2 On a family of $C^{1+bv}$ -undistorted diffeomorphisms

Here and in what follows, all maps we consider are supposed to preserve the orientation.

Recall that for a $C^{1+bv}$ diffeomorphism f of a compact 1-manifold, the asymptotic distortion was defined by Navas in [Reference Navas13] as

$$ \begin{align*}\mathrm{dist}_{\infty} (f) := \lim_{n \to \infty} \frac{\mathrm{var} (\log Df^n)}{n}.\end{align*} $$

By the subadditivity of $\mathrm {var} (\log D(\cdot ))$ , if f is a distorted element of the group of $C^{1+bv}$ diffeomorphisms, then $\mathrm {dist}_{\infty } (f) = 0$ .

The family of diffeomorphisms with positive asymptotic distortion studied in [Reference Navas14] is as follows: Start with a $C^{1+bv}$ diffeomorphism of $[0,1]$ with vanishing asymptotic distortion and no fixed point in $]0,1[$ . Let I be a fundamental domain for the action of f, that is, an open interval with endpoints $x_0$ and $x_1 := f (x_0)$ for a certain $x_0 \in \, ]0,1[$ . Let g be any non-trivial $C^{1+bv}$ diffeomorphism of $]0,1[$ supported on I. Then the diffeomorphism $\bar {f} := fg$ has positive asymptotic distortion and, in particular, it is undistorted in $\mathrm {Diff}_+^{1+bv} ([0,1])$ (hence in $\mathrm {Diff}_+^2([0,1])$ if f and g are of class $C^2$ ). This fact follows from [Reference Eynard-Bontemps and Navas5] (see Lemmas 2.2 and 7.2 therein) by using the relation between the asymptotic distortion and the Mather invariant. For the reader’s convenience, below we present a short and direct argument based on Kopell’s like estimates [Reference Kopell9, Reference Navas11].

Let us consider the product $\bar {f}^n f^{-n}$ . Since f has vanishing asymptotic distortion, if we show that $\mathrm {var} (\log D (\bar {f}^n f^{-n}) )$ has linear growth, the same will hold for $\mathrm {var} (\log D \bar {f}^n)$ . Now, notice that

$$ \begin{align*}\bar{f}^n f^{-n} = (fgf^{-1}) (f^2 g f^{-2}) (f^3 gf^{-3}) \cdots (f^{n} g f^{-n})\end{align*} $$

has support in the union of the intervals $f(I), f^2(I), \ldots , f^n (I)$ , and equals $f^k g f^{-k}$ on each such interval $f^k (I)$ . In particular, its derivative at the endpoints $x_k,x_{k+1}$ of each of these intervals equals 1. We claim that there is a constant $\lambda> 1$ such that, for all $k \geq 1$ , there is a point $y_k \in f^k (I)$ satisfying $D (f^k g f^{-k}) (y_k) \geq \lambda $ . Assuming this, we conclude

$$ \begin{align*}\mathrm{var} (\log D (\bar{f}^n f^{-n}) ) \geq \sum_{k=1}^{n} |\!\log D(f^kgf^{-k}) (y_k) - \log D (f^kgf^{-k}) (x_k)| \geq n \, \log (\lambda),\end{align*} $$

which yields the desired linear growth.

Now, to check the existence of $\lambda $ and the points $y_k$ , let $V := \mathrm {var} (\log Df)$ , and let $N \geq 1$ be such that $Dg^N (z)> e^{2V}$ holds for some $z \!\in \! I$ . We claim that $\lambda := e^{V/N}$ works. Assume otherwise. Then, for a certain $K \geq 0$ , one would have $ \| D (f^K g f^{-K} ) \|_{\infty } \leq e^{V / N}$ , which by the chain rule would yield $ \| D (f^K g^N f^{-K} ) \|_{\infty } \leq e^{V}$ . However, at the point $z_K \!:=\! f^K (z)$ , we have $D (f^K g^N f^{-K}) (z_K)> e^V$ . Indeed,

$$ \begin{align*} \log (D (f^K g^N f^{-K}) (z_K)) &= \log Df^K (g^N (z) ) + \log Dg^N (z) - \log Df^K (z) \\ &\geq \log Dg^N (z) - |\!\log Df^K (g^N (z) ) - \log Df^K (z) | \\ &> 2V - \sum^{K-1}_{k=0} |\!\log Df (f^k (g^N (z))) - \log Df (f^k (z)) |. \end{align*} $$

Since both z and $g^N (z)$ lie in the fundamental domain I of f,

$$ \begin{align*}\sum^{K-1}_{k=0} |\!\log Df (f^k (g^N (z))) - \log Df (f^k(z)) | \leq \mathrm{var} (\log Df) \leq V.\end{align*} $$

We thus conclude that $\log (D (f^K g^N f^{-K}) (z_K))> V,$ as announced.

3 Distortion in class $C^{1+\alpha }$ for $\alpha < 1/2$

In this section, we start by briefly recalling the construction of the group $\Gamma $ with a distorted element $\bar {f}$ considered in [Reference Navas14]. Next, we proceed to smooth the action of $\Gamma $ in order to achieve any differentiability class $C^{1+\alpha }$ for $\alpha < 1/2$ . Upgrading $\alpha $ to any number less than 1 will require the introduction of an extra element plus a tricky new computation, and will be carried out in the next section.

Start with the vector fields $\hat {\mathcal {X}}$ and $\mathcal {X}$ on the real line whose time-1 maps are, respectively,

$$ \begin{align*}\hat{F} := \hat{\mathcal{X}}^1 : x \mapsto 2x \quad { \mathrm{and} } \quad F := \mathcal{X}^1 : x \mapsto x + 1.\end{align*} $$

Let $\varphi \! : \mathbb {R} \to \, ]0,1[$ be a $C^{\infty }$ diffeomorphism such that $\hat {\mathcal {Y}} := \varphi _* (\hat {\mathcal {X}})$ and $\mathcal {Y} := \varphi _* (\mathcal {X})$ extend to the endpoints of $[0,1]$ as infinitely flat vector fields. Denote $\hat {f} := \hat {\mathcal {Y}}^1$ and $f := \mathcal {Y}^1$ , which we view as diffeomorphisms of $[-1,2]$ that coincide with the identity outside $[0,1]$ . The affine relation $\hat {f} f \hat {f}^{-1} = f^2$ yields that $\| f^n \| = O (\log (n))$ ; in particular, f has vanishing asymptotic distortion.

Let $x_0 := \varphi (0)$ and, for each $k \!\in \! \mathbb {Z}$ , let $x_k := f^k (x_0) = \varphi (k)$ . Denote also $x_{-1/2} := \varphi (-1/2)$ and $x_{-3/4} := \varphi (-3/4)$ . Let $\varphi _0$ be the affine diffeomorphism sending $I := [x_0,x_1]$ onto $[0,1]$ , and let $g := \varphi _0^{-1} f \varphi _0$ . This can be extended to $[-1,2]$ by the identity outside I.

We next define two diffeomorphisms $\hat {h}$ and h as follows.

1. (i) They act by the identity outside $[0,1]$ .

2. (ii) On each interval $I_k := f^k (I)$ , the diffeomorphism $\hat {h}$ (respectively h) coincides with the $s_k$ -time map (respectively $t_k$ -time map) of the flow of the vector field $f^k_* (\varphi _0^* (\hat {\mathcal {Y}} ) )$ (respectively $f^k_* (\varphi _0^* (\mathcal {Y})) $ ).

   Here, $s_k$ and $t_k$ are sequences of real numbers such that as follows.

3. (iii) If $2^{i-1} \leq k < 2^{i}$ for a certain positive even integer i, then

   $$ \begin{align*}s_k := \log_2 \bigg( 1 - \frac{1}{\sqrt{\ell_{i/2}}} \bigg) \quad \mbox{ and } \quad t_k := \frac{1}{ \sqrt{\ell_{i/2}} },\end{align*} $$
   where $\ell _j$ is a prescribed sequence of positive integers diverging to infinity to be fixed below.

4. (iv) Otherwise, $s_k = t_k := 0$ .

   Finally, we let $\psi $ be a $C^{\infty }$ diffeomorphism of $[-1,2]$ such that:

5. (v) $\psi $ coincides with the identity on $[x_{-1/2} , x_0];$

6. (vi) $\psi (x_{-3/4}) = 0$ and $\psi (x_1) = 1$ .

The group we consider is $\Gamma := \langle \hat {f}, f, g, \hat {h}, h, \psi \rangle $ . The computations of [Reference Navas14] show the following relation for certain powers of $\bar {f} := fg$ (which justifies the construction):

(3.1) $$ \begin{align} (\bar{f})^{2^{i-1}} = f^{1+n/2} \hat{f}^i [\hat{f}^{-i}f^{-n}hf^n\hat{f}^i , \psi \hat{f}^{-i}f^{-n}\hat{h}f^n\hat{f}^i\psi^{-1} ]^{\ell_{i/2}} \hat{f}^{-i} f^{-1}, \end{align} $$

where i is an even (positive) integer and $n:= 2^i$ . Roughly, this works as follows: set

$$ \begin{align*}a_n:=\hat{f}^{-i}f^{-n}hf^n\hat{f}^i \quad \mbox{ and } \quad b_n:= \psi \hat{f}^{-i}f^{-n}\hat{h}f^n\hat{f}^i\psi^{-1}.\end{align*} $$

One easily checks that

$$ \begin{align*}&\text{supp} (a_n) \subset[0,x_{-3/2}]\cup[x_{-1/2},x_0]\cup[x_1,1] \quad \mbox{ and }\\ &\text{supp} (b_n) \subset[-1,0]\cup[x_{-1/2},x_0]\cup[1,2].\end{align*} $$

Thus, the commutator $c_n:=[a_n,b_n] = a_n b_n a_n^{-1} b_n^{-1}$ is supported on $[x_{-1/2},x_0]$ , so the conjugate $\hat {f}^{i}c_n\hat {f}^{-i}$ is supported on $ [x_{-2^{i-1}},x_0] $ . Besides, on each $[x_k,x_{k+1}]\subset [x_{-2^{i-1}},x_0]$ , this conjugate $\hat {f}^{i}c_n\hat {f}^{-i}$ coincides with the time- ${1}/{\ell _{i/2}}$ map of the flow of $f^k_\ast (\varphi _0^\ast (\mathcal {Y}))$ . Moreover, the restriction of the map

$$ \begin{align*}h_{n/2} := (f^{-n/2}gf^{n/2}) \cdots (f^{-2} g f^2) (f^{-1}gf) = f^{-n/2} (f^{-1} \bar{f}^{n/2} f)\end{align*} $$

to each $[x_k,x_{k+1}]\subset [x_{-2^{i-1}},x_0]$ equals the time-1 map of the flow of $f^k_\ast (\varphi _0^\ast (\mathcal {Y}))$ . This implies that

$$ \begin{align*}h_{n/2} = (\hat{f}^i c_n \hat{f}^{-i})^{\ell_{i / 2}},\end{align*} $$

which corresponds to identity (3.1).

Since $\| f^n \| = O (\log (n))$ , identity (3.1) implies that $\| (\bar {f})^{2^{i-1}}\| = O (i \, \ell _{i/2})$ . Therefore, $\bar {f}$ is distorted provided $\ell _{j}$ grows to infinity in such a way that

(3.2) $$ \begin{align} \lim_{n \to \infty} \frac{\log (n) \, \ell_i}{n} = \lim_{i \to \infty} \frac{i \, \ell_i }{2^i} = 0. \end{align} $$

The maps $\hat {f}, f, g, \psi $ above are obviously smooth. However, regularity for $\hat {h}, h$ is more subtler. Indeed, their $C^1$ smoothness is ensured by the conditions $s_n \to 0$ and $t_n \to 0$ as $|n| \to \infty $ (which are equivalent to $\ell _j \to \infty $ as $j \to \infty $ ) together with the next lemma. This is strongly inspired on the work of Kopell (a proof together with a discussion appears in the Appendix of [Reference Navas14]). Below, by $C^{1+\mathrm {Lip}}$ we refer to maps with Lipschitz derivative.

Unfortunately, as we will see in the Appendix, this fails to extend to class $C^{1+\alpha }$ . Because of this, we need to go into more explicit computations for our example. Although these are difficult to handle, the following key elementary lemma taken from the work of Pixton [Reference Pixton15] and Tsuboi [Reference Tsuboi17] will be enough for us.

Lemma 3.2. Given a $C^2$ vector field $\mathcal {X}$ on an interval $[0,a]$ , denote $C_1 := \| D \mathcal {X} \|$ and $C_2 := \| D^2 \mathcal {X} \| $ . If $f^t$ denotes its flow, then, for all $t \geq 0$ ,

$$ \begin{align*}\| D \log Df^t \| \leq \frac{C_2}{C_1} \, ( e^{C_1 t} -1 ).\end{align*} $$

Proof Taking derivatives on the equality $d f^t / dt = \mathcal {X} \circ f^t$ , we deduce

$$ \begin{align*}\frac{d}{dt} D f^t = D \mathcal{X} (f^t) \cdot Df^t,\end{align*} $$

so

$$ \begin{align*}\frac{d}{dt} \log Df^t = \frac{({d}/{dt}) Df^t}{Df^t} = D \mathcal{X} (f^t).\end{align*} $$

Since $f^0 = \mathrm {Id}$ , we conclude

$$\begin{align*}\log Df^t = \int_0^t D \mathcal{X} (f^s) \, ds. \end{align*}$$

Since $\| D \mathcal {X} \| \leq C_1$ , this yields $|\!\log Df^t | \leq C_1 t$ , so $|D f^t| \leq e^{C_1 t}$ . Moreover,

$$ \begin{align*}D \log Df^t = D \bigg( \int_0^t D \mathcal{X} (f^s) \, ds \bigg) = \int_0^t D^2 \mathcal{X} (f^s) \cdot Df^s \, ds.\end{align*} $$

Since $\| D^2 \mathcal {X} \| \leq C_2$ , we conclude

$$ \begin{align*}| D \log Df^t | \leq C_2 \int_0^t |Df^s| \, ds \leq C_2 \int_0^t e^{C_1 s} ds = \frac{C_2}{C_1} \, ( e^{C_1 t} -1 ),\end{align*} $$

as announced.

We now turn to very long computations that will allow us to ensure that the resulting maps $h, \hat {h}$ built via the procedure above are $C^{1+\alpha }$ diffeomorphisms for well chosen $\varphi $ and $\ell _i$ that respect all the properties we have imposed. This will close the proof of our theorem.

In order to simplify these computations, let us remind the chain rules for different derivatives of maps between one-dimensional spaces, namely logarithmic (L), affine (A), and Schwarzian (S):

$$ \begin{align*}L (f) &:= \log Df, \quad A(f) := DL(f) = \frac{D^2 f}{D f} ,\\ S (f) &:= DA(f) - \frac{A(f)^2}{2} = \frac{D^3 f}{D f} - \frac{3}{2} \bigg( \frac{D^2 f}{D f} \bigg)^2.\end{align*} $$

These are listed below:

(3.3) $$ \begin{align} L(fg) = L(g) + L(f) \circ g, \end{align} $$

(3.4) $$ \begin{align} A (fg) = A (g) + A(f) \circ g \cdot Dg, \end{align} $$

(3.5) $$ \begin{align} S (fg)= S (g) + S(f) \circ g \cdot (Dg)^2. \end{align} $$

We let $\varphi \! : (0,1) \to \mathbb {R}$ be a $C^{\infty }$ diffeomorphism such that, for a sufficiently small $\delta> 0$ ,

$$ \begin{align*}\varphi(x)= \begin{cases} - \text{exp}{(\text{exp}{(1/x)})} \quad & \mbox{if } 0<x\leqslant \delta, \\ \text{exp}{(\text{exp}{(1/(1-x))})}\quad & \mbox{if } 1-\delta \leqslant x <1. \end{cases} \end{align*} $$

If we denote $\mathcal {Z} := \varphi _0^* (\hat {\mathcal {Y}} )$ , then we need to control $f^n_* (\mathcal {Z})$ .

3.1 An estimate for lengths of fundamental domains

Let us come back to the group $\Gamma = \langle \hat {f}, f, g, \hat {h}, h, \psi \rangle $ . Recall that I denotes the interval $[x_0,x_1]$ . We claim that, for a certain constant ${C> 0}$ ,

(3.6) $$ \begin{align} |f^n (I)| = O \bigg( \frac{C}{n \log(n) \, (\log (\log (n)))^2} \bigg). \end{align} $$

This is checked via a direct computation. Namely, for a sufficiently large n,

$$ \begin{align*}| f^n (I)| = \varphi^{-1} (n+1) - \varphi^{-1} (n) = \frac{1}{\log \log (n)} - \frac{1}{\log \log (n+1)}.\end{align*} $$

Since the derivative of $x \mapsto 1/\log \log (x)$ is $1 / [x \, \log (x) \, (\log \log (x))^2]$ , a direct application of the mean value theorem yields the desired estimate (3.6).

3.2 Estimates for the vector field and its derivative

Notice that

$$ \begin{align*}f^n_* (\mathcal{Z}) (x) = [Df^n(f^{-n}(x))] \, \mathcal{Z}(f^{-n}(x)), \hspace{0,5cm} x\in I_n := f^n (I) .\end{align*} $$

Taking derivatives, we obtain

$$ \begin{align*}D (f^n_\ast(\mathcal{Z}))(x) = \frac{D^2f^n(f^{-n}(x))}{Df^{n}(f^{-n}(x))} \, \mathcal{Z}(f^{-n}(x)) + D \mathcal{Z}(f^{-n}(x)).\end{align*} $$

Now, using the chain rule (3.4), this yields

$$ \begin{align*}D (f^n_\ast(\mathcal{Z}))(x) = \sum_{i=0}^{n-1}\bigg( \frac{D^2f(f^{i-n}(x))}{Df(f^{i-n}(x))}\bigg)Df^{i}(f^{-n}(x)) \, \mathcal{Z}(f^{-n}(x)) + D \mathcal{Z}(f^{-n}(x)).\end{align*} $$

Thus, letting

$$ \begin{align*}C' := \bigg\| \frac{D^2 f}{Df} \bigg\| \| \mathcal{Z} \|, \quad C" := \| D \mathcal{Z} \|,\end{align*} $$

we obtain

$$ \begin{align*}| D (f^n_\ast (\mathcal{Z}))(x) | \leq C' \sum_{i=0}^{n-1} Df^{i}(f^{-n}(x)) + C".\end{align*} $$

We claim that the sum above is uniformly bounded (independently of n and $x \in I_n$ ), so that

(3.7) $$ \begin{align} | D (f^n_\ast (\mathcal{Z}))(x) | \leq C \end{align} $$

for a certain constant C. Indeed, a standard control of distortion argument yields that $Df^{i}(f^{-n}(x))$ is of the order of $|I_i| / |I_0|$ , so

$$ \begin{align*}\sum_{i=0}^{n-1} Df^{i}(f^{-n}(x)) \sim \sum_{i=0}^{n-1} |I_i| \leq 1.\end{align*} $$

3.3 Estimates for the second derivative

We now claim that, for a certain constant ${C>0}$ and all $x \in I_n$ ,

(3.8) $$ \begin{align} | D^2 (f^n_\ast(\mathcal{Z})) (x) | \leq C \, n \log (n) \, (\log \log (n))^2. \end{align} $$

To show this, notice that, from

$$ \begin{align*}D (f^n_\ast(\mathcal{Z}))(x) = A(f^n) (f^{-n}(x)) \, \mathcal{Z}(f^{-n}(x)) + D \mathcal{Z}(f^{-n}(x)),\end{align*} $$

we obtain

$$ \begin{align*}D^2 (f^n_\ast(\mathcal{Z}))(x) = [ DA(f^n) \cdot \mathcal{Z} + A(f^n) \, D \mathcal{Z} + D^2 \mathcal{Z} ] \circ f^{-n} (x) \cdot Df^{-n}(x).\end{align*} $$

which is equal to

$$ \begin{align*}\big[ \big( S(f^n) + \tfrac{1}{2} A(f^n)^2\big) \cdot \mathcal{Z} + A(f^n) \, D \mathcal{Z} + D^2 \mathcal{Z} \big] \circ f^{-n} (x) \cdot Df^{-n}(x).\end{align*} $$

Let us analyze each term entering in this expression. First, by equation (3.6) and the control of distortion argument above,

$$ \begin{align*}Df^{-n}(x) = 1 / D f^n (f^{-n}(x)) = O ( n \log (n) \, (\log \log (n))^2 ).\end{align*} $$

We next claim that $A(f^n)$ is uniformly bounded on $I_0$ . Indeed, letting $C:= \| A (f) \|$ , the chain rule (3.4) yields

$$ \begin{align*}A (f^n) = \sum_{i=0}^{n-1} A (f) \circ f^i \cdot Df^i \leq C \sum_{i=0}^{n-1} Df^i.\end{align*} $$

The control of distortion argument above shows that the last sum is bounded from above by a constant, which is the claim.

Since $\mathcal {Z}$ , $D \mathcal {Z}$ and $D^2 \mathcal {Z}$ are obviously uniformly bounded, to show equation (3.8), it remains to check that $S (f^n) (f^{-n} (x))$ is uniformly bounded. To see this, we use the chain rule (3.5) to obtain

$$ \begin{align*}Sf^n(f^{-n}(x)) = \sum_{i=0}^{n-1}Sf(f^{i-n}(x))(Df^{i}(f^{-n}(x)))^2.\end{align*} $$

This implies

$$ \begin{align*}| Sf^n(f^{-n}(x)) | \leq C \, \sum_{i=0}^{n-1} ( Df^i (f^{-n}(x)) )^2,\end{align*} $$

and the last sum can be estimated as it was done before. (The sum here is even smaller since it involves the squares of the derivatives.)

3.4 Estimates for the maps

We are now in a position to check that the group $\Gamma $ is made of $C^{1+\alpha }$ diffeomorphisms for $\alpha < 1/2$ and $\ell _i$ of order $n / \log (n)^2$ . (Notice that, according to equation (3.2), the element $\bar {f}$ is distorted in $\Gamma $ for this choice.) Notice that this is obvious for all the generators except h and $\hat {h}$ . The estimates for these two elements are similar, so that we only deal with h. Besides, we may deal with $\log Dh$ instead of $Dh$ , since the condition ‘ $Dh$ is of class $C^{\alpha }$ ’ is equivalent to that ‘ $\log Dh$ is of class $C^{\alpha }$ ’.

We need to check that there exists a uniform bound B for expressions of type

$$ \begin{align*}\frac{|\!\log Dh(y) - \log Dh(x)|}{|y-x|^{\alpha}}\end{align*} $$

for all points $x< y$ in the same interval $I_n$ . Indeed, having such an estimate, one can easily treat the case of arbitrary pairs $x < y$ just by noticing that at each endpoint of an interval of the form above, the derivative of h equals 1. Namely, letting $z_1$ (respectively $z_2$ ) be such an endpoint that is immediately to the right of x (respectively to the left of y), one has

$$ \begin{align*} \frac{|Dh(y) - Dh(x)|}{|y-x|^{\alpha}} &\leq \frac{|Dh(y) - Dh(z)|}{|y-x|^{\alpha}} + \frac{|Dh(z) - Dh(x)|}{|y-x|^{\alpha}} \\ &\leq \frac{|Dh(y) - Dh(z)|}{|y-z|^{\alpha}} + \frac{|Dh(z) - Dh(x)|}{|z-x|^{\alpha}} \\ &\leq 2 \, B. \end{align*} $$

Now, for all $ z \in I_n$ (with $n \geq 0$ ), Lemma 3.2 and estimate (3.7) yield, for $t_n$ small enough,

$$ \begin{align*}D (\log Dh) (z) \leq 2 \, \| D^2 f^n_* (\mathcal{Z}) \| \, t_n.\end{align*} $$

By estimate (3.8), this implies, for a certain constant $C>0$ ,

(3.9) $$ \begin{align} D (\log Dh) (z) \leq 2 \, C \, n \log (n) \, (\log(\log(n)))^2 \, t_n. \end{align} $$

Moreover, for $x,y$ in $I_n$ ,

$$ \begin{align*}\frac{ |\!\log Dh (y) - \log Dh (x)| }{|y-x|^{\alpha}} &= \frac{ |\!\log Dh (y) - \log Dh (x)| }{|y-x|} \, |y-x|^{1-\alpha}\\ &= D (\log Dh (z)) \, |y-x|^{1-\alpha}\end{align*} $$

for a certain point $z \in I_n$ . By equation (3.9), this yields

(3.10) $$ \begin{align} &\frac{ |\!\log Dh (y) - \log Dh (x)| }{|y-x|^{\alpha}}\nonumber\\ &\quad\leq 2 \, C \, n \log (n) \, (\log(\log(n)))^2 \, t_n \, \bigg[ \frac{C}{ n \log (n) (\log (\log (n)))^2}, \bigg] ^{1-\alpha}. \end{align} $$

Since $t_n = 1/ \sqrt {\ell _{i/2}} \leq C \, \log (n) / \sqrt {n}$ , we finally obtain

$$ \begin{align*}&\frac{ |\!\log Dh (y) - \log Dh (x)| }{|y-x|^{\alpha}}\\ &\quad\leq 2 \, C' \, n \log (n) \, (\log(\log(n)))^2 \, \frac{\log(n)}{n^{1/2}} \, \frac{1}{[n \log (n) (\log(\log(n)))^2]^{1-\alpha}}.\end{align*} $$

To get the desired upper bound B, it suffices that the total exponent of n in the expression above is negative. Since this exponent equals $1 - 1/2 - (1-\alpha ) = \alpha - 1/2$ , this condition reduces to $\alpha < 1/2$ , which is our hypothesis.

4 Distortion in class $C^{1+\alpha }$ for $\alpha < 1$

It is unclear whether the previous action can be smoothed beyond the class $C^{3/2}$ (compare [Reference Castro, Jorquera and Navas2, Reference Deroin, Kleptsyn and Navas3, Reference Kleptsyn and Navas8, Reference Navas12]). To achieve a larger differentiability class, we will need to accelerate the distorted behavior of $\bar {f}$ , which will allow us to consider smaller integration times for the flows of vector fields (in concrete terms, we will increase the sequence $\ell _i$ ). This will be crucial to improve the regularity from $\alpha < 1/2$ to any $\alpha < 1$ .

4.1 Adding an extra element

We consider the map $\tilde {h}$ acts by the identity outside the intervals $I_k$ , and that each such interval coincides with the $r_k$ -time of the time flow of the vector field $f^k_* (\varphi _0^* (\hat {\mathcal {Y}}))$ , where $r_k := 1 / \sqrt {\ell _{i/2}}$ for $2^{i-1} \leq k < 2^{i}$ and $r_k := 0$ otherwise. Notice that $\tilde {h}$ is very similar to $\hat {h}$ . (Actually, we could perform the computations that follow using $\hat {h}$ instead of $\tilde {h}$ , but this would become much harder.)

Then we let $d_n := \hat {f}^{-i}f^{-n}\tilde {h}f^n\hat {f}^{i}$ for $n = 2^i$ , where i is an even integer. We have $\mathrm {supp} (d_n) \subset [0,x_{-3/2}]\cup [x_{-1/2},x_0]\cup [x_1,1]$ . Since $\mathrm {supp} (c_n) \subset [x_{-1/2},x_0]$ , for every integer $L_i \geq 1$ , the support of $d_n^{L_i} c_n d_n^{-L_i}$ is also contained in $[x_{-1/2}, x_0]$ , thus the support of

$$ \begin{align*}\hat{f}^{-i} d_n^{L_i} c_n d_n^{-L_i} \hat{f}^i = (\hat{f}^{-i} d_n^{L_i} \hat{f}^{i}) (\hat{f}^{-i} c_n \hat{f}^{i}) (\hat{f}^{-i} d_n^{-L_i} \hat{f}^{i})\end{align*} $$

is contained in $[x_{-2^{i-1}},x_0]$ .

Now recall that, on each $[x_k,x_{k+1}]\subset [x_{-2^{i-1}},x_0]$ , the conjugate $\hat {f}^{i}c_n\hat {f}^{-i}$ coincides with the time- ${1}/{\ell _{i/2}}$ map of the flow of $f^k_\ast (\varphi _0^\ast (\mathcal {Y}))$ . Moreover, by construction, on the same interval, the conjugate $\hat {f}^{i} d_n\hat {f}^{-i}$ coincides with the time- ${1}/{\sqrt {\ell _{i/2}}}$ map of the flow of $f^k_\ast (\varphi _0^\ast (\hat {\mathcal {Y}}))$ . By the affine relation, still on the same interval, the map $\hat {f}^{-i} d_n^{L_i} c_n d_n^{-L_i} \hat {f}^i$ lies in the flow of $f^k_\ast (\varphi _0^\ast (\mathcal {Y}))$ , and arises at time

$$ \begin{align*}\frac{2^{({L_i}/{\sqrt{\ell_{i/2}}})}}{\ell_{i/2}}.\end{align*} $$

If $L_i := \sqrt {\ell _{i/2}} \, \log _2 (\ell _{i/2})$ (which will be chosen to be an integer number), then this quantity equals 1. Therefore, for this choice, $\hat {f}^{-i} d_n^{L_i} c_n d_n^{-L_i} \hat {f}^i$ coincides with $h_{n/2}$ .

4.2 The distortion estimate

The identity $h_{n/2} = \hat {f}^{-i} d_n^{L_i} c_n d_n^{-L_i} \hat {f}^i$ implies that, in the new group $\tilde {\Gamma } := \langle \hat {f}, f, g, \hat {h}, h, \tilde {h}, \psi \rangle $ , we have the estimate

$$ \begin{align*}\lVert h_{n/2} \rVert \leqslant 2\lVert \hat{f}^i\rVert + 2 \, L_i \lVert d_n\rVert+\lVert c_n\rVert\leqslant 2i + 2 \, L_i \, (2i+1+2\lVert f^n\rVert) + 8 \, (1+i+\lVert f^n\rVert).\end{align*} $$

Since $\| f^n \| = O (\log (n)) = O (i)$ , we conclude that

$$ \begin{align*}\lVert h_{n/2} \rVert = O ( i \, \sqrt{\ell_{i/2}} \, \log (\ell_{i/2}) ).\end{align*} $$

Since $h_{n/2} = f^{-n/2} (f^{-1} \bar {f}^{n/2} f) $ and $\| f^{n/2} \| = O (\log (n))$ , this yields

$$ \begin{align*}\| \bar{f}^{n/2} \| = 2 + \| f^{n/2} \| + \| h_{n/2} \| = O ( i \, \sqrt{\ell_{i/2}} \, \log (\ell_{i/2}) ).\end{align*} $$

Notice that the last estimate is much better than what we had in the group $\Gamma $ of the previous section. There, $\| \bar {f}^{n/2}\|$ was of the order $O (i \, \ell _{i/2})$ , so $\bar {f}$ was distorted provided the growth of $\ell _j$ was smaller than exponential. In the new setting, that is, in the modified group $\tilde {\Gamma }$ , the diffeomorphism $\bar {f}$ is distorted whenever the condition below is satisfied (recall that $n = 2^i$ ):

(4.1) $$ \begin{align} \lim_{n \to \infty} \frac{i \, \sqrt{\ell_{i/2}} \, \log (\ell_{i/2})}{2^i} = 0. \end{align} $$

4.3 Checking regularity

We thus choose a new sequence $\ell _i$ so that condition (4.1) holds and $\sqrt {\ell _{i/2}} \, \log _2 (\ell _{i/2})$ is an integer number. This can be achieved for a sequence of type

$$ \begin{align*}\sqrt{\ell_{i/2}}\sim \frac{n}{\log (n)^{3}}\end{align*} $$

that we fix from now on. With such a choice, we claim that $\tilde {\Gamma }$ is a group of $C^{1+\alpha }$ diffeomorphisms. Again, this is obvious for all generators except $h,\hat {h},\tilde {h}$ , and for these three elements the computations are the exact same, because each of the sequences $r_n, s_n, t_n$ is equivalent to $1/ \sqrt {\ell _{i/2}}$ . We thus write everything only for h. Remind estimate (3.10):

$$ \begin{align*}&\frac{ |\!\log Dh (y) - \log Dh (x)| }{|y-x|^{\alpha}}\\ &\quad\leq 2 \, C \, n \log (n) \, (\log(\log(n)))^2 \, t_n \, \bigg[ \frac{C}{ n \log (n) (\log (\log (n)))^2}, \bigg] ^{1-\alpha}.\end{align*} $$

With the new estimate for $t_n$ , this becomes

$$ \begin{align*}&\frac{ |\!\log Dh (y) - \log Dh (x)| }{|y-x|^{\alpha}}\\ &\quad\leq 2 \, C \, n \log (n) \, (\log(\log(n)))^2 \, \frac{\log (n)^{3}}{n} \, \bigg[ \frac{C}{ n \log (n) (\log (\log (n)))^2}, \bigg] ^{1-\alpha}.\end{align*} $$

The expression on the right is of order

$$ \begin{align*}O \bigg( \frac{\log(n)^{3+\alpha} \, (\log \log(n))^{2\alpha}}{n^{1-\alpha}} \bigg),\end{align*} $$

which converges to $0$ as n goes to infinity. This allows us to show that h is a $C^{1+\alpha }$ diffeomorphism as it was done in the previous section.