The Koopman Operator

The Koopman Operator

Given(X,B, µ,T), we can take the space of complex-valued square-integrable observables L²(µ). This is a Hilbert space, equipped with inner product

hf,gi= Z

X f(x)·g(x) dµ.

TheKoopman operatoris dened as

U_T :L²(µ)→L²(µ), U_Tf =f ◦T.

Exercise 8.1: Show that the Koopman operator is linear and positive.

By T -invariance ofµ, it is a unitary operator (it preserves the inner product). Indeed

hU_Tf,U_Tgi = Z

X f ◦T(x)·g ◦T(x) dµ

= Z

X(f ·g)◦T(x) dµ= Z

Xf ·g dµ=hf,gi, and therefore U_T^∗U_T =U_TU_T^∗ =I .

Remark: This has several consequences, common to all unitary operators. The spectrumσ(U_T) of U_T is a closed subset of the unit circle, and it is in fact a group under multiplication.

The Transfer Operator

Denition: ThePerron-Frobenius or transfer operator of a transformation T :X →X is the dual of the Koopman operator:

Z

XP_Tf ·g dµ= Z

X f ·U_Tg dµ= Z

X f ·g ◦T dµ.

Note that, although U_T is independent of the measure, P_T is not.

Often it will be important to specify the measure explicitly, and this measure need not be invariant.

Exercise 8.2: Show that the Perron-Frobenius operator has the following properties:

1. PT is linear;

2. P_T is positive: f ≥0 implies P_Tf ≥0.

3. R

P_Tf dµ=R f dµ. 4. P_Tk = (P_T)^k.

Lemma: Let T : [0,1]→[0,1]be a piecewise C¹ interval map.

Then the Perron-Frobenius operator P_T w.r.t. Lebesgue measure λ has the pointwise formula

P_Tf(x) = X

y∈T⁻¹(x)

f(y)

|T⁰(y)|.

Specically, a xed point of P is an invariant density for T . Proof Let 0=a₀<a₁ <· · ·<a_N =1 be such that T is C¹ monotone on each(a_i−1,a_i). Let y_i =T⁻¹(x)∩(a_i−1,a_i).

The Transfer Operator

We obtain (P_Tf)(x) =

Fund. Th. Calc|{z}

d dx

Z _x

0 P_Tf(s) ds= d dx

Z ₁

0 (P_Tf)(s)·1[0,x](s)ds

=

def. P|{z}T

d dx

Z ₁

0 f ·1[0,x]◦T(s) ds =

s=|{z}T(u)

d dx

Z

T⁻¹[0,x]

f(u) du

= X

T|

(ai−1,ai)increasing

x∈T((ai−1,ai))

d dx

Z _yi(x) ai−1

f(u) du

+ X

T|(ai−1,ai)decreasing

x∈T((ai−1,ai))

d dx

Z _ai

yi(x)

f(u) du

+ X

T((ai−1,ai))⊂[0,x]

d dx

Z _ai

ai−1

f(u) du

Continuing this:

(P_Tf)(x) = X

T|(ai−1,ai)increasing

x∈T((ai−1,ai))

f(y_i)

T⁰(y_i) + X

T|(ai−1,ai)decreasing

x∈T((ai−1,ai))

− f(y_i) T⁰(y_i) +0

= X

i

f(yi)

|T⁰(y_i)|, as required.

The Transfer Operator

There is also a Perron-Frobenius operator with respect to a measureµLebesgue, instead of Lebesgue measure itself:

Lemma: If dµ=hdx, then the operator P_T,µf = P_T(f ·h)

h

acts as the Perron-Frobenius operator on(X,B,T, µ). Viewed dierently, if a function h≥0 is xed by P_T (w.r.t.

Lebesgue) then dµ=h dx is an invariant measure.

Proof of the Lemma: Let A beµ-measurable and f ∈L¹(µ). Then

Z

AP_T,µf dµ = Z

A

P_T(f ·h) h h dx

= Z

AP_T(f ·h)dx

= Z

XPT(f ·h)1Adµ= Z

X f ·h·(1A◦T)dx

= Z

T⁻¹Af ·h dx = Z

T⁻¹Af dµ.

Because A is arbitrary, the lemma is proved.

Mixing

Denition: A probability measure preserving dynamical system (X,B, µ,T) ismixing (orstrong mixing) if

µ(T⁻ⁿ(A)∩B)→µ(A)µ(B) as n→ ∞ for every A,B ∈ B.

This says that the events A and B are asymptotically independent.

Exercise: Show that Lebesgue measureµis mixing for the doubling map. In fact, the n-thcorrelation coecient

Corn(A,B) :=µ(T⁻ⁿ(A)∩B)−µ(A)µ(B) =0 for every n≥1.

Proposition: A probability preserving dynamical system (X,B,T, µ) is mixing if and only if

Z

X f ◦Tⁿ(x)·g(x) dµ→ Z

X f(x) dµ· Z

X g(x) dµas n→ ∞ for all f,g ∈L²(µ), or written in the notation of the Koopman operator U_Tf =f ◦T :

hU_Tⁿf,gi → hf,1ih1,gi as n→ ∞.

Proof: The if-direction follows by taking indicator functions f =1_A and g =1_B. For the only if-direction, general f,g ∈L²(µ) can be approximated by linear combinations of indicator functions.

Exponential Mixing

Assume without loss of generality thatR

g dµ=h1,gi=0. Then mixing means that

hU_Tⁿf,gi=hf,P_Tⁿgi →0 as n→ ∞.

If the operator normkP_Tkrestricted to {g ∈L²(µ) :R

g dµ=0} is strictly less than 1, then

Corn(f,g) =hf,P_Tⁿgi ≤ kfk kgk kP_Tkⁿ→0 exponentially fast.

Thusµis exponentially mixing: the events f and g are exponentially independent.

This motivates the spectral analysis of the transfer operator.

Denition: An operator P is called quasi-compact ifλ=1 is an eigenvalue of multiplicity one, and the rest of the spectrum (i.e., all λ∈Csuch that (P−λI)⁻¹ is not a wel-dened bounded

operator), lies in a disk{|λ| ≤σ} for someσ <1.

I Multiplicity one means that the space of eigenfunctions satisfying Pf =λf (forλ=1) is one-dimensional.

I Thisσ is called theessential spectral radius. The actual spectral radius is 1.

I Perron-Frobenius operators are seldom quasi-compact on L²(µ). The (non-trivial) task is to nd a Banach spaceB restricted to which P is quasi-compact.

Exponential Mixing

TheLasota-Yorke(orDoeblin-Fortet or Tulcea-Ionescu-Marinescu) inequality holds if there are two Banach spaces(B_s,k k_s) and (B_w,k k_w) (for strong and weak) and σ∈(0,1), L≥1 such that

I B_s ⊂B_w and the unit ball {f ∈B_s :kfk_s ≤1}is a compact subset of B_w.

I kPfk_s ≤σkfk_s+Lkfk_w for all f ∈B_s.

I kPfk_w ≤ kfk_w for all f ∈B_w.

Under these conditions, P:Bs →Bs is quasi-compact, with essential spectral radius≤σ.

Theorem: If the Perron-Frobenius operator for T : [0,1]→[0,1]is quasi-compact with Bs ⊂L¹(m)for Lebesgue measure m, then there is an absolutely continuous invariant measureµ which is exponentially mixing.

Exercise 8.3: Show that kP_Tfk_L1 ≤ kfk_L1.

Exercise 8.4: Suppose we have two sequences(a_n)_n∈Nand(b_n)_n∈N

satisfying: there areσ ∈(0,1) and L>0 such that:

a_n+1 ≤a_n and b_n≤σb_n+La_n.

Show that(b_n)_n∈N is bounded. Show that lim sup_nb_n≤L/(1−σ). The Lasota-Yorke inequality got reinvented several times (Andrej Lasota and James Yorke being the most recent), and they applied this to expanding interval maps with specic choices of Banach spaces.

Exponential Mixing

Denition: Let g : [a,b]→R. The variation of g is dened to be Var[a,b]g =supXⁿ

i=1

|g(x_i)−g(x_i−1)|,

where the supremum runs over all nite partitions generated by points a=x₀ <x₁<· · ·<x_n=b.

Note that Var is aseminorm, i.e.,Var(f) =06⇒f ≡0. In fact, Var f =Var(f +C) for every constant C. However,

BV :={f ∈L¹:Var(f)<∞}with normkfk_BV =Var(f) + Z

|f|dx is a Banach space. Its unit ball is compactly embedded in L¹.

Lasota-Yorke Theorem: If T : [0,1]→[0,1]is a piecewise C² map with inf_x∈[0,1]|T⁰(x)|>1, then T preserves an invariant measure µm=Lebesgue, and the density h= ^d_dx^µ is of bounded variation. Alsoµis exponentially mixing.

I Compared to the Folklore Theorem, there is an important dierence that the branches need not be onto.

I For instant, β-transformations:

Tβ : [0,1]→[0,1], Tβ(x) =βx mod 1

have an absolutely continuous measures for allβ >1, not just the integers β ≥2.

Exponential Mixing

The technical details are in the notes, but the key-steps are:

I Take an iterate Tⁿsuch that inf_x|(Tⁿ)⁰(x)|>2, so σ :=2/infx|(Tⁿ)⁰(x)|<1.

I For B_s =BV and B_w =L¹(m) the Lasota-Yorke ineq. holds:

Var_[0,1]P_T,mg ≤σVar_[0,1]g +Lkgk_L1(m).

I The limit µof Césaro means ¹_nP_n−1

k=0m◦T^−k is invariant.

I The invariant densityh = ^d_dx^µ =lim_nP_Tⁿ_,m1∈BV.

I The eigenspace of P_T,m for λ=1 is spanned by h.

I For all f ∈L¹(m)and g ∈BV : Cor_n(f,g) =

Z

f ·P_Tⁿ,mg h dx− Z

f dµ Z

g dµ because P_Tⁿ_,mh=h =

Z

f ·P_T,m(g − Z

g dµ)h dx

≤ kfk_L1kgk_BVkPⁿ|_h⊥k

≤ kfk_L1kgk_BVσⁿ.