2.1 Some generalities

The map $\rho $ is then an increasing homeomorphism.

When the foliation is bi-Lipschitz, we will prove that $\rho $ is a bi-Lipschitz homeomorphism (see Proposition 6.1). We recall a well-known result concerning the link between invariant measures and semi-conjugacies for orientation-preserving homeomorphisms of  $\mathbb {T}$ .

Proposition 2.1. Let $g:\mathbb {T}\rightarrow \mathbb {T}$ be an orientation-preserving homeomorphism. Assume that $\mu $ is a non-atomic Borel invariant probability measure by g. Then, if the rotation number $\rho $ of g is irrational or g is $C^0$ conjugate to a rotation, the map $h:\mathbb {T}\rightarrow \mathbb {T}$ defined by $h(\theta )=\int _0^\theta d\mu $ is a semi-conjugacy between g and the rotation of angle $\rho $ , that is,

$$ \begin{align*}h(g(\theta))=h(\theta)+\rho.\end{align*} $$

Proof. Let $\tilde \mu $ be the pullback measure of $\mu $ to $\mathbb {R}$ and let $\tilde g:\mathbb {R}\rightarrow \mathbb {R}$ be a lift of g to $\mathbb {R}$ . Then we have, for every $\Theta \in [0, 1]$ lift of $\theta \in \mathbb {T}$ ,

$$ \begin{align*}\tilde \mu([0, \Theta])=\tilde \mu([\tilde g(0), \tilde g(\Theta)])=\tilde \mu_c([ \lfloor \tilde g(0)\rfloor, \tilde g(\Theta)])-\tilde \mu([\lfloor \tilde g(0)\rfloor, \tilde g(0)]),\end{align*} $$

where $\lfloor \tilde g(0)\rfloor $ is the integer part of $\tilde g(0)$ . This implies

$$ \begin{align*}h(\theta)=h(g(\theta))-\tilde\mu([0, g(0)])=h(g(\theta))-\hat r.\end{align*} $$

(Recall that if $f : \mathbb {T} \to \mathbb {T}$ is an orientation-preserving homeomorphism then either $\rho (f)$ is irrational, f is semi-conjugate (by h) to the rotation $R_{\rho (f)}$ and the only invariant measure is the pullback of the Lebesgue measure by h; or $\rho (f)$ is rational and the ergodic invariant measures are supported on periodic orbits. When $\rho (f)$ is irrational or when f is $C^0$ conjugate to a rotation and $\rho (f)\in [0, 1[$ , then for any invariant measure $\mu $ and $x\in \mathbb {T}$ , $\mu ([x,f(x)[) = \rho (f)$ .) Moreover, as we assumed that $\mu $ is non-atomic, h is continuous.

Before entering the core of the proof, let us mention a useful fact about iterates of $C^0$ -integrable symplectic twist diffeomorphisms.

This is specific to the integrable case: in general, an iterated twist diffeomorphism is not a twist diffeomorphism, as can be seen in the neighborhood of an elliptic fixed point.

Proof. We argue by induction on $n>0$ . The initialization being trivial, let us assume the result true for some $n>0$ . Let $F : \mathbb {R}^2 \to \mathbb {R}^2$ be a lift of f. For any $c\in \mathbb {R}$ , using the notation given at the beginning of §2, we have

$$ \begin{align*}\text{for all } \theta \in \mathbb{T},\quad \text{for all } m>0, \quad f^m(\theta, \eta_c(\theta)) = (g_c^m(\theta), \eta_c\circ g_c^m(\theta)).\end{align*} $$

Observe that if $f^m$ satisfies the twist condition and $c_1<c_2$ are two real numbers, denoting by $\sim $ the lifts of the considered functions, then we have

$$ \begin{align*}\tilde g^m_{c_1}(t)=\pi_1\circ F^m(t, \eta_{c_1}(t))<\pi_1\circ F^m(t, \eta_{c_2}(t))=\tilde g^m_{c_2}(t)\end{align*} $$

and ${\lim _{t\rightarrow \pm \infty }\tilde g_{c_1}(t)=\pm \infty }$ .

Let $c_1<c_2$ and $t\in \mathbb {R}$ . We obtain that

$$ \begin{align*}& \pi_1(F^{n+1}(t,\eta_{c_2}(t))) -\pi_1(F^{n+1}(t,\eta_{c_1}(t)))\\ &\quad = \tilde g_{c_2}\circ \tilde g^{n}_{c_2}(t) -\tilde g_{c_1}\circ \tilde g^{n}_{c_1}(t)> \tilde g_{c_2}\circ \tilde g^{n}_{c_1}(t) -\tilde g_{c_1}\circ \tilde g^{n}_{c_1}(t) ,\end{align*} $$

where we have used the induction hypothesis, $\tilde g^{n}_{c_2}(t)> \tilde g^{n}_{c_1}(t)$ , and the fact that $\tilde g_{c_2}$ is increasing. It follows that $c\mapsto \pi _1(F^{n+1}(t,\eta _{c}(t))) $ is an increasing diffeomorphism onto its image. Observe also that this inequality implies that

$$ \begin{align*}\displaystyle{\lim_{c_2\rightarrow+\infty} \pi_1(F^{n+1}(t,\eta_{c_2}(t))) =+\infty}\end{align*} $$

because ${\lim _{c_2\rightarrow +\infty }\tilde g_{c_2}(s) =+\infty }$ . Moreover,

$$ \begin{align*}& \pi_1(F^{n+1}(t,\eta_{c_2}(t))) -\pi_1(F^{n+1}(t,\eta_{c_1}(t)))\\ &\quad = \tilde g_{c_2}\circ \tilde g^{n}_{c_2}(t) -\tilde g_{c_1}\circ \tilde g^{n}_{c_1}(t) \leq \tilde g_{c_2}\circ \tilde g^{n}_{c_2}(t) -\tilde g_{c_1}\circ \tilde g^{n}_{c_2}(t) \end{align*} $$

implies that ${\lim _{c_1\rightarrow -\infty } \pi _1(F^{n+1}(t,\eta _{c_1}(t))) =-\infty }$ because ${\lim _{c_1\rightarrow -\infty }\tilde g_{c_1}(s) =-\infty }$ . So finally $c\mapsto \pi _1(F^{n+1}(t,\eta _{c}(t))) $ is an increasing diffeomorphism onto $\mathbb {R}$ .

2.2 Differentiability and conjugacy along rational curves

It is proved in [Reference Arnaud1] that for every $r={p}/{q}\in \mathbb {Q}$ , $\eta _c=\eta _{\rho ^{-1}(r)}$ is $C^k$ and the restriction of f to the graph $\Gamma _c$ of $\eta _c$ is completely periodic: $f^q_{\vert \Gamma _c}={\textrm {Id}}_{\Gamma _c}$ . Moreover, along these particular curves, the two Green bundles (see Appendix B for definition and results) are equal:

$$ \begin{align*}G_-(\theta, \eta_c(\theta))=G_+(\theta, \eta_c(\theta)).\end{align*} $$

Proof of the first point of Theorem 2.1. We fix $A\in \mathbb {R}$ such that $\rho (A)={p}/{q}\in \mathbb {Q}$ . Replacing f by $f^q$ , we can assume that $\rho (A)\in \mathbb {Z}$ . Observe that because of the $C^0$ -integrability of f, $f^q$ is also a $C^k$ symplectic twist diffeomorphism that is $C^0$ -integrable with the same invariant foliation (Proposition 2.2).

We define $G_A: \mathbb {A}\rightarrow \mathbb {A}$ by

(1) $$ \begin{align} G_A(\theta, r)=(\theta, r+\eta_A(\theta)).\end{align} $$

Then $G_A^{-1}\circ f^q\circ G_A$ is also a $C^0$ -integrable $C^k$ symplectic twist diffeomorphism and $\mathbb {T}\times \{ 0\}$ is filled with fixed points.

We finally have to prove our theorem in this case and we use the notation f instead of $G_A^{-1}\circ f^q\circ G_A$ . We can assume that $\rho (A)=0$ instead of $\rho (A)\in \mathbb {Z}$ .

If $\varepsilon>0$ is arbitrarily small, because of the semi-continuity of the two Green bundles $G_-=\mathbb {R}(1, s_-)$ and $G_+=\mathbb {R}(1, s_+)$ , we have that for any point $x=(\theta , r)$ sufficiently close to $\mathbb {T}\times \{ 0\}$ : $\max \{\lvert s_-(x)\rvert , \lvert s_+(x)\rvert \} <\varepsilon $ is small.

Now we fix c small and consider for every $\theta \in \mathbb {T}$ the small triangular domain ${\mathcal T}(\theta )$ that is delimited by the following three curves:

• • the graph of $\eta _c$ ;

• • the vertical ${\mathcal V}_\theta =\{\theta \}\times \mathbb {R}$ ;

• • the image $f({\mathcal V}_\theta )$ of the vertical at $\theta $ .

To be more precise, ${\mathcal T}(\theta )$ is ‘semi-open’ in the following sense; it contains its entire boundary except for the image $f({\mathcal V}_\theta )$ of the vertical at $\theta $ .

We assume that $c>0$ . The case $c<0$ is similar.

As the slope of $\eta _c$ is almost $0$ (because between the slope of the two Green bundles; see Proposition B.1) and the slope of the side of the triangle that is in $f({\mathcal V}_\theta )$ is almost ${1}/{s(\theta )}$ where $s(\theta )>0$ is the torsion that is defined by

(2) $$ \begin{align} Df(\theta, 0)=\begin{pmatrix} 1&s(\theta)\\ 0& 1\end{pmatrix},\end{align} $$

the area of this triangle is

(3) $$ \begin{align}\lambda({\mathcal T}(\theta))=\tfrac{1}{2}( \eta_c(\theta))^2( s(\theta)+\varepsilon (\theta, c)),\end{align} $$

where

$$ \begin{align*}\text{uniformly for }\ \theta \in \mathbb{T},\quad \lim_{c\rightarrow 0}\varepsilon (\theta, c)=0.\end{align*} $$

Let $\lambda $ be the Lebesgue measure restricted to the invariant sub-annulus

$$ \begin{align*}{\mathcal A}_c=\bigcup_{\theta\in\mathbb{T}} \{ \theta\}\times [0, \eta_c(\theta)].\end{align*} $$

Being symplectic, f preserves $\lambda $ . Moreover, every ergodic measure $\mu $ for f with support in ${\mathcal A}_c$ is supported on one curve $\Gamma _A$ with $A\in [0, c]$ . But $f_{\lvert \Gamma _A}$ is semi-conjugate to a rotation with an angle $\rho (A)$ that is in $[0, 1[$ . Hence every interval in $\Gamma _A$ that is between some $(\theta , \eta _A(\theta ))$ and $f(\theta , \eta _A(\theta ))$ has the same $\mu $ -measure, which is just given by the rotation number $\rho (A)\in [0, 1]$ on the graph of $\eta _A$ . This implies that $\theta \mapsto \mu ({\mathcal T}(\theta ))$ is constant. Hence for every $\theta , \theta '\in \mathbb {T}$ and for every ergodic measure $\mu $ with support in ${\mathcal A}_c$ , we have $\mu ({\mathcal T}(\theta ))=\mu ({\mathcal T}(\theta '))$ .

We now use the ergodic decomposition of invariant measures (see, for example, [Reference Mañé14]) applied to the Lebesgue measure restricted to ${\mathcal A}_c$ . Let $\delta _x$ be the notation for the Dirac measure at x. For every $a\in \mathbb {A}$ , we denote ${\lim _{n\rightarrow \infty }({1}/{n})\sum _{k=1}^{n-1}\delta _{f^n(a)}}$ by $\lambda _a$ (such a limit exists and is ergodic as all Birkhoff averages converge to an ergodic measure for orientation-preserving circle homeomorphisms). Then we have $\lambda =\int _{{\mathcal A}_c}\!\lambda _a d\lambda (a)$ . Denoting the rotation number of a point $a\in \mathbb {A}$ by ${\mathcal R}(a)$ , we deduce that,

(4) $$ \begin{align}\text{for all } \theta, \theta'\in \mathbb{T},\quad \lambda({\mathcal T}(\theta))=\lambda({\mathcal T}(\theta'))=\int_{{\mathcal A}_c}\!\! {\mathcal R}(a)\, d\lambda(a).\end{align} $$

This last inequality is again a consequence of the fact that for all $a\in {\mathcal A}_c$ , $\lambda _a({\mathcal T}(\theta )) = {\mathcal R}(a)$ as explained above.

We deduce from equation (3) that

$$ \begin{align*}\text{uniformly for }\theta, \theta'\in\mathbb{T},\quad \lim_{c\rightarrow 0} \frac{\eta_c(\theta')}{\eta_c(\theta)}=\sqrt{\frac{s(\theta)}{s(\theta')}}.\end{align*} $$

Integrating with respect to $\theta '$ and recalling that $\int _{\mathbb {T}} \eta _c = c$ , we deduce that uniformly in $\theta $ , we have

$$ \begin{align*}\lim_{c\rightarrow 0} \frac{c}{\eta_c(\theta)}=\sqrt{s(\theta)}\int_{\mathbb{T}} \frac{dt}{\sqrt{s(t)}}.\end{align*} $$

This implies that

(5) $$ \begin{align} \frac{\partial \eta_c(\theta)}{\partial c}_{\lvert c=0} =\bigg( {\int_{\mathbb{T}} \frac{dt}{\sqrt{s(t)}}}\bigg)^{-1}\frac{1}{\sqrt{s(\theta)}}, \end{align} $$

and even

(6) $$ \begin{align} \eta_c(\theta)=c\bigg( {\int_{\mathbb{T}} \frac{dt}{\sqrt{s(t)}}}\bigg)^{-1}\bigg(\frac{1}{\sqrt{s(\theta)}}+\varepsilon(\theta,c)\bigg), \end{align} $$

where

$$ \begin{align*}\text{uniformly for }\theta \in \mathbb{T},\quad \lim_{c\rightarrow 0}\varepsilon (\theta, c)=0.\end{align*} $$

Observe that $ {\partial \eta _c}/{\partial c}=( \int _{\mathbb {T}} ({dt}/{\sqrt {s(t)}}))^{-1}({1}/{\sqrt {s(\cdot )}})$ is a $C^{k-1}$ function of $\theta $ . This proves the first point of Theorem 2.1. $\square $

Proof of the second point of Theorem 2.1. We deduce from the first point that for any c such that $\rho (c)$ is rational, the function ${\partial \eta _c}/{\partial c} $ is continuous and positive. Moreover, its integral on $\mathbb {T}$ is 1. Hence ${\partial \eta _c}/{\partial c} $ is the density of a Borel probability measure that is equivalent to Lebesgue. We now introduce some notation.

Then f preserves $\Lambda _{c, c'}$ . Observe that for any measurable $I\subset \mathbb {T}$ , we have

(7) $$ \begin{align}\Lambda_{c, c'} (\{ (\theta, r); \theta \in I, \ r\in [\eta_c(\theta), \eta_{c'}(\theta)]\})=\frac{1}{c-c'}\int_I (\eta_c(\theta)-\eta_{c'}(\theta))\, d\theta. \end{align} $$

Lemma 2.1. If $\rho (c)$ is rational, then ${\lim _{c'\rightarrow c}\Lambda _{c, c'}}$ is a measure supported on the graph of $\eta _c$ whose projected measure $\mu _c$ has density ${\partial \eta _c}/{\partial c}$ with respect to the Lebesgue measure of $\mathbb {T}$ . Hence if $h_c(\theta )=\int _0^\theta ({\partial \eta _c}/{\partial c})(t)\,dt$ , we have

$$ \begin{align*}h_c\circ \pi_1\circ f(\theta, \eta_c(\theta))=h_c(\theta)+\rho(c).\end{align*} $$

Proof. Using equation (6), we can take the limit in equation (7), or more precisely for any $\psi \in C^0(\mathbb {A}, \mathbb {R})$ in

$$ \begin{align*}\int \psi(\theta, r)\, d\Lambda_{c, c'}(\theta, r)=\int_{\mathbb{T}}\frac{1}{c-c'}\bigg( {\int_{\eta_{c'}(\theta)}^{\eta_c(\theta)}\psi(\theta, r)\,dr}\bigg)\, d\theta,\end{align*} $$

and obtain that the limit is an invariant measure supported in the graph of $\eta _c$ whose projected measure $\mu _c$ has a density with respect to Lebesgue (hence has no atoms) that is equal to ${\partial \eta _c}/{\partial c}$ . Moreover, by [Reference Arnaud1] (this fact was recalled at the beginning of §2.2), $g_c$ is $C^0$ conjugate to a rotation. We then use Proposition 2.1 to conclude that $h_c$ is the desired conjugacy.

Proof of the third point of Theorem 2.1. We notice that when $\rho (c)$ is irrational, there is only one invariant Borel probability measure that is supported on the graph of $\eta _c$ . This implies the continuity of the map $c\mapsto \mu _c$ at such a c. Let us look at what happens when $\rho (c)$ is rational.

Proposition 2.3. For every $c_0\in \mathbb {R}$ such that $\rho (c_0)$ is rational, for every $\theta \in [0, 1]$ , we have

$$ \begin{align*}\lim_{c\rightarrow c_0} \mu_c([0, \theta])=\mu_{c_0}([0, \theta])\end{align*} $$

and the limit is uniform in $\theta $ .

Together with the continuity of $h_{c_0}$ , this implies the continuity of $(\theta , c)\mapsto h_c(\theta )$ at $(\theta , c_0)$ .

Proof. In this proof, we will use different functions $\varepsilon _i(\tau , c)$ and, uniformly in $\tau $ , all these functions will satisfy

$$ \begin{align*}\lim_{c\rightarrow 0}\varepsilon_i(\tau, c)=0.\end{align*} $$

As in the proof of the first point of Theorem 2.1, we can assume that $\eta _{c_0}=0$ (and then $c_0=0$ ) and $\rho (0)=0$ .

We fix $\varepsilon>0$ . Because of the continuity of $\rho $ , we can choose $\alpha $ such that if $\lvert c\rvert <\alpha $ , then $\lvert \rho (c)\rvert <\varepsilon $ .

Let us introduce the notation $N_c=\lfloor {1}/{\rho (c)}\rfloor $ for $c\not =0$ . Let us assume that $c>0$ and $\theta \in (0, 1]$ . We also denote by $ \tilde g_c : \mathbb {R} \to \mathbb {R}$ the lift of $g_c$ such that $ \tilde g_c(0)\in [0, 1)$ and by $M_c(\theta )$ ,

$$ \begin{align*}M_c(\theta)=\sharp \{ j\in \mathbb{N};\ \tilde g _{c}^j(0)\in [0, \theta]\}.\end{align*} $$

Hence, $M_c(\theta )$ is the number of points of the orbit of $0$ under $\tilde g_c$ that belong to $[0, \theta ]$ . Observe that $M_c(\theta )$ is non-decreasing with respect to $\theta $ .

As $\eta _c>0$ , any primitive ${\mathcal N}_c$ of $\eta _c$ is increasing, hence $M_c(\theta )$ is also the number of $\tilde g^k(0)$ such that ${\mathcal N}_c(\tilde g^k(0))$ belongs to $[{\mathcal N}_c(0), {\mathcal N}_c(\theta )]$ , that is,

(8) $$ \begin{align} M_c(\theta)&=\sharp \bigg\{\displaystyle j\in \mathbb{N};\ \int_0^{\tilde g_{c}^j(0)}\eta_{c}(t)\,dt\leq \int_0^{\theta}\eta_{c}(t)\,dt\bigg\}\nonumber\\ & =\sup\bigg\{\displaystyle j\in \mathbb{N};\ \int_0^{\tilde g_{c}^j(0)}\eta_{c}(t)\,dt\leq \int_0^{\theta}\eta_{c}(t)\,dt\bigg\}. \end{align} $$

Note that $M_c(1)=N_c$ because $g_c$ has rotation number $\rho (c)$ and that we have, for all $\theta \in (0, 1]$ , $M_c(\theta )\leq N_c$ as $M_c$ is non-decreasing. We also have

$$ \begin{align*}\mu_c([0, \theta])=\sum_{j=0}^{M_c(\theta)-1}\mu_c([\tilde g_c^j(0), \tilde g_c^{j+1}(0)[)+\mu_c([\tilde g^{M_c(\theta)}(0), \theta])\end{align*} $$

and thus $\mu _c([0, \theta ])=M_c(\theta )\rho (c) +\Delta \rho (c)$ with $\Delta \in [0, 1]$ because $[\tilde g^{M_c(\theta )}(0), \theta ]\subset [\tilde g^{M_c(\theta )}(0), \tilde g^{M_c(\theta )+1}(0)]$ .

Hence

(9) $$ \begin{align}\mu_c([0, \theta])\in [ M_c(\theta)\rho(c), M_c(\theta)\rho(c)+\rho(c)]\subset \bigg[ \frac{M_c(\theta)}{N_c+1}, \frac{M_c(\theta)+1}{N_c}\bigg].\end{align} $$

Hence to estimate the measure $\mu _c([0, \theta ])$ we need a good estimate of the number of j such that $\tilde g_{c}^j(0)$ belongs to $[0, \theta ]$ . We have proved in equation (6) that

(10) $$ \begin{align} \eta_{c}(\tau)=\bigg( {\int_{\mathbb{T}} \frac{dt}{\sqrt{s(t)}}}\bigg)^{-1}\frac{c(1+\varepsilon_0(\tau,c))}{\sqrt{s(\tau)}}. \end{align} $$

We deduce from equation (2) that $\tilde g_{c}(\tau )= \tau +(s(\tau )+\varepsilon _1(\tau , c))\eta _c(\tau )$ where, uniformly in $\tau $ , we have ${\lim _{c\rightarrow 0}\varepsilon _1(\tau , c)=0}$ and then, by equation (10),

(11) $$ \begin{align}\int_\tau^{\tilde g_{c}(\tau)}\eta_{c}(t)\,dt=\eta_c(\tau)^2(s(\tau)+\varepsilon_2(\tau, c)) =\frac{c^2(1+\varepsilon_3(\tau, c))}{\big(\!\int_{\mathbb{T}} ({dt}/{\sqrt{s(t)}})\big)^2}. \end{align} $$

This says that the area that is delimited by the zero section, the graph of $\eta _{c}$ and the vertical lines between ${\mathcal V}_\tau $ and ${\mathcal V}_{\tilde g_c(\tau )}$ is almost constant (that is, does not depend much on $\tau $ ).

We deduce from equation (8) that

$$ \begin{align*}\int_0^{\tilde g_c^{M_c(\theta)}(0)}\eta_c(t)\,dt\leq\int_0^\theta\eta_c(t)\,dt<\int_0^{\tilde g_c^{M_c(\theta)+1}(0)}\eta_c(t)\,dt.\end{align*} $$

Hence

$$ \begin{align*}\sum_{j=0}^{M_c(\theta)-1}\int_{\tilde g^j(0)}^{\tilde g^{j+1}(0)}\eta_c(t)\,dt\leq \int_0^\theta\eta_c(t)\,dt\leq \sum_{j=0}^{M_c(\theta)}\int_{\tilde g^j(0)}^{\tilde g^{j+1}(0)}\eta_c(t)\,dt.\end{align*} $$

Using equation (11), we deduce that

$$ \begin{align*}M_c(\theta)\frac{ c^2(1+\varepsilon_4(\theta, c))}{\big(\!\int_{\mathbb{T}}({dt}/{\sqrt{s(t)}})\big)^2}\leq\frac{c(1+\varepsilon_5(\theta, c))}{\int_{\mathbb{T}}({dt}/{\sqrt{s(t)}})}\int_0^\theta \frac{dt}{\sqrt{ s(t)}} <(M_c(\theta)+1)\frac{ c^2(1+\varepsilon_6(\theta, c))}{\big(\!\int_{\mathbb{T}}({dt}/{\sqrt{s(t)}})\big)^2}, \end{align*} $$

and then

(12) $$ \begin{align}M_c (\theta) =\bigg\lfloor \frac{1}{c}\bigg(\bigg( {\int_{\mathbb{T}} \frac{du}{\sqrt{s(u)}}}\bigg)\bigg( {\int_0^{\theta}\frac{dt}{\sqrt{s(t)}}\bigg)+\varepsilon_7(\theta, c)}\bigg)\bigg\rfloor.\end{align} $$

This implies that

(13) $$ \begin{align}N_c=M_c(1)=\bigg\lfloor \frac{1}{c}\bigg(\bigg({\int_{\mathbb{T}} \frac{dt}{\sqrt{s(t)}}}\bigg)^2+\varepsilon_8(1, c)\bigg)\bigg\rfloor,\end{align} $$

and by equations (9), (12) and (13),

(14) $$ \begin{align}\mu_c([0, \theta])=\frac{M_c(\theta)}{N_c}+\varepsilon_9(\theta, c)= \frac{\int_0^{\theta}({dt}/{\sqrt{s(t)}})}{\int_{\mathbb{T}} ({dt}/{\sqrt{s(t)}})}+\varepsilon_{10}(\theta, c)=\mu_0([0,\theta])+\varepsilon_{11}(\theta, c).\end{align} $$

As none of the measures $\mu _c$ has atoms, this implies that $c\mapsto \mu _c$ and all the maps $c\mapsto \mu _c([0, \theta ])=h_c(\theta )$ are continuous. As every map $h_c$ is non-decreasing in the variable $\theta $ , we deduce from the Dini–Polyá theorem [Reference Rudin19, Exercise 13.b, p. 167] that $c\mapsto h_c$ is continuous for the $C^0$ uniform topology.

Remark. If $\rho (A)={p}/{q}$ , then we proved that

$$ \begin{align*}({\partial \eta_c(\theta)}/{\partial c})_{\vert c=A}\! =\! \bigg( {\int_{\mathbb{T}} \frac{dt}{\sqrt{s_q(t, \eta_A(t))}}}\bigg)^{-1} \bigg(\frac{1}{\sqrt{s_q(\theta, \eta_A(\theta))}}\bigg),\end{align*} $$

where

$$ \begin{align*}Df^q(x)=\begin{pmatrix} a_q(x)&s_q(x)\\ c_q(x)&d_q(x) \end{pmatrix}.\end{align*} $$

Indeed, the term $s_q(t, \eta _c(t))$ does not change when we conjugate by the map $G_A$ where $G_A(\theta , r)=(\theta , r+\eta _A(\theta ))$ as we did in §2.2.

This gives for the conjugacy

$$ \begin{align*}h_A(\theta)=\mu_A([0, \theta])=\bigg( {\int_{\mathbb{T}} \frac{dt}{\sqrt{s_q(t, \eta_A(t))}}}\bigg)^{-1}\int_0^\theta \frac{1}{\sqrt{s_q(t, \eta_A(t))}}\,dt.\end{align*} $$

Observe that this $C^k$ depends on $\theta $ .

Observe too that equation (6) can be rewritten as

(15) $$ \begin{align}\eta_c(\theta)=\eta_A(\theta)+(c-A)\bigg[\bigg( {\int_{\mathbb{T}} \frac{dt}{\sqrt{s_q(t, \eta_A(t))}}}\bigg)^{-1}\frac{1}{\sqrt{s_q(\theta, \eta_A(\theta))}}+\varepsilon(\theta,c)\bigg], \end{align} $$

where

$$ \begin{align*}{\textrm{uniformly}}\ {\textrm{for}}\ \theta \in \mathbb{T}, \quad \lim_{c\rightarrow A}\varepsilon (\theta, c)=0.\end{align*} $$

Observe that the formula does not give any continuous dependence of $h_c$ or ${\partial \eta _c}/{\partial c}$ in the c variable, because q can become very large when c changes.

2.3 Generating function and regularity

To finish the proof of Theorem 1.1, we have to prove that u admits a derivative with respect to c everywhere and that

$$ \begin{align*}\text{for all } \theta\in\mathbb{T}, \text{ for all } c\in\mathbb{R} ,\quad h_c(\theta)=\theta+\frac{\partial u}{\partial c}(\theta, c).\end{align*} $$

Because we proved that $(\theta , c)\mapsto h_c(\theta )$ is continuous, we will deduce that u is $C^1$ .

Observe that for every $\theta $ , the map $c\mapsto u(\theta , c)+c\theta $ is increasing because every $ c\mapsto \eta _c(\theta )$ is increasing.

Let us restate Theorem 1.1 with the new notation.

Proof. The first point is a consequence of the definition of u.

Then $u(\cdot ,c)$ and ${\partial u}/{\partial \theta }=\eta _{c}-c$ continuously depend on $(\theta , c)$ .

Observe that with the notation (7), we have

$$ \begin{align*} &\Lambda_{c, c'} (\{ (\theta, r); \ \theta \in [\theta_1, \theta_2], r\in [\eta_{c}(\theta), \eta_{c'}(\theta)]\}) \\ &\quad=\frac{1}{c'-c}( (u(\theta_2, c')-u(\theta_1, c'))-(u(\theta_2,c)-u(\theta_1,c)) )+(\theta_2-\theta_1). \end{align*} $$

Moreover, if $\rho (c_0)\in \mathbb {Q}$ , we deduce from Lemma 2.1 that $u(\cdot , c)$ admits a derivative with respect to c at $c_0$ ,

$$ \begin{align*}\frac{\partial u}{\partial c}(\theta, c_0)=\lim_{c\rightarrow c_0}\frac{1}{c-c_0}((u(\theta,c)-u(0,c))-(u(\theta,c_0)-u(0, c_0))),\end{align*} $$

that is given by

$$ \begin{align*}\frac{\partial u}{\partial c}(\theta, c_0)=\mu_{c_0}([0, \theta])-\theta=h_{c_0}(\theta)-\theta\end{align*} $$

and this derivative continuously depends on $\theta $ .

Assume now that $\rho (c_0)$ is irrational and let c tend to $c_0$ . Every limit point of $\Lambda _{c, c_0}$ when c tends to $c_0$ is a Borel probability measure that is invariant by f and supported on the graph of $\eta _{ c_0}$ . As there exists only one such measure, whose projection was denoted by $\mu _{c_0}$ , we deduce that

$$ \begin{align*}{ \pi_{1*}}(\lim_{c\rightarrow c_0}\Lambda_{c, c_0}) =\mu_{c_0}.\end{align*} $$

As $\mu _{c_0}$ has no atom, we have, for all $\theta _0\in [0, 1)$ ,

$$ \begin{align*} h_{c_0}(\theta_0)&=\ \mu_{c_0}([0, \theta_0]) =\lim_{c\rightarrow c_0}\Lambda_{c_0, c} (\{ (\theta, r); \theta \in [0, \theta_0], r\in [\eta_{c_0}(\theta), \eta_{c}(\theta)]\}) \\ & =\lim_{c\rightarrow c_0}\frac{1}{c-c_0}((u(\theta_0,c)-u(0,c))-(u(\theta_0, c_0)-u(0,c_0)))+\theta_0 \\ &=\frac{\partial u}{\partial c}(\theta_0,c_0)+\theta_{0}. \end{align*} $$

Let us explain the above equality. If $\varepsilon>0$ , as $\mu _{c_0}$ has no atoms, there exist two continuous functions $\chi ^\pm $ such that $\chi ^-\leq \chi _{[0,\theta ]}\leq \chi ^+$ and $\int _{\mathbb {T}} (\chi ^+-\chi ^-)\,d \mu _{c_0} <\varepsilon $ . Setting $\nu _c = { \pi _{1*}}(\Lambda _{c, c_0})$ , we have $\int _{\mathbb {T}} \chi ^- \,d\nu _c \leq \nu _c([0,\theta ]) \leq \int _{\mathbb {T}} \chi ^+ \,d\nu _c $ . As $\int _{\mathbb {T}} \chi ^\pm \, d\nu _c \to \int _{\mathbb {T}} \chi ^\pm \, d\mu _{c_0}$ when $c\to c_0$ , we conclude that $\lvert \nu _c([0,\theta ]) - \mu _{c_0}([0,\theta ])\rvert <2\varepsilon $ for c close to $c_0$ .

Hence u admits a derivative with respect to c and

$$ \begin{align*}h_{c_0}(\theta)=\mu_{c_0}([0, \theta])=\theta+\frac{\partial u}{\partial c}(\theta, c_0).\end{align*} $$

Because of Theorem 2.1, $(\theta , c)\mapsto ({\partial u}/{\partial c})(\theta ,c)=h_c(\theta )-\theta $ is continuous. As the two partial derivatives ${\partial u}/{\partial \theta }$ and ${\partial u}/{\partial c}$ are continuous in $(\theta , c)$ , we conclude that u is  $C^1$ . This finishes the proof of Theorem 1.1.

4.1 Proof of the first implication

Let $\Phi :{\mathcal U}\rightarrow {\mathcal V}$ be an exact symplectic homeomorphism that maps the standard horizontal foliation onto a foliation ${\mathcal F}$ that is transverse to the vertical one and that preserves the orientation of the leaves. (When ${\mathcal V}=\mathbb {A}$ , the exactness condition implies that $\Phi $ maps the circle $\mathbb {T}\times \{c\}$ to the graph of $\eta _c$ . In the other cases, as the domains are simply connected, the exactness condition is empty.) Assume that $\Phi $ extends to a homeomorphism $\overline {\Phi } : \overline {{\mathcal U}}\to \overline {{\mathcal V}}$ and that $\overline {\Phi }(\partial ^\pm {\mathcal U})=\partial ^\pm {\mathcal V}$ . Then the endpoints of the leaves of ${\mathcal F}$ are not in $\partial ^\pm {\mathcal V}$ . As the foliation ${\mathcal F}$ is transverse to the ‘vertical’ foliation ${\mathcal G}_0$ into $\pi _1^{-1}(\{\theta \})$ for $\theta \in \pi _1({\mathcal V})$ , all the leaves of ${\mathcal F}$ are graphs above $(\alpha , \beta )$ .

Another result of the transversality of ${\mathcal F}$ and the vertical foliation ${\mathcal G}_0$ is that the foliation ${\mathcal G}=\Phi ^{-1}({\mathcal G}_0)$ is a foliation of ${\mathcal U}$ that is transverse to the standard (‘horizontal’) foliation ${\mathcal F}_0=\Phi ^{-1}({\mathcal F})$ . This means precisely that the foliation ${\mathcal G}$ is a foliation by graphs of maps $\zeta _\theta : I=\pi _2({\mathcal U})\rightarrow \pi _1({\mathcal U})$ . Hence there exists a continuous function $v:{\mathcal W}\rightarrow \mathbb {R}$ that admits a continuous derivative with respect to r such that the foliation ${\mathcal G}$ is the foliation by graphs $\Phi ^{-1}(\pi _1^{-1}(\{\theta \}))$ of $\zeta _\theta : r\mapsto \theta +({\partial v}/{\partial r})(\theta , r)$ . Observe that by definition of $\zeta _\theta $ , we have $\Phi (\zeta _\theta (c), c)=(\theta ,\eta _c(\theta ))$ . As a result, every map $\theta \mapsto \zeta _\theta (c)$ is a homeomorphism onto its image.

We now use the preservation of the area. We fix $\theta _1<\theta _2$ in $\pi _1({\mathcal V})$ and $c_1<c_2$ in $\pi _2({\mathcal U})$ such that the domain $D_1$ delimited by the horizontals $\pi _2^{-1}( \{ c_1\})$ , $\pi _2^{-1}( \{ c_2\})$ , the graph of $c\in [c_1, c_2]\mapsto \zeta _{\theta _1}(c)$ and the graph of $c\in [c_1, c_2]\mapsto \zeta _{\theta _2}(c)$ is contained in ${\mathcal U}$ . Because $\Phi $ is a symplectic homeomorphism, $\Phi $ preserves the area and so $D_1$ and $\Phi (D_1)$ have the same area. Observe that $\Phi (D_1)$ is the domain delimited by the graphs of $\eta _{c_1}$ , $\eta _{c_2}$ and the verticals $\pi _1^{-1}(\{\theta _1\})$ and $\pi _1^{-1}(\{ \theta _2\})$ . This can be written

$$ \begin{align*} &\int_{c_1}^{c_2}\bigg( \bigg(\theta_2+\frac{\partial v}{\partial c} (\theta_2,c)\bigg)-\bigg(\theta_1+\frac{\partial v}{\partial c} (\theta_1,c)\bigg)\bigg)\, dc\\ &\quad=\int_{\theta_1}^{\theta_2}\bigg(\bigg(c_2+\frac{\partial u}{\partial \theta} (\theta,c_2)\bigg) - \bigg(c_1+\frac{\partial u}{\partial \theta} (\theta,c_1)\bigg)\bigg)\,d\theta .\end{align*} $$

It follows that

$$ \begin{align*} &u(\theta_2,c_2)-u(\theta_1,c_2)-u(\theta_2,c_1)+u(\theta_1,c_1) \\ &\quad=v(\theta_2,c_2) -v(\theta_2,c_1) -v(\theta_1,c_2) +v(\theta_1,c_1). \end{align*} $$

Evaluating for $\theta _1=0$ we find

$$ \begin{align*}u(\theta_2,c_2)-u(\theta_2,c_1) = v(\theta_2,c_2) -v(\theta_2,c_1) -v(0,c_2) +v(0,c_1).\end{align*} $$

Finally, as v admits a continuous partial derivative with respect to c, we conclude that $({\partial u}/{\partial c})(\theta , c) = ({\partial v}/{\partial c})(\theta , c)-({\partial v}/{\partial c})(0, c)$ exists and is continuous. Hence u is $C^1$ . Moreover, every map $\theta \mapsto \theta +({\partial u}/{\partial c})(\theta , c)= \zeta _c(\theta )-({\partial v}/{\partial c})(0, c)$ is a homeomorphism onto its image and we have established the first implication.

4.2 Proof of the second implication

We assume that there exists a $C^1$ map $u : {{\mathcal W}} \to \mathbb {R}$ such that:

• • $u(0,c) = 0$ for all $c\in I$ where $I=(c_-, c_+)$ or $I=\mathbb {R}$ ;

• • $\eta _c(\theta ) = c+ ({\partial u}/{\partial \theta })(\theta ,c)$ for all $(\theta , c) \in {\mathcal W}$ ;

• • for all $c\in I$ , the map $h_c :\theta \mapsto \theta + ({\partial u}/{\partial c})(\theta , c)$ is increasing.

Then we can define a unique homeomorphism $\Phi $ by

$$ \begin{align*}\Phi\bigg(\theta + \frac{\partial u}{\partial c}(\theta , c), c\bigg)=\bigg(\theta, c+ \frac{\partial u}{\partial \theta}(\theta,c)\bigg).\end{align*} $$

We first prove the result when ${\mathcal U}=\mathbb {A}$ . Let $v:\mathbb {R}^2\rightarrow \mathbb {R}_+$ be the $C^\infty $ function with support in $B(0, 1)$ defined by $v(\theta , c) = a \exp ((1-\|(\theta , c)\|)^{-2})$ for $(\theta , c) \in B(0,1)$ and where a is such that $\int \!v=1$ . We denote by $v_\varepsilon $ the function $v_\varepsilon (x)=({1}/{\varepsilon ^2})v({x}/{\varepsilon })$ .

For every $\varepsilon>0$ , we define

(18) $$ \begin{align}U_\varepsilon(\theta, c)=(u*v_\varepsilon) (\theta, c),\end{align} $$

where we recall the convolution formula

$$ \begin{align*}u*v(x)= \int u(x-y)v(y)\,dy.\end{align*} $$

Then when $\varepsilon $ tends to $0$ , the functions $U_\varepsilon $ tend to U in the $C^1$ compact-open topology. Moreover, $U_\varepsilon $ is $1$ -periodic in $\theta $ and smooth. Observe that for every $\theta $ , the function $c\mapsto c+({\partial u}/{\partial \theta })(\theta ,c)$ is increasing. We deduce that the convolution $c\mapsto c+({\partial U_\varepsilon }/{\partial \theta })(\theta ,c)$ is a $C^\infty $ diffeomorphism as it is a mean of $C^\infty $ diffeomorphisms thanks to Lemma 4.1. Finally, the maps $F_\varepsilon : (\theta , c)\mapsto (\theta , c+({\partial U_\varepsilon }/{\partial \theta })(\theta , c))$ define $C^\infty $ foliations that converge to the initial foliation $F_0: (\theta , c)\mapsto (\theta , c+({\partial u}/{\partial \theta })(\theta , c))$ for the $C^0$ compact-open topology when $\varepsilon $ tends to $0$ .

Observe that the $h_c : \theta \mapsto \theta +({\partial u}/{\partial c})(\theta , c)$ are assumed to be increasing. We deduce that the maps $G_\varepsilon : (\theta , c)\mapsto (\theta +({\partial U_\varepsilon }/{\partial c})(\theta ,c), c)$ are $C^\infty $ diffeomorphisms of $\mathbb {A}$ that converge for the $C^0$ compact-open topology to $G_0:(\theta , c)\mapsto (\theta +({\partial u}/{\partial c})(\theta , c), c)$ .

Finally, the maps ${\mathcal H}_\varepsilon = F_\varepsilon \circ G_\varepsilon ^{-1}$ are $C^\infty $ diffeomorphisms of $\mathbb {A}$ that converge for the $C^0$ compact-open topology to $F_0\circ G_0^{-1}=\Phi $ . Moreover, because they are defined via generating functions $U_\varepsilon $ in the classical sense, we know that the diffeomorphisms ${\mathcal H}_\varepsilon $ are all exact symplectic.

This means precisely that $\Phi $ is an exact symplectic homeomorphism.

We now deal with the local case when ${\mathcal U}\neq \mathbb {A}$ . The computation from the first implication (with $u=v$ ) shows that $\Phi $ is area preserving. We conclude with the next proposition.

Proof. We only prove the first point, the second one being very similar. As f is smooth and compactly supported, $f*g$ is well defined and smooth. As f is even, $f'$ is odd. Just notice that

$$ \begin{align*}(f*g)'(x) = \int_{\mathbb{R}} f'(y)g(x-y)\, dy = \int_0^{+\infty} f'(y) (g(x-y)-g(x+y))\, dy.\end{align*} $$

Then $f*g$ is an increasing diffeomorphism onto its image as $g(x-y)-g(x+y) <0$ and $f'(y) \leq 0$ and $f'$ is not the zero function.

Let us now assume that $\lim _{x\to +\infty } g(x) = +\infty $ . We will prove that $\lim _{x\to +\infty } f*g(x) = +\infty $ . Let $M>0$ such that the support of f is included in $[-M,M]$ . Let $A>0$ and $B\in \mathbb {R}$ such that $A< g(x)$ as soon as $x>B$ . Then if $x>M+B$ ,

$$ \begin{align*}f*g(x) = \int_{-M}^M f(y)g(x-y)\,dy>A\int_{\mathbb{R}} f(y)\,dy.\end{align*} $$

This proves our result as $\int _{\mathbb {R}} f(y)\,dy>0$ .

5.1 Proof of Proposition 1.1: bi-Lipschitz foliations with $C^1$ generating functions are straightenable

Let $u:\mathbb {A}\rightarrow \mathbb {R}$ be the $C^1$ generating function of a continuous foliation of $\mathbb {A}$ into graphs. We recall the following result due to Minguzzi [Reference Minguzzi16].

5.1.1 Proof of the first implication

We assume that the invariant foliation is K-Lipschitz on a compact ${\mathcal K} =\{ (\theta , \eta _c(\theta )); \theta \in \mathbb {T}, c\in [a, b]\}$ , which means

(19) $$ \begin{align} \text{for all } \theta \in \mathbb{T},\text{ for all } c_1, c_2\in [a,b], \quad \frac{\lvert c_1-c_2\rvert}{K}\leq \lvert\eta_{c_1}(\theta)-\eta_{c_2}(\theta)\rvert\leq K\lvert c_1-c_2\rvert. \end{align} $$

As $\eta _c(\theta )=c+({\partial u}/{\partial \theta })(\theta , c)$ , this means that $({\partial u}/{\partial \theta })(\theta , \cdot )$ is locally Lipschitz, locally uniformly with respect to $\theta $ . Hence, by Minguzzi’s theorem, for every $c\in (a, b)$ , $({\partial u}/{\partial c})(\cdot , c)$ is locally Lipschitz, locally uniformly with respect to c, and at almost all $(\theta , c)\in \mathbb {T}\times (a, b)$ we have that ${\partial ^2u}/{\partial c\partial \theta }$ and ${\partial ^2u}/{\partial \theta \partial c}$ exist and are equal and uniformly bounded.

Hence $h_c=Id_{\mathbb {T}}+ ({\partial u}/{\partial c})(\cdot ,c )$ is locally uniformly Lipschitz and because of Inequality (19), we have Lebesgue almost everywhere

(20) $$ \begin{align} \frac{\partial^2u}{\partial \theta\partial c}(\theta, c)= \frac{\partial^2u}{\partial c\partial \theta}(\theta, c)=\frac{\partial \eta}{\partial c}(\theta_0, c_0)-1\in \bigg[\!-1+\frac{1}{K}, -1+K\bigg]=[k_-, k_+].\end{align} $$

This implies that $h_c(\theta )=\theta +({\partial u}/{\partial c})(\theta , c)$ defines a $({1}/{K}, K)$ -bi-Lipschitz homeomorphism of $\mathbb {T}$ for almost every $c\in [a, b]$ and then for all $c\in [a, b]$ by continuity. By Theorem 1.3, we deduce that u is the generating function of an exact symplectic homeomorphism $\Phi :\mathbb {A}\rightarrow \mathbb {A}$ that maps the invariant foliation onto the standard one.

5.2 Proof of Corollary 1.4

Let $k\geq 1$ and $r\mapsto f_r$ be a $C^k$ -foliation in graphs and let u be its generating function. As $k\geq 1$ , the foliation is Lipschitz when restricted to every compact set. Hence we can use Proposition 1.1. In this case, u is the generating function of an exact symplectic homeomorphism $\Phi :\mathbb {A}\rightarrow \mathbb {A}$ that maps the standard foliation onto the invariant one and we have

$$ \begin{align*}\Phi\bigg(\theta + \frac{\partial u}{\partial c}(\theta , c), c\bigg)=\bigg(\theta, c+ \frac{\partial u}{\partial \theta}(\theta,c)\bigg).\end{align*} $$

Moreover, u is $C^k$ , hence $F_0(\theta , c)=(\theta , c+({\partial u}/{\partial \theta })(\theta , c))$ defines a $C^{k-1}$ homeomorphism that is locally bi-Lipschitz, hence a $C^{k-1}$ diffeomorphism.

Also $h_c(\theta )=\theta +({\partial u}/{\partial c})(\theta , c)$ is $C^{k-1}$ in $(\theta , c)$ . Observe that every $h_c$ is a bi-Lipschitz homeomorphism that is $C^{k-1}$ , hence $G_0: (\theta , c)\mapsto (h_c(\theta ), c)$ is also a $C^{k-1}$ diffeomorphism and then $\Phi =F_0\circ G_0^{-1}$ is a $C^{k-1}$ symplectic diffeomorphism (where a $C^0$ -diffeomorphism is a homeomorphism).

6.1 Proof of Proposition 1.2

Let $f:\mathbb {A} \to \mathbb {A}$ be an exact symplectic homeomorphism. We assume that f has an invariant foliation ${\mathcal F}$ into $C^0$ graphs that is symplectically homeomorphic (by $\Phi ^{-1} : \mathbb {A} \to \mathbb {A}$ ) to the standard foliation ${\mathcal F}_0=\Phi ^{-1}({\mathcal F})$ . Then the standard foliation is invariant by the exact symplectic homeomorphism $g=\Phi ^{-1}\circ f\circ \Phi $ . Hence we have

$$ \begin{align*}g(\theta, r)=(g_1(\theta ,r), r).\end{align*} $$

As g is area preserving, for every $\theta \in [0, 1]$ and every $r_1<r_2$ , the area of $[0, \theta ]\times [r_1, r_2]$ is equal to the area of $g( [0, \theta ]\times [r_1, r_2])$ , that is,

$$ \begin{align*}\theta(r_2-r_1)=\int_{r_1}^{r_2} (g_1(\theta, r)-g_1(0, r))\,dr.\end{align*} $$

Dividing by $r_2-r_1$ and taking the limit when $r_2$ tends to $r_1$ , we obtain

$$ \begin{align*}g_1(\theta, r_1)=\theta+g(0, r_1).\end{align*} $$

This proves the proposition for $\rho =g_1(0, \cdot )$ .

6.2 Proof of Corollary 1.5

The ‘if’ part is obvious by Proposition 1.2.

Let us prove the ‘only if’ part, that is, we assume f is $C^0$ -integrable with the dynamics on each leaf conjugate to a rotation. We denote by $u : \mathbb {A} \to \mathbb {R}$ the map given by Theorem 1.1 and which enjoys the properties of Theorem 2.1. Hence $h_c:\theta \mapsto \theta +({\partial u}/{\partial c})(\theta ,c)$ is a semi-conjugation between the projected dynamics $g_c: \theta \mapsto \pi _1\circ f(\theta , c+({\partial u}/{\partial \theta })(\theta ,c))$ and the rotation $R_{\rho (c)}$ of $\mathbb {T}$ and is even a conjugation when $\rho (c)$ is rational.

If $\rho (c)$ is irrational, it follows from the hypothesis that $g_c$ is conjugate to a rotation. As the dynamics is minimal, there is up to constants a unique (semi)-conjugacy and then $h_c$ is a true conjugation. We then conclude by using Theorem 1.3.

6.3 Proof of Corollary 1.6

6.3.1 Arnol’d–Liouville coordinates for f

Let $f:\mathbb {A}\rightarrow \mathbb {A}$ be a symplectic twist diffeomorphism that is Lipschitz integrable with generating function u of its invariant foliation.

By Theorem 1.1 and Proposition 1.1, u is the generating function of an exact symplectic homeomorphism $\Phi :\mathbb {A}\rightarrow \mathbb {A}$ that maps the standard foliation onto the invariant one, and for every compact subset ${\mathcal K}\subset \mathbb {A}$ there exist two constants $k_+>k_->-1$ such that ${\partial ^2 u}/{\partial \theta \partial c}\in [k_-, k_+]$ Lebesgue almost everywhere in ${\mathcal K}$ .

By Proposition 1.2, we have

$$ \begin{align*}\text{for all } (x, c)\in \mathbb{A},\quad \Phi^{-1}\circ f\circ \Phi(x, c)=(x+\rho(c), c),\end{align*} $$

where $\rho :\mathbb {R}\rightarrow \mathbb {R}$ is continuous. Moreover, because of the twist condition, $\rho $ is an increasing homeomorphism of $\mathbb {R}$ .

6.3.2 Proof that $\rho :\mathbb {R}\rightarrow \mathbb {R}$ is a bi-Lipschitz homeomorphism

Proposition 6.1. Assume that a $C^1$ symplectic twist diffeomorphism $f:\mathbb {A}\rightarrow \mathbb {A}$ has an invariant locally Lipschitz continuous foliation by graphs $c\in \mathbb {R}\mapsto \eta _c\in C^{0}(\mathbb {T}, \mathbb {R})$ . Then the map $\rho : c\in \mathbb {R}\mapsto \rho (c)$ is a locally bi-Lipschitz homeomorphism.

We will use the following lemma.

Lemma 6.1. Let $f,g : \mathbb {R} \to \mathbb {R}$ be lifts of homeomorphisms of $\mathbb {T}$ that preserve orientation (implying $f(\cdot +1) = f(\cdot ) +1$ and $g(\cdot +1) = g(\cdot ) +1$ ). Assume that:

• • either f or g is conjugate to a translation $t_\alpha : x \mapsto x+\alpha $ by a homeomorphism h that is a lift of a homeomorphism of $\mathbb {T}$ that preserves orientation;

• • h and $h^{-1}$ are K-Lipschitz.

Then:

1. (1) if there exists $d>0$ such that $f<g+d$ , then $\rho (f)\leq \rho (g)+Kd$ ;

2. (2) if there exists $d>0$ such that $f+d <g$ then $\rho (f) +{d}/{K}\leq \rho (g)$ .

Proof. Let us say that $h\circ g \circ h^{-1} = t_\alpha $ , hence $\rho (g) = \alpha $ (the proof when f is conjugate to a translation is the same).

(1) By hypothesis, $ f\circ h^{-1}< g\circ h^{-1} +d $ . Using that h is increasing and K-Lipschitz, it follows that for all $ x\in \mathbb {R}$ ,

$$ \begin{align*}\quad h\circ f\circ h^{-1}(x)<h(g\circ h^{-1}(x) +d)\leq h\circ g\circ h^{-1}(x)+Kd=x+\alpha+Kd.\end{align*} $$

Finally, as $\rho (f) = \rho ( h\circ f\circ h^{-1})$ , we conclude that

$$ \begin{align*}\rho(f)\leq \alpha+Kd = \rho(g)+Kd.\end{align*} $$

(2) By hypothesis, $ f\circ h^{-1}+d< g\circ h^{-1} $ . Using that h is increasing, it follows that

$$ \begin{align*}\text{for all } x\in \mathbb{R}, \quad h( f\circ h^{-1}(x)+d)<h\circ g\circ h^{-1}(x)=x+\alpha.\end{align*} $$

Because $h^{-1}$ is K-Lipschitz and increasing, observe that

$$ \begin{align*} d&=h^{-1}(h(f\circ h^{-1}(x)+d))-h^{-1}(h\circ f\circ h^{-1}(x))\\ &\leq K( h(f\circ h^{-1}(x)+d)-h\circ f\circ h^{-1}(x)). \end{align*} $$

Then

$$ \begin{align*}h\circ f\circ h^{-1}(x)\leq h(f\circ h^{-1}(x)+d)-\frac{d}{K}<x+\alpha -\frac{d}{K}.\end{align*} $$

Hence $\rho (f)+{d}/{K}\leq \rho (g)$ .

Proof of Proposition 6.1

The proof is now a direct application of the previous lemma. Indeed, we have seen that when the foliation is K-Lipschitz, if c varies in a compact set $\mathcal K$ , the dynamics $g_c$ are all conjugate to rotations. We have, moreover, proven there exists a constant $\widetilde K$ such that the conjugating functions $h_c$ may be chosen equi-bi-Lipschitz (for $c\in \mathcal K$ ) of constant $\widetilde K$ .

We denote the minimum and maximum torsions on ${\mathcal K}$ by

$$ \begin{align*}b_{\min}=\min_{x\in{\mathcal K}}\frac{\partial f_1}{\partial \theta}(x)\quad\text{and}\quad b_{\max}=\max_{x\in{\mathcal K}}\frac{\partial f_1}{\partial \theta}(x).\end{align*} $$

For $c_1<c_2$ in $[a, b]$ , we have

$$ \begin{align*}\tilde g_{c_2}(\theta)-\tilde g_{c_1}(\theta)=F_1(\theta, \eta_{c_2}(\theta))-F_1(\theta, \eta_{c_1}(\theta))\end{align*} $$

and so

$$ \begin{align*}\tilde g_{c_2}(\theta)-\tilde g_{c_1}(\theta)\in [b_{\min}(\eta_{c_2}(\theta)-\eta_{c_1}(\theta)), b_{\max}(\eta_{c_2}(\theta)-\eta_{c_1}(\theta))]\end{align*} $$

and

$$ \begin{align*}\tilde g_{c_2}(\theta)-\tilde g_{c_1}(\theta)\in \bigg[\frac{b_{\min}}{K}(c_2-c_1), K.{b_{\max}} (c_2-c_1)\bigg].\end{align*} $$

We deduce from Lemma 6.1 that

$$ \begin{align*}{K.\tilde K}.b_{\max}(c_2-c_1)\geq \rho(g_{c_2})-\rho(g_{c_1})\geq \frac{b_{\min}}{K.\tilde K}(c_2-c_1).\\[-42pt]\end{align*} $$

7.1 Proof of Corollaries 1.1–1.3

As u is not $C^1$ , we deduce from Theorems 1.1 and 1.3 that this foliation cannot be globally straightenable by a symplectic homeomorphism and also that it cannot be invariant by a symplectic twist diffeomorphism.

Proof of Corollary 1.3

We now work in $\mathbb {R}^2$ and assume by contradiction that there are two open sets ${\mathcal U}$ and ${\mathcal V}$ as in Theorem 1.3, and a symplectic homeomorphism $\Phi : {\mathcal U} \to {\mathcal V}$ that maps the horizontal lines of ${\mathcal U}$ to the leaves of the foliation in ${\mathcal V}$ preserving the orientation of the leaves. Let us denote $(c_-,c_+) = \pi _2 ({\mathcal U})$ and $J\subset \mathbb {R}$ such that ${\mathcal V} =\{ (\theta , \eta _c(\theta )), \ \theta \in (\alpha ,\beta ), c\in J\} $ . By hypothesis, there exists an increasing homeomorphism ${\mathfrak h : (c_-,c_+ ) \to J}$ such that if $c\in (c_-,c_+)$ , the horizontal line $\mathbb {R}\times \{c\} \cap {\mathcal U}$ is mapped to the leaf of the foliation $\{ (\theta , \eta _{\mathfrak h(c)}(\theta )), \ \theta \in (\alpha ,\beta )\}$ by $\Phi $ .

Returning to the generating function, if $c\in (c_-,c_+)$ we have $u(0,c)=0$ and $c+ ({\partial u}/{\partial \theta })(c,\theta ) = \eta _{\mathfrak h(c)}(\theta )$ from which we deduce that

$$ \begin{align*}\text{for all } (\theta,c)\in {\mathcal W},\quad u(\theta,c) = \theta(\mathfrak h(c) -c) + \frac{\varepsilon\circ \mathfrak h(c)}{2\pi} \sin(2\pi\theta).\end{align*} $$

Now let $\theta _1,\theta _2 \in (\alpha ,\beta )$ such that the matrix

$$ \begin{align*}A=\begin{pmatrix} \theta_1 & {\sin(2\pi \theta_1)}/{2\pi} \\ \theta_2 & {\sin(2\pi \theta_2)}/{2\pi} \end{pmatrix}\end{align*} $$

is invertible. It follows that

$$ \begin{align*}\text{for all } c\in (c_-,c_+), \quad \begin{pmatrix} \mathfrak h(c) \\ \varepsilon\circ \mathfrak h(c) \end{pmatrix} = A^{-1} \begin{pmatrix} u(\theta_1,c)+\theta_1c \\u(\theta_2,c)+\theta_2c \end{pmatrix}. \end{align*} $$

As by Theorem 1.3 the function u is $C^1$ , we deduce that $\mathfrak h$ and $\varepsilon \circ \mathfrak h$ are $C^1$ functions. It follows that $\mathfrak h'$ must vanish at all points c such that $\varepsilon $ is not differentiable at $h(c)$ . As such points are dense, we conclude that $\mathfrak h$ is constant which contradicts it being a homeomorphism onto its image.

7.2 An exact symplectic twist map that leaves the strange foliation invariant

Let us prove, however, that this foliation, for a simple choice of $\varepsilon $ , can be invariant by a certain $C^1$ exact symplectic (weakly) twist map as defined below.

Let us now assume that $\varepsilon $ is a $C^2$ function away from $c=0$ and that at $0$ it has left and right derivatives up to order $2$ . For the sake of simplicity, let us assume also that $\varepsilon (0)=0$ so that $\mathbb {T}\times \{0\}$ is a leaf of the foliation and that $\varepsilon $ restricted to $[0,+\infty )$ (respectively, $(-\infty , 0]$ ) is the restriction of a $C^2$ periodic function.

The proof of Theorem 1.4 gives us two $C^1$ functions

$$ \begin{align*}\Phi^\pm : (\theta,r ) \mapsto (h^\pm(\theta,r) , \eta( h^\pm(\theta,r), r) )\end{align*} $$

where $\Phi ^+$ is a $C^1$ exact symplectic diffeomorphism of $\mathbb {A}^+ = \mathbb {T} \times [0,+\infty )$ to itself (up to the boundary) and $\Phi _-$ is a $C^1$ exact symplectic diffeomorphism of $\mathbb {A}^- = \mathbb {T} \times (-\infty , 0]$ to itself (up to the boundary). Note that here $\Phi ^+$ and $\Phi ^-$ do not coincide on $\mathbb {T}\times \{0\}$ . explaining why the foliation is not straightenable.

Let $\rho : \mathbb {R} \to \mathbb {R}$ be an increasing, $C^1$ homeomorphism such that $\rho (0) = \rho '(0) = 0$ . We denote $f_\rho : (\theta ,r) \mapsto (\theta +\rho (r) , r)$ . The function $f = \Phi ^\pm \circ f_\rho \circ (\Phi ^\pm )^{-1}$ is well defined on $\mathbb {A}$ ; it is the identity on $\mathbb {T} \times \{0\}$ . It is clearly an area-preserving homeomorphism that is $C^1$ away from $\mathbb {T}\times \{0\}$ .

If $r> 0$ and $\theta \in \mathbb {T}$ , let us set $(\Theta ,R) = \Phi ^+(\theta ,r) $ . Then one finds that

$$ \begin{align*} D f(\Theta,R) &= D\Phi^+(\theta + \rho(r),r) \cdot Df_\rho ( \theta,r)\cdot D\Phi^+(\theta,r)^{-1} \\ &= D\Phi^+(\theta + \rho(r),r) \cdot \begin{pmatrix}1&\rho'(r)\\0&1 \end{pmatrix} \cdot D\Phi^+(\theta,r)^{-1}. \end{align*} $$

It follows from the properties on $\Phi ^+$ and $\rho (0)=\rho '(0) = 0$ that as $R\to 0$ , $Df(\Theta ,R)$ uniformly converges to the identity. As the same holds for $R<0$ , we deduce that f is in fact $C^1$ with a differential on $\mathbb {T} \times \{0\}$ being identity.

It remains to choose $\rho $ in such a way that the map obtained is a twist map. We construct it on $[0,+\infty )$ . The twist condition we aim at is: for every $\Theta \in \mathbb {R}$ , the map $r\mapsto h^+((h^{+}_r)^{-1}(\Theta )+\rho (r),r)$ is an increasing homeomorphism of $\mathbb {R}$ where $h^+_r = h^+(\cdot ,r)$ .

After computation, if we denote $h^+(\theta _r, r)=\Theta $ , the derivative of the above function is the following for $r>0$ (the inequality is our goal):

$$ \begin{align*} &\frac{\partial h^+}{\partial r}(\theta_r+\rho(r),r)-\bigg( \frac{\partial h^+}{\partial \theta}(\theta_r,r)\bigg)^{-1}\frac{\partial h^+}{\partial \theta}(\theta_r+\rho(r),r)\frac{\partial h^+}{\partial r}(\theta_r,r) \\ &\quad+ \frac{\partial h^+}{\partial \theta}(\theta_r+\rho(r),r)\rho'(r)>0. \end{align*} $$

The first line above is smaller in absolute value than $M_1\rho (r)$ where (recall that by hypothesis, all the functions at play are continuous periodic, hence bounded)

$$ \begin{align*}M_1 = \bigg \|\frac{\partial^2 h^+}{\partial r\partial \theta}\bigg \|_\infty +\bigg \|\bigg(\frac{\partial h^+}{\partial \theta}\bigg)^{-1}\bigg \|_\infty. \bigg \|\frac{\partial^2 h^+}{\partial \theta^2}\bigg \|_\infty \bigg \|\frac{\partial h^+}{\partial r}\bigg \|_\infty .\end{align*} $$

On the other hand, the second line is greater than $M_2 \rho '(r)$ , where we set $ M_2 = \min ({\partial h^+}/{\partial \theta })>0$ . If $\rho (t) = t^2 e^{Mt}$ with $M= 2M_1/M_2$ , then we have $\rho (0)=\rho '(0)=0$ and $\rho '(t)>({M_1}/{M_2})\rho (t)$ which implies the twist condition.