DECAY OF CORRELATIONS FOR PIECEWISE

SMOOTH MAPS WITH INDIFFERENT FIXED POINTS

Huyi Hu

Abstract. We consider a piecewise smooth expanding map f on the unit interval that has the form f(x) = x + x^{1+γ} + o(x^{1+γ}) near 0, where 0 < γ < 1. We prove that the density function h of an absolutely continuous invariant probability measure μ has order x^{−γ} as x → 0, and that the decay rate of correlations with respect to μ is polynomial for Lipschitz functions. Perron-Frobenius operators are the main tool used for proofs.

0. Introduction

Let f : I → I be a piecewise smooth map on the unit interval I. It is well known that if f is uniformly expanding, then it admits an absolutely continuous invariant probability measure μ, and (f, μ) has exponential decay of correlations. If f has indifferent fixed points, then f still admits an absolutely continuous invariant measure μ. In addition, if f is C^{1+γ}, 0 < γ < 1, then the measure μ is finite (See e.g. [P]). The purpose of this paper is to show that such systems has polynomial decay of correlations.

We assume that f has an indifferent fixed point 0, and fx = x + x^{1+γ} + o(x^{1+γ}) near 0. We use Perron-Frobenius operator L to get the density function h. The fact Lh = h implies that as x → 0, h(x) goes to infinite just like x^{−γ} multiplied by a constant related to the value of f′ and h at f^{−1}(0). Then we use η(x) = \frac{h(x)}{h(fx)f′(x)}, instead of \frac{1}{f′(x)}, to define a different operator L̃. This operator preserves L^{1} norms and leaves constant functions invariant. So L̃^{n}g → μ(g) for any continuous function g. Moreover, if higher order terms are ignored, then near 0, L̃g(x) ≈ (1−x_{1}^{γ})g(x_{1})+ x_{1}^{γ} ḡ(x_{1}), where x_{1} is the preimage of x near 0, and ḡ is the average of g with weight η at the rest of preimages (see (4.3) for details). Since restricted to a neighborhood of 0, all backward orbits approach to 0 in a polynomial rate, the rate of the convergence L̃^{n}g → μ(g), both in L^{1}(I, μ) and in measure, is polynomial. Therefore the rate of decay of correlations is polynomial as well.

We state assumptions and the main results in §1. Theorem A, which is concerning existence and properties of density functions of invariant measures, is proved in §3. Theorem B and its corollary, which deal with decay rate of correlations of Lipschitz functions and mixing rate of sets respectively, are proved in §7. To obtain Theorem

Typeset by AMS-TEX

B, we prove Proposition 5.2 in §5 and §6, which asserts that the rate of convergence L̃^{n}g → μ(g) is polynomial.

1. Assumptions, Statements of Results and Notations

Let I = [0, 1] be the unit interval and f : I → I be a piecewise smooth map. A fixed point p of f is called indifferent if fp = p and lim_{x→p} f′(x) = 1.

Assumptions. Let f : I → I such that

(I) There is a finite partition ξ = {I_{0}, I_{1}, ⋯, I_{Q}} into subintervals such that for each q, restricted to I_{q}, f|_{int I_{q}} is twice differentiable and f|_{int I_{q}} maps int I_{q} to (0, 1) diffeomorphically.

(II) 0 is an indifferent fixed point of f.

(III) f′ > 1 on (0, 1], and f′′ is bounded on [τ, 1] ∀τ > 0. Moreover, we need the following assumption for technical reasons.

(IV) Near x = 0, f and its derivative have the form

f(x) = x + x^{1+γ} + x^{1+γ}δ_{0}(x),  (1.1)
f′(x) = 1 + (1 + γ)x^{γ} + x^{γ}δ_{1}(x),  (1.2)
f′′(x) = \frac{γ(1 + γ) + δ_{2}(x)}{x^{1−γ}},  (1.3)

where δ_{i}(x) → 0 as x → 0 for i = 0, 1, 2.

The last assumption says that f is equal to x + x^{1+γ} plus higher order terms, and the first and the second derivative of the higher order terms are still of higher orders.

We denote by I_{0} the element of the partition ξ that contains 0.

Theorem A. Suppose f : I → I satisfies Assumption (I)−(IV). Then f has an absolutely continuous invariant probability measure μ whose density function h(x) satisfies

i) 0 < h(x) < ∞ ∀x ∈ (0, 1]; ii) h is Lipschitz on [τ, 1] ∀0 < τ < 1; iii) ∃R > 0 such that

|x^{γ}h(x) − σ_{0}| ≤ R max{x^{γ}, δ_{1}(x)},

where σ_{0} = lim_{x→0} ∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{h(x̄_{1})}{f′(x̄_{1})} is a constant. In particular

lim_{x→0} x^{γ}h(x) = σ_{0}.

DECAY OF CORRELATIONS FOR MAPS WITH INDIFFERENT FIXED POINTS 3

The part of existence of absolutely continuous invariant measures was proved by Pianigiani ([P]) in more general setting by using the first return map. For Part iii), a similar result can be seen in [CF1] and [CF2] for a map with the form fx = \frac{x}{1−x} as 0 ≤ x ≤ \frac{1}{2}, which admits σ-finite absolutely continuous invariant measure.

For a Lipschitz function F, denote by ∥F ∥ the C^{0} norm.

Theorem B. Suppose f : I → I satisfies Assumption (I)−(IV). Let μ be the absolutely continuous invariant probability measure and let β = γ^{−1}. Then

i) for any Lipschitz function G, there is a constant C = C(G) > 0 such that for any Lipschitz function F,

|∫ (F ∘ f^{n})Gdμ − ∫ F dμ ∫ Gdμ| ≤ \frac{C}{n^{β−1}} ∥F ∥ ∀n > 0;

ii) there exist Lipschitz functions G and F, and a constant C′ > 0 such that

|∫ (F ∘ f^{n})Gdμ − ∫ F dμ ∫ Gdμ| ≥ \frac{C′}{n^{β−1}} ∀n > 0.

A result similar to Part i) has been proved by L.-S. Young recently in more general setting (see [Y]). However, her method is quite different with ours. She uses tails of tower, and we use Perron-Frobenius operators. Earlier, M. Mori proved polynomial decay of correlations for piecewise linear maps (see [M]).

Remark. By the proof of the theorem, we can see that Part i) still holds if we use L^{∞}(I, μ) function F and the L^{∞}(I, μ) norm ∥F ∥_{∞} instead of Lipschitz function and C^{0} norm respectively. On the other hand, we can find C^{∞} functions F and G satisfying the inequality in Part ii).

Denote ξ_{m} = ⋀_{i=0}^{m−1} f^{−i}ξ. So if E ∈ ξ_{m}, then E = ⋂_{i=0}^{m−1} f^{−i}I_{q_{i}} for some q_{0}, ⋯, q_{m−1}.

Also we denote by E^{(m)} the element of ξ_{m} containing 0.

Corollary. Under the supposition of Theorem B, there exist constants C > C′ > 0 and l > 0 such that for any m ≥ 0, E ∈ ξ_{m}, and for any measurable set E′ ⊂ [0, 1],

|μ(f^{−n−m}E′ ∩ E) − μE · μE′| ≤ \frac{Cm^{β−1}}{(n − l)^{β−1}} μE · μE′ ∀n > l,

and if in addition m ≥ l and E = E^{(m)}, then

|μ(f^{−n−m}E′ ∩ E) − μE · μE′| ≥ \frac{C′m^{β−1}}{(n + m)^{β−1}} μE · μE′ ∀n > 0.

We introduce some notations.

Let I_{0} be the element of the partition ξ containing 0. For x ∈ I_{0}, we denote x_{0} = x and x_{i+1} = f^{−1}x_{i} ∩ I_{0} ∀i > 0. Choose a small neighborhood P_{0} ⊂ I_{0} of 0.

For any function g, if x ∈ I_{0}, then we denote

σ_{g}(x) = ∑_{x̄_{1}∈f^{−1}(fx)\I_{0}} \frac{g(x̄_{1})}{f′(x̄_{1})}.  (1.4)

We should note that σ_{g}(x) depends on values of g at f^{−1}(fx) but at x itself.

Take nondecreasing functions ρ_{±}(x) ≥ 0 and denote B(x, ρ(x)) = (x−ρ_{−}(x), x+ ρ_{+}(x)). We require that ρ_{±}(x) are chosen in such a way that ρ_{±}(x) = O(x^{1+γ}) on P_{0}, and fB(x, ρ(x)) ⊃ B(fx, ρ(fx)) ∀x ∈ I, and y ∈ B(x, ρ(x)) if and only if x ∈ B(y, ρ(y)). The latter implies ρ_{+}(x) > ρ_{−}(x) on P_{0}. So B(x, ρ(x)) is not a ball in Euclid metric. Since ρ_{±}(x) are nondecreasing, we have ρ(x) ≥ ρ̄ for some ρ̄ > 0 on I\I_{0}.

For any n ≥ 0, denote B_{n}(x, ρ) = {y ∈ I : d(f^{i}y, f^{i}x) ≤ ρ(f^{i}x) ∀0 ≤ i ≤ n}. We always denote β = γ^{−1}. Choose β_{−} < β < β_{+} such that β_{+} − β and β − β_{−}

are small, for example, less than 0.1 and 0.1(β − 1).

2. Preliminary

Lemma 2.1. Let x ∈ I_{0}. For any θ ≥ 0,

\frac{x^{θ}}{x_{1}^{θ}} · \frac{d(x_{1}, y_{1})}{d(x, y)} = \frac{x^{θ}}{x_{1}^{θ}} · \frac{1}{f′(x)} + o(x^{γ}) = 1 − (1 + γ − θ)x^{γ} + o(x^{γ}), x → 0,

where y ∈ B(x, ρ(x)) and y_{1} ∈ B(x_{1}, ρ(x_{1})).

Proof. This is because by Assumption (IV), x = fx_{1} = x_{1}(1 + x_{1}^{γ} + o(x^{γ}), and d(x, y) = (f′(x_{1}) + o(x^{γ}))d(x_{1}, y_{1}) = (1 + (1 + γ)x^{γ} + o(x^{γ}))d(x_{1}, y_{1}). □

Lemma 2.2. Given β_{−} < β < β_{+}, we can choose P_{0} small enough such that for any x ∈ P_{0},

i) if x = x_{0} ≥ (\frac{β_{−}}{r} )^{β} for some r > 0, then x_{n} ≥ (\frac{β_{−}}{r + n})^{β};

ii) if x = x_{0} ≤ (\frac{β_{+}}{r} )^{β} for some r > 0, then x_{n} ≤ (\frac{β_{+}}{r + n})^{β}.

Proof. If x is small, then we can find 1 < λ < β/β_{−} such that f(x) ≤ x(1 + λx^{γ}).

Suppose x ≤ (\frac{β_{−}}{r} )^{β}. We have

f(x) ≤ (\frac{β_{−}}{r} )^{β}(1 + \frac{λβ_{−}}{r} ) = β_{−}^{β} \frac{r + λβ_{−}}{r^{β+1}} .

Note that λβ_{−} < β. If r is large enough, then

(1 − \frac{1}{r})^{β}(1 + \frac{λβ_{−}}{r} ) ≤ 1  or  (r − 1)^{β}(r + λβ_{−}) ≤ r^{β+1}.

So we get that

f(x) ≤ (\frac{β_{−}}{r − 1})^{β}.

This implies the result in (i).

Part (ii) can be proved similarly. □

Define

Δ(x, y) = \{ \begin{matrix}1 + \frac{J_{0}}{x} d(x, y), & ∀x ∈ P_{0}, y ∈ B(x, ρ(x)); \\ 1 + Jd(x, y), & ∀x ∈ I\P_{0}, y ∈ B(x, ρ(x)), \end{matrix}

where J, J_{0} > 0 are constants satisfying the proposition below.

Proposition 2.3. (Distortion Estimates) There exist constants J, J_{0} > 0 such that for all x ∈ I, y ∈ B(x, ρ(x)),

i) if x_{1} ∈ f^{−1}x, y_{1} ∈ f^{−1}y ∩ B(x_{1}, ρ(x_{1})), then

Δ(x_{1}, y_{1}) · \frac{f′(x_{1})}{f′(y_{1})} ≤ Δ(x, y);

ii) for all n > 0, if x_{n} ∈ f^{−n}x, y_{n} ∈ f^{−n}y ∩ B_{n}(x_{n}, ρ), then

\frac{(f^{n})′(x_{n})}{(f^{n})′(y_{n})} ≤ Δ(x, y).

Proof. i) First we suppose x ∈ P_{0}. By (1.3) and the fact f′(y) > 1, there is a constant c > 0 such that

\frac{f′(x)}{f′(y)} < 1 + (f′(x) − f′(y)) ≤ 1 + cx^{γ} \frac{d(x, y)}{x}.

Note that x^{−1}d(x, y) is of order x^{γ}. So by Lemma 2.1 with θ = 1 we have

Δ(x_{1}, y_{1}) · \frac{f′(x_{1})}{f′(y_{1})} ≤ (1 + J_{0} \frac{d(x_{1}, y_{1})}{x_{1}}) · (1 + cx_{1}^{γ} \frac{d(x_{1}, y_{1})}{x_{1}})

=1 + J_{0}[1 + \frac{cx_{1}^{γ}}{J_{0}} + O(x_{1}^{2γ})]\frac{d(x_{1}, y_{1})}{x_{1}}
=1 + J_{0}(1 + \frac{cx_{1}^{γ}}{J_{0}} + O(x_{1}^{2γ}))(1 − γx^{γ} + o(x^{γ}))\frac{d(x, y)}{x}

If J_{0} is large enough, then the right side is less than 1 + J_{0}x^{−1}d(x, y). For the case x∉ P_{0}, the result is clear since f is uniformly expanding outside P_{0}. ii) can be obtained from i) by induction. □

Remark. The ratio \frac{(f^{n})′(x_{n})}{(f^{n})′(y_{n})} only depends on preimages of x and y. So if f^{n−1}x_{n} ∈

I\I_{0}, then we still have

\frac{(f^{n})′(x_{n})}{(f^{n})′(y_{n})} ≤ 1 + Jd(x, y)

for some J > 0 even if x ∈ P_{0}.

Recall the definition (1.4) of σ_{g}.

Corollary 2.4. Let x, y ∈ P_{0}. If g satisfies g(ȳ_{1}) ≤ g(x̄_{1})Δ(x̄_{1}, ȳ_{1}) for all x̄_{1} ∈ f^{−1}x\I_{0}, ȳ_{1} ∈ f^{−1}y ∩ B(x̄_{1}, ρ̄), then

σ_{g}(y_{1}) ≤ σ_{g}(x_{1})(1 + Jd(x, y)).

Proof. By (1.4) and Proposition 2.3.i),

\frac{σ_{g}(y_{1})}{σ_{g}(x_{1})} = \frac{∑_{ȳ_{1}∈f^{−1}y\I_{0}} g(ȳ_{1})/f′(ȳ_{1})}{∑_{x̄_{1}∈f^{−1}x\I_{0}} g(x̄_{1})/f′(x̄_{1})} ≤ max\{\frac{g(ȳ_{1})}{g(x̄_{1})} · \frac{f′(x̄_{1})}{f′(ȳ_{1})}\}≤ max\{Δ(x̄_{1}, ȳ_{1})\frac{f′(x̄_{1})}{f′(ȳ_{1})}\}

where max is taken over all pairs x̄_{1} ∈ f^{−1}x\I_{0} and ȳ_{1} ∈ f^{−1}y ∩ B(x̄_{1}, ρ̄). Since

x̄_{1}, ȳ_{1} ∉ I_{0}, Δ(x̄_{1}, ȳ_{1})\frac{f′(ȳ_{1})}{f′(x̄_{1})} ≤ (1 + Jd(x̄_{1}, ȳ_{1}))\frac{f′(ȳ_{1})}{f′(x̄_{1})} ≤ 1 + Jd(x, y). □

3. The Density Function

In this section we prove Theorem A, and then prove a result (Lemma 3.5) which implies that decreasing rate of h is arbitrarily large as x goes to 0.

Proof of Theorem A.

Define Perron-Frobenius Operator L = L_{− log f′} from the set of continuous functions on (0, 1] to itself by

Lg(x) = ∑_{x̂_{1}∈f^{−1}x} \frac{g(x̂_{1})}{f′(x̂_{1})}.

Let υ denote the Lebesgue measure on I. Clearly, υ(Lg) = υ(g) for any integrable function g on (0, 1].

Also it is well known that for any fixed point h of L, a measure μ given by μ(g) = υ(g · h) is an invariant measure of f. In fact, we can check directly that L(h · (g ∘ f)) = g · (Lh), then we have μ(g ∘ f) = υ(h · (g ∘ f)) = υ(L(h · (g ∘ f))) =

υ((Lh) · g) = υ(h · g) = μ(g). (See e.g. [B] for more details.)

Let B denote the set of continuous functions g on (0, 1] with the norm

∥g∥ = sup_{x∈(0,1]}{xg(x)}.  (3.1)

It is easy to check that B is a Banach space and L is a Linear operator on B. Lemma 3.1 below implies that the operator L is continuous.

Put

\begin{aligned}G = \{g ∈ B : & g > 0, υ(g) = 1, g(y) ≤ g(x)Δ(x, y) ∀x ∈ I, y ∈ B(x, ρ(x)), \\ & x^{γ}g(x) ≤ H_{0} ∀x ∈ P_{0}\}. \end{aligned}

where H_{0} is a constant to be determined later.

G is not empty since (1 − γ)x^{−γ} ∈ G. It is clear that G is a convex set. By Lemma 3.2 and 3.3, G is compact and LG ⊂ G if H_{0} is large enough. So by Schauder-Tychonoff fixed point theorem (see e.g. [DS]), L has a fixed point h ∈ G, and therefore, i) and ii) follows from the definition of G. Part iii) can be obtained from Lemma 3.4 and the fact ϕ(x) = (1 + γ)x^{γ} + o(x^{γ}). □

Lemma 3.1. L is a bounded linear operator.

Proof. Since L is a positive operator and x^{−1} is the maximal element in the unit ball with respect to the norm in (3.1), we only need to prove that xL(x^{−1}) is bounded.

Note f′(x) ≥ 1. We have

L(\frac{1}{x}) < ∑_{x̂_{1}∈f^{−1}x} \frac{1}{x̂_{1}} ≤ \frac{1}{x_{1}} + ∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{1}{x̄_{1}} ≤ \frac{1}{x} max_{z∈[0,1]}\{f′(z)\} + ∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{1}{x̄_{1}} ,

where the last inequality follows from the fact x = f(x_{1}) ≤ x_{1} max_{z∈[0,1]}{f′(z)}. Since

the second term is bounded, ∥L∥ = sup_{x∈(0,1]}{xL(x^{−1})} is finite. □

Lemma 3.2. The set G is compact.

Proof. First, G is a bounded set. In fact, for any g ∈ G, if x ∉ P_{0}, then

1 ≥ ∫_{B(x,ρ(x))} g(y)dy ≥ g(x)\frac{1}{1 + Jρ(x)} · 2ρ(x).

That is,

xg(x) ≤ \frac{x}{2ρ(x)}(1 + Jρ(x)) ≤ sup_{x∉P_{0}}\{\frac{x(1 + Jρ(x))}{2ρ(x)}\}.

If x ∈ P_{0}, then xg(x) ≤ H_{0}x^{1−γ} ≤ H_{0}.

Using the facts that g(y) ≤ Δ(x, y)g(x) ∀y ∈ B(x, ρ(x)) and xg(x) ≤ H_{0}x^{1−γ} ∀x ∈ P_{0}, we know that G is also an equicontinuous set. □

Lemma 3.3. If H_{0} is large enough, then LG ⊂ G.

Proof. Take g ∈ G. We prove Lg ∈ G.

It is clear that Lg > 0 and υ(Lg) = υ(g) = 1.

If x, y ∈ I with d(x, y) ≤ ρ(x), then Proposition 2.3.i) and the same arguments as in the proof of Corollary 2.4 give

\frac{Lg(y)}{Lg(x)} = \frac{∑_{ŷ_{1}∈f^{−1}y} g(ŷ_{1})/f′(ŷ_{1})}{∑_{x̂_{1}∈f^{−1}x} g(x̂_{1})/f′(x̂_{1})} ≤ max\{Δ(x̂_{1}, ŷ_{1})\frac{f′(ŷ_{1})}{f′(x̂_{1})}\} ≤ Δ(x, y),

where max is taken over all pairs x̂_{1} ∈ f^{−1}x and ŷ_{1} ∈ f^{−1}y ∩ B(x̂_{1}, ρ(x̂_{1})).

Suppose x ∈ P_{0}. Using Lemma 2.1 with θ = γ and using the fact x^{γ}g(x) ≤ H_{0} ∀x ∈ P_{0}, we get

\begin{matrix}x^{γ}Lg(x) = & x_{1}^{γ}g(x_{1})\frac{x^{γ}}{x_{1}^{γ}} \frac{1}{f′(x_{1})} + x^{γ} ∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{g(x̄_{1})}{f′(x̄_{1})} \\ ≤ & H_{0}[1 − x^{γ} + o(x^{γ}) + \frac{x^{γ}}{H_{0}} ∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{g(x̄_{1})}{f′(x̄_{1})} ]. \end{matrix}

Since all element g in G are uniformly bounded on I\I_{0}, the summation in the second term are bounded. So if we take H_{0} large enough, then the right side of the inequality is less than H_{0}. □

Lemma 3.4. There exists R > 0 such that

|h(x)ϕ(x) − (1 + γ)σ_{0}| ≤ R max{x^{γ}, δ_{1}(x)},

where ϕ(x) = f′(x) − 1 = (1 + γ)x^{γ} + x^{γ}δ_{1}(x).

Proof. Denote α(x) = max{x^{γ}, |δ_{1}(x)|} for x ∈ P_{0}. We may assume that α(x) is nondecreasing on P_{0}, otherwise we use max_{0≤y≤x}{α(y)} instead.

First we claim that there exist R > 0 such that if

h(x)ϕ(x) ≥ (1 + γ + c + Rα(x_{1}))σ_{0}

for some c ≥ 0 and x ∈ P_{0}, then

h(x_{1})ϕ(x_{1}) ≥ (1 + γ + c(1 + \frac{1}{2}x_{1}^{γ}) + Rα(x_{2}))σ_{0}.

In fact, since Lh = h, we have that for x ∈ P_{0},

h(x_{1}) = (1 + ϕ(x_{1}))(h(x) − σ_{h}(x_{1})) ≥ (1 + ϕ(x_{1}))(h(x) − σ_{0} − Jσ_{0}x_{1}), (3.2)

where σ_{h}(x_{1}) ≤ σ_{0}(1 + Jx_{1}) follows from Corollary 2.4. Also, it is easy to check by (1.1) and the definition of δ_{1}(x) that

(1 + ϕ(x_{1})) · \frac{ϕ(x_{1})}{ϕ(x)} = 1 + x_{1}^{γ} + x_{1}^{γ}δ_{1}^{∗}(x_{1}).

for some δ_{1}^{∗}(x) which is bounded by δ_{1}(x) multiplied by a constant coefficient. So by (3.2) we get

\frac{ϕ(x_{1})h(x_{1})}{σ_{0}} ≥ \frac{ϕ(x)h(x)}{σ_{0}} · (1 + ϕ(x_{1}))\frac{ϕ(x_{1})}{ϕ(x)} − (1 + Jx_{1})ϕ(x_{1})(1 + ϕ(x))

≥(1 + γ + c + Rα(x_{1}))(1 + x_{1}^{γ} + x_{1}^{γ}δ_{1}^{∗}(x_{1}))
− ((1 + γ)x_{1}^{γ} + x_{1}^{γ}δ_{1}(x_{1}))(1 + ϕ(x_{1})) − Jx_{1}ϕ(x_{1})(1 + ϕ(x_{1}))
=(1 + γ + c + Rα(x_{1})) + (c + Rα(x_{1}))(x_{1}^{γ} + x_{1}^{γ}δ_{1}^{∗}(x_{1})) + (1 + γ)x_{1}^{γ}δ_{1}^{∗}(x_{1})
− (1 + γ)x_{1}^{γ}ϕ(x_{1}) − x_{1}^{γ}δ_{1}(x_{1})(1 + ϕ(x_{1})) − Jx_{1}ϕ(x_{1})(1 + ϕ(x_{1})).

If P_{0} is small enough, then |δ_{1}^{∗}(x_{1})| ≤ \frac{1}{2} and therefore cx_{1}^{γ}(1 + δ_{1}^{∗}(x_{1})) ≥ \frac{1}{2}cx_{1}^{γ}. Note that α(x) is greater than or equal to δ_{1}(x) and x^{γ}. So

\frac{1}{2}Rα(x_{1}) + (1 + γ)δ_{1}^{∗}(x_{1}) − (1 + γ)ϕ(x_{1}) − (δ_{1}(x_{1}) + Jx_{1}^{1−γ}ϕ(x_{1}))(1 + ϕ(x_{1})) > 0.

if R is sufficiently large. Hence we have

\frac{ϕ(x_{1})h(x_{1})}{σ_{0}} ≥ (1 + γ + c + Rα(x_{1})) + \frac{1}{2}cx_{1}^{γ} ≥ (1 + γ + c + Rα(x_{2})) + \frac{1}{2}cx_{1}^{γ}.

It means that the claim is true.

Using this claim we can see that

ϕ(x)h(x) ≤ (1 + γ)σ_{0} + 2Rσ_{0}α(x_{1}) ∀x ∈ P_{0}.

Otherwise we may have

ϕ(x)h(x) ≥ (1 + γ)σ_{0} + 2Rσ_{0}α(x_{1}) = (1 + γ)σ_{0} + cσ_{0} + Rσ_{0}α(x_{1})

for some x ∈ P_{0}, where c = Rα(x_{1}) > 0. Then by using the claim repeatedly, and

using the fact c · (1 + \frac{1}{2} ∑_{i=1}^{n−1} x_{i}^{γ})(1 + \frac{1}{2}x_{n}^{γ}) ≥ c · (1 + \frac{1}{2} ∑_{i=1}^{n} x_{i}^{γ}) we get that

ϕ(x_{n})h(x_{n}) ≥ (1 + γ)σ_{0} + Rσ_{0}α(x_{n+1}) + cσ_{0} · (1 + \frac{1}{2} ∑_{i=1}^{n} x_{i}^{γ}).

By Lemma 2.2, x_{i}^{γ} ≥ \frac{β_{−}}{r + i} ∀i ≥ 0 for some r > 0 and therefore ∑_{i=1}^{∞} x_{i}^{γ} diverges.

This contradicts to the fact that x^{γ}h(x) is bounded for all x ∈ P_{0}.

By using ϕ(x)h(x) > 0, the inequality of the other direction can be proved similarly. □

Lemma 3.5. For any γ′ > 0, we can choose P_{0} small enough such that

h(y) ≥ h(x) · (1 + \frac{J′_{0}}{x^{1−γ′}} d(x, y)) ∀x ∈ P_{0}, x − ρ(x) ≤ y ≤ x

for some J′_{0} > 0.

Proof. Denote τ = inf_{x∈(0,1]}\{\frac{1 − γ}{x^{γ}h(x)}\}. By Lemma 2.1, there exists c > 0 such that

\frac{x^{γ}}{x_{1}^{γ}} · \frac{1}{f′(x)} = 1 − x^{γ} + o(x^{γ}) > 1 − cx^{γ} for all x ∈ P_{0}. We take H′_{0} ≤ min_{x∈P_{0}}\{\frac{τσ_{h}(x)}{c}\}

and then define

\begin{aligned}G_{1} = \{g ∈ G : & g(x) ≥ τh(x) ∀x ∈ (0, 1], x^{γ}g(x) ≥ H′_{0} ∀x ∈ P_{0}, \\ & g(y) ≥ g(x)(1 + \frac{J′_{0}}{x^{1−γ′}} d(x, y)) ∀x ∈ P_{0}, x − ρ(x) ≤ y ≤ x\}. \end{aligned}

G_{1} is not empty because (1 − γ′)x^{−γ′} ∈ G_{1}. Clearly, G_{1} is compact since it is closed in G. We will prove LG_{1} ⊂ G_{1}. Then we can take h as a fixed point of L in G_{1}, and therefore h has the required property.

Let g ∈ G_{1}. First, we have

Lg(x) ≥ Lτh(x) = τLh(x) = τh(x).

Secondly, since σ_{g}(x) ≥ τσ_{h}(x) and H′_{0} ≤ c^{−1}τσ_{h}(x) ∀x ∈ P_{0}, we get

x^{γ}Lg(x) = x_{1}^{γ}g(x_{1})\frac{x^{γ}}{x_{1}^{γ}} \frac{1}{f′(x_{1})} + x^{γ}σ_{g}(x_{1}) ≥ H′_{0}(1 − cx^{γ} + \frac{x^{γ}}{H′_{0}} τσ_{h}(x_{1})) ≥ H′_{0}.

Now it remains to check Lg(y) ≥ Lg(x)(1 + \frac{J′_{0}}{x^{1−γ′}} d(x, y)). That is,

\frac{g(y_{1})}{f′(y_{1})} + σ_{g}(y_{1}) ≥ (\frac{g(x_{1})}{f′(x_{1})} + σ_{g}(x_{1}))(1 + \frac{J′_{0}}{x^{1−γ′}} d(x, y)).  (3.3)

By (1.3), if x ∈ P_{0}, then \frac{f′(x)}{f′(y)} ≥ 1 + f′(x) − f′(y) ≥ 1 + \frac{c′}{x^{1−γ}} d(x, y) for some

c′ > 0. Also, using Lemma 2.1 for θ = 1 − γ′ we have

\begin{matrix}\frac{g(y_{1})}{f′(y_{1})} ≥ & \frac{g(x_{1})}{f′(x_{1})}(1 + \frac{J′_{0}}{x_{1}^{1−γ′}} d(x_{1}, y_{1}))(1 + \frac{c′}{x^{1−γ}} d(x, y)) \\ ≥ & \frac{g(x_{1})}{f′(x_{1})} (1 + J′_{0}(1 − (γ + γ′)x^{γ} + o(x^{γ}))\frac{d(x, y)}{x^{1−γ′}} + c′ \frac{d(x, y)}{x^{1−γ}}) \\ ≥ & \frac{g(x_{1})}{f′(x_{1})}(1 + J′_{0} \frac{d(x, y)}{x^{1−γ′}} + \frac{c′}{2} \frac{d(x, y)}{x^{1−γ}}), \end{matrix}

if P_{0} is small enough. Therefore, using Corollary 3.4 and interchanging the roles of x and y, we can see that (3.3) holds if we show

\frac{g(x_{1})}{f′(x_{1})} \frac{c′}{2} \frac{d(x, y)}{x^{1−γ}} ≥ Jσ_{g}(y_{1})d(x, y) + σ_{g}(x_{1})\frac{J′_{0}}{x^{1−γ′}} d(x, y),

or

x^{γ−γ′} \frac{c′g(x_{1})}{2f′(x_{1})} ≥ Jx^{1−γ′}σ_{g}(y_{1}) + J′_{0}σ_{g}(x_{1}).

However, this is true if P_{0} is small, because x^{γ−γ′}g(x) ≥ x^{γ−γ′}x^{−γ}H′_{0} = x^{−γ′}H′_{0} → ∞, while all other quantities are bounded as x → 0. □

4. The Operator L̃

Take η(x) = \frac{h(x)}{f′(x)h(fx)} if x > 0 and η(0) = 1. By Lemma 4.4 below, η(x) is

continuous on each I_{q}.

Define a new Perron-Frobenius Operator L̃ = L_{log η} from the set of continuous functions on [0, 1] to itself by

L̃g(x) = ∑_{x̂_{1}∈f^{−1}x} η(x̂_{1})g(x̂_{1}),

or equivalently,

L̃g(x) = \frac{1}{h}L(hg) = \frac{1}{h(x)} ∑_{x̂_{1}∈f^{−1}x} \frac{h(x̂_{1})}{f′(x̂_{1})}g(x̂_{1}).

Recall that the measure μ, defined by μ(g) = υ(hg), is an f invariant measure, where υ is the Lebesgue measure on I.

DECAY OF CORRELATIONS FOR MAPS WITH INDIFFERENT FIXED POINTS 11

Lemma 4.1. The operator L̃ has the following properties.

i) L̃c = c for any constant function c.

ii) μ(L̃g) = μ(g) for any integrable function g. Proof. i) Using Lh = h, we get L̃c = \frac{1}{h}L(ch) = \frac{c}{h}Lh = \frac{c}{h}h = c.

ii) Since υ(Lg) = υ(g), μ(L̃g) = υ(h · \frac{1}{h}L(hg)) = υ(L(hg)) = υ(hg) = μ(g). □

Lemma 4.2. lim_{w→0} \frac{μ[0, w]}{w^{1−γ}} = \frac{σ_{0}}{1 − γ}. Consequently, there exists a > \frac{σ_{0}}{1 − γ} > a′ > 0 such that

a′w^{1−γ} ≤ μ[0, w] ≤ aw^{1−γ}.

Proof. Use the fact that μ[0, w] = ∫_{0}^{w} h(x)dx and then use Theorem A.iii). □ Lemma 4.3. Let w ∈ I_{0}. Then

∫_{0}^{w} ∏_{i=1}^{n} η(x_{i})dμ(x) = μ[0, w_{n}].

Proof. Note that f^{n} : [0, w_{n}] → [0, w] is a one to one map. We have L̃^{n}χ_{[0,w_{n}]}(x) = ∏_{i=1}^{n} η(x_{i}) as x ∈ [0, w] and L̃^{n}χ_{[0,w_{n}]}(x) = 0 as x ∈ [w, 1]. So by Lemma 4.1.ii),

∫_{0}^{w} ∏_{i=1}^{n} η(x_{i})dμ(x) = μ(L̃^{n}χ_{[0,w_{n}]}) = μ[0, w_{n}].  □

Take ψ(x) such that η(x) = 1 − ψ(x). Recall the definition (1.4) of σ_{g}. Lemma 4.4. η and ψ have the following properties:

i) ψ(x) = \frac{1}{h(fx)}σ_{h}(x) if x ∈ I_{0};

ii) lim_{x→0} \frac{ψ(x)}{x^{γ}} = 1;

iii) lim_{x→0} η(x) = 1, and therefore η is continuous on each I_{q};

iv) 0 ≤ η(x) ≤ 1, and η(x) = 1 if and only if x = 0;

v) ψ(x) is strictly increasing and η(x) is strictly decreasing on P_{0}; vi) ∀x ∈ P_{0} and x̄ ∈ I\P_{0}, η(x) > η(x̄), if P_{0} is small enough.

Proof. Since Lh = h, h(fx) = \frac{h(x)}{f′(x)} + σ_{h}(x) for x ∈ P_{0}. So η(x) = \frac{h(x)}{f′(x)h(fx)} =

1 − \frac{1}{h(fx)}σ_{h}(x). This implies i).

Part ii) follows from Part i) and Theorem A.iii).

By Part ii), η = 1 − x^{γ} + o(x^{γ}). So η is continuous at 0. By the definition, it is continuous at all other points.

By Lemma 4.1.i), ∑_{x̂_{1}∈f^{−1}x} η(x̂_{1}) = 1, so 0 ≤ η(x̂_{1}) ≤ 1. Then Part iv) is clear.

To get Part v), we use Part i) and then compare Lemma 3.5 and Corollary 2.4, from which we see that h(x) changes in a faster rate than σ_{h}(x).

Part vi) simply follows from Part iv) and v). □

By part i) of the lemma, for x ∈ I_{0} we can write L̃g(x) as

L̃g(x) = η(x_{1})g(x_{1}) + ψ(x_{1})ḡ(x_{1}) = (1 − ψ(x_{1}))g(x_{1}) + ψ(x_{1})ḡ(x_{1}),  (4.1)

where ḡ(x_{1}) is the average of {g(x̄_{1})} with weight {η(x̄_{1})}, x̄_{1} ∈ f^{−1}x\I_{0}, i.e.

ḡ(x_{1}) = \frac{∑_{x̄_{1}∈f^{−1}x\I_{0}} η(x̄_{1})g(x̄_{1})}{∑_{x̄_{1}∈f^{−1}x\I_{0}} η(x̄_{1})} = \frac{∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{h(x̄_{1})}{f′(x̄_{1})}g(x̄_{1})}{∑_{x̄_{1}∈f^{−1}x\I_{0}} \frac{h(x̄_{1})}{f′(x̄_{1})}}.  (4.2)

The second part of the lemma says that if higher order terms are ignored, then ψ(x_{1}) ≈ x_{1}^{γ} and therefore L̃ has the form

L̃g(x) ≈ (1 − x_{1}^{γ})g(x_{1}) + x_{1}^{γ} ḡ(x_{1}).  (4.3)

Lemma 4.5. Let g_{n}(x) = L̃^{n}g(x). Then for any x ∈ P_{0},

g_{n}(x) = g(x_{n}) ∏_{i=1}^{n} η(x_{i}) + g_{n}^{∗}(x),

where

g_{n}^{∗}(x) = ∑_{j=1}^{n} ḡ_{n−j}(x_{j})ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}).

Proof. Use induction. By (4.1) the result is true for n = 1. Suppose it is true for some n. Then

L̃g_{n}(x) = η(x_{1})g(x_{n+1}) ∏_{i=2}^{n+1} η(x_{i}) + η(x_{1})g_{n}^{∗}(x_{1}) + ψ(x_{1})ḡ_{n}(x_{1}).

Since

\begin{matrix}η(x_{1})g_{n}^{∗}(x_{1}) = η(x_{1}) ∑_{j=1}^{n} ḡ_{n−j}(x_{j+1})ψ(x_{j+1}) ∏_{i=2}^{j} η(x_{i}) \\ = & η(x_{1}) ∑_{j=2}^{n+1} ḡ_{n+1−j}(x_{j})ψ(x_{j}) ∏_{i=2}^{j−1} η(x_{i}) = ∑_{j=2}^{n+1} ḡ_{n+1−j}(x_{j})ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}), \end{matrix}

we get

η(x_{1})g_{n}^{∗}(x_{1}) + ψ(x_{1})ḡ_{n}(x_{1}) = ∑_{j=1}^{n+1} ḡ_{n+1−j}(x_{j})ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}),

which is equal to g_{n+1}^{∗}(x). This completes the proof. □

Lemma 4.6. Let x, y ∈ P_{0} with x > y. If ḡ_{i}(x) ≥ 0 ∀0 ≤ i ≤ n − 1, then

\begin{matrix}L̃^{n}g(x) − L̃^{n}g(y) ≥ & g(x_{n}) ∏_{i=1}^{n} η(x_{i}) − g(y_{n}) ∏_{i=1}^{n} η(y_{i}) \\ + & ∑_{j=1}^{n}(ḡ_{n−j}(x_{j}) − ḡ_{n−j}(y_{j}))ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}). \end{matrix}

Proof. By Lemma 4.5, we only need prove

g_{n}^{∗}(x) − g_{n}^{∗}(y) ≥ ∑_{j=1}^{n}(ḡ_{n−j}(x_{j}) − ḡ_{n−j}(y_{j}))ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}).

Note that

\begin{matrix}g_{n}^{∗}(x) − g_{n}^{∗}(y) = & ∑_{j=1}^{n}(ḡ_{n−j}(x_{j}) − ḡ_{n−j}(y_{j}))ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) \\ + & ∑_{j=1}^{n} ḡ_{n−j}(y_{j})(ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) − ψ(y_{j}) ∏_{i=1}^{j−1} η(y_{i})). \end{matrix}

We only need prove that

ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) = \frac{σ_{h}(x_{j})}{h(x_{j−1})} · \frac{h(x_{j−1})}{(f^{j−1})′(x_{j−1})h(x)} = \frac{σ_{h}(x_{j})}{(f^{j−1})′(x_{j−1})h(x)}

is increasing, where the first inequality follows from Lemma 4.4.i) and the definition of η(x). By Proposition 2.3 and Corollary 2.4, both (f^{j−1})′(y_{j−1})/(f^{j−1})′(x_{j−1}) and σ_{h}(x_{j})/σ_{h}(x_{j}) are bounded by 1 + Jd(x, y). Hence by Lemma 3.5, we see that h(x) decreasing faster than σ_{h}(x_{j}) and (f^{j−1})′(x_{j−1}) if P_{0} is small enough. □

Proposition 4.7. Given β_{−} < β < β_{+}, we can choose P_{0} sufficiently small such that for any x ∈ P_{0},

i) if x = x_{0} ≤ (\frac{β}{r} )^{β} for some r > 0, then ∏_{i=1}^{n} η(x_{i}) ≥ (\frac{r}{r + n})^{β_{+}};

ii) if x = x_{0} ≥ (\frac{β}{r} )^{β} for some r > 0, then ∏_{i=1}^{n} η(x_{i}) ≤ (\frac{r}{r + n})^{β_{−}}.

Proof. Take β_{+} > β′_{+} > β′′_{+} > β. Let P_{0} be small enough such that for any

x = (\frac{β′_{+}}{r} )^{β} ∈ P_{0}, 1 − ψ(x) ≥ 1 − \frac{β′_{+}}{β′′_{+}} x^{γ} and 1 − (\frac{β′_{+}}{r} ) ≥ (1 − \frac{1}{r})^{β_{+}}. Hence, using

Lemma 2.2 for β′′_{+}, we have

1 − ψ(x_{i}) ≥ 1 − \frac{β′_{+}}{β′′_{+}} x_{i}^{γ} ≥ 1 − \frac{β′_{+}}{r + i} ≥ (1 − \frac{1}{r + i})^{β_{+}} = (\frac{r + i − 1}{r + i})^{β_{+}}.

Taking product we get the result of Part i).

Part ii) can be proved in a similar way. □

We denote

Δ̃(x, y) = \{ \begin{matrix}1 + \frac{J̃_{0}}{x} d(x, y), & ∀x ∈ P_{0}, y ∈ B(x, ρ(x)); \\ 1 + J̃d(x, y), & ∀x ∈ I\P_{0}, y ∈ B(x, ρ(x)), \end{matrix}

where J̃, J̃_{0} > 0 are constants to be determined by the following lemma.

Lemma 4.8. There exist constants J̃, J̃_{0} > 0 such that ∀x ∈ I, y ∈ B(x, ρ(x)), i) if x_{1} ∈ f^{−1}x, y_{1} ∈ f^{−1}y ∩ B(x_{1}, ρ(x_{1})), then

Δ̃(x_{1}, y_{1}) · \frac{η(y_{1})}{η(x_{1})} ≤ Δ̃(x, y);

ii) if x_{n} ∈ f^{−n}x, y_{n} ∈ f^{−n}y ∩ B_{n}(x_{n}, ρ), then

∏_{i=1}^{n} \frac{η(f^{i}y_{n})}{η(f^{i}x_{n})} ≤ Δ̃(x, y) ∀n > 0;

iii) if a function g satisfies g(ŷ_{1}) ≤ g(x̂_{1})Δ̃(x̂_{1}, ŷ_{1}) ∀x̂_{1} ∈ f^{−1}x, ŷ_{1} ∈ f^{−1}y ∩ B(x̂_{1}, ρ(x̂_{1})), then L̃g(y) ≤ L̃g(x) · Δ̃(x, y).

Proof. Notice that \frac{η(y)}{η(x)} = \frac{h(y)}{h(x)} · \frac{f′(x)}{f′(y)} · \frac{h(fx)}{h(fy)}. So if we take J̃, J̃_{0} > 0 such that

Δ̃(x, y) ≥ Δ(x, y)^{2}Δ(fx, fy),

then the rest is the same as in the proof of Proposition 2.3 and Lemma 3.3. □

Remark. Recall the remark after Proposition 2.3. We also have that if f^{n−1}x_{n} ∈ I\I_{0}, then

∏_{i=1}^{n} \frac{η(f^{i}y_{n})}{η(f^{i}x_{n})} ≤ 1 + J̃d(x, y)

for some J̃ > 0 even if x ∈ P_{0}.

Recall the definition of ḡ(x) in (4.2).

Lemma 4.9. There exists a constant J̄ > 0 such that for all x, y ∈ I_{0}, with d(x, y) ≤ ρ̄, if g(ȳ) ≤ g(x̄)Δ(x̄, ȳ) ∀x̄ ∈ f^{−1}(fx)\I_{0}, ȳ ∈ f^{−1}(fy) ∩ B(x̄, ρ(x̄)), then

ḡ(y) ≤ ḡ(x)(1 + J̄d(x, y)).

Proof. Clearly, η(ȳ)g(ȳ) ≤ η(x̄)g(x̄)Δ̃(x̄, ȳ)^{2}. Hence, by (4.2),

ḡ(y)= \frac{∑_{ȳ∈f^{−1}(fy)\I_{0}} η(ȳ)g(ȳ)}{∑_{ȳ∈f^{−1}(fy)\I_{0}} η(ȳ)} ≤ \frac{∑_{x̄∈f^{−1}(fx)\I_{0}} η(x̄)g(x̄)Δ̃(x, y)^{2}}{∑_{x̄∈f^{−1}(fx)\I_{0}} η(x̄)Δ̃(x̄, ȳ)^{−1}} =ḡ(x) max\{Δ̃(x̄, ȳ)^{3}\},

where max is taken over all pairs x̄ ∈ f^{−1}(fx)\I_{0} and ȳ ∈ f^{−1}(fy) ∩ B(x̄, ρ(x̄)). So the result follows by choosing J̄ > 0 such that 1 + J̄d(x, y) ≥ (1 + J̃d(x̄, ȳ))^{3}. □

5. Convergent Rate

The main result in this section is Proposition 5.2, which shows that the rate of convergence L̃^{n}g → μ(g) is polynomial. This proposition plays a key role for the proof of Theorem B. Since the proof is long, we put some lemmas in next section.

From now on we denote g_{n}(x) = L̃^{n}g(x).

For any b_{+} ∈ (0, 1), define a function Γ(x, y) = Γ_{b_{+}}(x, y) by

Γ(x, y) = \{ \begin{matrix}1 + \frac{K_{0}}{x} d(x, y) & ∀x ∈ P_{0}, y ∈ B(x, ρ(x)); \\ 1 + Kd(x, y), & ∀x ∈ I\P_{0}, y ∈ B(x, ρ(x)), \end{matrix}

where K, K_{0} > 0 are constants chosen as in the following lemma.

Lemma 5.1. There exist constants K, K_{0} > 0 such that for any x ∈ I, y ∈ B(x, ρ(x)),

i) if g(x) ≤ b_{+}, g(x) ≤ g(y)Δ̃(y, x) and g(y) ≤ g(x)Δ̃(x, y), then

1 − g(y) ≤ (1 − g(x))Γ(x, y);

ii) if 1 − g(ȳ_{1}) ≤ (1 − g(x̄_{1}))Γ(x̄_{1}, ȳ_{1}) ∀x̄_{1} ∈ f^{−1}x, ȳ_{1} ∈ f^{−1}y ∩ B(x̄_{1}, ρ(x̄_{1})),

then

1 − L̃g(y) ≤ (1 − L̃g(x))Γ(x, y);

iii) there exist constant K̄ > 0 such that for all x, y ∈ I_{0} with d(x, y) ≤ ρ̄, if 1 − g(ȳ)≤(1 − g(x̄))Γ(x̄, ȳ) ∀x̄∈f^{−1}(fx)\I_{0}, ȳ∈f^{−1}(fy) ∩ B(x̄, ρ(x̄)), then

1 − ḡ(y) ≤ (1 − ḡ(x))(1 + K̄d(x, y)).

Proof. Since g(x) ≤ g(y)Δ̃(y, x) = g(y) + g(y)(Δ̃(y, x) − 1), we have

1 − g(y) ≤ 1 − g(x) + g(y)(Δ̃(y, x) − 1)≤(1 − g(x))(1 + \frac{g(y)}{1 − b_{+}} (Δ̃(y, x) − 1))

≤(1 − g(x))(1 + Δ̃(y, x) − 1)^{max\{1, \frac{g(y)}{1−b_{+}} \}} ≤ (1 − g(x))(Δ̃(y, x))^{max\{1, \frac{g(y)}{1−b_{+}} \}}.

Note that for x ∈ P_{0}, d(x, y) ≤ ρ(x) = O(x^{1+γ}), and g(y) ≤ g(x)Δ̃(x, y) ≤ b_{+}Δ̃(x, y). So g(y) is bounded. Hence, it is clear that K_{0} and K exist. This is Part i). Part ii) and iii) follow from the same arguments as in the proof of Lemma 4.8 and Lemma 4.9. □

Proposition 5.2. For any 0 < b_{−} ≤ b_{+} < 1, we can find arbitrarily small v ∈ P_{0} such that for any continuous functions g_{+} ≥ g_{−} > 0 of the form

g_{±}(x) = \{ \begin{matrix}A_{±} ∏_{i=0}^{k−1} η(x_{i}), & x ∈ [0, v]; \\ b_{±}, & x ∈ [v, 1], \end{matrix}  (5.1)

where A_{+} ≥ A_{−} > 1 and k > 0 are constants that make μ(g_{+}) ≥ 1 and μ(g_{−}) ≤ 1, if a function g satisfies

(a) g(x) ≤ g_{+}(x) ∀x ∈ I, and g(x) ≥ g_{−}(x) ∀x ≤ v,

(b) μ(g) = 1,

(c) g(y) ≤ g(x)Δ̃(x, y) ∀x ∈ I, y ∈ B(x, ρ(x)), and

(d) g is decreasing on [0, v], then for all n ≥ 0,

i) 1 − g_{n}(x) ≥ \frac{D′A_{−}}{(n + k)^{β−1}} ∀x ∈ I\I_{0},

ii) 1 − g_{n}(x) ≤ \frac{DA_{+}}{n^{β−1}} ∀x ∈ I,

iii) \frac{D̄′A_{−}}{(n + k)^{β−1}} ≤ ∫ |g_{n}(x) − 1|dμ(x) ≤ \frac{D̄A_{+}}{n^{β−1}} ,

where D, D′, D̄, D̄′ > 0 are constants only depending on f.

Proof. We divide the proof into three steps.

Step I. We choose v and construct functions g_{±}(x).

Take 0 < b_{−} ≤ b_{+} < 1.

Take u ∈ I_{0} with u ≤ ρ̄, where ρ̄ = inf{ρ(x) : x ∈ I\I_{0}}, such that for all x > u, η(x) ≤ η(u), and for all x ∈ [u, fu], y ∈ B(x, ρ(x)), Γ(x, y) ≥ 1 + 3K̄d(x̄, ȳ) ∀x̄, ȳ ∈ I\I_{0} with d(x̄, ȳ) ≤ ρ(x̄). This is possible because of the definition of Γ(x, y).

Take v = u_{m} ∈ P_{0} for some m > 0, and write v = (\frac{β}{s} )^{β}. We assume first that s ≥ m, otherwise we can choose a smaller u. Then we assume that m is large enough such that

∏_{i=1}^{m} η(x_{i}) ≤ \frac{1}{2}  ∀x ∈ I_{0}\[0, u].  (5.2)

DECAY OF CORRELATIONS FOR MAPS WITH INDIFFERENT FIXED POINTS 17

Since ∏_{i=1}^{n} η(x_{i}) + ∑_{j=1}^{n} ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) = 1, it implies that for any n ≥ m,

∑_{j=1}^{n} ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) ≥ \frac{1}{2}  ∀x ∈ I_{0}\[0, u].  (5.3)

Lastly we assume that s is large enough such that

c′s^{\frac{β−1}{β_{+}−β+1}} ≥ max\{(\frac{2^{β_{+}}}{b_{−}} )^{\frac{1}{β_{−}}} , (\frac{2aβ^{β−1}}{a′b_{−}β_{−}^{β−1}} )^{\frac{1}{β_{−}−β+1}} \},  (5.4)

and

s^{β_{−}−β+1} ≥ \frac{4C_{1}C_{2}b_{+}}{C′_{3}b_{−}} ,  (5.5)

where c′ is given in (5.13), a and a′ are given in Lemma 4.2, C_{2} and C′_{3} are as in Lemma 6.1 and 6.2 respectively, and C_{1} ≥ (1 − b_{+})^{−1} and satisfies that if

1 − g(y) ≤ 2(1 − g(x))(1 + K̄d(x, y)) ∀x ∈ [u, fu], 0 < y ≤ x;

1 − g(y) ≤ (1 − g(x))Γ(x, y) ∀x ≥ u, y ∈ B(x, ρ(x)),

then

max{1 − g(x), x ∈ I} ≤ C_{1} min{1 − g(x), x ≥ u}.  (5.6)

Now we choose A_{+} ≥ A_{−} ≥ 1 and k > 0 such that

A_{±} ∏_{i=0}^{k−1} η(v_{i}) = b_{±},  (5.7)
A_{+}μ[0, v_{k}] + b_{+}μ[v, 1] ≥ 1 and A_{−}μ[0, v_{k}] + b_{−}μ[v, 1] ≤ 1. (5.8)

This is possible. In fact, by Lemma 4.7 and 4.2 we have

(\frac{s}{s + k} )^{β_{+}} ≤ ∏_{i=1}^{k} η(v_{i}) ≤ (\frac{s}{s + k} )^{β_{−}},  (5.9)
a′(\frac{β_{−}}{s + k} )^{β−1} ≤μ[0, v_{k}] ≤ a(\frac{β_{+}}{s + k} )^{β−1}.  (5.10)

These imply lim_{k→0} \frac{∏_{i=0}^{k−1} η(v_{i})}{μ[0, v_{k}]} = 0. So we can take k such that

\frac{1}{b_{+}} − μ[v, 1] ≤ \frac{μ[0, v_{k}]}{∏_{i=0}^{k−1} η(v_{i})} ≤ \frac{1}{b_{−}} − μ[v, 1]  (5.11)

and then take A_{±} such that (5.7) is satisfied.

Now we define g_{±}(x) by using (5.1). Lemma 4.3 and (5.8) give

μ(g_{+}) = A_{+}μ[0, v_{k}] + b_{+}μ[v, 1] ≥ 1, μ(g_{−}) = A_{−}μ[0, v_{k}] + b_{−}μ[v, 1] ≤ 1.

Note that by (5.9)−(5.11) we can obtain

\frac{s^{β_{+}}}{aβ_{+}^{β−1}} (\frac{1}{b_{+}} − μ[v, 1]) ≤ (s + k)^{β_{+}−β+1}, (s + k)^{β_{−}−β+1} ≤ \frac{s^{β_{−}}}{a′β_{−}^{β−1}} (\frac{1}{b_{−}} − μ[v, 1]),

and therefore,

c′s^{\frac{β_{+}}{β_{+}−β+1}} ≤ k + s,  k ≤ cs^{\frac{β_{−}}{β_{−}−β+1}} ,  (5.12)

where c > c′ > 0 are constants satisfying

c′^{β_{+}−β+1} = \frac{1}{aβ_{+}^{β−1}} (\frac{1}{b_{+}} − μ[v, 1]), c^{β_{−}−β+1} = \frac{1}{a′β_{−}^{β−1}} (\frac{1}{b_{−}} − μ[v, 1]). (5.13)

Hence by (5.4) we have

k + s ≥ c′s^{\frac{β−1}{β_{+}−β+1}} · s ≥ (\frac{2^{β_{+}}}{b_{−}} )^{\frac{1}{β_{−}}} · s ≥ 2s, i.e. k ≥ s.  (5.14)

Moreover, by (5.7) and (5.9),

b_{±}(\frac{s + k}{s} )^{β_{−}} ≤ A_{±} ≤ b_{±}(\frac{s + k}{s} )^{β_{+}},  (5.15)

and therefore by (5.14),

A_{−} ≥ b_{−}(\frac{s + k}{s} )^{β_{−}} ≥ 2^{β_{+}} > 1.  (5.16)

Step II. We prove that any function g satisfying condition (a)-(d) has the following property: (A_{n}) 1 − g_{n}(x) > 0 ∀x ≥ u; ( B_{n}) max{1 − g_{n}(x), x ∈ I} ≤ C_{1} min{1 − g_{n}(x), x ≥ u}. First we consider the case 0 ≤ n ≤ m. Since 1 − g(x) ≥ 1 − b_{+} > 0 ∀x ≥ v = u_{m},

by Lemma 4.1.i), 1 − g_{n}(x) ≥ 1 − b_{+} > 0 ∀x ≥ u ≥ f^{n}v. We get (A_{n}). Since C_{1} ≥ (1 − b_{+})^{−1}, ( B_{n}) follows.

Now we consider the cases n > m. We only need prove the following: (A_{n}^{∗}) 1 − g_{n}(x) > 2g(x_{n}) ∏_{i=1}^{n} η(x_{i}) ∀x ∈ [u, fu]; (B′_{n}) 1 − g_{n}(y) ≤ 2(1 − g_{n}(x))(1 + K̄d(x, y)) ∀x ∈ [u, fu], 0 < y ≤ x; (B′′_{n}) 1 − g_{n}(y) ≤ (1 − g_{n}(x))Γ(x, y) ∀x ≥ u, y ∈ B(x, ρ(x)).

In fact, by the definition of L̃, we know that (A_{n−1}) and (A_{n}^{∗}) imply (A_{n}). Also, by (B′_{n}), (B′′_{n}) and (5.6), we can get (B_{n}).

To prove (A_{n}^{∗}), (B′_{n}) and (B′′_{n}), we use induction. Assume (B_{j}) are true for all 0 ≤ j ≤ n − 1. Then by Lemma 6.3 and the choice of C_{1}, (A_{n}^{∗}) is true.

Note that (B′′_{j}) holds for any j = 0, 1, ⋯, m because of Lemma 5.1.i), ii) and the fact that 1 − g(x) ≥ 1 − b_{+} > 0 ∀x ≥ v. So we may assume (A_{j}) and (B′′_{j}) for all j = 0, 1, ⋯, n − 1. Hence, if we assume (A_{n}^{∗}) in addition, then by Lemma 6.4 and Lemma 6.5, (B′_{n}) and (B′′_{n}) hold respectively.

Step III. We prove that g_{n} satisfies i)-iii).

Since μ(g_{n}) = μ(g) = 1,

∫_{\{g_{n}>1\}}(g_{n}(x)−1)dμ(x) = ∫_{\{g_{n}<1\}}(1−g_{n}(x))dμ(x) = \frac{1}{2} ∫ |g_{n}(x)−1|dμ(x). (5.17)

Then the first inequality of Part iii) follows immediately from Lemma 6.2 with D̄′ = 2C′_{3}. By using (A_{n}) ∀n > 0, we get that the upper bound estimate in Part iii) follows from Lemma 6.6 with D̄ = 2C_{4}.

If we use (B_{n}), then

∫_{{g_{n}<1}}(1 − g_{n}(x))dμ(x) ≤ max{1 − g(x) : x ∈ I}
≤C_{1} min\{1 − g(x) : x ≥ u\} ≤ \frac{C_{1}}{μ(I\I_{0})} ∫_{\{g_{n}<1\}}(1 − g_{n}(x))dμ(x). (5.18)

Considering (5.17) and the results in Part iii), we get i) with D′ = (2C_{1})^{−1}D̄′ = C_{1}^{−1}C′_{3}, and get ii) with D = (2μ(I\I_{0}))^{−1}C_{1}D̄ = (μ(I\I_{0}))^{−1}C_{1}C_{4}. □

6. Some Supplementary Lemmas

In this section we prove lemmas which are used for the proof of Proposition 5.2.

Lemma 6.1. There exists C_{2} > 0 such that for any x > u,

∏_{i=1}^{k+n} η(x_{i}) ≤ C_{2} \frac{1}{k^{β_{−}−β+1}} · \frac{1}{(n + k)^{β−1}} ∀n, k > 0.

Proof. By Lemma 4.7, for x^{∗} = (\frac{β_{−}}{r^{∗}} )^{β} ∈ P_{0} fixed, ∏_{i=1}^{k+n} η(x_{i}^{∗}) ≤ (\frac{r^{∗}}{r^{∗} + k + n})^{β_{−}}.

So the result is clear for this x^{∗}. Since by Lemma 4.4.iv) η(x) is smaller outside P_{0} than inside P_{0}, the result holds for all x ∈ I_{0}\P_{0} as well. □

20 HUYI HU

Lemma 6.2. Let C′_{3} = 2^{−β}a′β_{−}^{β−1}, where a′ is given in Lemma 4.2. Then

∫_{\{g_{n}>1\}}(g_{n}(x) − 1)dμ(x) ≥ C′_{3}A_{−} \frac{1}{(n + k)^{β−1}} ∀n > 0.

Proof. Take t > 0 such that

(\frac{t}{t + k} )^{β_{+}} = \frac{1}{b_{−}} (\frac{s}{s + k} )^{β_{−}}.  (6.1)

Clearly, s ≤ t. Also, by (5.16) the right side is no more than \frac{1}{2^{β_{+}}} . So \frac{t}{t + k} ≤ \frac{1}{2}. We get t ≤ k.

Take

z^{(n)} = (\frac{β}{t(1 + \frac{n}{k})})^{β}.  (6.2)

We claim

[0, z^{(n)}] ⊂ {x : g_{n}(x) ≥ 1} ∀n ≥ 0. (6.3)

In fact, for any x ≤ z^{(n)}, by Proposition 4.7 and (5.9),

∏_{i=1}^{k+n} η(x_{i}) ≥ (\frac{t(1 + \frac{n}{k})}{t(1 + \frac{n}{k}) + k + n})^{β_{+}} = (\frac{t}{t + k} )^{β_{+}} = \frac{1}{b_{−}} (\frac{s}{s + k} )^{β_{−}} ≥ \frac{1}{b_{−}} ∏_{i=1}^{k} η(v_{i}).

Then by (5.1), (4.1) and (5.7), g_{n}(x) ≥ A_{−} ∏_{i=1}^{k+n} η(x_{i}) ≥ \frac{A_{−}}{b_{−}} ∏_{i=1}^{k} η(v_{i}) = 1.

Now using Lemma 4.3 we have

∫_{{g_{n}>1}}(g_{n}(x) − 1)dμ(x) ≥A_{−} ∫_{0}^{z^{(n)}} ∏_{i=1}^{k+n} η(x_{i})dμ(x) − μ[0, z^{(n)}]
=A_{−}μ[0, z_{n+k}^{(n)}] − μ[0, z^{(n)}].

Since k > t, by (6.2) and Lemma 2.2, z_{n+k}^{(n)} ≥ (\frac{β_{−}}{t(1 + \frac{n}{k}) + k + n})^{β} ≥ (\frac{β_{−}}{2(k + n)})^{β}.

Then by Lemma 4.2,

A_{−}μ[0, z_{n+k}^{(n)}] ≥ A_{−}a′(z_{n+k}^{(n)})^{β−1} ≥ \frac{A_{−}a′β_{−}^{β−1}}{2^{β−1}} (\frac{1}{k + n})^{β−1}.

So the result follows if we show A_{−}μ[0, z_{n+k}^{(n)}] ≥ 2μ[0, z^{(n)}].

Note that \frac{z_{n+k}^{(n)}}{z^{(n)}} ≥ (\frac{β_{−}t}{β(t + k)})^{β}. Using (5.16) and the fact t > s, we can get

A_{−} \frac{μ[0, z_{n+k}^{(n)}]}{μ[0, z^{(n)}]} ≥ b_{−}(\frac{s + k}{s} )^{β_{−}} · \frac{a′}{a} (\frac{β_{−}t}{β(t + k)})^{β−1} ≥ \frac{a′b_{−}}{a} (\frac{β_{−}}{β} )^{β−1}(\frac{s + k}{s} )^{β_{−}−β+1}.

The right side is greater than or equal to 2 because by (5.12) and (5.4),

(\frac{k + s}{s})^{β_{−}−β+1} ≥ (c′s^{\frac{β−1}{β_{+}−β+1}} )^{β_{−}−β+1} ≥ \frac{2a}{a′b_{−}} (\frac{β}{β_{−}} )^{β−1}.  □

DECAY OF CORRELATIONS FOR MAPS WITH INDIFFERENT FIXED POINTS 21

Lemma 6.3. Let n > m. Suppose for all 0 ≤ j ≤ n − 1,

max{1 − g_{j}(x), x ∈ I} ≤ C_{1} min{1 − g_{j}(x), x ≥ u}.

Then 1 − g_{n}(x) > 2g(x_{n}) ∏_{i=1}^{n} η(x_{i}) ∀x ∈ [u, fu].

Proof. By (5.17), (5.18) and Lemma 6.2, we have that for all 1 ≤ j ≤ n,

\begin{matrix}1 − g_{j}(x) ≥ \frac{1}{C_{1}} ∫_{\{g_{j}>1\}}(g_{j}(x) − 1)dμ(x) \\ ≥ & \frac{C′_{3}A_{−}}{C_{1}} \frac{1}{(k + j)^{β−1}} ≥ \frac{C′_{3}A_{−}}{C_{1}} \frac{1}{(k + n)^{β−1}} ∀x ≥ u. \end{matrix}

So the same inequality is true for 1 − ḡ_{j}(x). By Lemma 4.5 and (5.3),

1 − g_{n}(x) > ∑_{j=1}^{n}(1 − ḡ_{n−j}(x_{j}))ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) ≥ \frac{C′_{3}A_{−}}{2C_{1}} \frac{1}{(k + n)^{β−1}} .

On the other hand, since n > m, g(x_{n}) ≤ A_{+} ∏_{i=1}^{k} η(x_{i}). So by Lemma 6.1,

g(x_{n}) ∏_{i=1}^{n} η(x_{i}) ≤ A_{+} ∏_{i=1}^{k+n} η(v_{i}) ≤ A_{+}C_{2} \frac{1}{k^{β_{−}−β+1}} \frac{1}{(k + n)^{β−1}} .

Now, considering (5.7) and (5.14) we have

\frac{1 − g_{n}(x)}{2g(x_{n}) ∏_{i=1}^{n} η(x_{i})} ≥ \frac{C′_{3}A_{−}}{2^{2}C_{1}C_{2}A_{+}} k^{β_{−}−β+1} ≥ \frac{C′_{3}b_{−}}{4C_{1}C_{2}b_{+}} s^{β_{−}−β+1}.

By (5.5) it is greater than or equal to 1. □ Lemma 6.4. Let n > m. Suppose g(x) is decreasing on [0, v]. Suppose further

(i) 1 − g_{j}(x) > 0 ∀0 ≤ j ≤ n − 1, x ≥ u, (ii) 1 − g_{j}(y) ≤ (1 − g_{j}(x))Γ(x, y), ∀0 ≤ j ≤ n − 1, x ≥ u, y ∈ B(x, ρ(x)), and

(iii) 1 − g_{n}(x) ≥ 2g(x_{n}) ∏_{i=1}^{n} η(x_{i}) ∀x ∈ [u, fu].

Then for all x ∈ [u, fu] with 1 − g_{n}(x) > 0,

1 − g_{n}(y) ≤ 2(1 − g_{n}(x))(1 + K̄d(x, y)) ∀0 < y ≤ x.

Proof. By Supposition (ii) and Lemma 5.1.iii),

(1 − ḡ_{n−j}(x_{j})) − (1 − ḡ_{n−j}(y_{j})) ≥ −K̄d(x, y) · (1 − ḡ_{n−j}(x_{j})).  (6.4)

Using Lemma 4.5 for the function 1−g(x) and then using Supposition (iii), we have

∑_{j=1}^{n}(1 − ḡ_{n−j}(x_{j}))ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) = 1 − g_{n}(x) − (1 − g(x_{n})) ∏_{i=1}^{n} η(x_{i})

<1 − g_{n}(x) + g(x_{n}) ∏_{i=1}^{n} η(x_{i}) ≤ \frac{3}{2}(1 − g_{n}(x)) ≤ 2(1 − g_{n}(x)).  (6.5)

Therefore, by using Lemma 4.6 for the function 1 − g(x), we obtain

(1 − g_{n}(x)) − (1 − g_{n}(y))
≥(1 − g(x_{n})) ∏_{i=1}^{n} η(x_{i}) − (1 − g(y_{n})) ∏_{i=1}^{n} η(y_{i}) − 2K̄(1 − g_{n}(x))d(x, y). (6.6)

If 1 − g(y_{n}) ≤ 0, then either 1 − g(x_{n}) ≥ 0 or 0 ≥ 1 − g(x_{n}) ≥ 1 − g(y_{n}). Since η(x) is decreasing, (6.6) becomes

(1 − g_{n}(x)) − (1 − g_{n}(y)) ≥ −2K̄(1 − g_{n}(x))d(x, y).  (6.7)

Then the result follows.

If 1 − g(y_{n}) ≥ 0, then 0 ≤ 1 − g(y_{n}) ≤ 1 − g(x_{n}) ≤ 1 − g_{n}(x). Since η(x_{i}) > 0 and η(y_{i}) < 1, (6.6) becomes

(1 − g_{n}(x)) − (1 − g_{n}(y)) ≥ −(1 − g_{n}(x)) − 2K̄(1 − g_{n}(x))d(x, y).

This is the result of the lemma. □

Lemma 6.5. Suppose all conditions in Lemma 6.4 are satisfied. Then

1 − g_{n}(y) ≤ (1 − g_{n}(x))Γ(x, y) ∀x ∈ [u, fu], y ∈ B(x, ρ(x)).

Proof. First we assume y ≤ x, The same argument as in the proof of above lemma tells that (6.5) holds. Further, if 1 − g(y_{n}) ≤ 0, then (6.7) follows as well and therefore the result is true. So we consider the case 1 − g(y_{n}) ≥ 0. Note that g(y_{n}) ≥ g(x_{n}) ≥ g_{n}(x). By Lemma 4.8.ii) and (5.2),

(1 − g(x_{n})) ∏_{i=1}^{n} η(x_{i}) − (1 − g(y_{n})) ∏_{i=1}^{n} η(y_{i}) ≥ (1 − g(y_{n}))(∏_{i=1}^{n} η(x_{i}) − η(y_{i}))

≥ −(1 − g(y_{n})) ∏_{i=1}^{n} η(x_{i})(Δ̃(x, y) − 1) ≥ −(1 − g_{n}(x))\frac{Δ̃(x, y) − 1}{2}.

So by (6.6),

(1 − g_{n}(x)) − (1 − g_{n}(y)) ≥ − (1 − g_{n}(x))(\frac{Δ̃(x, y) − 1}{2} + \frac{3K̄d(x, y)}{2})
≥ − (1 − g_{n}(x))(Γ(x, y) − 1),

where the last step follows from the choice of u. This is the result.

Now we assume y ≥ x. We use Lemma 4.6 for the function g(x), while interchange the roles of x and y, and replace ḡ_{n−j}(x_{j}) − ḡ_{n−j}(y_{j}) by (1 − ḡ_{n−j}(y_{j})) − (1 − ḡ_{n−j}(y_{j})), to get

g_{n}(x) − g_{n}(y) ≤ g(x_{n}) ∏_{i=1}^{n} η(x_{i}) − g(y_{n}) ∏_{i=1}^{n} η(y_{i})

+ ∑_{j=1}^{n}[(1 − ḡ_{n−j}(y_{j})) − (1 − ḡ_{n−j}(x_{j}))]ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}).  (6.8)

Since g(y_{n}) ≤ g(x_{n}) and η(y_{n}) ≤ η(x_{n}), by Lemma 4.8.i) and Supposition (iii),

\begin{matrix}g(x_{n}) & ∏_{i=1}^{n} η(x_{i}) − g(y_{n}) ∏_{i=1}^{n} η(y_{i})≤[g(y_{n}) ∏_{i=1}^{n} η(y_{i})](Δ̃(x, y) − 1) \\ ≤ & [g(x_{n}) ∏_{i=1}^{n} η(x_{i})](Δ̃(x, y) − 1) ≤ \frac{1 − g_{n}(x)}{2}(Γ(x, y) − 1). \end{matrix}

Note that the arguments for (6.4) and (6.5) still hold. So (6.8) becomes

\begin{matrix}(1 − g_{n}(y)) − (1 − g_{n}(x)) = g_{n}(x) − g_{n}(y) \\ ≤ & (1 − g_{n}(x))(\frac{Γ(x, y) − 1}{2} + \frac{3K̄d(x, y)}{2}) ≤ (1 − g_{n}(x))(Γ(x, y) − 1). \end{matrix}

This completes the proof. □

Lemma 6.6. Let C_{4} = aβ_{+}^{β−1}. Suppose 1 − g_{j}(x) > 0 ∀0 ≤ j ≤ n, x ≥ u. Then

∫_{\{g_{n}>1\}}(g_{n}(x) − 1)dμ(x) ≤ \frac{C_{4}A_{+}}{n^{β−1}} .

Proof. The supposition implies that ∀x ∈ I_{0},

∑_{j=1}^{n}[(ḡ_{n−j}(x_{j}) − 1)]ψ(x_{j}) ∏_{i=1}^{j−1} η(x_{i}) ≤ 0.

If g_{n}(x) ≥ 1 > b_{+}, then g(x_{n}) ≥ 1 > b_{+} and therefore x_{n} ≤ v. So by Lemma 4.5,

g_{n}(x) − 1 ≤ (g(x_{n}) − 1) ∏_{i=1}^{n} η(x_{i}) < g(x_{n}) ∏_{i=1}^{n} η(x_{i}) ≤ A_{+} ∏_{i=1}^{n+k} η(x_{i}).

Note that {x : g_{n}(x) > 1} ⊂ [0, u]. Also note that k ≥ s ≥ m and therefore u_{n+k} = v_{n+k−m} ≤ v_{n}. We have

∫_{{g_{n}>1}}(g_{n}(x) − 1)dμ(x) ≤ A_{+} ∫_{0}^{u} ∏_{i=1}^{n+k} η(x_{i})dμ(x) = A_{+}μ[0, u_{n+k}]

≤ A_{+}μ[0, v_{n}] ≤ A_{+} · av_{n}^{1−γ} ≤ aA_{+}(\frac{β_{+}}{s + n})^{β−1} ≤ aA_{+}(\frac{β_{+}}{n} )^{β−1}.  □

7. Proofs of Theorem B and its Corollary

Proposition 7.1. There exist B, B̄ > 0 such that for any Lipschitz function g with μ(g) = 1, and for all n > 0,

i) |1 − L̃^{n}g(x)| ≤ \frac{B}{ϵn^{β−1}} ∀x ∈ I\I_{0},

ii) ∫ |L̃^{n}g(x) − 1|dμ(x) ≤ \frac{B̄}{ϵn^{β−1}} ,

where ϵ > 0 only depends on the Lipschitz constant of g.

Proof. Take 0 < b_{−} < b_{+} < 1, and take v ∈ P_{0}, k > 0, and functions g_{±} with μ(g_{+}) > 1 and μ(g_{−}) < 1 as in Proposition 5.2 . Then we choose A and b such that

A ∏_{i=1}^{k} η(v_{i}) = b and such that the function ĝ defined by

ĝ(x) = \{ \begin{matrix}A ∏_{i=0}^{k−1} η(x_{i}), & x ∈ [0, v]; \\ b, & x ∈ [v, 1] \end{matrix}

satisfies μ(ĝ) = 1. Then we write

1 − L̃^{n}g = \frac{1}{2ϵ}[1 − L̃^{n}(ĝ − ϵ[1 − g)])] − \frac{1}{2ϵ}[1 − L̃^{n}(ĝ + ϵ[1 − g])].

Suppose we can find ϵ > 0 such that both functions ĝ(x) + ϵ[1 − g(x)] and ĝ(x) − ϵ[1 − g(x)] satisfy the requirements (a), (c) and (d) in Proposition 5.2. By using the proposition for these functions, we can get

|1 − L̃^{n}g(x)| ≤ \frac{1}{2ϵ} · \frac{DA_{+}}{n^{β−1}} + \frac{1}{2ϵ} · \frac{DA_{+}}{n^{β−1}} = \frac{DA_{+}}{ϵn^{β−1}} ∀x ∈ I\I_{0},

and

∫ |L̃^{n}g(x) − 1|dμ(x) ≤ \frac{1}{2ϵ} · \frac{D̄A_{+}}{n^{β−1}} + \frac{1}{2ϵ} · \frac{D̄A_{+}}{n^{β−1}} = \frac{D̄A_{+}}{ϵn^{β−1}} .

Therefore the result follows with B = DA_{+} and B̄ = D̄A_{+}.

Clearly we can find ϵ > 0 such that (a) and (d) in Proposition 5.2 hold for functions ĝ(x) ± ϵ(1 − g(x)). It remains to show that there exists ϵ > 0 such that

\frac{ĝ(y) ± ϵ(1 − g(y))}{ĝ(x) ± ϵ(1 − g(x))} ≤ Δ̃(x, y) ∀x ∈ I, y ∈ B(x, ρ(x)).

That is, we need

\frac{Δ̃(x, y)ĝ(x) − ĝ(y)}{|Δ̃(x, y)(1 − g(x)) − (1 − g(y))|} ≥ ϵ > 0  (7.1)

for all x ∈ I and y ∈ B(x, ρ(x)).

First, we consider the case x ∈ [0, v].

Recall the definition of Δ̃(x, y) and Lemma 4.8.i), we have

(1 + \frac{J̃_{0}d(x_{k}, y_{k})}{x_{k}}) · \frac{∏_{i=0}^{k−1} η(y_{i})}{∏_{i=0}^{k−1} η(x_{i})} = Δ̃(x_{k}, y_{k}) · \frac{∏_{i=0}^{k−1} η(y_{i})}{∏_{i=0}^{k−1} η(x_{i})} ≤ Δ̃(x, y).

Hence, by the definition of ĝ,

\begin{matrix}Δ̃(x, y)ĝ(x) − ĝ(y) ≥ A(Δ̃(x, y) · \frac{∏_{i=0}^{k−1} η(x_{i})}{∏_{i=0}^{k−1} η(y_{i})} − 1) ∏_{i=0}^{k−1} η(y_{i}) \\ ≥ & A(1 + \frac{J̃_{0}d(x_{k}, y_{k})}{x_{k}} − 1) ∏_{i=0}^{k−1} η(y_{i}) = AJ̃_{0} \frac{d(x_{k}, y_{k})}{x_{k}} ∏_{i=0}^{k−1} η(y_{i}). \end{matrix}

Also, we have

\begin{matrix}|Δ̃(x, y)(1 − g(x)) − (1 − g(y))| \\ ≤ & (Δ̃(x, y) − 1)|1 − g(x)| + |g(y) − g(x)| ≤ (|1 − g(x)| + \frac{xL_{g}}{J̃_{0}} ) · \frac{J̃_{0}d(x, y)}{x}, \end{matrix}

where L_{g} is a Lipschitz constant of g.

Now we get that the left side of (7.1) is greater than or equal to

\frac{A ∏_{i=0}^{k−1} η(y_{i})}{|1 − g(x)| + xL_{g}J̃_{0}^{−1}} · \frac{d(x_{k}, y_{k})}{d(x, y)} · \frac{x}{x_{k}} .

It is bounded from below for all x ∈ [0, v] and y ∈ B(x, ρ(x)) because (f^{k})′(x) → 1 and η(y) → 1 as x → 0.

The case x ∈ [v, 1] can be considered similarly. □

Proof of Theorem B.

First, we note that by the definition of L̃, for any functions F and G defined on I, L̃((F ∘ f) · G) = F · (L̃G). Hence

L̃^{n}((F ∘ f^{n}) · G) = F · (L̃^{n}G).

So, by using Lemma 4.1.ii) we have that

μ((F ∘ f^{n}) · G) − μ(F)μ(G) = μ(L̃^{n}((F ∘ f^{n}) · G)) − μ(F · μ(G))
=μ(F · (L̃^{n}G)) − μ(F · μ(G)) = μ(F · (L̃^{n}G − μ(G))).  (7.2)

To prove Part i), we take Lipschitz functions F and G on [0, 1]. Above formula gives

μ((F ∘ f^{n}) · G) − μ(F)μ(G) ≤ ∥F ∥μ(|L̃^{n}G − μ(G)|).

By Proposition 7.1, there exist B̄ > 0 and ϵ = ϵ(G) > 0 such that

μ(|L̃^{n}G − μ(G)|) = μ(|L̃^{n}(G − μ(G) + 1) − 1|) ≤ \frac{B̄}{ϵn^{β−1}} .

So we can take C = B̄ϵ^{−1}.

Now we prove Part ii). Let G be any Lipschitz function satisfying the requirements (a)-(d) in Proposition 5.2 for some functions g_{−}(x) ≤ g_{+}(x). In particular, μ(G) = 1. Then we know that there exists D′ > 0 such that for all n > 0,

1 − L̃^{n}G(x) ≥ \frac{D′A_{−}}{(n + k)^{β−1}} ∀x ∈ I\I_{0},

where A_{−} and k are described in the same proposition.

Take a Lipschitz function F (x) ≥ 0 such that F (x) = 0 on I_{0} and μ(F) > 0. Then by (7.2) we have

\begin{matrix}|μ((F ∘ f^{n}) · G) − μ(F)μ(G)| = |μ(χ_{I\I_{0}} · F · (L̃^{n}G − 1))| \\ ≥ & μ(F) min_{x∈I\I_{0}}\{1 − L̃^{n}G(x)\} ≥ μ(F)\frac{D′A_{−}}{(n + k)^{β−1}} . \end{matrix}

Now the result follows with C′ = (k + 1)^{−(β−1)}D′A_{−}μ(F). □

Recall that E^{(j)} is the element of ξ_{j} containing 0.

Lemma 7.2. There exist l > 0 such that for all j ≥ l, if a function g satisfies (a) g(x) > 0 as x ∈ E^{(j)} and g(x) = 0 as x ∉ E^{(j)},

(b) ∫_{E^{(j)}} gdμ = 1, and

(c) g(y) ≤ g(x)(1 + J̃d(x, y)) ∀x, y ∈ E^{(j)}, then ∀n > 0,

i) 1 − L̃^{n+j}g(x) ≥ \frac{D′A_{−}}{(n + j)^{β−1}} ∀x ∈ I\I_{0},

ii) 1 − L̃^{n+j}g(x) ≤ \frac{DA_{+}}{n^{β−1}} ∀x ∈ I,

where D, D′ are as in Proposition 5.2, and A_{+} = sup{g(x) : x ∈ E^{(j)}} and A_{−} = inf{g(x) : x ∈ E^{(j)}}.

Proof. Take 0 < b_{−} ≤ b_{+} < 1 such that \frac{b_{+}}{b_{−}} = \frac{A_{+}}{A_{−}}. Let v = (\frac{β}{s} )^{β} be the point

given in Proposition 5.2.

For each j > 0, consider the function g_{j}(x) = L̃^{j}g(x). Since f^{j} : E^{(j)} → I is a one to one map,

g_{j}(x) = g(x_{j}) ∏_{i=1}^{j} η(f^{i}x_{j}) ≤ A_{+} ∏_{i=1}^{j} η(f^{i}x_{j}),

for all x ∈ I, where x_{j} = f^{−j}x ∩ E^{(j)}.

Note that if x ≤ y, then

\frac{g(x)η(x)}{g(y)η(y)} ≥ \frac{1}{1 + J̃d(x, y)} · \frac{1 − ψ(x)}{1 − ψ(y)} ≥ \frac{1 + ψ(y) − ψ(x)}{1 + J̃d(x, y)}.

It is easy to see by Lemma 4.4.i), Lemma 3.5 and 2.4 that the right side is greater than or equal to 1 if x is small. It means that g(x)η(x) and therefore g_{j}(x) is decreasing on [0, v] if E^{(j)} is small enough.

By Lemma 2.2, the length of E^{(j)} is between (\frac{β_{−}}{r + j} )^{β} and (\frac{β_{+}}{r + j} )^{β} for some

r > 0. So if j is large enough, then by (c), g(y) ≤ 2g(x) for any x, y ∈ E^{(j)}. Hence by (b) and Lemma 4.2, we have

\frac{(r + j)^{β−1}}{2aβ_{+}^{β−1}} ≤ \frac{1}{2μE^{(j)}} ≤ A_{−} ≤ A_{+} ≤ \frac{2}{μE^{(j)}} ≤ \frac{2(r + j)^{β−1}}{a′β_{−}^{β−1}} .  (7.3)

On the other hand, by Lemma 4.7, ∏_{i=1}^{j} η(v_{i}) ≤ (\frac{s}{s + j} )^{β}. So if j is large enough,

then g(v) ≤ b_{+} and therefore g(x) ≤ b_{+} ∀x > v.

Now we see that g_{j} satisfies all conditions in Proposition 5.2, with j = k. Therefore the results of the lemma follow. □

Lemma 7.3. There exist C > 0 and l > 0 such that for any m ≥ 0, if E ∈ ξ_{m}, then for all n > 0,

|μE − L̃^{n+m+l}χ_{E}(x)| ≤ \frac{Cm^{β−1}}{n^{β−1}} μE ∀x ∈ I\I_{0}.

Proof. Note that f^{m} : E → I is a one to one map and f^{m−1}E = I_{q} for some q. First we consider the case f^{m−1}E = I_{q} ≠ I_{0}. Put

g(x) = \frac{1}{μE} L̃^{m}χ_{E}(x) = \frac{1}{μE} ∏_{i=1}^{m} η(f^{i}x_{m}),

where x_{m} = f^{−m}x ∩ E. By the remark after Lemma 4.8, we know g(y) ≤ g(x)(1 + J̃d(x, y)) for any x ∈ I, y ∈ B(x, ρ̄). Since μ(g) = 1, by similar arguments as in the proof of Lemma 3.2 we know that g is bounded and the bounds is independent of m and E provided f^{m−1}E ≠ I_{0}. Consequently, g is a Lipschitz function and the Lipschitz constant is independent of m and E. So by Proposition 7.1, we have

|μE − L̃^{n+m}χ_{E}(x)| ≤ \frac{C}{n^{β−1}} μE ≤ \frac{Cm^{β−1}}{n^{β−1}} μE ∀x ∈ I\I_{0}

for all n > 0, where C ≥ Bϵ^{−1}.

Secondly, we consider the case that there exists l ≤ j ≤ m such that f^{m−j}E = E^{(j)} ⊂ I_{0}, where l is as in Lemma 7.2. We may assume that j is the largest number with this property. Take

g(x) = \frac{1}{μE} L̃^{m−j}χ_{E}(x) = \frac{1}{μE} ∏_{i=1}^{m−j} η(f^{i}x_{m−j}),

where x_{m−j} = f^{−m+j}x ∩ E. Clearly g(x) satisfies all requirements in Lemma 7.2. So we get that for all n > 0

1 − \frac{1}{μE} L̃^{n+m}χ_{E}(x) = 1 − L̃^{n+j}g(x) ≤ \frac{DA_{+}}{n^{β−1}} ∀x ∈ I\I_{0}.

Recall (7.3), and note that r only depends on f. We may assume l > r. Since

j ≤ m, we have A_{+} ≤ \frac{2^{β}m^{β−1}}{a′β_{−}^{β−1}} . So the result follows with C ≥ \frac{2^{β}D}{a′β_{−}^{β−1}} .

Lastly, we consider the case that f^{m−j}E = E^{(j)} ⊂ I_{0} hold only for j < l. We take a partition E = E_{l−j} ⋃(∪_{i=1}^{l−j−1} ∪_{q=1}^{Q} E_{i,q}) such that f^{m+i−1}E_{i,q} = I_{q} and f^{m+l−j−1}E_{l−j} = I_{0}. For each E_{i,q}, we use the argument similar to the first case for the function g(x) = (μE_{i,q})^{−1}L̃^{m+i}χ_{E_{i,q}}(x) to get that for all n > 0,

|μE_{i,q} − L̃^{n+m+i}χ_{E_{i,q}}(x)| ≤ \frac{Cm^{β−1}}{n^{β−1}} μE_{i,q}  ∀x ∈ I\I_{0}.  (7.4)

Also, we have f^{m−j}E_{l−j} = E^{(l)}. So by taking g(x) = (μE_{l−j})^{−1}L̃^{m−j}χ_{E_{l−j}}(x), the same reasons as in the second case imply that for all n > 0,

μE_{l−j} − L̃^{n+m+l−j}χ_{E_{l−j}}(x) ≤ \frac{Cm^{β−1}}{n^{β−1}} μE_{l−j}  ∀x ∈ I\I_{0}.  (7.5)

Since i ≤ l and l−j ≤ l, (7.4) and (7.5) still hold if we use L̃^{n+m+l} instead of L̃^{n+m+i} and L̃^{n+m+l−j} respectively. Hence the result follows if we take summation. □

Proof of Corollary of Theorem B.

Use (7.2) and take F = χ_{E′} and G = χ_{E}, we get

μ(f^{−n−m}E′ ∩ E) − μE′ · μE = μ(χ_{E′} · (L̃^{n+m}χ_{E} − μE)).

Since E′ ⊂ I\I_{0},

\begin{matrix}μE′ · min_{x∈I\I_{0}}(μE − L̃^{n+m}χ_{E}(x)) \\ ≤ & μ(χ_{E′} · (μE − L̃^{n+m}χ_{E})) ≤ μE′ · max_{x∈I\I_{0}}(μE − L̃^{n+m}χ_{E}(x)). \end{matrix}

Therefore the first inequality follows from Lemma 7.3 with n + l replaced by n. For the second one, we take g(x) = (μE)^{−1}χ_{E}(x) and then apply Lemma 7.2.i) with j = m to get

μE − L̃^{n+m}χ_{E}(x) ≥ \frac{D′A_{−}}{(n + m)^{β−1}} μE ∀x ∈ I\I_{0}

for all n > 0. By (7.3), A_{−} ≥ (2a)^{−1}β_{+}^{1−β}m^{β−1}. So we get the inequality by taking C′ ≤ (2a)^{−1}β_{+}^{1−β} · D′. □

Acknowledgment. It is my pleasure to thank Professor Lai-Sang Young for introducing me this problem. I also wish to thank Professor Sheldon Newhouse for his valuable suggestion and helpful conversation.

References

[B] R. Bowen, Equilibrium states and the ergodic theory of Anosov diffeomorphisms, Lecture Notes in Math., vol. 470, Springer, New York, 1975.

[CF1] P. Collet and P. Ferrero, Absolutely continuous invariant measure for expanding pick maps of the interval expect at a marginal fixed point, Instabilities and nonequilibrium structures, E. Tirapehui and D. Villaroel Eds., Kluwer Academic Publishers, New York, 1989.

[CF2] P. Collet and P. Ferrero, Some limit ratio theorem related to a real endomorphism in case of neutral fixed point, Ann. Inst. Henri Poicaré 52 (1990), 283-301.

[DS] N. Dunford and J. Schwartz, Linear Operator Part I, Interscience Publishers, New York, 1958.

[M] M. Mori, On the intermittency of a piecewise linear map, Tokyo J. Math. 16 (1993), 411-428.

[P] G. Pianigiani, First return map and invariant measures, Israel J. Math. 35 (1980), 32-48.

[Y] L.-S. Young, Recurrence time and rate of mixing, preprint.

Department of Mathematics, Michigan State University, East Lansing, MI 48824, USA

Current address: Department of Mathematics, University of Alabama at Birmingham, Birmingham, AL 35294, USA