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On non-uniformly hyperbolicity assumptions in one-dimensional dynamics

We give an essentially equivalent formulation of the backward contracting property, defined by Juan Rivera-Letelier, in terms of expansion along the orbits of critical values, for complex polynomials of degree at least two which are at most finitely renormalizable and have only hyperbolic periodic points, as well as for all smooth interval maps with non-flat critical points.


Introduction
In the context of one-dimensional dynamics, widely adopted non-uniform hyperbolicity conditions involve expansion along the orbits of critical values, such as the Collet-Eckmann condition, the summability conditions and the large derivatives condition, see [3,11,4,2,1] among others. Recently, Rivera-Letelier [13] introduced a new notion called backward contraction which serves as a different type of non-uniform hyperbolicity condition. This condition is more convenient to use as it follows immediately that the first return maps to suitably chosen small neighborhoods of critical points have good combinatorial and geometric properties. For instance, this notion plays an important role in the work [1].
It has been realized that sufficient expansion along the orbits of critical points often implies backward contraction, see [13, Theorem A] and [1,Theorem 1]. In this paper, we study further the relation between these two types of non-uniformly hyperbolicity conditions. For notational definiteness, we shall mainly work on complex maps and leave the argument for interval maps in Appendix B.
Given a complex polynomial f , let Crit(f ) denote the set of critical points of f in C, let J(f ) denote the Julia set of f , and let For every z ∈ Crit(f ) and δ > 0 we denote by B(f (z), δ) the Euclidean ball of radius δ and centered at f (z). Moreover, we denote by B(z, δ) the connected component of f −1 (B(f (z), δ)) that contains z. δ 0 > 0 such that for every c ∈ Crit ′ (f ), every 0 < δ ≤ δ 0 , every integer n ≥ 1 and every component W of f −n ( B(c, rδ)), we have that dist(W, CV (f )) ≤ δ ⇒ diam(W ) < δ where CV (f ) = f (Crit(f )). If f ∈ BC(r) for all r > 1, we will say that f ∈ BC(∞).
In [13,Theorem A], the author showed that for a complex polynomial (or more generally a rational map), if ∞ n=0 1 |Df n (f (c))| < ∞ holds for all c ∈ Crit ′ (f ), then f satisfies BC(∞).
Definition 2. We say that a polynomial f satisfies the large derivative condition with constant K (f ∈ LD(K) in short) if there exists a neighborhood V of Crit ′ (f ) such that for each c ∈ Crit ′ (f ) and n ≥ 1 with f n (c) ∈ V , we have |Df n (f (c))| ≥ K. If f ∈ LD(K) for all K > 0, we will say that f ∈ LD(∞).
Obviously, given a polynomial f , if for every c ∈ Crit ′ (f ), we have This definition was given first in [1] for C 3 interval maps with non-flat critical points and with all periodic points hyperbolic repelling, where it was proved that for such maps, f ∈ LD(∞) implies f ∈ BC(∞), where the properties LD(K) and BC(r) are defined as above except that we use the standard metric on the interval and use Crit ′ (f ) = Crit(f ). The proof uses a special tool in real one-dimensional dynamics, namely the one-sided Koebe principle, which has no complex analogy.
We shall prove the following in §2: Theorem A (Large derivative implies backward contraction). For each integer d ≥ 2, there exists K 0 = K 0 (d) > 0 such that if f is a polynomial of degree d which is at most finitely renormalizable and has only hyperbolic periodic points and if f satisfies LD(K 0 r) for some r > 1, then f satisfies BC(r).
Recall that a renormalization of f is a map f s : U → V , where s is a positive integer and V ⋑ U are Jordan disks, such that the following hold: • f s : U → V is proper; • U contains a critical point in J(f ); • the following set (called the filled Julia set of f s : U → V ) is connected: {z ∈ U : f sn (z) ∈ U for all n = 1, 2, . . .}; • for each c ∈ Crit(f ), there exists at most one j ∈ {0, 1, . . . , s − 1} with c ∈ f j (U ); We say that f is infinitely renormalizable if there exists a sequence of renor- As a consequence of the Schwarz lemma, we shall also prove the following converse statement in §3: Theorem B (Backward contraction implies large derivative). Let f be a polynomial of degree at least 2. There exists a constant r 0 depending on the maximal critical order of f such that if f satisfies BC(Kr 0 ) for some K > 1, then f satisfies LD(K).
Combining these two theorems, we obtain that Corollary 1. Let f be a polynomial which is at most finitely renormalizable and has only hyperbolic periodic points. Then f satisfies BC(∞) if and only if f satisfies LD(∞).
Our proof of the Theorem A is based on the following complex bounds established in [6], which depends heavily on the recent analytic result [5] and the enhanced nest construction [7]. See also [12] for the case of Cantor Julia sets. Since the precise form is not stated explicitly in [6], we include a proof of the proposition in the Appendix A for completeness.
Proposition 2 (Complex bounds). Assume that f is a complex polynomial of degree d ≥ 2 which is at most finitely renormalizable and has only hyperbolic periodic points. Then there exists ρ 0 = ρ 0 (d) > 0 such that for each c ∈ Crit ′ (f ), there exists an arbitrarily small ρ 0 -nice topological disk which contains c.
Recall that an open set V ⊂ C is called nice if f n (∂V ) ∩ V = ∅ for all n ≥ 0. We say that V is ρ-nice if for each return domain U of V , there is an annulus A ⊂ V \ U such that U is contained in the bounded component of C \ A and such that mod(A) ≥ ρ.
In appendix B, we shall prove similar results for interval maps.

Terminology and notation:
A topological disk means a simply connected domain in C. An annulus A is a doubly connected domain in C, and the modulus mod(A) of A is defined to be log R/r, where A is conformal isomrophic to the round annulus {r < |z| < R}. Given an open set V and a set E with E ⊂ V , let where the supremum is taken over all annuli A with the property that A ⊂ V \ E and E is contained in the bounded component of C \ A.
Given a nice set V, let The first entry map Acknowledgement. We would like to thank Juan Rivera-Letelier for reading carefully a first version of the manuscript and several valuable discussions.

Large derivative implies backward contraction
The goal of this section is to prove Theorem A. So let us fix an integer d ≥ 2 and let f denote a polynomial of degree d. We assume throughout this section that f is at most finitely renormalizable and has only hyperbolic periodic points. For each critical point c, let ℓ c denote the order of c, and let ℓ max (f ) = max{ℓ c : c ∈ Crit ′ (f )}.
In the following we shall not state explicitly dependence of constants on the degree d. So a constant depending only on d will be called universal.
2.1. Preparation. We shall use the following variation of the Koebe principle.
Lemma 3. For any ρ > 0 and N ≥ 1, there exists A 0 = A 0 (ρ, N ) > 0 such that the following holds. Let V ⊃ D, U ⊃ E be bounded topological disks and let s be a positive integer with the following properties: provided that diam(D) is small enough.
Proof. Certainly we only need to prove the lemma in the case ρ < 1. By considering a suitable restriction of the map f s : U → V we may assume that mod(V ; D) = ρ.
Let ϕ : f (U ) → D, ψ : V → D be Riemann mappings with ϕ(f (x)) = 0 and ψ(f s (x)) = 0, where D denotes the unit disk in C. Then F := ψ • f s−1 • ϕ −1 is a holomorphic map from D into itself and F (0) = 0. Thus by the Schwarz lemma, |F ′ (0)| ≤ 1. Since F is proper and deg(F ) ≤ N , we have By the Koebe distortion theorem, it follows that Again by the Koebe distortion theorem, we obtain that where C 2 = C 2 (ρ). Finally, provided that diam(D) is small enough, we have where C 3 > 0 is a constant depending only on the degree of f . Combining all these estimates, we obtain
We shall use the following procedure to construct nice sets with bounded shape. Given a nice topological disk V , a point z 0 ∈ V and a constant λ > 0, let B V (z 0 , λ) be the hyperbolic ball centered at z 0 and of radius λ (in the hyperbolic Riemann surface V ), let V * [z 0 , λ] be the union of B V (z 0 , λ) and all the return domains of V that intersect this set, and let V [z 0 , λ] be the filling of V * [z 0 , λ], i.e. the union of V * [z 0 , λ] and the bounded components of C \ V * [z 0 , λ]. Clearly, for each n ≥ 1, Since V is ρ-nice, the hyperbolic diameters of return domains of V in V are uniformly bounded from above by a constant depending only on ρ. It follows that the hyperbolic diameter of V * [z 0 , λ], hence that of V [z 0 , λ] in V is bounded from above by a constant depending only on ρ. Consequently, mod(V ; U ) is bounded away from zero. By the Riemann mapping theorem and the Koebe distortion theorem, we obtain that Shape(U, z 0 ) is bounded from above.
Let us prove that U is ρ ′ -nice. Indeed, each return domain W of U is contained in a return domain W ′ of V . The first return map R V to V maps W into either U or a return domain of V . In both cases, we have that mod(V ; R V (W )) is bounded away from zero. Since R V |W ′ has bounded degree and since W ′ ⊂ U , we obtain that mod(U ; W ) is bounded away from zero.
We say that (V ′ , V) is an admissible pair of neighborhoods of a set A ⊂ Crit ′ (f ) if the following hold: We shall need the following lemma which is essentially [7, Lemma 6.3].
Lemma 5. Let c 0 ∈ Crit ′ (f ) and let (V ′ , V) be an admissible pair of neighborhoods of Back(c 0 ). Assume that max c∈Back(c 0 ) V ′ c is sufficiently small. Let U be an entry domain of V with entry time s, let c 1 ∈ Back(c 0 ) be such that

2.2.
Nested nice sets. In this section, we shall prove the following proposition, which is a crucial step of the proof of Theorem A.
Proposition 6 (Nested nice sets). There exist universal constants K * > 0, κ * > 0 and C * > 0 such that if f satisfies LD(K) with K ≥ K * , then for each c 0 ∈ Crit ′ (f ), there exists an infinite sequence of nice topological disks that contain c 0 such that the following hold: To prove this proposition, we need a few lemmas.
Lemma 7. Given any ρ > 0 and M > 0, there exist Proof. Let us first construct an admissible pair ( V, V) of neighborhoods of Back(c 0 ) as follows. Let D be the collection of return domains of V outside V and V itself. Let c 1 , c 2 , . . . , c m be the critical points in Back(c 0 ) \ {c 0 }.
For each i = 1, 2, . . . , m, the degree of the first entry map Provided that diam( V ) is small enough, by the assumption that f satisfies LD(K), we have Thus, by Lemma 3, we have For each return domain U of V with return time s and with f s (U ) = V i , let U and U be the components of f −s ( V i ) and f −s ( V i ) that contain U respectively. By Lemma 5, the degree of f s : depends only on ρ and M . This proves the claim.
Since each return domain of V is contained in a return domain of V, the statement (ii) follows from the claim. Moreover, the claim implies that diam(f L c 0 (V )) ≤ CK −1 diam(f (V )), which together with (1) implies (i).
, the lemma follows.
Proof. Let U be a return domain of V with return time s. We need to find a universal bound for mod has uniformly bounded degree, hence mod(V ; U ) is bounded from below by a positive constant. So assume that (2) fails for some c ∈ Crit ′ (f ). Then clearly c ∈ Back(c 0 ) \ {c 0 } and there exists 1 ≤ s 1 < s such that U ′ s 1 ⊂ L 4 c (V ′ ). Therefore, there exists a minimal integer ) which contains f j (U ). By the minimality of s ′ , we have that for each c ∈ Crit ′ (f ),   1]. By Lemma 4, V 1 is (ρ 1 , M 1 )-bounded for some universal constants ρ 1 and M 1 . The estimate on the diameter of diam(f V 1 ) follows from the construction.
Lemma 11. There exist universal constants K * > 0 andκ * > 0 such that if f satisfies LD( K * ), then for each c 0 ∈ Crit ′ (f ), there exists an infinite sequence of (ρ 1 , M 1 )-bounded nice topological disks that contain c 0 such that for each k ≥ 1, Proof. It suffices to prove existence of an arbitrarily small (ρ 1 , M 1 )-bounded nice topological disk V 0 ∋ c 0 under the assumption that f satisfies LD(K) with a large K, since then we may apply Lemma 10 successively to obtain the desired sequence. By Proposition 2, there exists an arbitrarily small ρ 0 -nice topological disk V ∋ c 0 . Let V = V [c 0 , 1]. By Lemma 4, V is uniformly bounded, so by Lemma 10, we obtain existence of V 0 .
Proof of Proposition 6. Let K * = max( K * , K 1 (ρ 1 )). Assume that f satisfies LD(K) with K ≥ K * . Let V k be as in Lemma 11 and let V k = V k [c 0 , 1]. Then the first, second and third statements hold with suitable choices of C * and κ * . By Lemma 4, Shape(V k , c 0 ), hence Shape(f (V k ), f (c 0 )), are uniformly bounded. By Lemma 7, the last statement holds.

Proof of Theorem A.
In [13,Section 6], the author introduced another notion called univalent pull back condition, which is closely related to backward contraction. (A similar notion, BC * (r), was used in [1].) Given δ ′ > δ > 0, we say that f satisfies the (δ, δ ′ )-univalent pull back condition if for every z ∈ C and every integer n ≥ 1 such that • for each j = 1, 2, . . . , n − 1, f j (z) ∈ c∈Crit ′ (f ) B(c, δ) • for some c ∈ Crit ′ (f ), we have f n (z) ∈ B(c, δ ′ ), then f n maps a neighborhood of z conformally onto B(c, δ ′ ). Given r > 1, we say that f satisfies the univalent pull back condition with constant r if for all δ > 0 sufficiently small, f satisfies the (δ, rδ)-univalent pull back condition.
Lemma 12. There exists a constant r 0 > 1 such that if f satisfies the univalent pull back condition with constant rr 0 , where r > 1 is a constant, then f satisfies the backward contracting condition with constant r.
Fix r > 1 and assume that f satisfies LD(K) with K = rK 0 . Let us prove that f satisfies the univalent pull back condition with constant rr 0 which implies that f satisfies the backward contraction condition with constant r by Lemma 12. To this end, it suffices to prove that for each c 0 ∈ Crit ′ (f ) and each δ > 0 small enough, if U is a pull back of B(c 0 , δ) that intersects Crit ′ (f ), then diam(f (U )) ≤ (rr 0 ) −1 δ. Let V k be as given in proposition 6. For each δ ∈ (0, (2C * ) −1 diam(V 1 )), there exists a maximal integer k ≥ 1 such that B(c 0 , δ) ⊂ V k . By part 2 and 3 of the proposition 6, we have . Thus by part 4 of the proposition 6, we obtain The proof is completed.

Backward contraction implies large derivatives
In this section, we shall prove Theorem B. Let f be a polynomial which satisfies the backward contraction condition with constant r > 4. So there exists δ 0 > 0 such that for each δ ∈ (0, δ 0 ] and each c, c ′ ∈ Crit ′ (f ), if U ∋ c is a component of f −n ( B(c ′ , rδ)) then diam(f (U )) < δ. We continue to use ℓ max to denote the maximal order of critical points in the Julia set.
Lemma 13. For each n ≥ 1 and c ∈ Crit ′ (f ), if s is a positive integer with f s (c) ∈ V n \ V n+1 and with f j (c) ∈ V n+1 for all 1 ≤ j < s, then where κ > 0 is a constant depending only on ℓ max (f ).

Lemma 14.
There exists r * > 4 depending only on ℓ max (f ) such that if f satisfies BC(r * ) then the following holds. Let x ∈ V n+1 and let s be a positive integer with f s (x) ∈ V n \ V n+1 and with f j (x) ∈ V n+1 for all j = 1, 2, . . . , s − 1. Then |Df s (f (x))| ≥ 1.
Proof. Assume that f satisfies BC(r * ) with r * sufficiently large so that Let c, c ′ ∈ Crit ′ (f ) be such that x ∈ V c n+1 and f s (x) ∈ V c ′ n . For each j = 0, 1, . . . , s, let U j be the connected component of f −(s−j) ( B(c ′ , r * δ n+1 )) which contains f j (x). Since f j (x) ∈ B(Crit ′ (f ), δ n+1 ) for all 1 ≤ j < s, f s−1 : U 1 → U s is conformal. Moreover, diam(U 1 ) ≤ δ n+1 . Provided that r * is large enough, U s contains a ball centered at f s (x) and of radius at least diam(U s )/3. By the Schwarz lemma, we have Proof of Theorem B. Let r 0 = max(r * , κ −1 ), where κ is as in Lemma 13 and r * is as in Lemma 14.
Assume that f satisfies BC(Kr 0 ) with K ≥ 1. We shall prove that for any S ≥ 1 and c ∈ Crit Given S and c as above, let us define inductively non-negative integers n 1 > n 2 > · · · > n m and S 1 < S 2 < · · · < S m = S as follows. First, (Observe that the backward contracting assumption on f implies that f has no critical relation, so n 1 , and hence S 1 is well-defined.) If S 1 = S then we stop. Otherwise, let Repeating the argument, we must stop within finitely many steps.
By Lemma 13, |Df S 1 (f (c))| ≥ κKr 0 ≥ K. By Lemma 14, for each i = 1, 2, . . . , m − 1 we have Thus The goal of this section is to prove Proposition 2. Besides the results of [6], we shall also use some arguments in [7,Section 6]. Throughout this section, assume that f is a polynomial, at most finitely renormalizable and having only hyperbolic periodic points.
A.1. The puzzle construction. We shall now describe a puzzle partition which will provide nice topological disks as required. The construction given below is a modification of that in [6, Section 2] and makes use of equipotential curves for the Green function and all bounded periodic Fatou components, external rays and internal rays. The modification is necessary since we use the usual definition of renormalization rather than the one used in [6], and since we want to have an arbitrarily small puzzle piece for each c ∈ Crit ′ (f ).
Let G denote the Green function of f . A smooth external ray is a smooth gradient line of G which starts from infinity and tends to the Julia set of f . A smooth external ray has a well-defined angle t ∈ R/Z at which the ray goes to ∞. An external ray is either a smooth external ray, or else a limit of such rays. So for each t ∈ R/Z there is exactly one smooth external rays R t , or two (non-smooth) external rays R t,+ and R t,− . Let P bad be the set of periodic points which are contained in the forward orbit of either a critical point or the landing point of a non-smooth external ray. Note that P bad is a finite set. Let P sep be the set of periodic points which are the common landing point of at least 2 external rays. (Here "sep" stands for "separable".) Observation 1. Let K ′ be a periodic component of the filled Julia set of f that contains a critical point but no attracting periodic points. Then K ′ ∩P sep is an infinite set.
In fact by [8], we only need to consider the case that f has a connected Julia set. So ∂K ′ = J(f ) and f has only repelling periodic points. Let N n (resp. N ′ n ) be the number of periodic points (resp. external rays) of f which has period n. Then N 1 = N ′ 1 + 1 and for all n > 1, N n = N ′ n . Let us construct a sequence of integers s 1 = 1 < s 2 < ..., such that for each j ≥ 1, f has a periodic point p j of period s j which is not the landing point of an external ray of the same period. Since N 1 > N ′ 1 , there exists a fixed point of f which is not the landing point of an external ray fixed by f . Suppose now that p j and s j have been defined. Let γ j be an external ray landing at p j and let s j+1 be the period of γ j . Clearly, s j+1 > s j ≥ 1. Since N s j+1 = N ′ s j+1 , f has a periodic point p j+1 of period s j+1 which is not the landing point of an external ray of period s j+1 . This proves the existence of {s j } ∞ j=1 and {p j } ∞ j=1 . Since p j ∈ P sep for all j ≥ 1, the statement follows. Let K be the collection of all periodic components of the filled Julia set of f that contain a critical point. For each K ′ ∈ K, choose a K ′ ∈ (K ′ ∩P sep )\P bad and let Ξ K ′ be the union of the orbit of a K ′ and all external rays landing on this orbit. Moreover, let Now we shall define for each bounded Fatou component B, an equipotential curve Γ B and internal rays γ θ B , θ ∈ R/Z. We start by choosing bounded periodic Fatou components B 1 , B 2 , . . . , B m , such that the orbits of B i 's are pairwise disjoint and such that the grand orbit of m i=1 B i covers the interior of the filled Julia set. By assumption, f has an attracting periodic point p i ∈ B i with period s i ≥ 1. We choose Jordan curves Γ B i ⊂ B i such that B i \ Ω i is disjoint from the orbit of all critical points and such that Γ ′ where Ω i is the topological disk bounded by Γ B i . Then Γ ′ B i is also a Jordan curve which bounds a topological disk Ω ′ i . Denote by d i the degree of the map f s i : . Given a bounded Fatou component B there exists a unique i ∈ {1, 2, . . . , m} and a minimal non-negative integer n B ≥ 0 such that f n B (B) which is also a Jordan curve, and define internal rays There exists θ 0 ∈ R/Z which is periodic under the map θ → d i θ, such that γ θ 0 B i converges to a periodic point in ∂B i \ P bad . In fact, as in [6, Lemma 2.1], we can prove that for each θ ∈ R/Z which is periodic under the map θ → d i θ, γ θ B i converges to a periodic point in ∂B i . Moreover, for two distinct θ, γ θ B i converges to different points. Since P bad is finite, the statement follows.
For each i = 1, 2, . . . , m, let us fix i be the union of the orbit of a i and the external rays landing on this orbit, and let Now we are ready to construct the puzzle. Let ε > 0 be such that the equipotential set Θ 0 := {G(z) = ε} is disjoint from the grand orbit of all critical points. Let Θ = Θ 0 ∪Θ 1 ∪Θ 2 . This is a finite union of smooth curves and periodic points, and each of these periodic points is the landing point of two or more smooth (external or internal) rays. Let P 0 be the collection of all components of C \ Θ which intersect the Julia set of f and for n ≥ 1, let P n be the collection of components of f −n (P ), where P runs over all elements of P 0 . An element of P n is called a puzzle piece of depth n.
Since Θ is disjoint from the orbit of critical points, for each c ∈ Crit ′ (f ) and each n ≥ 0 there is a puzzle piece P n (c) of depth n that contains c. If for each c ∈ Crit ′ (f ) and each integer s ≥ 1, there exists n ≥ 0 such that f s (c) ∈ P n (c), then the arguments in [6] show that diam(P n (c)) → 0 as n → ∞ for each c ∈ Crit ′ (f ).
Otherwise, there exists c ∈ Crit(f ) and a minimal integer s ≥ 1 such that f s (c) ∈ P n (c) for all n ≥ 0. The degree of f s : P n+s (c) → P n (c) is non-increasing, hence eventually constant. So there exists n 0 such that all the critical points of f s : P n 0 +s (c) → P n 0 (c) do not escape P n 0 +s (c) under forward iteration of this map. Using the "thickening" technique [10], we can find topological disks P n 0 +s (c) ⊃ P n 0 +s (c) and P n 0 (c) ⊃ P n 0 (c) such that f s : P n 0 +s (c) → P n 0 (c) is a renormalization of f . Then arguing as in Observation 1, there exists a period point q ∈ P n 0 +s (c) ∩ (P sep \ P bad ). We add the orbit of q and the external rays landing on the orbit of q in the set Θ and construct a new puzzle. Then either for each c ∈ Crit ′ (f ) we obtain an arbitrarily small new puzzle pieces containing c, or we obtain a new renormalization of f . The new renormalization either has a larger period, or has a smaller degree. Repeating the argument, if f is not infinitely renormalizable, then we must stop within finitely steps.
In conclusion, if f is at most finitely renormalizable and has only hyperbolic periodic points, then we can construct a puzzle such that for each c ∈ Crit ′ (f ) there exists an arbitrarily small puzzle pieces containing c. In the following we fix such a puzzle.
A.2. The bounds. Given a recurrent critical point c of f and a nice topological disk V ∋ c, we say that a topological disk U is a child of V if there exist c ′ ∈ [c] and s ≥ 1 such that U ∋ c ′ and such that f s : U → V is a proper map with a unique critical point, where We say that c is persistently recurrent if for each c ′ ∈ [c], each nice topological disk V ∋ c ′ has only finitely many children. Otherwise, we say that the recurrent critical point c is reluctantly recurrent.
Let V be a nice topological disk which contains a critical point c ∈ Crit ′ (f ). We say that V is essentially ρ-nice if for each return domain U of V which intersects orb(c), we have mod(V; U ) ≥ ρ.
Lemma 15. For each c ∈ Crit ′ (f ) which is persistently recurrent, there exists an arbitrarily small puzzle piece that contains c and is essentially ρ 1nice, where ρ 1 > 0 is a constant depending only on the degree of f . Proof. Given a small puzzle piece W that contains c, we can construct a complex box mapping as in Lemma 2.2 of [6], but we take V to be the union of the components of the domain of the first entry map to W which intersect [c]. Then the box mapping has only persistently recurrent critical points. Then by the proposition 10.1 of [6], the I n constructed in §8 of [6] is essentially ρ 1 -nice for some universal ρ 1 > 0 when n is large enough.
Let us say that an open set V is a puzzle neighborhood of a set A ⊂ Crit ′ (f ) if V is nice and each component of V is a puzzle piece containing exactly one point of A. As before, If V ′ ⊃ V are two puzzle neighborhoods of A such that f n (∂V a ) ∩ V ′ a = ∅, holds for all n ≥ 1 and a ∈ A, where V a (resp. V ′ a ) denotes the component of V (resp. V ′ ) which contains a, then we say that (V ′ , V) is an admissible pair. We say that a ∈ A is special for the pair (V ′ , V) if For any ρ > 0 there exists ρ ′ > 0 such that the following holds. Let c ∈ Crit ′ (f ) and let (V ′ , V) be an ρ-bounded admissible pair of puzzle neighborhoods of Back(c). Assume that V ′ is sufficiently small. Then for each x ∈ D(V), we have Proof. Take x ∈ D(V). Let s be the entry time of x into V, let U = L x (V). Let a ∈ Back(c) be such that f s (x) ∈ V a and let U ′ be the component of f −s (V ′ a ) which contains x. By Lemma 5, the degree of f s : U ′ → V ′ a is bounded from above by a constant N 0 . Thus mod( This proves the last statement. Lemma 17. For any ρ > 0 there exists ρ ′ > 0 such that the following holds. Let c ∈ Crit ′ (f ) and let (V ′ , V ) be a ρ-bounded admissible pair of puzzle neighborhoods of c. Suppose that for each x ∈ c ′ ∈Back(c) orb(c ′ ), either If diam(V ′ ) is sufficiently small, then (i) there exists a ρ ′ -bounded admissible pair (V ′ , V) of puzzle neighborhoods of Back(c) such that V ′ c = V ′ and V c = V and such that c is special for this admissible pair; is an admissible pair of puzzle neighborhoods of Back(c), and c is a special critical point. Since the maps f t i : V ′ i → V ′ has uniformly bounded degree, the admissible pair (V ′ , V) is ρ ′ -bounded, where ρ ′ > 0 is a constant.
(ii) By Lemma 16 and redefining ρ ′ > 0, we obtain that for each x ∈ D(V), (iii) Since c is special for the pair (V ′ , V), applying the last statement of Lemma 16 proves the result.
Lemma 18. If c ∈ Crit ′ (f ) is reluctantly recurrent. Then for any ρ > 0, there exists an arbitrarily small puzzle neighborhood V k of c which is ρ-nice. In particular, V k is essentially ρ-nice.
Proof. By [7, Lemma 6.5], there exists a puzzle piece V ∋ c, a positive integer N and a sequence of integers s k → ∞ with the following properties: • V is λ-nice for some λ > 0; • f s k (c) ∈ V , and letting V k = Comp c (f −s k V ), we have • f s k : V k → V has degree at most N . By replacing V with some pull back of V , we may also assume f j (V k ) ∋ c, for all 1 ≤ j < s k , which implies that (5) f j (V k ) ∩ V k = ∅ for all 1 ≤ j < s k . Let where L y (V k ) = V k if y ∈ V k and L y (V k ) = L y (V k ) otherwise. Since L y (V k ) ⊂ L y (V ) for each y ∈ V ∩ D(V k ), we have λ k ≥ λ. Claim. Each V k is λ k /N -nice.
To prove this claim, take x ∈ V k ∩ D(V k ). By (5), the return time of x into V k is at least s k and hence f s k (L x (V k )) = L y (V k ), where y = f s k (x). Therefore mod(V k ; L x (V k )) ≥ N −1 mod(V ; L y (V k )) ≥ λ k /N.
Letting U k = L c (V k ), and applying Lemma 17, we obtain that for each x ∈ D(U k ), mod(L x (V k ); L x (U k )) ≥ ρ 1 , where ρ 1 > 0 is a constant. It follows that λ k → ∞. The lemma follows.
Proof of Proposition 2. By Lemmas 15 and 18, there exists a universal constant ρ 1 > 0 such that any c ∈ Crit ′ (f ), there exists an arbitrarily small puzzle piece that contains c and is essentially ρ 1 -nice (note that by the definition of essentially ρ 1 -nice, the statement is trivial for non-recurrent c ∈ Crit ′ (f )). By Lemma 17 (i), it follows that there exists an arbitrarily small ρ 2 -bounded admissible pair (V ′ , V) of [c] for which c is special , where ρ 2 > 0 is a constant.
Theorem A'. For each ℓ > 1, there exists K 0 = K 0 (ℓ) > 0 such that if f is a C 3 interval maps with #Crit ′ (f ) = N and with ℓ max ≤ ℓ, and if f satisfies LD(K 0 r) for some r > 1, then f satisfies BC(r). This is proved in [1, Theorem 1], although the statement here is slightly more general. We observe that [1, Proposition 1]) remains true under the more general assumption here, if we require that f s (T ) is contained a small neighborhood of Crit ′ (f ) (which is given by [15,Theorem C]). The dependence of constants follows from the proof.
We also have the following Theorem B'. Let f be a C 3 interval maps with non-flat critical points. There exists a constant r 0 > 1 depending only on ℓ max such that if f satisfies BC(Kr 0 ) for some K > 1, then f satisfies LD(K).
The proof of this theorem follows the same outline as that of Theorem B, replacing the Schwarz lemma by the following real version.
Real Schwarz Lemma. Let f be a C 3 interval maps with non-flat critical points. There exists η = η(f ) > 0 and a universal constant θ ∈ (0, 1) such that if f n : U → V is a diffeomorphism between intervals, V ⊂ B(c, η) for some c ∈ Crit ′ (f ) and x ∈ U is such that f n (x) is the middle point of V , then provided that η is sufficiently small. Thus the lemma holds with θ = 1/(2K).