoretically isomorphic to Λl by a topological homeomorphism, we may assume that the transformation Tl on ([−1, 1], μ) satisfies all the conditions of Lemma 2.

Proposition 1. For the measure preserving transformation Tl on [−1, 1] defined by l-th Chebyshev polynoimal, if an ℝ-valued function f(x) is a step function with finite discontinuity points and f(x) = g(Tlx)−g(x)+c with a constant c, then g(x) is also a step function with finite discontinuity points.

Proof. Recall that the measure preserving transformation Tl on ([−1, 1], μ) and the measure preseving transformation Λl on ([0, 1], ν) are measure theoretically isomorphic via the topological homeomorphism arccos(x) by Lemma 1. As in Lemma 1, let ϕ(y) = cos(πy) be the inverse function of ψ(x). Note that f(x) is a step function with finite discontinuity points if and only if f( ϕ(y)) is a step function with finite discontinuity points. Furthermore the functional equation

f(x) = g(Tlx) − g(x) + c

has a solution if and only if the functional equation

f(ϕ(y)) = g(Tl(ϕ(y))) − g(ϕ(y)) + c

has a solution. Let v be the variation of f(x), and Note that the number of discontinuity points of f(x) is equal to the number of discontinuity points of F(y) and if the functional equation

f(ϕ(y)) = g(Tl(ϕ(y))) − g(ϕ(y)) + c

has a solution then the functional equation

F(y)G(Λly) = G(y)

has a solution. So it is enough to show that G(y) is also a step function with finite discontinuity points, because if g(ϕ(y)) is a bounded variation function and G(y) = is a step function with finite discontinuity points, then g(ϕ(y)) also has to be a step function with finite discontinuity points. For the notational simplicity, we will prove the proposition in the case l = 2.

Let P be a partition of [0, 1] defined by , and Let D = {z | F(y) is discontinuous at y = z}, m be the cardinality of discontinuity D and D∊ be the ∊-neighborhood of D, i.e., . Then there exists ∊0 such that for all 0 < ∊ < ∊0, ν (D∊) = 2m∊. Now choose an integer N such that and

If I ∈ PN and if I ∩ D ≠ ϕ, then I ⊂ D∊ for . Hence the totality of I ∈ PN with I ∩ D ≠ ϕ measures at most . By the similar argument, the totality of I ∈ PN+j, j ≥ 0 such that I ∩ D ≠ ϕ measures at most

Fix L > 0 and consider the collection of I ∈ PN+L having the property that TjI ∩ D ≠ ϕ for some 0 ≤ j ≤ L − 1. Since TjI ∈ PN+L−j for these j, and Λ2 is Lebesgue measure preserving, these intervals have the total Lebesgue measure at most

Let Q(N,L) be the sub-collection of PN+L such that TjI ∩ D = ϕ for all 0 ≤ j ≤ L−1. Then for each I ∈ Q(N, L), F(y)F(Λ2y) ··· is constant, say λI,L with |λI,L| = 1 . Since G(y) = F(y)G(Λ2y), G(y) = F(y)F(Λ2y) ··· . Hence holds almost everywhere on I. Letting the map : I → J is bijective and it is easily shown that

Since Q(N, L) measures at least , the set of y which is interior to some I ∈ Q(N, L) for an infinitely number of L must also measures at least . Fixing such an y, we have that (1) holds. We may assume that y is also a Lebesgue point of G. Since PN is finite, it can be assumed J is always the same on the right side of (1). By the Lebesgue density theorem[4], we can assume that the left side of (1) tends to G(y). Hence

Since |G(z)| = 1 for all z ∈ [0, 1], G(z) has to be constant on J. Since F(y) is a step function with finite discontinuity and G(y) is also a step function with finite discontinuity. Hence the conclusion follows.

Theorem 1. Let Tl be a measure preserving transformation on defined by Chebyshev polynomial of degree l (l ≥ 2). If f(x) is a nonconstant step function with finite k-discontinuity points with k < l − 1 then it satisfies the Central Limit Theorem, i.e.,

and, for every α,

where Snf(x) = , dμ = and μ(f) = f(x) dμ(x).

Proof. It is enough to show that the functional equation

f(x) = g(Tlx) − g(x) + c

has no solution. Suppose it is not, by Proposition 1, g(x) is also a step function with finite discontinuity points. Hence g(x) can be expressed as

where −1 = a0 < a1 < ⋯ < am = 1. Since g(x) has m − 1 discontinuity points, g(Tlx) has at least l(m − 1) discontinuity points and g(Tlx) − g(x) + k has at least (l − 1)(m − 1) discontinuity points. Since f(x) has k discontinuity points, we have

So if k < l − 1 then m has to be 1 and g(x) has to be a constant function. It is a contradiction to the assumption that f(x) is not a constant function.

Theorem 2. Let Tl be a measure preserving transformation on defined by Chebyshev polynomial of degree l (l ≥ 2). If f(x) is a nonconstant step function with finite discontinuity points and f(x) is constant on the interval then it satisfies the Central Limit Theorem.

Proof. Letting J = [−1, ], we have Tl(J) = [−1, 1]. Suppose there exists an function g(x) which satisfies the functional equation, f(x) = g(Tlx) − g(x) + c.

By Proposition 1, there exists x1 such that g(x) is constant on [−1, Tl(x1)] ⊃ [−1, x1]. If we take any x ∈ [−1, x1], then both x and Tl(x) are in [−1, x1] and g(Tl(x)) = g(x). Since f(x) = g(Tlx)−g(x)+c, we have f(x) = c for all x ∈ J. Therefore g(Tl(x)) = g(x) for all x ∈ J, and g(x) = g(−1) for all x ∈ [−1, Tl(x1)]. If Tl([−1, x1]) = [−1, 1], then g(x) has to be a constant function and f(x) also has to be constant. it completes the proof. Otherwise, letting x2 = Tl(x1), we have g(x) is a constant on Tl([−1, x2)) by exactly the same argument by using x2 in the place of x1. Iterating this argument if we need it, we get g(x) is constant and the conclusion follows.

In the following Proposition, we give an explicit method how to check whether it satisfies the Central Limit Theorem or not in the cases of general step functions with finite discontinuity points. For the simplicity, we consider the case l = 2 and f(x) is a step function with 1 or 2 discontinuity points.

Proposition 2. Let T2 be a measure preserving transformation on ([−1, 1], ) defined by Chebyshev polynomial of degree 2. If f(x) is a step function with finite k-discontinuity points with k ≤ 2, then it satisfies the Central Limit Theorem with respect to T2 except for the functions of the form

with some constants b, c.

Proof. As in the proof of Proposition 1, let ϕ(y) = cos(πy). Then the functional equation

f(x) = g(Tlx) − g(x) + c

has a solution if and only if the functional equation

f(ϕ(y)) = g(Tl(ϕ(y)) − g(ϕ(y)) + c

has a solution. Furthermore F(y) = f(ϕ(y)) has the same discontinuity points as f(x).

Case 1) Suppose that f(x) has 1-discontinuity point, i.e., F(y) = f(ϕ(y)) has the form of

F(y) = b0 · 1[0,d](y) + b1 · 1[d,1](y)

and G(y) = g(ϕ(y)) is the solution of the functional equation F(y) = G(Λ2y) − G(y) + c, then G(y) can be expressed as

where 0 = a0 < a1 < ⋯ < am = 1. By exactly the same argument as in the proof of Theorem 1, we have m = 1 or 2. When m = 1, G(y) has to be constant, and f(x) also has to be constant. It is a contradiction. When m = 2, then G(y) has the form of G(y) = c01[0,a](y)+c11[a,1](y) with some constants c0, c1 and constant 0 < a < 1. Since both G(y) and G(Λ2y) have the same value on the interval [0, a/2], we have b0 = c. Integrating the functional equation

F(y) = G(Λ2y) − G(y) + c,

we get a equation dc + (1 − d)b1 = c. Hence b1 = c and f(x) is constant. It is a contradiction to the assumption of f(x). Thus if f(x) has 1-discontinuity point, then it satisfies the Central Limit Theorem.

Case 2) Suppose that f(x) has 2-discontinuity points. Then F(y) = f(ϕ(y)) has the form of

As in the case 1, letting G(y) be a solution of the functional equation F(y) = G(Λ2y) − G(y) + c, we have with m = 2 or 3. When m = 2, the discontinuity points of G(Λ2y)− G(y)+c are Hence we have and Thus

and F(y) has to be in the form of

and with some constants b, c.

When m = 3, by the similar argument as in the case m = 2, we have a1 = , a2 = and

Hence for G(Λ2y) − G(y) having 2-discontinuity points, we have c2 − c1 = 0. It contradicts the assumption that G(y) has 2-discontinuity points.

Remark 1. By exactly the same argument as in the proof of the case in Proposition 2, if f(x) has only 1-discontinuity point, then it satisfies the Central Limit Theorem with respect to any Chebyshev polynomials of degree l ≥ 2.