A modified Katok entropy formula

The following definitions of ‘strongly separated’ and ‘almost spanning’ sets are inspired by Pfister and Sullivan and designed for use in the setting of maps with the almost specification property.

Definition 6.3.1. LetZ ⊆X. For m < n, a set S is (n,−m, ε) separated forZ ifS ⊂Z and for every Λ ∈ I(n,−m), we have dΛ(x, y) > ε for every x, y ∈ S. We define a set S to be (g;n, ε)

separated if it is(n,−g(n, ε), ε)separated. Equivalently,S is(g;n, ε)separated if for everyx, y∈S

#{j∈ {0, . . . n−1}:d(fjx, fjy)> ε}> g(n, ε).

We think of an (n,−m, ε) separated set to be ‘a set which remains (n, ε) separated when you permitm mistakes’. In particular, a setS which is(g;n, ε) separated is (n, ε) separated in the usual sense. We define the natural dual notion of a(g;n, ε) spanning set.

Definition 6.3.2. Form < n, a setS ⊂Z is(n,−m, ε)spanning forZ if for allx∈Z, there exists

y∈S andΛ∈I(n,−m)such thatdΛ(x, y)≤ε. Note thatΛ depends onxand an(n, ε) spanning

set is always(n,−m, ε)spanning. We define a set S to be(g;n, ε) spanning if it is (n,−g(n, ε), ε) spanning.

We think of an(n,−m, ε) spanning set to be ‘a set which requires up tom mistakes to be

(n, ε)spanning’. Let

sn(g;Z, ε) = sup{#S : S is(g;n, ε) separated for Z}, rn(g;Z, ε) = inf{#S : S is(g;n, ε) spanning forZ},

sn(Z, ε) = sup{#S : S is(n, ε) separated forZ}, rn(Z, ε) = inf{#S : S is(n, ε)spanning for Z}. Lemma 6.3.1. We have

(1)rn(g;Z, ε)≤sn(g;Z, ε)≤sn(Z, ε),

(2)sn(2g;Z,2ε)≤rn(g;Z, ε)≤rn(Z, ε)≤sn(Z, ε).

Proof. Suppose that S is a(g;n, ε) separated set forZ of maximum cardinality such that S is not

(g;n, ε) spanning. We can find

z∈Z\ [ Λ∈I(g;n,ε) [ x∈S BΛ(x, ε) = \ Λ∈I(g;n,ε) Z\ [ x∈S BΛ(x, ε) ! .

SincedΛ(x, z)> εfor all x∈S andΛ∈I(g;n, ε), thenS∪ {z} is a(g;n, ε) separated set, which

contradicts the maximality ofS. Thus, every(g;n, ε)separated set of maximal cardinality is(g;n, ε)

spanning.

For (2), supposeE is(2g;n,2ε)separated andF is(g;n, ε)spanning forZ. Defineφ:E 7→

F by choosing for eachx∈E someφ(x)∈F and someΛx ∈I(g;n, ε)such that dΛx(x, φ(x))≤ε.

Suppose x 6= y. Let Λ = Λx∩Λy. Since Λ ∈ I(2g;n, ε), we have dΛ(φ(x), φ(y)) > 0 and thus φ(x)6=φ(y) . Thusφ is injective and hence|E| ≤ |F|.

Theorem 6.3.1(Modified Katok entropy formula). Let(X, d)be a compact metric space,f :X 7→

X be a continuous map and µ be an ergodic invariant measure. For γ ∈ (0,1) and any mistake functiong, we have

hµ= lim

ε→0lim infn→∞

1

nlog(inf{rn(g;Z, ε) :Z ⊂X, µ(Z)≥1−γ})

The formula remains true if we replace thelim inf bylim supand/or rn(g;Z, ε) bysn(g;Z, ε). The value taken by thelim inf (or lim sup) is independent of the choice of mistake functiong.

Proof. Since rn(Z, ε) ≥ rn(g;Z, ε), it follows from the original Katok entropy formula that the

expression on the right hand side is less than or equal to hµ (this is the easier inequality to prove

directly any way). To prove the opposite inequality, we give a method inspired by the proof of theorem A2.1 of [Pes].

For any η > 0, there exists δ, 0 < δ ≤η, a finite Borel partition ξ = {C1, . . . , Cm} and a

finite open coverU ={U1, . . . , Uk} of X wherek≥m with the following properties:

(1) Diam(Ui)≤η, Diam(Cj)≤η for all i= 1, . . . , m,j= 1, . . . , k,

(2)Ui ⊂Ci for all i= 1, . . . , m,

(4)2δlogm≤η.

This is a consequence of the regularity of the measure µ. Fixη so 1−γ > η >0 and take the corresponding numberδ, covering U and partitionξ. FixZ ⊂X with µ(Z)>1−γ. Lettn(x)

denote the number of l, 0 ≤ l ≤ n−1 for which fl(x) ∈ Sk

i=m+1Ui. Let ξn = Wni=0−1f−iξ and Cξn(x) denote the member of the partitionξn to which xbelongs.

Lemma 6.3.2. There exists a set A ⊂ Z and N > 0 with µ(A) ≥ µ(Z)−δ such that for every

x∈A andn≥N

(1)tn(x)≤2δn

(2)µ(Cξn(x))≤exp(−(hµ(f, ξ)−δ)n)

Proof. Let χ be the characteristic function ofSk_i=m+1Ui. We can writetn(x) =Pni=0−1χ(fix). By

Birkhoff’s ergodic theorem and Egorov’s theorem, we can find a setA1⊂Xwithµ(A1)≥µ(Z)−δ₂

such that forx∈A1, we have uniform convergence n−1tn(x) = 1 n n_X−1 i=0 χ(fix)→ Z χdµ=µ( k [ i=m+1 Ui)≤δ.

ChooseN1 such that if n≥N1 andx∈A1, then tn(x)≤2δn.

By Shannon-Mcmillan-Brieman theorem and Egorov’s theorem, we can find a set A2 ⊂ X

withµ(A2)≥µ(Z)−δ₂ such that forx∈A1, we have uniform convergence

−1

nlogµ(Cξn(x))→hµ(f, ξ)

There exists N2 such that if n ≥ N2 and x ∈ A2, then −n1 logµ(Cξn(x)) ≤ h(f, ξ) +δ. Set

A=A1∩A2 andN = max{N1, N2} and the lemma is proved.

Letξ_n∗ be the collection of elementsCξn of the partitionξnfor whichCξn∩A6=∅. Then for

n≥N, using property (2) of A, #ξ_n∗ ≥ X C∈ξ∗ n µ(C) exp{n(hµ(f, ξ)−δ)} ≥ µ(A) exp{n(hµ(f, ξ)−δ)}}.

Let 2ε be a Lebesgue number for U and let S be (g;n, ε) spanning for Z. We have Z ⊆

S

x∈SBΛx(x, ε) for suitably chosen Λx ∈ I(g;n, ε). Let us fix B = BΛx(x, ε). Let ξΛ be the

partitionW_i∈Λf−i_{ξ. We estimate the numberp(B, ξ}

Λx) of elements of the partitionξΛx which have

non-empty intersection withA∩B.

Since 2ε is a Lebesgue number for U, then B(fj_{x, ε) ⊂ U}

ij for some Uij ∈ U. If ij ∈ {1, . . . , m} then f−j(Uij) ⊂f

−j_(C

formf−j_(C

ij)may have non-empty intersection with f

−j_(U

ij). It follows, using property (1) of A,

that

p(B, ξΛx)≤m

2δn_{= exp(2δnlogm)}

The number p(B, ξn) of elements of the partitionξn which have non-empty intersection with both Aand B satisfies p(B, ξn)≤p(B, ξΛx)m g(n,ε)_{≤exp{(2δn+g(n, ε)) logm}.} It follows that #ξn∗ ≤ X x∈S p(BΛx(x, ε), ξn)≤#Sexp{(2δn+g(n, ε)) logm}. Rearranging, we have 1 nlog #S ≥hµ(f, ξ)−δ− 2δ+ g(n, ε) n logm.

Since2δlogm < η, Diam(ξ)< η,δ < η, g(n,ε_n ) →0, and η was arbitrary, we are done.

As a corollary, we have a version of theorem 6.3.1 for topological entropy (which we do not use).

Theorem 6.3.2. Let(X, d) be a compact metric space and f :X 7→X be a continuous map. We have

htop(f) = lim ε→0lim infn→∞

1

nlogrn(g;X, ε)

The formula remains true if we replace thelim inf bylim supand/or rn(g;X, ε) bysn(g;X, ε). The value taken by thelim inf (or lim sup) is independent of the choice of mistake functiong.