500

400

300

200

100

0

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

Figure 4.1: The histogram of iterates of L4 compared with σ.

90 CHAPTER 4. SPACE AND TIME AVERAGES

it follows that if T— ν = ν, then T— (h— ν) = h— ν. So to solve L4—µ = µ, we must

merely compute h— ν. According to (4.3) this is the measure with density

1 1

σ(y) = = .

π sin 2 cos πx

πx

|h (x)| 2

But since y = sin2 πx

this becomes

2

1

σ(y) =

y(1 ’ y)

π

as desired.

To prove (ii), it is enough to prove the corresponding result for the tent

transformation: that ρ = const. is the only continuous function satisfying (4.5).

To see this, let us consider the binary representation of T : Let

x = 0.a1 a2 a3 . . .

1

be the binary expansion of x. If 0 ¤ x < 2 , so a1 = 0, then T x = 2x or

T (0.0a2a3 a4 . . . ) = 0.a2 a3 a4 . . . .

1

If x ≥ 2 , so a1 = 1, then

T (x) = ’2x + 2 = 1 ’ (2x ’ 1) = 1 ’ S(x) = 1 ’ 0.a2 a3 a4 . . . .

Introducing the notation

¯ = 1, ¯ = 0,

0 1

we have

0.a2 a3 a4 · · · + 0.a1 a2 a3 · · · = 0.1111 · · · = 1

¯¯¯

so

T (0.1a2a3 a4 . . . ) = 0.a2 a3 a4 . . . .

¯¯¯

In particular, T ’1 (0.a1 a2 a3 . . . ) consists of the two points

0.0a1 a2 a3 . . . and 0.1a1 a2 a3 . . . .

¯¯¯

Now let us iterate (4.5) with ρ replaced by f , and show that the only solution

is f = constant. Using the notation x = 0.a1 a2 · · · = 0.a repeated application

of (4.5) gives:

1

f (x) = [f (.0a) + f (.1¯)]

a

2

1

= [f (.00a) + f (.01¯) + f (.10a) + f (.11¯)]

a a

4

1

[f (.000a) + f (.001¯) + f (.010a) + f (.011¯) + f (.100a) + · · · ]

= a a

8

’ f (t)dt.

4.2. THE HISTOGRAM OF L4 91

But this integral is a constant, independent of x. QED.

The third statement, (iii), about the limiting histogram for “generic” initial

seed, x0 , demands a more careful formulation. What do we mean by the phrase

“generic”? The precise formulation requires a dose of measure theory, the word

“generic” should be taken to mean “outside of a set of measure zero with respect

to µ”. The usual phrase for this is “for almost all x0 ”. Then assertion (iii)

becomes a special case of the famous Birkho¬ ergodic theorem. This theorem

asserts that for almost all points, p, the “time average”

n’1

1

φ(Lk p)

lim 4

n

k=0

equals the “space average”

φµ

for any integrable function, φ. Rather than proving this theorem, we will explain

a simpler theorem, von Neumann™s mean ergodic theorem, which motivated

Birkho¬ to prove his theorem.

Let F be a transformation with an invariant measure, µ. By this we mean

that F— µ = µ. We let H denote the Hilbert space of all square integrable

functions with respect to µ, so the scalar product of f, g ∈ H is given by

(f, g) = f g µ.

¯

The map F induces a transformation U : H ’ H by

Uf = f —¦ F

and

(f —¦ F )(g —¦ F )µ =

(U f, U g) = f gµ = (f, g).

¯

In other words, U is an isometry of H. The mean ergodic theorem asserts that

the limit of

n’1

1

U kf

n0

exists in the Hilbert space sense, “convergence in mean”, rather than the almost

everywhere pointwise convergence of the Birkho¬ ergodic theorem. Practically

ˆ ˆ ˆ

by its de¬nition, this limiting element f is invariant, i.e. satis¬es U f = f .

Indeed, applying U to the above sum gives an expression which di¬ers from

that sum by only two terms, f and U n f and dividing by n sends these terms

to zero as n ’ ∞. If, as in our example, we know what the possible invariant

ˆ

elements are, then we know the possible limiting values f

The mean ergodic theorem can be regarded as a smeared out version of the

Birkho¬ theorem. Due to inevitable computer error, the mean ergodic theorem

may actually be the version that we want.