Claim 4.1. If $1\le |\alpha |+|\beta |=\rho \le r-1$ , then

(29) $$ \begin{align} \partial_x^{\alpha} \partial_{y}^{\beta}(\mathcal{L}^n h)(x,y) = \sum_{j=1}^{N^n} \sum_{|a|+|b| \leq \rho} \partial_x^{a} \partial_{y_1}^{b_1}\cdots \partial_{y_d}^{b_d} h (H^{n,j}(x,y)) \cdot Q_{\alpha,\beta,a,b,n;j}(x,y), \end{align} $$

where $Q_{\alpha ,\beta ,a,b,n;j}(x,y)$ is an homogeneous polynomial on the derivatives of order up to $|\alpha |+|\beta |-|a|-|b|+1$ of the components of $H^{n,j}$ and $\Theta _n\circ H^{n,j}$ at the point $(x,y)$ , where $\Theta _n(x,y)=|\!\det DE^n(x)|^{-1}|\!\det D_yC^n(x,y)|^{-1}$ . Also, if $|a|+|b|=|\alpha |+|\beta |$ and $|b|\geq |\beta |$ , then $Q_{\alpha ,\beta ,a,b,n;j}(x,y)=p(x,y)\Theta _n( H^{n,j}(x,y))$ , where p is an homogeneous polynomial of degree $|\alpha |+|\beta |$ on the partial derivatives of order 1 of $H^{n,j}$ . In fact, ${ p=\sum _{|a|+|b|= \rho }c_{a,b} p_{a,b},} $ where $c_{a,b}$ are positive integers and

(30) $$ \begin{align} p_{a,b}=\frac{\partial{h^{n,j}}}{\partial x_{i_1}}\cdots \frac{{\partial h^{n,j}}}{\partial x_{i_{|a|}}}\frac{{\partial H^{n,j}}}{\partial x_{i_{|a|+1}}}\cdots \frac{{\partial H^{n,j}}}{\partial y_{i_{|a|+|\alpha|-|a|}}}\frac{{\partial H^{n,j}}}{\partial y_{i_{|\alpha|-|a|+1}}}\cdots \frac{{\partial H^{n,j}}}{\partial y_{i_\rho}}. \end{align} $$

Proof. The proof follows by induction on $\rho $ , and we describe here the main steps. Iterating ${\mathcal L}$ in equation (3) and using the chain rule, we can write ${\mathcal L}^n$ as

(31) $$ \begin{align} {\mathcal L}^n h(x,y)= \sum_{j=1}^{N^n} h( H^{n,j}(x,y) ) \cdot \Theta_n \circ H^{n,j}(x,y), \end{align} $$

which corresponds to the case $\rho =0$ for $Q_{0,0,0,0,n;j} = Q_{n,j} = \Theta _n \circ H^{n,j}$ .

Consider the inverse branches $H^{n,j}$ of $T^n$ , $j=1,\ldots , N^n$ , and denote by $H_k^{n,j}$ its kth component. Here, $H^{n,j}$ are locally written as $ H^{n,j}(x,y)=(h^{n,j}(x),C^{n,j}(x,y) ). $

Suppose that equation (29) is valid for some $\rho $ and let us differentiate it with respect to some variable $u_\ell \in \{x_1,\ldots ,x_u, y_1,\ldots , y_d\}$ . Denote by $u_k$ the kth variable of $(x,y)$ , that is, $x_k$ for $1\leq k \leq u$ and $y_{k-u}$ for $u+1 \leq k \leq u+d$ . By the rule of product and chain rule, $\partial _{u_\ell } \partial ^\alpha _x\partial _y^\beta ({\mathcal L}^nh)$ is a sum of terms of the type:

(32) $$ \begin{align} \partial_{u_k}\partial_x^{a} \partial_{y_1}^{b_1}\cdots \partial_{y_d}^{b_d} h (H^{n,j}(x,y)) \cdot [\partial_{u_\ell} H_k^{n,j}(x,y) \cdot Q_{\alpha,\beta,a,b,n;j}(x,y)] \end{align} $$

and

(33) $$ \begin{align} \partial_x^{a} \partial_{y_1}^{b_1}\cdots \partial_{y_d}^{b_d} h (H^{n,j}(x,y)) \cdot \partial_{u_\ell} Q_{\alpha,\beta,a,b,n;j}(x,y). \end{align} $$

All the terms in the bracket in equation (32) are the product of $Q_{\alpha ,\beta ,a,b,n;j}$ with some derivative of some component of $H_k^{n,j}$ , so are also homogeneous polynomials on the derivatives of the components of $H^{n,j}$ and $\Theta _n \circ H^{n,j}$ of order 1 or of the same order of $Q_{\alpha , \beta , a, b, n;j}$ . In the case $|\alpha |+|\beta |=|a|+|b|$ and $|b|\geq |\beta |$ , since we do not differentiate $Q_{\alpha ,\beta ,a,b,n;j}$ , no derivatives of $\Theta _n \circ H^{n,j}$ will appear. Since $\partial _{y_i}H^{n,j}_k=0$ for $k=1,\ldots , u$ , in the polynomials $p_{a,b}$ , the derivatives of $h^{n,j}$ with respect to $y_i$ -variables are null.

For the terms of the type in equation (33), notice that $\partial _{u_\ell } Q_{\alpha ,\beta ,a,b,n;j}(x,y)$ is also an homogeneous polynomial on the derivatives of the components of $H^{n,j}$ and $\Theta _n\circ H^{n,j}$ of order increased by $1$ . Notice also that a and b are the same, and $|\alpha |+|\beta |$ increases by $1$ . In particular, these terms do not correspond to the case $|a|+|b|=|\alpha |+|\beta |$ .

Proof of Proposition 3.2

(First Lasota–Yorke inequality). Using the formula of Claim 4.1 for $\psi \in \mathcal {S}$ , $\phi \in C^r(\psi )$ with $\| \phi \| _{C^\rho }\leq 1$ , and considering $\psi _1,\ldots ,\psi _{N^n}$ , $g_1,\ldots ,g_{N^n}$ such that $T^n(\psi _i(g_i(y)),g_i(y))=(\psi (y),y)$ , we have

(34) $$ \begin{align} &\int \phi(y)\cdot \partial_x^{\alpha} \partial_{y}^{\beta} (\mathcal{L}^n h)(\psi(y),y)\, dy \nonumber \\ &\quad= \sum_{i=1}^{N^n} \int \sum_{|a|+|b|\leq \rho} \phi(y) \partial_x^{a} \partial_{y_1}^{b_1}\cdots \partial_{y_d}^{b_d} h(\psi_i(g_i(y)),g_i(y) )\cdot Q_{\alpha,\beta,a,b,n;i}(\psi(y),y)\, dy \nonumber \\ &\quad= \sum_{i=1}^{N^n}\sum_{|a|+|b|\leq \rho} \int \Psi_{i,\alpha,\beta,a,b,n;i}(y') \cdot \partial_x^{a} \partial_{y_1}^{b_1}\cdots \partial_{y_d}^{b_d} h(\psi_i(y'),y')\, dy', \end{align} $$

where $\Psi _{\alpha ,\beta ,a,b,n;i}(y')\!=\!\phi (g_i^{-1}(y'))\cdot Q_{\alpha ,\beta ,a,b,n;i}((\psi \circ g_i^{-1})(y'),g_i^{-1}(y'))\cdot \! |\!\det Dg_{i}^{-1}(y')|.$

Note that $\!\Psi _{\alpha ,\beta ,a,b,n;i}$ has $C^{|a|+|b|}$ -norm uniformly bounded by some constant $K(n)$ depending on the constants $k_1,k_2,\ldots ,k_r$ on the definition of $\mathcal {S}$ but not on h. Actually, we can bound from above its $C^0$ norm by

$$ \begin{align*}K_0(n)= \underset{\begin{subarray}{c} 1\leq i \leq N^n \\ \alpha|,|\beta|,|a|,|b| \leq r-1 \\ (x,y) \in T^{-n}({\mathbb{T}^u} \times [-K,K]^d) \end{subarray}}{\sup} |Q_{\alpha,\beta,a,b,n;i}(x,y)| \cdot |\!\det D_y C(x,y)|^{-n}. \end{align*} $$

To bound its $C^\ell $ -norm, we can differentiate $\ell $ times $\Psi _{\alpha ,\beta ,a,b,n;i}$ , using the chain rule and product rule, to obtain an expression involving derivatives of $\phi $ , derivatives of $g_i^{-1}$ , and derivatives of $\psi $ , and all of them are uniformly bounded.

In particular, we have

(35) $$ \begin{align} \sum_{i=1}^{N^n}\sum_{|a|+|b|< \rho} \int {\Psi_{\alpha,\beta,a,b,n;i}(y') \cdot \partial_x^{a} \partial_{y_1}^{b_1}\cdots \partial_{y_d}^{b_d} h(\psi_i(y'),y')}\, dy' \leq K_0(n) \| h\| ^\dagger_{\rho-1}. \end{align} $$

To estimate the sum of the terms with $|a|+|b|=\rho $ , we integrate by parts if $|b|>0$ . For this, we rewrite the integral as a surface integral over $G_{\psi _i}$ , $\partial _{y_j}h$ as a combination of Lie derivatives of $\partial _x h$ and ${\mathcal L}_{V_j}h$ , where $V_j$ are tangent vector fields, and apply the following integration by parts: $ \int _S {\mathcal L}_X(f)g = - \int _S f {\mathcal L}_X(g) + \int _S f g \operatorname {div} X.$ Integrating by parts repeatedly when $|b|>0$ , only expressions with at most $\rho -1$ derivatives with respect to x and y or $\rho $ derivatives only in the variable x will remain.

Denote by $S_1$ the sum corresponding to $\rho $ derivatives only on x and by $S_2$ the sum of the remaining terms. Here, $S_2$ is a sum of terms of the form $ \int _{U_{\psi _i}} \tilde {\phi } (y') {\mathcal L}_{V_j^{(i)}}(\tilde {h}) (\psi _i(y'),y')\, dy', $ where $\tilde {h}$ has at most $\rho -1$ derivatives of h. Since $\tilde {h}$ has at most $\rho -1$ derivatives of h, we have that $S_2 \leq K(n) \| h \| ^\dagger _{\rho -1}$ .

So now we only need to consider the terms for $|b|=0$ and $|a|=\rho $ . To estimate $\label {eq.estimate2.2} \int \Psi _{\alpha ,\beta ,a,b,n;i_0}(y') \cdot \partial _x^a h(y')\, dy', $ we write

(36) $$ \begin{align} \Psi_{\alpha,\beta,a,b,n;i_0}(y') \cdot \partial^a_x h(y')=\frac{\phi_1(y')\phi_2(y')\phi_3(y')\phi_4(y') }{\det DE^n(x_{i_0})}\partial_x^{a}h(\psi_{i_0}(y'),y'), \end{align} $$

where $\displaystyle \phi _1(y')=\phi \circ g_{i_0}^{-1}(y')$ , $ \phi _2(y')={\det DE^n(x_{i_0})}/({\det DE^n\circ \psi _{i_0}(y')})$ , $ \phi _3(y')= \det Dg_{i_0}^{-1}/({\det D_yC^n\circ (\psi _{i_0},I)})$ , and $ \phi _4=p\circ (\psi ,I)\circ g_{i_0}^{-1}. $ We can notice that there exists a constant $K>0$ such that

(37) $$ \begin{align} \| p\circ (\psi,I)\circ g_{i_0}^{-1}\| _{C^{\rho}} &\le K n^{\rho^2} \underline{\mu}^{-\rho n},\ \ \,\kern3pt\qquad\qquad\end{align} $$

(38) $$ \begin{align} \bigg\| \frac{\det Dg_{i_0}^{-1}}{\det D_yC^n\circ (\psi_{i_0},I)}\bigg\| _{C^\rho} &\le K,\qquad\qquad\qquad\qquad\kern1pt\qquad \end{align} $$

(39) $$ \begin{align} \| [\det\circ (DE^n)\circ \psi_{i_0}]^{-1}\| _{C^\rho}&\le K n^{r-1} \| \det\circ [DE^n(x_{i_0})]^{-1}\|, \end{align} $$

for all $ x_{i_0}\in h_{i_0}^n(\psi (U_\psi )).$ Actually, equation (37) follows from the expression of $p_{a,b}$ in equation (30) noticing that $b=0$ and $|\alpha |=|a|$ , so only derivatives of $h^{n,j}$ with respect to x variables appear that are bounded by $\underline {\mu }^{-n}$ . Equations (38) and (39) follow from the bounded distortion property, noticing that $ [DE^n \circ \psi _{i_0}]^{-1}(y) = Dh^{n,j}_{i_0}(\psi (g^{-1}_{i_0}(y) ))$ and that $h_{n,j}$ is a composition of inverse branches of E.

Then the $C^{\rho }$ norms of $\phi _1$ , $\phi _2$ , and $\phi _3$ are bounded for some constant independent on n. The $C^\rho $ norm of $\phi _4$ is bounded from above by $K n^{\rho ^2} \underline {\mu }^{-\rho n} $ . Then $\| \phi _1\phi _2\phi _3\phi _4\| _{C^\rho }$ is bounded by $K n^{\rho ^2+r-1}\underline {\mu }^{-\rho n}$ . So $ S_1 \le K n^{\rho ^2+r-1}\underline {\mu }^{-\rho n}\| h\| _\rho ^{\dagger }. $

Consequently, we conclude that equation (34) is bounded by $ K n^{\rho ^2+r-1}\underline {\mu }^{-\rho n}\| h\| _\rho ^{\dagger }+K(n)\| h\| _{\rho -1}^{\dagger } $ for some constant K independent on n.

The estimate in equation (21) is analogous and easier.

Proof. Consider W a u-dimensional subspace containing $(\xi ,\eta )$ as described above such that $(DT_{(x,y)}^q)^*W\subset {\mathcal C}^*$ for all $(x,y)\in B(x_0,R) \times B(y_0,R) \supset {\mathcal R}({\textbf {c}}{\textbf {a}})$ and $(\xi ,\eta )\in W$ . We have $|{\mathcal F}({\mathcal L}^q(\chi \phi ))(\xi ,\eta )| (\| \xi \| ^{\rho _0} + \| \eta \| ^{\rho _0}) \leq K |{\mathcal F}(\partial ^{\rho _0}_{x_j}{\mathcal L}^q(\chi \phi ))(\xi ,\eta )|, $ where the $\rho $ derivatives are taken with respect to the variable $x_j$ ( $\xi _j$ or $\eta _j$ ) that has greatest absolute value. Since the support of ${\mathcal L}^q(\chi \phi )$ is contained in $D\cap ({\mathcal R}({\textbf {c}})\times {\mathbb {R}}^d)$ , we have

$$ \begin{align*} |{\mathcal F}(\partial^{\rho_0}_{x_j}{\mathcal L}^q(\chi\phi))(\xi,\eta)| \leq \int_\Gamma \int_{\sigma } | \partial^{\rho_0}_{x_j}{\mathcal L}^q(\chi\phi)) |\, dm_\sigma \, d\hat{m} \leq K \sup_{\sigma \in \Gamma}\int_{\sigma} |\partial^{\rho_0}_{x_j}{\mathcal L}^q(\chi\phi) |\,dm_\sigma. \end{align*} $$

For each $\sigma \in \Gamma $ , there is a unique $\widetilde {\sigma }$ contained in ${\mathcal R}({\textbf {c}}{\textbf {a}})\times {\mathbb {R}}^d$ such that $T^q(\widetilde {\sigma })=~\sigma $ . For $x\in \widetilde {\sigma }$ and $(u,v)$ tangent to $\sigma $ at $T^q(x,y)$ , we have $ 0=\langle (u,v) ,(w_1,w_2) \rangle =\langle (DT^q_{(x,y)})^{-1}(u,v) ,(DT^q_{(x,y)})^*(w_1,w_2) \rangle $ for all $(w_1,w_2)\in W$ . Since $(DT^q_{(x,y)})^*W$ is a u-dimensional subspace contained in ${\mathcal C}^*$ , we have $(DT^q_{x})^{-1}(u,v)\in {\mathcal C}$ . So, we conclude that $\hat {\sigma }:=T^{-q} \sigma \cap ({\mathcal R}({\textbf {c}}{\textbf {a}})\times {\mathbb {R}}^d)$ is the graph of some $\tilde {\psi }$ in ${\mathcal S}$ . Since $\chi $ is supported in $\mathcal {R}({\textbf {c}}{\textbf {a}})\times B$ , we have that ${\mathcal L}^q(\chi \phi ) = ({(\chi \phi )}/{|\!\det DT^q|})\circ g$ for the inverse branch $g:{\mathcal R}({\textbf {c}})\times {\mathbb {R}}^d \to {\mathcal R}({\textbf {c}}{\textbf {a}})\times {\mathbb {R}}^d$ of the restriction of $T^{q}$ to ${\mathbb {R}}({\textbf {c}}{\textbf {a}})\times {\mathbb {R}}^d$ . Then,

$$ \begin{align*} |\partial^{\rho_0} {\mathcal L}^q(\chi\phi)(x,y)| \leq K(\chi, g) \sum_{|\beta|\leq \rho_0} {|\partial^{\beta}\phi(g(x,y))|}. \end{align*} $$

Integrating and changing variables, we obtain

$$ \begin{align*} \int_{\sigma} |\partial^{\rho_0} {\mathcal L}^q(\chi \phi)(x,y)|\,dm_{\sigma} \leq K(\chi,q) \sum_\beta \int_\sigma |\partial^\beta \phi(g(x,y))|\, dm_{\sigma} \leq K(\chi,q) \| \phi\| ^\dagger_{\rho_0}. \\[-30pt] \end{align*} $$

Proof. The proofs are the same as [Reference Avila, Gouezel and Tsujii7, Lemmas 7 and 9].

Proof. Let us first consider the s integer. Recall that $T_{{\textbf {c}},{\textbf {a}}}^{-q}$ is an inverse branch defined over ${\mathcal R}({\textbf {c}}) \times {\mathbb {R}}^d$ by $T_{{\textbf {c}},{\textbf {a}}}^{-q}(x,y)=(E_{{\textbf {c}},{\textbf {a}}}^{-q}x,C^{-q}(x,y))$ . If we call by $g_1, g_2, \ldots , g_{u+d}$ the components of $T^{-q}_{{\textbf {c}},{\textbf {a}}}$ , then we may observe that $\| \partial ^{\sigma }g_j\| \le K \underline {\unicode{x3bb} }^{-q}$ for all $\sigma $ multi-index with $|\sigma | \le s $ . From this, we can write

$$ \begin{align*}\partial^{\sigma}[(\chi_{{\textbf{c}},{\textbf{a}},B}\phi)\circ T_{{\textbf{c}},{\textbf{a}}}^{-q}](z) = \sum_{|\sigma'|\le|\sigma|} \psi_{\sigma',\sigma}(z) (\partial^{\sigma'} \chi_{{\textbf{c}},{\textbf{a}},B}\phi)\circ T_{{\textbf{c}},{\textbf{a}}}^{-q}(z), \end{align*} $$

where $\psi _{\sigma ',\sigma }$ is a polynomial function of degree at most s in the variables $\partial ^{\gamma }g_j$ , and $\gamma $ goes through the multi-indexes with $|\gamma |\leq |\sigma '|$ and $j=1,2,\ldots ,u+d$ . Notice that $|\psi _{\sigma ',\sigma }(z)|\le K \underline {\unicode{x3bb} }^{-qs}$ for some constant K. So we have

$$ \begin{align*} \| {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B}\phi)\| ^2_{H^s}\!\leq\! K \!\!\!\sum_{ \substack{ |\sigma|{\le}s \\ |\sigma'|+|\sigma"|\le|\sigma|}} \!\! \int \bigg| \bigg( (\partial^{\sigma'}\! \chi_{{\textbf{c}},{\textbf{a}},B}\phi)\! \partial^{\sigma"}\bigg(\frac{1}{|\!\det DT^q| } \bigg) \bigg)\! \circ T_{{\textbf{c}},{\textbf{a}}}^{-q}(z)\! \psi_{\sigma',\sigma}(z) \bigg|\, dz. \end{align*} $$

Write this last sum above as $\mathcal {S}_1+\mathcal {S}_2$ , where $\mathcal {S}_1$ correspond to the terms with $\sigma "=0$ and $\mathcal {S}_2$ to terms with $|\sigma "|>0$ (which implies $|\sigma '|\leq s-1$ ). We deal with $\mathcal {S}_1$ making a change of variables and with $\mathcal {S}_2$ bounding it by the $H^{s-1}$ norm and using Young’s inequality to bound by the $H^s$ norm and $L^1$ norm. So

$$ \begin{align*} \mathcal{S}_1 &\leq K \underline{\unicode{x3bb}}^{-2qs} \sum_{|\sigma|{\le}s} \int \bigg(\sum_{|\sigma'|\le|\sigma|} \frac{|(\partial^{\sigma'} \chi_{{\textbf{c}},{\textbf{a}},B}\phi)\circ T_{{\textbf{c}},{\textbf{a}}}^{-q}(z)|}{|\!\det DT^q|\circ T_{{\textbf{c}},{\textbf{a}}}^{-q}(z)}\bigg)^2\, dz\\ &= K \underline{\unicode{x3bb}}^{-2qs} \inf|\!\det DT^q|^{-1} \| \chi_{{\textbf{c}},{\textbf{a}},B}\phi\| _{H^s}^2 \end{align*} $$

and

$$ \begin{align*}\mathcal{S}_2 \leq K_1(q) \| \chi_{{\textbf{c}},{\textbf{a}},B}\phi\| _{H^{s-1}} \leq \underline{\unicode{x3bb}}^{-2qs} \inf|\!\det DT^q|^{-1}\| \chi_{{\textbf{c}},{\textbf{a}},B}\phi\| _{H^{s}} + K(q) \| \chi_{{\textbf{c}},{\textbf{a}},B}\phi\| _{L^1} .\end{align*} $$

For non-integer values of s, we shall use interpolation of Hilbert spaces [Reference Chandler-Wilde, Hewett and Moiola20]. For $s=0$ , the estimative for $\mathcal {S}_1$ gives

$$ \begin{align*} \| \mathcal{L}^q (\chi_{{\textbf{c}},{\textbf{a}},B}\phi)\| ^2_{L^2} \leq \frac{K}{\inf |\!\det DT^q|} \| \chi_{{\textbf{c}},{\textbf{a}},B}\phi \| ^2_{L^2} =: {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {\chi_{{\textbf{c}},{\textbf{a}},B}\phi} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_{H_0}^2. \end{align*} $$

For every $s\leq r$ integer, since $\| h\| _{L^2} \leq K(D) \| u\| _{L^1}$ , we have

$$ \begin{align*} \| {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B}\phi)\| ^2_{H^s} \leq \frac{K }{ \inf |\!\det DT^q| \underline{\unicode{x3bb}}^{2qs} } \| \chi_{{\textbf{c}},{\textbf{a}},B}\phi\| ^2_{H^s} \kern1.3pt{+}\kern1.3pt K(q) \| \chi_{{\textbf{c}},{\textbf{a}},B} \phi\| ^2_{L^2} =: \kern-2pt{\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {\chi_{{\textbf{c}},{\textbf{a}},B}\phi} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_{H_1}^2. \end{align*} $$

This implies that

$$ \begin{align*} \| {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B}\phi)\| ^2_{H^{t}}\! \leq \! {\left\vert\kern-0.25ex\left\vert\kern-0.25ex\left\vert {\chi_{{\textbf{c}},{\textbf{a}},B}\phi} \right\vert\kern-0.25ex\right\vert\kern-0.25ex\right\vert}_{H_t}^2 = \frac{K }{ (\inf |\!\det DT^q| \underline{\unicode{x3bb}}^{2qt})} \| \chi_{{\textbf{c}},{\textbf{a}},B}\phi\| ^2_{H^{t}} + K(q) \| \chi_{{\textbf{c}},{\textbf{a}},B} \phi\| ^2_{L^2}. \end{align*} $$

Equation (49) follows using equation (47) for some small $\epsilon = \epsilon (q)$ .

Proof. Note that if $E^q(x_{\textbf {a}})=x$ for some $x_{\textbf {a}} \in E^{-q}_{{\textbf {c}},{\textbf {a}}}({\mathcal R}_{*}({\textbf {c}}))$ , then

$$ \begin{align*} (DT^q_{({\textbf{a}}(x),y) } )^*=\left( \begin{array}{@{}cc@{}} (DE^q_{{\textbf{a}}(x)})^* & (DE^q_{{\textbf{a}}(x)})^*(D_xC^q_{{\textbf{a}}(x)}(y))^* \\ 0 & (D_yC^q_{{\textbf{a}}(x)})^*\\ \end{array} \right)\!. \end{align*} $$

Supposing that $(DT^q_{({\textbf {a}}(x_{\textbf {a}}),y_{\textbf {a}})})^*(\xi ,\eta )\not \in {\mathcal C}_1^*$ for some ${\textbf {a}}(x_{\textbf {a}})\in E^{-q}_{{\textbf {c}},{\textbf {a}}}({\mathcal R}_{*}({\textbf {c}}))$ and $y \in B_{\textbf {a}}$ , then we claim that $(DT^q_{({\textbf {b}}(x_{\textbf {b}}),y_{\textbf {b}})})^*(\xi ,\eta )\in {\mathcal C}_1^*$ for all ${\textbf {b}}(x_{\textbf {b}})\in E^{-q}_{{\textbf {c}},{\textbf {b}}}({\mathcal R}_{*}({\textbf {c}}))$ and $y_{\textbf {b}}\in B_{\textbf {b}}$ . In fact, if both vectors are not in ${\mathcal C}_1^*$ , then

$$ \begin{align*}\| D_yC^q({\textbf{i}}(x_{\textbf{i}}),y_{\textbf{i}})^* \eta \|> \tfrac{9}{10}\alpha_0^{-1} \| DE^q({\textbf{i}}(x_{\textbf{i}}))^*\xi + D_xC^q({\textbf{i}}(x_{\textbf{i}}),y_{\textbf{i}})^*\eta\| \end{align*} $$

for ${\textbf {i}} \in \{{\textbf {a}}, {\textbf {b}}\}$ . Denoting $f_{\textbf {i}} = DE^q({\textbf {i}}(x_{\textbf {i}}))^*\xi + D_xC^q({\textbf {i}}(x_{\textbf {i}}),y_{\textbf {i}})^*\eta $ , it satisfies

$$ \begin{align*} \| (DE^q_{{\textbf{i}}(x_1)})^*)^{-1}f_{\textbf{i}} \| \leq \underline{\mu}^{-q} \| f_{\textbf{i}}\| < \tfrac{10}{9}\alpha_0 \underline{\mu}^{-q} \| D_yC^q({\textbf{i}}(x_{\textbf{i}}),y_{\textbf{i}})^*\eta\| \leq \tfrac{10}{9}\alpha_0 \theta^q \| \eta \|. \end{align*} $$

Reminding that $S(x,({\textbf {i}},y))= \pi _2 T^q (E_{\textbf {i}}^{-q}(x),y)$ , where $\pi _2(x,y)=y$ , we have

$$ \begin{align*} &\tfrac{20}{9} \alpha_0 \theta^q \| \eta\| \geq \| (DE^q_({\textbf{a}}(x_1))^*)^{-1}f_{\textbf{a}} - (DE^q_({\textbf{b}}(x_2))^*)^{-1}f_{\textbf{b}} \| \\ &\quad=\| (DE^q({\textbf{a}}(x_1))^*)^{-1} D_xC^q({\textbf{a}}(x_1),y_1)^*\eta - (DE^q({\textbf{b}}(x_2))^*)^{-1} D_xC^q({\textbf{b}}(x_2),y_2)^*\eta \| \\ &\quad=\| (DS(x_1,({\textbf{a}},y_1))-DS(x_2,({\textbf{b}},y_2)))^*\eta\| \geq 3 \alpha_0\theta^q\| \eta\| , \end{align*} $$

which is a contradiction.

Proof of Proposition 3.4

By Lemma 4.3, we have

$$ \begin{align*} \| {\mathcal L}^q \phi\| ^2_{H^s} &\leq K \sum_{{\textbf{c}} } \| \chi_{{\textbf{c}}} {\mathcal L}^q \phi\| ^2_{H^s} + K \| \phi\| ^2_{L^1} \leq K \sum_{{\textbf{c}}} \bigg\| \sum_{{\textbf{a}},B} {\mathcal L}^q (\chi_{{\textbf{c}},{\textbf{a}},B}\phi) \bigg\| ^2_{H^s} + K \| \phi\| ^2_{L^1} \\ &= K \sum_{{\textbf{c}}\in A^p,\,{\textbf{a}},{\textbf{b}} \in A^q, B_1,B_2 \in \mathcal{Y}_p} \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s} + K \| \phi\| ^2_{L^1}. \end{align*} $$

Denote ${\textbf {a}} \pitchfork _{\textbf {c}} {\textbf {b}}$ if ${\textbf {a}}\times B_1$ is transversal to $ {\textbf {b}} \times B_2$ in ${\textbf {c}}$ for every $B_1, B_2 \in \mathcal {Y}_p$ , and denote ${\textbf {a}} \times B_1 \pitchfork _{\textbf {c}}^g {\textbf {b}} \times B_2$ if ${\textbf {a}}\times B_1$ is geometrically transversal to $ {\textbf {b}} \times B_2$ in ${\textbf {c}}$ .

For ${\textbf {a}} \pitchfork _{\textbf {c}} {\textbf {b}}$ , notice that

$$ \begin{align*} &\sum_{{\textbf{a}} \pitchfork_{\textbf{c}} {\textbf{b}}, B_1,B_2}\! \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi)\! \rangle_{H^s} =\! \sum_{{\textbf{a}} \pitchfork_{\textbf{c}} {\textbf{b}}} \sum_{B_1, B_2 \in {\mathcal Y}_p}\! \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s}\\ &\quad\leq \sum_{{\textbf{a}}\times B_1 \pitchfork_{\textbf{c}}^g {\textbf{b}}\times B_2} \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s}\\ &\qquad + \sum_{d(B_1,B_2)> M^q } \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s}. \end{align*} $$

If the distance between $B_1$ and $B_2$ is greater than $M^q$ , then by equation (48),

$$ \begin{align*} \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s} &\leq K(q,M) \| \chi_{{\textbf{c}},{\textbf{a}},B_1} \phi\| _{L^1} \| \chi_{{\textbf{c}},{\textbf{b}},B_2} \phi\| _{L^1}\\ &\leq K(q,M) \| \phi\| ^\dagger_{\rho_0} \| \phi\| _{H^s}. \end{align*} $$

If ${\textbf {a}}\times B_1$ is geometrically transversal to $ {\textbf {b}} \times B_2$ in ${\textbf {c}}$ , then by Lemma 4.5, for every $(\xi ,\eta ) \neq (0,0)$ , we have either $(DT_{(x,y)}^q)^*(\xi ,\eta )\in {\mathcal C}_1^*$ for all ${(x,y)}\in {\mathcal R}({\textbf {c}}{\textbf {a}}) \times B_1$ or $(DT^q_{(x,y)})^*(\xi ,\eta )\in {\mathcal C}_1^*$ for all ${(x,y)}\in {\mathcal R}({\textbf {c}}{\textbf {b}}) \times B_2 $ . Denote by U the set of $(\xi ,\eta )$ such that the first occurs, and V the set such that the second occurs. Then, if $(\xi ,\eta )\in U$ , by Lemma 4.2, we have $ |{\mathcal F}({\mathcal L}^q(\chi _{{\textbf {c}},{\textbf {a}},B_1}\phi ))(\xi ,\eta )| \leq K ( 1+ |\xi |^2 + |\eta |^2)^{-{\rho _0}/{2}} \| \phi \| ^\dagger _{\rho _0}. $ So,

$$ \begin{align*} &\bigg| \sum_{\xi}\int_U (1+|\xi|^2+|\eta|^2)^s {\mathcal F}{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi)(\xi,\eta) \overline{{\mathcal F}{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi)}(\xi,\eta)\, d\eta \bigg| \\ &\quad \leq \bigg( \sum_{\xi}\int_U (1+|\xi|^2+|\eta|^2)^s |{\mathcal F}{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi)(\xi,\eta)|^2\, d\eta \bigg)^{1/2} \| {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi)\| _{H^s}\\ &\quad \leq K(q)\| \phi\| _{H^s} \sum_{\xi} \int (1+|\xi|^2+|\eta|^2)^{s-\rho_0} \| \phi\| ^\dagger_{\rho_0}\,d\,\eta \leq K(q) \| \phi\| ^\dagger_{\rho_0} \| \phi\| _{H^s}, \end{align*} $$

where we used that the integral is finite since $s-\rho _0<-(u+d)/2$ .

Summing it with the same integrals over V instead of U, we obtain that

(50) $$ \begin{align} \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s} \leq K(q) \| \phi\| ^\dagger_{\rho_0} \| \phi\| _{H^s}. \end{align} $$

For ${\textbf {a}} \not \pitchfork _{\textbf {c}} {\textbf {b}}$ , we use $\langle u_1,u_2 \rangle _{H^s} \leq \tfrac 12 (\| u_1 \|^2_{H^s}+ \| u_2 \| ^2_{H^s} ) $ together with the definition of $\tau _M(q)$ to obtain

(51) $$ \begin{align} \sum_{{\textbf{a}} \not\pitchfork_{\textbf{c}} {\textbf{b}}, B_1,B_2} \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s} \leq \tau_M(q) \sum_{{\textbf{a}}, B_1} \| {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi)\| ^2_{H^s}. \end{align} $$

Using it with Lemmas 4.3 and 4.4, we have

$$ \begin{align*} \sum_{{\textbf{c}}\in A^p,\,{\textbf{a}}\not\pitchfork_{\textbf{c}} {\textbf{b}}} \langle {\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{a}},B_1}\phi),{\mathcal L}^q(\chi_{{\textbf{c}},{\textbf{b}},B_2}\phi) \rangle_{H^s} \leq \frac{K \tau_M(q,p)}{(\inf |\!\det DT^q| \underline{\unicode{x3bb}}^{2sq}) } \| \phi \| ^2_{H^s} + K(q) \| \phi\| _{L^1}^2. \end{align*} $$

Then

$$ \begin{align*} \| \mathcal{L}^q \phi\| _{H^s}^2 &\leq \frac{K \tau_M(q,p)}{\inf |\!\det DT^q| \underline{\unicode{x3bb}}^{2sq}} \| \phi \| ^2_{H^s} +K \| \phi\| ^2_{L^1}+ K(q,M) \| \phi\| ^\dagger_{\rho_0} \| \phi\| _{H^s} \\ &\leq \frac{K \tau_M(q)}{\inf |\!\det DT^q| \underline{\unicode{x3bb}}^{2sq}} \| \phi\| ^2_{H^s} + K(q,M) \| \phi\| ^\dagger_{\rho_0} \| \phi\| _{H^s}.\\[-3.6pc] \end{align*} $$

Proof of Theorem A $^{\prime }$

The transversality condition implies that we can consider q sufficiently large such that $\omega = ({K_2 \tau _M(q)}/{(|\!\det DT|{\underline {\unicode{x3bb} }}^{2s})^q}) <1$ . Consider $\rho _0 = r-1$ and $\rho _1=0$ . Since $s<r-u/2-d/2-1$ , we have that $s+u/2+d/2<\rho _0 $ , so we can apply Proposition 3.5 for some $\zeta $ between $\omega $ and $1$ .

Fix some non-negative function $\psi _0 \in C^r(D)$ with $\| \psi _0\| _{L^1}=1$ , $\nu _0=\psi _0 m$ , $\psi _n = ({1}/{n}) (\psi _0 + {\mathcal L}\psi _0 + \cdots + {\mathcal L}^{n-1}\psi _0)$ , and $\nu _n=\psi _n m$ , then $\nu _n = ({1}/{n}) \sum _{j=0}^{n-1}T^j_{*}\nu _0 \overset {*}{\to } \mu $ since $\mu $ is the SRB measure for T. It is easy to check that $\mu =\psi _{\infty }m$ , where $\psi _\infty $ is a weak* accumulation point of $\psi _n$ in $L^2$ , so $\mu $ is an absolutely continuous invariant probability.

The openness in $T{\kern-1.3pt}={\kern-1.3pt}T(E,{\kern-1pt}C)$ follows from the fact that $\tau _M(q)$ is upper semi-continuous on $(E,C)\in C^r({\mathbb {T}^u}) \times C^r({\mathbb {T}^u} \times {\mathbb {R}}^d, {\mathbb {R}}^d)$ , from the openness of the condition $K_{2}({\tau _M(q)}/ {\inf |\!\det DT^q|\underline {\unicode{x3bb} }^{2sq}})<1$ and because we are considering a $C^r$ neighborhood of maps T where the bounds in equations (7), (8), (9), and (10) hold with the same constants. This concludes the proof of Theorem A $^{\prime }$ .

Proof. A proof of this theorem can be found in [Reference Sarig53] (see also [Reference Demers, Kiamari and Liverani22, Theorem B.14 B]). In [Reference Sarig53, Appendix A.3], it is proved that conditions (1)–(4) imply that the operator is quasi-compact. In [Reference Sarig53, Exercise 2.2], it is proved that if L is quasi-compact and satisfies condition (5), then L has a spectral gap.

Proof. It is standard to verify that $\mu $ is T-invariant, since for every bounded continuous function g, it holds

$$ \begin{align*} \int g\circ T d\mu &= \int (g\circ T) hd\nu = \unicode{x3bb}^{-1} \int (g \circ T) h\, d(\mathcal{L}^*\nu) \\ &= \unicode{x3bb}^{-1}\int \mathcal{L}((g\circ T) h)\, d\nu = \unicode{x3bb}^{-1} \int g \mathcal{L}(h)\, d\nu = \int g h\, d\nu = \int g \, d\mu. \end{align*} $$

Since ${\mathcal L}$ has spectral gap in $\mathcal {B}$ , for each $\phi \in \mathcal {B}$ , we write $\phi =a(\phi )\phi _0+ \phi _1$ with $\| \mathcal {L}^n\phi _1\| _{\mathcal {B}}\le \zeta ^n\| \phi _1\| _{\mathcal {B}}$ . Then the properties $\int {\mathcal L}^n u \, d\nu = \unicode{x3bb} ^n \int u \, d\nu $ and ${\mathcal L}^n \phi _0 = \unicode{x3bb} ^n \phi _0$ imply that for every n,

$$ \begin{align*} \bigg|\!\int \phi_1 \,d\nu\bigg| &=\bigg|\!\int \unicode{x3bb}^{-n}{\mathcal L}^n(\phi_1) \,d\nu\bigg| \leq C \unicode{x3bb}^{-n}\| {\mathcal L}^n(\phi_1)\| _{L^2(\nu)} \\ &\leq C\unicode{x3bb}^{-n}\| {\mathcal L}^n(\phi_1)\| _{{\mathcal B}} \leq C(\zeta \unicode{x3bb}^{-1})^n\| \phi_1\| _{{\mathcal B}} \to 0. \end{align*} $$

So it follows that $\int \phi _1\, d\nu =0$ . Since $\int \phi _0\, d\nu =1$ , we have $a(\phi )=\int \phi \,d\nu $ . In particular, $\unicode{x3bb} ^{-n}{\mathcal L}^n(\phi )=(\int \phi \,d\nu ) \phi _0 + \unicode{x3bb} ^{-n} {\mathcal L}^n\phi _1 $ converges to $(\int \phi \,d\nu ) \phi _0$ exponentially fast in ${\mathcal B}$ .

We also have the property ${\mathcal L}^n(\phi \cdot \psi \circ T^n)=\psi \cdot \mathcal {L}^n\phi $ and $\phi _1 = \mathcal {N}(\phi )$ implies that $\| \phi _1\| _{\mathcal {B}} \leq K \| \phi \| _{\mathcal {B}}$ for $K= \| \mathcal {N}\| $ .

Given $\psi \in L^\infty (\mu )$ and $\phi \in \tilde{\mathcal{B}}$ , it follows that

$$ \begin{align*} \bigg|\! \int\kern-1.3pt \phi(\psi\circ T^n)\,d\mu \kern1.3pt{-} \int\kern-1.5pt \phi\, d\mu \int\kern-1.5pt \psi\, d\mu \bigg| &=\bigg|\! \int \phi(\psi\circ T^n)\phi_0\,d\nu - \int \phi\phi_0 d\nu \int \psi\phi_0\, d\nu \bigg| \\&= \bigg|\! \int \bigg[\dfrac{{\mathcal L}^n (\phi\phi_0)}{\unicode{x3bb}^n} - \bigg(\!\int \phi \phi_0\,d\nu\bigg) \phi_0\bigg] \psi\, d\nu \bigg|\\ &= \bigg|\! \int \unicode{x3bb}^{-n}{\mathcal L}^n \bigg[ (\phi\phi_0)_1 \bigg] \psi\, dm \bigg|\\ &\le \unicode{x3bb}^{-n}\| \psi\| _{L^2(m)} \| {\mathcal L}^n (\phi\phi_0)_1 \| _{L^2(m)} \\ &\le K \unicode{x3bb}^{-n}\| \psi\| _{L^2(m)} \| {\mathcal L}^n (\phi\phi_0) \| _{\mathcal{B} } \\ &\le K \unicode{x3bb}^{-n}\| \psi\| _{L^2(m)} \| (\phi\phi_0)_1 \| _{\mathcal{B} } \zeta^n \kern1.2pt{=}\kern1.2pt K(\phi,\psi) \bigg(\kern-0.2pt\dfrac{\zeta}{\unicode{x3bb}}\kern-0.2pt\bigg)^{\kern-1pt n}\kern-0.1pt. \end{align*} $$

So $(T,\mu )$ has exponential decay of correlations in $\tilde{\mathcal{B}}$ .

Proof of Theorem C $^{\prime }$

Consider the smallest integer $\rho _0$ and the greatest integer $\rho _1$ such that $ \rho _1+ u/2 < s < \rho _0 - u/2-d/2. $ Consider $t \in ( \rho _1+u/2,s)$ and an integer q such that $K_2 \tau _M(q) < (\inf |\!\det DT|{\underline {\unicode{x3bb} }}^{2s})^q$ . Since $\rho _0 - \rho _1 \leq u + ({d}/{2}) +2 $ and $\sum _{j=1}^{n}1/j\le 1+\log (n-1)$ , we have $\nu \le 1+\log (u+({d}/{2})+1) := a$ .

Consider $\mathcal {B}$ the completion of $C^r(D)$ with respect to the norm $\| \cdot \| _{\rho _0,s}$ and $\mathcal {B}'$ the completion of $C^r(D)$ with respect to the norm $\| \cdot \| ^\dagger _{\rho _1}$ .

If $\zeta \in ( \max \{\underline {\mu }^{{-1}/{a}},({(K_2 \tau _M(q))^{1/q}}/{|\!\det DT| \underline {\unicode{x3bb} }^{2s}})\} , 1 )$ , then $\zeta $ is in the interval in equation (27).

We will verify that the conditions of Hennion’s theorem are satisfied. Obviously, $\| \cdot \| _{\rho _1} \leq \| \cdot \| _{\rho _0} \leq \| \cdot \| $ , which implies condition (1). Condition (3) is an immediate consequence of Lemma 3.2. Condition (4) follows from Lemma 3.5 with $r=\zeta < 1$ .

Condition (5) follows from the fact that $T|\Lambda $ is topologically mixing and that the SRB measure is a mixing absolutely continuous invariant probability (see e.g. [Reference Sarig53, Exercise 2.3]).

To verify condition (2), we notice that the embedding of $\mathcal {B}$ in $\mathcal {B}'$ is compact. Actually, the embedding of $H^s(D)$ in $H^t(D)$ is compact and the embedding of $H^t(D)$ in $\mathcal {B}'$ is continuous, as we notice in the claim below.

So we can apply the theorem of Hennion and Proposition 4.8 to conclude the part of spectral gap and decay of correlations in Theorem C $^{\prime }$ .

It remains to notice that $C^{r-1}(D) \subset \mathcal {B}$ . The definition of $\| \cdot \| ^\dagger _\rho $ gives that

$$ \begin{align*}\bigg| \int_{U_\psi} \phi(y) \partial^\alpha_x \partial_y^\beta h(\psi(y),y)\, dy\bigg| \leq \int_{U_\psi} \bigg|\partial_x^\alpha \partial_y^\beta h (\psi(y),y)\bigg|\,dy \leq \operatorname{vol}(D) \| h\| _{C^{r-1}}\end{align*} $$

whenever $|\alpha |+|\beta | \leq \rho _0$ , $\psi \in \mathcal {S}$ , $\phi \in C^{|\alpha |+|\beta |}(U_\psi )$ , and $\| \phi \| _{C^{|\alpha |+|\beta |}}\leq 1$ . This implies immediately that $\| h\| _{\rho _0}^\dagger \leq K \| h\| _{C^{r-1}}$ and so $C^{r-1}(D) \subset \mathcal {B}$ . Since $C^{r-1}(D)$ is a Banach algebra, it follows that $C^{r-1}(D) \subset \hat {\mathcal {B}}$ .

Claim 4.9. The embedding of $H^t(D)$ in $\mathcal {B}'$ is continuous, that is, there exists a constant $K>0$ such that

(55) $$ \begin{align} \| u\| _{\rho_1}^{\dagger} \le K \| u\| _{H^t}. \end{align} $$

Proof of Claim 4.9

From the definition of $\| .\| _{\rho _1}^{\dagger }$ , it follows that $ \| u\| _{\rho _1}^\dagger $ is bounded by $ K \max _{|\gamma |\le \rho _1}\sup _{\psi \in {\mathcal S}}\| \partial ^{\gamma }u(\psi (\cdot ),\cdot )\| _{L^2({\mathbb {R}}^d)}. $ Define $v(x,y)=u(x+\psi (y),y)$ for ${y\in U_\psi }$ and $v(x,y)=0$ if $y\not \in U_{\psi }$ , then the norm of $\| \partial ^{\gamma }u(\psi (\cdot ),\cdot )\| _{L^2({\mathbb {R}}^d)}$ is bounded by $K\| v(0,\cdot )\| _{H^{\rho _1}}$ . Due to [Reference Adams1, Theorem 7.58(iii)] applied with $p=q=2$ , $\tilde {s}=\rho _1+u/2$ , $\chi =\rho _1$ , $k=d$ , $n=u+d$ , we have

(56) $$ \begin{align} \| v(0,\cdot)\| _{H^{\rho_1}} \leq K \| v\| _{H^{\tilde{s}}({\mathbb{R}}^{u+d})}. \end{align} $$

For any $0< t < r$ , it is easy to see that $ \| v\| _{H^t({\mathbb {R}}^{u+d})}\le K \| u\| _{H^t({\mathbb {R}}^{u+d})}, $ where K depends on only $k_1, k_2, \ldots , k_r$ . So, for $t>\rho _1 + ({u}/{2}) = \tilde {s}$ , we conclude that $ \| v\| _{H^{\tilde {s}}({\mathbb {R}}^{u+d})} \le K \| v\| _{H^t({\mathbb {R}}^{u+d})}\le K \| u\| _{H^t({\mathbb {R}}^{u+d})}. $

Proof. Without loss of generality, we can suppose that U, V, and $\tilde {V}$ are bounded open sets of ${\mathbb {R}}^{u+d}$ . Let $F: C_b^{r}(U,\tilde {V}) \times C_b^{r-k}(V,U) \to C_b^{r-k}(V,\tilde {V})$ be given by $ F(T,S)=T\circ S. $ The same proof of [Reference Franks23, Theorem 3.1] shows that F is $C^k$ . Moreover, for $k\ge 1$ , $\partial _SF(T,S):C_b^{r-k}(V,{\mathbb {R}}^{u+d}) \to C_b^{r-k}(V,{\mathbb {R}}^{u+d})$ is given by

$$ \begin{align*} \partial_SF(T,S)H=(DT\circ S)\cdot H. \end{align*} $$

So, if $T\in \mathcal {V}$ and $S=\xi (T)$ , then $\partial _SF(T,S)$ is a continuous linear isomorphism with inverse $\partial _SF(T,S)^{-1}:C_b^{r-k}(V,{\mathbb {R}}^{u+d}) \to C_b^{r-k}(V,{\mathbb {R}}^{u+d})$ given by $ \partial _SF(T,S)^{-1} H=(DT\circ S)^{-1}\cdot H. $ Since $F(T,\xi (T))=\mathrm {id}_{V}$ , the implicit function theorem implies that $\xi $ is $C^k$ .

Proof. It is enough to show that $\gamma (T)=DT\circ \xi (T)$ is $C^{k-1}$ . Consider

$$ \begin{align*}F: C_b^{r}(U,\tilde V) \times C_b^{r-k}(V,U)\to C_b^{r-1}(U,L({\mathbb{R}}^{u+d},{\mathbb{R}}^{u+d}))\times C_b^{r-k}(V,U)\end{align*} $$

given by $ F(T,S)=(DT,S) $ and

$$ \begin{align*}G:C_b^{r-1}(U,L({\mathbb{R}}^{u+d},{\mathbb{R}}^{u+d}))\times C_b^{r-k}(V,U) \to C_b^{r-1}(V,L({\mathbb{R}}^{u+d},{\mathbb{R}}^{u+d}))\end{align*} $$

given by $ G(A,S)=A\circ S. $ Notice that F is a continuous linear mapping and, by the same argument of the previous lemma, G is of class $C^{k-1}$ . So, $\beta (T,S)=G\circ F(T,S)= DT\circ S$ is $C^{k-1}.$ Since $\gamma (T)=\beta (T,\xi (T))$ , it follows that $\gamma $ is $C^{k-1}$ .

Proof. Consider $\tilde H=(H_1,\ldots ,H_j)\in (C_b^{a}(V,{\mathbb {R}}^{u+d})^j$ and write

$$ \begin{align*} &[(\omega_j(S+H)-\omega_j(S)).\tilde H](\varphi)(z) = (D^{j}\varphi(S(z)+H(z))-D^{j}\varphi(S(z))).\tilde H(z). \end{align*} $$

For each multi-index $\alpha $ with $|\alpha |\le a-k$ , we have

$$ \begin{align*} \partial^{\alpha}\kern-1pt[(\omega_j(S\kern1.3pt{+}\kern1.3ptH)\kern1.3pt{-}\kern1.3pt\omega_j(S)).\tilde{H}(\varphi)](z)\kern1pt{=}\kern1.3pt\sum_{\gamma_1+\tilde\gamma=\alpha}\![\partial^{\gamma_1}(D^{j} \varphi(S(z)\kern1.3pt{+}\kern1.3ptH(z))\kern1.3pt{-}\kern1.3ptD^{j}\varphi(S(z))).\partial^{\tilde \gamma}\tilde{H}(z)], \end{align*} $$

where $\tilde \gamma =\gamma _2+\cdots +\gamma _{j+1}$ and $\partial ^{\tilde \gamma }\tilde {H}(z)=(\partial ^{\gamma _2}H_1(z),\ldots ,\partial ^{\gamma _{j+1}}H_j(z))$ .

We can write

$$ \begin{align*} &\partial^{\gamma_1}(D^{j}\varphi(S(z)+H(z))-D^{j}\varphi(S(z))\\ &\quad=\sum_{1\le |\sigma|\le |\gamma_1|}(\partial^{\sigma}(D^{j}\varphi)(S(z)+H(z))-\partial^{\sigma}(D^{j}\varphi)(S(z))) Q_{\sigma,\gamma_1}(S+H;z)\\ &\qquad+ \partial^{\sigma}(D^{j}\varphi)(S(z))(Q_{\sigma,\gamma_1}(S+H;z)-Q_{\sigma,\gamma_1}(S;z)), \end{align*} $$

where $Q_{\sigma ,\gamma _1}(S;z)$ is a polynomial of degree $|\sigma |$ in the derivatives of $S_1, \ldots , S_{u+d}$ of order at most $|\gamma _1|$ .

For $\gamma _1\le \alpha $ and $|\sigma |\le |\gamma _1|$ , define

$$ \begin{align*} A_{S,H,\gamma_1,\sigma}(z)=(\partial^{\sigma} (D^{j}\varphi)(S(z)+H(z))-\partial^{\sigma}(D^{j}\varphi)(S(z))) Q_{\sigma,\gamma_1}(S+H;z) \end{align*} $$

and

$$ \begin{align*} B_{S,H,\gamma_1,\sigma}(z)=\partial^{\sigma}(D^{j}\varphi)(S(z)) (Q_{\sigma,\gamma_1}(S+H;z)-Q_{\sigma,\gamma_1}(S;z)). \end{align*} $$

Using the mean value theorem, we have

$$ \begin{align*} A_{S,H,\gamma_1,\sigma}(z)=\int_{0}^{1} (\partial^{\sigma}(D^{j+1}\varphi)(S(z)+tH(z))H(z))Q_{\sigma,\gamma_1}(S+H;z)\,dt. \end{align*} $$

Then, we can write

$$ \begin{align*} \partial^{\alpha}[(\omega_j(S+H)-\omega_j(S)).\tilde{H}(\varphi)](z) =\sum_{\gamma_1+\tilde\gamma=\alpha}\sum_{1\le |\sigma|\le |\gamma_1|} \mathcal{P}_1(z)+\mathcal{P}_2(z) \end{align*} $$

for $ \mathcal {P}_1(z)=(A_{S,H,\gamma _1,\sigma }(z)).\partial ^{\tilde \gamma }\tilde {H}(z) $ and $ \mathcal {P}_2(z)=(B_{S,H,\gamma _1,\sigma }(z)).\partial ^{\tilde \gamma }\tilde {H}(z). $

Now suppose that $|\alpha |\le \rho -k$ and let $\psi :D_\psi \subset {\mathbb {R}}^d \to {\mathbb {R}}^u$ in $\mathcal {S}$ (as in the definition of $\| \cdot \| _{\rho -k}^{\dagger }$ ) and $\eta :D_\psi \to {\mathbb {R}}$ a $C^{|\alpha |}$ function with compact support and $\| \eta \| _{|\alpha |}\le 1$ . Then

$$ \begin{align*} \int\eta(y)P_1(\psi(y),y)\,dy &=\int_0^1\int[(\eta(y)Q_{\sigma,\gamma_1}(S+H;(\psi(y),y)))\\&\qquad\times([\partial^{\sigma}(D^{j+1}\varphi)(S(\psi(y),y)+tH(\psi(y),y))]H(\psi(y),y))\\&\qquad \times (\partial^{\gamma_2}H_1(\psi(y),y),\ldots,\partial^{\gamma_{j+1}}H_j(\psi(y),y))]\,dy\,dt. \end{align*} $$

Recall that $S(\psi (y),y)+tH(\psi (y),y)=(\tilde \psi (g_t(y)),g_t(y))$ for some diffeomorphism $g_t:D_\psi \to \pi _2\circ (S+tH)(G_{\psi })$ . So, making the change of variable $y=g_t^{-1}(\tilde y)$ and writing $z(\tilde y)=(\psi (g_t^{-1}(\tilde y)),g_t^{-1}(\tilde y))$ , we obtain

$$ \begin{align*} \int\eta(y)P_1(\psi(y),y)\,dy \le K\| \varphi\| _{\rho}^{\dagger}\| H \| _{a}\| H_1\| _{a}\cdots\| H_j\| _{a}. \end{align*} $$

Since $Q_{\sigma ,\gamma _1}(S+H;z)-Q_{\sigma ,\gamma _1}(S;z)$ is bounded by $K \| H \| _a$ , we similarly prove that

$$ \begin{align*} \int\eta(y)P_2(\psi(y),y)\,dy \le K\| \varphi\| _{\rho}^{\dagger}\| H \| _{a}\| H_1\| _{a}\cdots\| H_j\| _{a}. \end{align*} $$

These estimates prove that

(59) $$ \begin{align} \| [(\omega_j(S+H)-\omega_j(S)).\tilde H](\varphi)\| _{\rho-k}^{\dagger}\le K\| \varphi\| _{\rho}^{\dagger}\| H \| _{a}\| H_1\| _{a}\cdots\| H_j\| _{a}. \end{align} $$

Now, for $|\alpha |\le s-k$ , we have

$$ \begin{align*} \| \mathcal{P}_1\| _{L^{2}}^2&=\int\bigg|\int_{0}^{1}D^{j+1} (\partial^{\sigma}\varphi)(S(z)+tH(z))\cdot\tilde H(z)Q_{\sigma,\gamma_1}(S+tH;z)\,dt\bigg|^2 \, dz. \end{align*} $$

Recall $S+tH$ is a diffeomorphism over V for all $t\in [0,1]$ . So, making the changing of variable $\tilde z=(S+tH)(z) $ and observing that $|Q_{\sigma ,\gamma _1}(S+tH;z)|$ is uniformly bounded, we conclude that