Digit and orbit frequencies via the Birkhoff theorem

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The measure μ\mu is assumed to be finite and to satisfy the normalization condition μ(X)=1\mu (X)=1 (such measures are called probability measures). Let TT be a transformation of the set XX. A set A⊂XA \subset X is called invariant if T−1(A)=AT^{-1}(A)=A. A transformation TT preserving the measure μ\mu is called ergodic if for every invariant set A⊂XA \subset X one of the following two holds: either μ(A)=0\mu (A)=0, or μ(X\A)=0\mu (X \backslash A)=0.

Consider the irrational shift

S(x)=(x+α) mod 1(x∈[0,1)=X,α∈R\Q). S(x)=(x+\alpha ) \bmod 1 \quad (x \in [0,1)=X, \alpha \in \mathbb {R} \backslash \mathbb {Q}) .

It is ergodic by 2.21a). By the ergodic theorem, for any summable function ff on XX and for almost all x∈Xx \in X the equality

lim⁡1n∑0⩽k<nf(Sk(x))=∫01f(t)dt(1) \lim \frac{1}{n} \sum _{0 \leqslant k<n} f\left(S^{k}(x)\right)=\int _{0}^{1} f(t) d t \tag {1}

holds.

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(a)

Prove that if f∈C([0,1])f \in C(\left[0,1\right]), then for the shift SS one can assert more, namely that equality (1) holds for every x∈[0,1)x \in [0,1). (First assume that f(0)=f(1)f(0)=f(1), and then consider the general case.)

(b)

Prove that if Δ=⟨a,b⟩⊂[0,1),χΔ(x)=1\Delta =\langle a, b\rangle \subset [0,1), \chi_{\Delta }(x)=1 for x∈Δx \in \Delta, χΔ(x)=0\chi_{\Delta }(x)=0 for x∉Δx \notin \Delta, then

lim⁡1n∑0⩽k<nχΔ(Sk(x))=b−a(2) \lim \frac{1}{n} \sum _{0 \leqslant k<n} \chi _{\Delta }\left(S^{k}(x)\right)=b-a \tag {2}

for all x∈[0,1)x \in [0,1).

If {Sk(x)}\left\{ S^{k}(x)\right\} is replaced by an arbitrary sequence {xk}\left\{ x_{k}\right\}, then property b) is called uniform distribution of this sequence (see [PS], [KN]).

For a uniformly distributed sequence {xk}\left\{ x_{k}\right\} and a function ff continuous on [0,1]\left[0,1\right], a relation analogous to (1) holds:

1n∑k=1nf(xk)⟶n→∞∫01f(t)dt \frac{1}{n} \sum _{k=1}^{n} f\left(x_{k}\right) \underset {n \rightarrow \infty }{\longrightarrow } \int _{0}^{1} f(t) d t

In particular,

1n∑k=1ne2πimxk⟶n→∞0 for any integer m≠0.  \frac{1}{n} \sum _{k=1}^{n} e^{2 \pi i m x_{k}} \underset {n \rightarrow \infty }{\longrightarrow } 0 \text{ for any integer } m \neq 0 \text{. }

This condition is not only necessary but also sufficient for uniform distribution. The latter statement is called Weyl's criterion (cf. problem I.3.9).

Remark.
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The Birkhoff--Khinchin ergodic theorem (see [Bi], [KSF]): let TT be a transformation of a set XX preserving a probability measure μ\mu, and let ff be a measurable function such that ∫X∣f(x)∣dμ(x)<+∞\int_{X}\left|f(x)\right| d \mu (x)<+\infty. Then for almost every (with respect to the measure μ\mu) x∈Xx \in X the limit

lim⁡1n∑0⩽k<nf(Tk(x))=fˉ(x) \lim \frac{1}{n} \sum _{0 \leqslant k<n} f\left(T^{k}(x)\right)=\bar{f}(x)

exists, called the mean value of the function ff along the trajectory of the point xx. Moreover the function fˉ\bar{f} is invariant and ∫Xfˉ(x)dμ(x)=∫Xf(x)dμ(x)\int_{X} \bar{f}(x) d \mu (x)=\int_{X} f(x) d \mu (x). If TT is ergodic, then fˉ\bar{f} turns out to be a constant (see problem 2.20), equal to ∫Xf(x)dμ(x)\int_{X} f(x) d \mu (x) (i.e. the mean value of ff on XX).

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(a)

Two points move along a circle, jumping once a second by the same angle, which is incommensurable with π\pi. What fraction of a sufficiently long time interval will both points simultaneously lie on some (fixed in advance) semicircle, if at the initial moment the angle between the points was equal to α\alpha?

(b)

Answer the same question if every second one point moves by an angle β1\beta_{1}, and the other by an angle β2\beta_{2}, where the numbers β1,β2\beta_{1}, \beta_{2}, π\pi are linearly independent over Q\mathbb {Q} (see the definition at the beginning of 4 of Chapter VII).

Let x=∑xk2k,xk∈{0,1}x=\sum \frac{x_{k}}{2^{k}}, x_{k} \in \left\{ 0,1\right\} be the binary expansion of a number x∈[0,1)x \in [0,1). Using the ergodicity of the doubling transformation (see problem 2.21b)), prove that for almost all x∈[0,1)x \in [0,1) the limit lim⁡1n∑0⩽k<nxk\lim \frac{1}{n} \sum_{0 \leqslant k<n} x_{k} exists (the frequency of occurrence of the digit one), and find it. Solve the analogous problem for expansions in a base-pp fraction (cf. problems IX.2.14, IX.2.16).

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Let aka_{k} be the first digit of the decimal representation of the number 2k(k=0,1,…)2^{k} (k=0,1, \ldots ), i.e., a0=1,a1=2,a2=4,a3=8,a4=1,…a_{0}=1, a_{1}=2, a_{2}=4, a_{3}=8, a_{4}=1, \ldots What is the frequency of occurrence of the digit p∈{1,…,9}p \in \left\{ 1, \ldots , 9\right\} in this sequence?

As is well known, every number xx from the interval (0,1](0,1] can be uniquely represented as a continued fraction

x=1a1+1a2+⋯ x=\frac{1}{a_{1}+\frac{1}{a_{2}+\cdots }}

which is also written as x=[a1,a2,…]x=\left[a_{1}, a_{2}, \ldots \right], where a1=a1(x),a2=a2(x)a_{1}=a_{1}(x), a_{2}=a_{2}(x), a3=a3(x),…a_{3}=a_{3}(x), \ldots are natural numbers. Such a fraction contains finitely many terms only when xx is rational. The finite continued fraction [a1,…,an]\left[a_{1}, \ldots , a_{n}\right] is uniquely written as an irreducible fraction pnqn\frac{p_{n}}{q_{n}}. The value of an infinite continued fraction is understood as the limit of the finite ones:

[a1,a2,…]=lim⁡[a1,…,an]=lim⁡pnqn, \left[a_{1}, a_{2}, \ldots \right]=\lim \left[a_{1}, \ldots , a_{n}\right]=\lim \frac{p_{n}}{q_{n}},

the number pnqn=pn(x)qn(x)\frac{p_{n}}{q_{n}}=\frac{p_{n}(x)}{q_{n}(x)} is called the convergent. It is easily verified that

∣x−pn(x)qn(x)∣⩽1qn(x)qn+1(x) \left|x-\frac{p_{n}(x)}{q_{n}(x)}\right| \leqslant \frac{1}{q_{n}(x) q_{n+1}(x)}

A variety of information about continued fractions can be gleaned from the book [X][\mathrm{X}]. Associated with the expansion of numbers into continued fractions is the Gauss transformation

T(x)=1x mod 1(x∈(0,1]). T(x)=\frac{1}{x} \bmod 1(x \in (0,1]) .

Indeed, let x=[a1,a2,…],y=[a2,a3,…]x=\left[a_{1}, a_{2}, \ldots \right], y=\left[a_{2}, a_{3}, \ldots \right]. Then x=1a1+yx=\frac{1}{a_{1}+y}, from which it follows that y=T(x)y=T(x). Using TT, the numbers ana_{n} are written as follows:

a1(x)=[1x],a2(x)=a1(T(x)),…,an(x)=a1(Tn−1(x)). a_{1}(x)=\left[\frac{1}{x}\right], a_{2}(x)=a_{1}(T(x)), \ldots , a_{n}(x)=a_{1}\left(T^{n-1}(x)\right) .

The Gauss transformation preserves the measure dμ=c1+xdxd \mu =\frac{c}{1+x} d x (see problem 2.5). The measure μ\mu is a probability measure for c=1ln⁡2c=\frac{1}{\ln 2}. The reader can learn about interesting applications of ergodic theory to number theory and continued fractions in [Po].

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References

[МГЛ+04] Макаров Б.М., Голузина М.Г., Лодкин А.А., Подкорытов А.Н. Избранные задачи по вещественному анализу. 2-е изд., Невский Диалект; БХВ-Петербург, 2004.