Let
T_{l}
be a transformation on the interval [−1, 1] defined by Chebyshev polynomial of degree
l
(
l
≥ 2), i.e.,
T_{l}
(cos
θ
) = cos(
lθ
). In this paper, we consider
T_{l}
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.
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
which equals the number of times that the points
T^{k}x
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 onesided derivative at the discontinuities. Then it is wellknown 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
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
T_{l}
be the transformation on the interval [−1, 1] defined by Chebyshev polynomial of degree
l
. In this paper, we consider
T_{l}
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
T_{l}
be the Chebyshev polynomial of degree
l
(
l
≥ 2). Recall that
T_{l}
is defined by
on [−1, 1]. Chebyshev polynomials are orthogonal in the Hilbert space
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 Λ
_{l}
be a map on [0, 1] defined by
for
k
= 0, 1, ⋯ ,
l
− 1.
It is wellknown that Λ
_{l}
preserves the Lebesgue measure
ν
and it is weakly mixing.
Lemma 1.
Let T_{l} be the lth Chebyshev polynomial of order l
≥ 2.
Then T_{l} 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
) =
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}
=
T_{l}
○
ϕ
holds. Hence
ψ
○
T_{l}
= 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
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
T_{l}
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
such that T is continuously differentiable on each
[
a
_{i−1}
,
a_{i}
] (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
where c is constant, has no solution g
(
x
)
of bounded variation. Then
and, for every α,
where
and
Since
T_{l}
is measure theoretically isomorphic to Λ
_{l}
by a topological homeomorphism, we may assume that the transformation
T_{l}
on ([−1, 1],
μ
) satisfies all the conditions of Lemma 2.
Proposition 1.
For the measure preserving transformation
T_{l} on
[−1, 1]
defined by lth Chebyshev polynoimal, if an
ℝ
valued function f
(
x
)
is a step function with finite discontinuity points and f
(
x
) =
g
(
T_{l}x
)−
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
T_{l}
on ([−1, 1],
μ
) and the measure preseving transformation Λ
_{l}
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
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
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_{∊}
) = 2
m∊
. Now choose an integer
N
such that
and
If
I
∈
P_{N}
and if
I
∩
D
≠
ϕ
, then
I
⊂
D_{∊}
for
. Hence the totality of
I
∈
P_{N}
with
I
∩
D
≠
ϕ
measures at most
. By the similar argument, the totality of
I
∈
P_{N+j}
,
j
≥ 0 such that
I
∩
D
≠
ϕ
measures at most
Fix
L
> 0 and consider the collection of
I
∈
P_{N+L}
having the property that
T^{j}
I
∩
D
≠
ϕ
for some 0 ≤
j
≤
L
− 1. Since
T^{j}
I
∈
P_{N+L−j}
for these
j
, and Λ
_{2}
is Lebesgue measure preserving, these intervals have the total Lebesgue measure at most
Let
Q
(
N,L
) be the subcollection of
P_{N+L}
such that
T^{j}
I
∩
D
=
ϕ
for all 0 ≤
j
≤
L
−1. Then for each
I
∈
Q
(
N, L
),
F
(
y
)
F
(Λ
_{2}
y
) ···
is constant, say λ
_{I,L}
with λ
_{I,L}
 = 1 . Since
G
(
y
) =
F
(
y
)
G
(Λ
_{2}
y
),
G
(
y
) =
F
(
y
)
F
(Λ
_{2}
y
) ···
. 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
P_{N}
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 T_{l} 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 kdiscontinuity points with k
<
l
− 1
then it satisfies the Central Limit Theorem, i.e.,
and, for every α
,
where
S_{n}f
(
x
) =
,
dμ
=
and μ
(
f
) =
f
(
x
)
dμ
(
x
).
Proof
. It is enough to show that the functional equation
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 =
a
_{0}
<
a
_{1}
< ⋯ <
a_{m}
= 1. Since
g
(
x
) has
m
− 1 discontinuity points,
g
(
T_{l}x
) has at least
l
(
m
− 1) discontinuity points and
g
(
T_{l}x
) −
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 T_{l} 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
T_{l}
(
J
) = [−1, 1]. Suppose there exists an function
g
(
x
) which satisfies the functional equation,
f(x) = g(T_{l}x) − g(x) + c.
By Proposition 1, there exists
x
_{1}
such that
g
(
x
) is constant on [−1,
T_{l}
(
x
_{1}
)] ⊃ [−1,
x
_{1}
]. If we take any
x
∈ [−1,
x
_{1}
], then both
x
and
T_{l}
(
x
) are in [−1,
x
_{1}
] and
g
(
T_{l}
(
x
)) =
g
(
x
). Since
f
(
x
) =
g
(
T_{l}x
)−
g
(
x
)+
c
, we have
f
(
x
) =
c
for all
x
∈
J
. Therefore
g
(
T_{l}
(
x
)) =
g
(
x
) for all
x
∈
J
, and
g
(
x
) =
g
(−1) for all
x
∈ [−1,
T_{l}
(
x
_{1}
)]. If
T_{l}
([−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}
=
T_{l}
(
x
_{1}
), we have
g
(
x
) is a constant on
T_{l}
([−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.
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],
)
defined by Chebyshev polynomial of degree
2.
If f
(
x
)
is a step function with finite kdiscontinuity points with k
≤ 2,
then it satisfies the Central Limit Theorem with respect to T
_{2}
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
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 1discontinuity 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
(Λ
_{2}
y
) −
G
(
y
) +
c
, then
G
(
y
) can be expressed as
where 0 =
a
_{0}
<
a
_{1}
< ⋯ <
a_{m}
= 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
(Λ
_{2}
y
) have the same value on the interval [0,
a
/2], we have
b
_{0}
=
c
. Integrating the functional equation
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 1discontinuity point, then it satisfies the Central Limit Theorem.
Case 2) Suppose that
f
(
x
) has 2discontinuity 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
(Λ
_{2}
y
) −
G
(
y
) +
c
, we have
with
m
= 2 or 3. When
m
= 2, the discontinuity points of
G
(Λ
_{2}
y
)−
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
a
_{1}
=
,
a
_{2}
=
and
Hence for
G
(Λ
_{2}
y
) −
G
(
y
) having 2discontinuity points, we have
c
_{2}
−
c
_{1}
= 0. It contradicts the assumption that
G
(
y
) has 2discontinuity points.
Remark 1.
By exactly the same argument as in the proof of the case in Proposition 2, if
f
(
x
) has only 1discontinuity point, then it satisfies the Central Limit Theorem with respect to any Chebyshev polynomials of degree
l
≥ 2.
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