Growth Gap vs. smoothness for diffeomorphisms of the interval

Given a diffeomorphism of the interval, consider the uniform norm of the derivative of its n-th iteration. We get a sequence of real numbers called the growth sequence. Its asymptotic behavior is an invariant which naturally appears both in smooth dynamics and in geometry of the diffeomorphisms groups. We find sharp estimates for the growth sequence of a given diffeomorphism in terms of the modulus of continuity of its derivative. These estimates extend previous results of Polterovich and Sodin, and Borichev.


Introduction and main results
Denote by Diff 0 [0, 1] the group of all C 1 -smooth diffeomorphisms of the interval [0, 1] fixing the end points 0 and 1. For any f ∈Diff 0 [0, 1], we define the growth sequence of f by Γ n (f ) = max{ (f n ) ′ (x) ∞ , (f −n ) ′ (x) ∞ }, for all n ∈ N, where . ∞ stands for the uniform norm.
We say that a subgroup G ⊆Diff 0 [0, 1] admits a growth gap if there exists a sequence of positive numbers γ n (G) that grows sub-exponentially to +∞, such that for any f ∈ G, either Γ n (f ) tends exponentially to +∞, or Γ n (f ) ≤ C(f ) · γ n (G), for all n ∈ N.
From a viewpoint of dynamics, growth sequence of an element reflects how the length changes asymptotically under iterations. At the same time, geometrically, growth sequence indicates how an element is distorted with respect to the multiplicative norm. In [DG], D'Ambra and Gromov suggested to study growth sequences of various classes of diffeomorphisms.
As it is shown in Theorem 1, decrease of smoothness assumptions leaves more room for exponential growth, i.e., the growth sequence Γ n (f ) becomes bigger, or in other words, "the growth gap" is smaller. Therefore, smaller subgroups of Diff 0 [0, 1] should be considered in order to discover a growth gap. In [PS] a growth gap was found for the subgroup of C 2 -diffeomorphisms of Diff 0 [0, 1]. Namely, Theorem 2. Let f ∈Diff 0 [0, 1] be a C 2 -diffeomorphism with γ(f ) = 1.
One can substitute ω(δ) = δ and ω(δ) = δ α into Theorem 4 for achieving Theorems 2 and 3. In the following corollary, we consider two toy models related to case(a) in order to test how Theorem 4 provides a growth gap. In the case when the modulus of continuity ω(δ) is close to the identity, the second term on the right hand side of ( * ) can be absorbed into the first one. Namely, .
The proofs easily follow by substituting the relevant assumptions into Theorem 4.
The drawback of Theorem 4 is that it does not provide a growth gap , then an attempt to apply Theorem 4 for this diffeomorphism yields only a trivial estimate Our second theorem mends this disadvantage. It shows that in case (b) (under additional regularity assumption imposed on ω) one can discard the first term on the right hand side of ( * ) : Theorem 5. Let ω(x) : [0, 1] → R + be a modulus of continuity such that for some 0 < α < 1, ω(x) x α is a decreasing function on (0, a(α)), where 0 < a(α) < 1. Then for f ∈ Diff ω 0 [0, 1], such that γ(f ) = 1, we have The next set of theorems present a sufficient sharpness for the estimates of the bounds in Theorems 4 and 5 respectively.

Growth gap: Proofs of theorems 4 and 5
The following lemma (see [EF, Dz]) states that every modulus of continuity admits an equivalent concave modulus of continuity: Lemma 1. For any modulus of continuity ω there exists a concave modulus of continuity ω * such that ω ≤ ω * ≤ 2ω everywhere on [0, 1].
Due to this lemma, we assume in the proofs of Theorems 4 and 5 that ω is a concave modulus of continuity.

Claim 1. (a) For any
In the same way, Therefore, for all k ∈ N : By summing the integrals we obtain the desirable inequality.
Proof of Claim 3. Denote by s = Ω(z), and notice that s · ω(s) = z. Thus, and we are done.
We turn now to the following two lemmas, on which the proof of Theorem 4 will be based. .
Proof of Lemma 2. We split the proof to 2 cases.
Case 1: a. x n+1 ∈ I x 1 and φ(x 1 ) < φ(x n+1 ). In this case, . We have two possibilities: Hence, this case is completed due to Claim 3 in the last inequality we have used Claim 1, hence we are done due to In the last inequality we have used Claim 1, hence we are done due to Claim 3.
Let us apply the above estimates for bounding our initial expressions: with C(f ) = 10 + A + 9 1+a . By Jensen's inequality completing the proof of Lemma 2.
Combining Lemmas 1 and 2, we get Corollary 2. Suppose that x 1 , ..., x n ∈ (0, δ). Then, At last, we turn to the details of the proof of Theorem 4, we shall show that estimate ( * * ) holds for each x ∈ (0, 1). Consider the decomposition of the interval into a union of open intervals [0, 1]\Fix(f ) = ∪ i∈I (a i , b i ). Let x ∈ (0, 1) be an arbitrary point, then x ∈ (a i , b i ) for some i ∈ I. If |b i − a i | ≤ δ, then the proof is complete by Corollary 2. There are only finitely many intervals such that |b i − a i | > δ. We take one of them and divide it into 3 subintervals: We denote by n 1 , n 2 , n 3 the length of the trajectory of the sequence x n in each of the 3 subintervals respectively. It is evident that n 2 is bounded by some constant N(f ). If n 3 = 0 or n 1 = 0, then we are done due to Corollary 2. Otherwise, n = n 1 + n 2 + n 3 , we continue using Lemma 2, Note that Moreover, we have the following estimate: Now we are going to find an upper bound for | log φ(xn) φ(x 1 ) |. As before, we split into two cases: a. φ(x n ) > φ(x 1 ). By using Claim 1 and the choice of δ 0 , we have the last inequality is due to Claim 2. Thus by Claim 3, we have In the same way, by Claim 3 it follows that .
Lemma 4. The following inequality is satisfied Proof of Lemma 4. We will prove it only for x ′ 0 , ..., x ′ n ∈ (0, ε), the general case is proved similarly. Denote by s := s(n) = log( . We wish to show that s ≤ C · nω( 1 n ). We have the following inequality, which is valid due to the Claim 1: We are going to change variables in this integral until we bound n(s) from above by some function of s. .
We obtain that n ≥ C ′ · s 0 dt ω(φ −1 0 (e −t ·φ 0 (x 0 )) . The inequality s ≤ C ·nω( 1 n ) will be proved if we find an absolute constant C such that Denote by Λ(x) = x · ω( 1 x ), for x ≥ 1 and continue Λ(x) in an arbitrary monotonically and smooth way on [0, 1], such that Λ(0) = 0. Let ξ := Λ(n), n = Λ −1 (ξ). Note that Thus, we have to show that, for ξ → ∞, Making once again change of variables [t = C · s], we get This is equivalent to In order to prove that inequality let us show that the expression inside the integral is always non-negative. I.e., we have to show that , x , where 0 < t < 1, we have to prove the following inequality: 0 ≥ (log( ω(t) t 1−c ) ′ Using the assumption that ω(t) t α is a decreasing function and choosing c = 1 − α we obtain the desired inequality, proving Claim 4. Now we can apply Claim 4 to the desirable inequality: We can choose C in a way that C · C ′ · c ∈ N. Notice that )) due to the sub-additivity property of ω. Thus , it is enough to show that In other words, Since φ 0 (x 0 ) < 1 and φ −1 0 is monotonically increasing, we can drop this multiplier and prove that .
We have to prove the following inequality: or, an equivalent form is Here, we used the monotonicity of ω(x). Hence, it is enough to show that: Equivalently, Since we can choose a constant C such that C·C ′ ·c·V 2 ≥ V, it is enough to prove that: x α is decreasing in some [0, a(α)], thus there exists a constant c(ω), such that ω(x) ≥ c(ω) · x α . Substituting it in the above inequality, we have to show that Certainly, such a C can be picked, since the left expression tends to ∞ when V → 0, while the right one tends to 0.
The next lemma allows us to complete the proof of Theorem 5 for a general function f (x).

Remark:
The same proof will work when we deal with the case log φ(xn) φ(x 0 ) = | log φ(xn) φ(x 0 ) |. The proof of Lemma 5 is based on the following Lemma 6, whose proof will be provided in the end of this section.
Proof of Claim 5. We have n ≥ c · We have obtained that N ≤ C · n. This proves Claim 5. Now we provide details, which complete the proof of Lemma 5. We may reformulate the previous claim as follows. Let m ∈ N some fixed number .
The proof of Lemma 5 is completed.
Now, let us complete the proof of Theorem 5 (modulo Lemma 6). Combining Lemmas 4 and 5 we have that , since we have ω(x) x → ∞, x → 0, due to the assumptions of the theorem. Therefore, due to Lemma 3, The proof of Theorem 5 is completed up to the proof of Lemma 6 which will be provided now.
Proof of Lemma 6. The proof is based on the following series of claims: . Integrate both sides and obtain the following: . Pick a = s and t = s − s 0 (s > s 0 , since φ 0 (x) ≥ φ(x)), we obtain that: Proof of Claim 7. Define g(x) = φ −1 0 (φ(x)) and notice that this follows from Claim 6. Now we conclude that g(x) is a Lipschitz, Proof of Claim 8. We will split both integrals to the sums: hence it is enough to check that Using Claim 7 we obtain that The claim is proved.
It completes the proof of Theorem 6.
We use this observation in the following form: We make change of variables [s = 1 t ] and get that , we obtain using integration by parts, )ds.
For α very close to 1, we have , hence, we have ). Therefore, It completes the proof of Claim 9.