CENTRAL LIMIT THEOREM ON CHEBYSHEV POLYNOMIALS
CENTRAL LIMIT THEOREM ON CHEBYSHEV POLYNOMIALS
The Pure and Applied Mathematics. 2014. Nov, 21(4): 271-279
• Received : August 14, 2014
• Accepted : October 10, 2014
• Published : November 30, 2014 PDF e-PUB PubReader PPT Export by style
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YOUNG-HO AHN

Abstract
Let Tl be a transformation on the interval [−1, 1] defined by Chebyshev polynomial of degree l ( l ≥ 2), i.e., Tl (cos θ ) = cos( ). 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 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.
Keywords
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 −1 E ) = 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 1 E be the characteristic function of a set E and consider the behavior of the sequence PPT Slide
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which equals the number of times that the points Tkx visit E . The Birkhoff Ergodic Theorem applied to the ergodic transformation PPT Slide
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Bernoulli shift on PPT Slide
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gives the Laws of the Large Numbers.
Let T be a transformation which is piecewise expanding on the unit interval X = [0, 1) and PPT Slide
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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 )  .
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 PPT Slide
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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 PPT Slide
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is log l for each l ≥ 2  .
2. PROPERTIES OF CHEBYSHEV POLYNOMIALS
Let Tl be the Chebyshev polynomial of degree l ( l ≥ 2). Recall that Tl is defined by
• Tl(cosx) = cos(lx)
on [−1, 1]. Chebyshev polynomials are orthogonal in the Hilbert space
• H=L2([−1, 1],ρ(x)dx)
where PPT Slide
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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: PPT Slide
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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 PPT Slide
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Definition 1. For each l ∈ ℕ , let Λ l be a map on [0, 1] defined by PPT Slide
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for PPT Slide
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k = 0, 1, ⋯ , l − 1.
It is well-known that Λ l 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]) a nd is measure theoretically isomorphic to the transformation Λ l on ([0, 1], ν ) by a topological homeomorphism ψ ( x ) = PPT Slide
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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 ○Λ l = Tl ϕ holds. Hence ψ Tl = Λ l 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 PPT Slide
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ϕ is a isomorphism and the following diagram commutes. PPT Slide
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Hence Tl is a measure preserving transformation on ([−1, 1], μ ) and weakly mixing. PPT Slide
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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  . 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
such that T is continuously differentiable on each [ a i−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 PPT Slide
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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 PPT Slide
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and, for every α, PPT Slide
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where PPT Slide
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and PPT Slide
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Since Tl is measure theoretically isomorphic to Λ l 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 Λ l on ([0, 1], ν ) are measure theoretically isomorphic via the topological homeomorphism PPT Slide
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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 ), PPT Slide
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and PPT Slide
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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 ) = PPT Slide
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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 PPT Slide
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, and PPT Slide
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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., PPT Slide
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. Then there exists 0 such that for all 0 < < 0 , ν ( D ) = 2 m∊ . Now choose an integer N such that PPT Slide
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and PPT Slide
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If I PN and if I D ϕ , then I D for PPT Slide
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. Hence the totality of I PN with I D ϕ measures at most PPT Slide
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. By the similar argument, the totality of I PN+j , j ≥ 0 such that I D ϕ measures at most PPT Slide
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Fix L > 0 and consider the collection of I PN+L having the property that Tj I D ϕ for some 0 ≤ j L − 1. Since Tj I PN+L−j for these j , and Λ 2 is Lebesgue measure preserving, these intervals have the total Lebesgue measure at most PPT Slide
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Let Q ( N,L ) be the sub-collection of PN+L such that Tj I D = ϕ for all 0 ≤ j L −1. Then for each I Q ( N, L ), F ( y ) F 2 y ) ··· PPT Slide
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is constant, say λ I,L with |λ I,L | = 1 . Since G ( y ) = F ( y ) G 2 y ), G ( y ) = F ( y ) F 2 y ) ··· PPT Slide
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. Hence PPT Slide
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holds almost everywhere on I . Letting PPT Slide
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the map PPT Slide
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: I J is bijective and it is easily shown that PPT Slide
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Since Q ( N, L ) measures at least PPT Slide
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, the set of y which is interior to some I Q ( N, L ) for an infinitely number of L must also measures at least PPT Slide
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. 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  , we can assume that the left side of (1) tends to G ( y ). Hence PPT Slide
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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 PPT Slide
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G ( y ) is also a step function with finite discontinuity. Hence the conclusion follows. PPT Slide
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Theorem 1. Let Tl be a measure preserving transformation on PPT Slide
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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., PPT Slide
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and, for every α , PPT Slide
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where Snf ( x ) = PPT Slide
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, = PPT Slide
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and μ ( f ) = PPT Slide
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f ( x ) ( 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 PPT Slide
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where −1 = a 0 < a 1 < ⋯ < 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 PPT Slide
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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. PPT Slide
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Theorem 2. Let Tl be a measure preserving transformation on PPT Slide
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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 PPT Slide
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then it satisfies the Central Limit Theorem.
Proof. Letting J = [−1, PPT Slide
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], 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 x 1 such that g ( x ) is constant on [−1, Tl ( x 1 )] ⊃ [−1, x 1 ]. If we take any x ∈ [−1, x 1 ], then both x and Tl ( x ) are in [−1, x 1 ] 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 ( x 1 )]. If Tl ([−1, x 1 ]) = [−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 x 2 = Tl ( x 1 ), we have g ( x ) is a constant on Tl ([−1, x 2 )) by exactly the same argument by using x 2 in the place of x 1 . Iterating this argument if we need it, we get g ( x ) is constant and the conclusion follows. PPT Slide
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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 T 2 be a measure preserving transformation on ([−1, 1], PPT Slide
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) 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 T 2 except for the functions of the form PPT Slide
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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 2 y ) − G ( y ) + c , then G ( y ) can be expressed as PPT Slide
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where 0 = a 0 < a 1 < ⋯ < 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 ) = c 0 1 [0,a] ( y )+ c 1 1 [a,1] ( y ) with some constants c 0 , c 1 and constant 0 < a < 1. Since both G ( y ) and G 2 y ) have the same value on the interval [0, a /2], we have b 0 = c . Integrating the functional equation
• F(y) =G(Λ2y) −G(y) +c,
we get a equation dc + (1 − d ) b 1 = c . Hence b 1 = 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 PPT Slide
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As in the case 1, letting G ( y ) be a solution of the functional equation F ( y ) = G 2 y ) − G ( y ) + c , we have PPT Slide
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with m = 2 or 3. When m = 2, the discontinuity points of G 2 y )− G ( y )+ c are PPT Slide
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Hence we have PPT Slide
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and PPT Slide
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Thus PPT Slide
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and F ( y ) has to be in the form of PPT Slide
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and PPT Slide
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with some constants b , c .
When m = 3, by the similar argument as in the case m = 2, we have a 1 = PPT Slide
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, a 2 = PPT Slide
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and PPT Slide
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Hence for G 2 y ) − G ( y ) having 2-discontinuity points, we have c 2 c 1 = 0. It contradicts the assumption that G ( y ) has 2-discontinuity points. PPT Slide
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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.
References