Assumption US.

1. (i) The distributions $E^{u}$ and $E^{s}$ are continuous.

2. (ii) (U-boundary control) There exist $r_0^u, \beta ^u>0, \gamma ^u>{\max } \{1, \beta ^u / \alpha \}$ , $0<\kappa ^u<1$ , and $C^u>1$ such that for any $x \in O$ satisfying $d(x, \partial O) \leqslant r_0^u$ , we have

   $$ \begin{align*} m(D f|_{E^{u}(x)})^{\kappa^u}-1 \geqslant C^u {\max} \bigg\{d(x, \partial O)^{\beta^u},\bigg(\frac{d(f(x), \partial O)}{d(x, \partial O)}\bigg)^{\gamma^u}-1\bigg\}.\end{align*} $$

3. (iii) (S-boundary control) There exist $r^s_0, \beta ^s>0, \gamma ^s>{\max } \{1, \beta ^s / \alpha \}$ , $0<\kappa ^s<1$ , and $C^s>1$ such that for any $x \in O$ satisfying $d(x, \partial O) \leqslant r^s_0$ , we have

   $$ \begin{align*} m(D f^{-1}|_{E^{s}(x)})^{\kappa^s}-1 \geqslant C^s {\max} \bigg\{d(x, \partial O)^{\beta^s},\bigg(\frac{d(f^{-1}(x), \partial O)}{d(x, \partial O)}\bigg)^{\gamma^s}-1\bigg\}. \end{align*} $$

Assumption R. $\alpha>({\beta }/{\gamma }) $ , where $\gamma =\gamma ^u\wedge \gamma ^s$ , $\beta =\beta ^u\vee \beta ^s$ .

Proof. The proof follows the arguments in [Reference Sarig22, Theorem 1.1], we provide some details.

Write $\widehat {\Sigma }=\widehat {\Sigma }(\widehat {\mathscr {G}})$ and let $\mu $ be an ergodic measure of maximal entropy for $f|_O$ . Proceeding as in [Reference Sarig22, §13], we can lift $\mu $ to an ergodic measure of maximal entropy $\widehat {\mu }$ for $\widehat {\sigma }$ via

(2.2) $$ \begin{align} \widehat{\mu}=\int_{O} \frac{1}{|\widehat{\pi}^{-1}(x)|}\bigg(\sum_{\underline{v} \in \widehat{\pi}^{-1}(x)} \delta_{\underline{v}}\bigg)\, d \mu(x), \end{align} $$

such that $h_{\widehat {\mu }}(\widehat {\sigma })=h_{\mu }(f|_O)$ .

By [Reference Gurevich11, Reference Gurevich12], $\widehat {\mu }$ is supported on a subset $\widehat {\Sigma }(\widehat {\mathscr {G}}^\prime )\subset \widehat {\Sigma }(\widehat {\mathscr {G}})$ , where the smaller TMS $(\widehat {\Sigma }(\widehat {\mathscr {G}}^\prime ),\widehat {\sigma })$ is topologically transitive. For a topologically transitive TMS with countable many vertexes, Gurevič [Reference Gurevich11, Reference Gurevich12] showed a good estimate on $\mathrm {Per}_n(\widehat {\sigma })$ : a topologically transitive TMS $(\widehat {\Sigma }(\widehat {\mathscr {G}}^\prime ),\widehat {\sigma })$ admits at most one measure of maximal entropy $h_{\max }(\widehat {\Sigma }(\widehat {\mathscr {G}}^\prime ))$ , and such a measure exists if and only if $\text {there exists } p\in \mathbb {N}$ such that for every vertex $v\in \widehat {\mathscr {G}}^\prime $ , $\text {there exists } C_v\geq 1$ , such that

$$ \begin{align*} C_v^{-1}\leq\frac{\mathrm{Per}_{pn}(\widehat{\sigma}|_{\widehat{\Sigma}(\widehat{\mathscr{G}}^\prime)},v)}{e^{pnh_{\max}( \widehat{\Sigma}(\widehat{\mathscr{G}}^\prime))}}\leq C_v \end{align*} $$

for all $n\in \mathbb {N}$ , where

$$ \begin{align*} &h_{\max}(\widehat{\Sigma}(\widehat{\mathscr{G}}^\prime)):=h_{\max}(\widehat{\sigma})\\ &\quad={\max}\{h_{\nu}(\widehat{\sigma}): \nu \text{ is } \widehat{\sigma}\text{-invariant Borel probability measure on } \widehat{\Sigma}(\widehat{\mathscr{G}}^\prime)\}, \end{align*} $$

and

$$ \begin{align*} \mathrm{Per}_{pn}(\widehat{\sigma}|_{\widehat{\Sigma}(\widehat{\mathscr{G}}^\prime)},v):=|\{\underline{v}\in\widehat{\Sigma}(\widehat{\mathscr{G}}^\prime): v_0=v, \widehat{\sigma}^{pn}(\underline{v})=\underline{v}\}|. \end{align*} $$

From Theorem 2.1(1) and (3), every periodic point in $\{\underline {v}\in \widehat {\Sigma }(\widehat {\mathscr {G}}^\prime ): v_0=v, \widehat {\sigma }^n(\underline {v})=\underline {v}\}\subset \widehat {\Sigma }^{\#}$ is mapped by $\widehat {\pi }$ to a periodic point of the same period in O. By Theorem 2.2, this mapping is at most $N(v)^2$ -to-one. Therefore, for any $v\in \widehat {\mathscr {G}}^\prime $ and all $n\geq 1$ ,

(2.3) $$ \begin{align} P_{pn}(f)\geq P_{pn}(f|_O)\geq \frac{1}{N(v)^2}\mathrm{Per}_{pn}(\widehat{\sigma}|_{\widehat{\Sigma}(\widehat{\mathscr{G}}^\prime)},v). \end{align} $$

Finally, by assumption and construction, ${h_{\max }(\widehat {\Sigma }(\widehat {\mathscr {G}}^\prime ){\kern-1pt}){\kern-1.2pt}={\kern-1.2pt}h_{\widehat {\mu }}(\widehat {\sigma }){\kern-1.2pt}={\kern-1.2pt}h_{\mu }(f|_O{\kern-1pt}){\kern-1.2pt}={\kern-1.2pt}h_{\mathrm {top}} (f{\kern-1pt},{\kern-1.5pt}O)}$ . Therefore, we have

$$ \begin{align*} P_{pn}(f)\geq \frac{1}{C_vN(v)^2}e^{pnh_{\max}( \widehat{\Sigma}(\widehat{\mathscr{G}}^\prime))}=\frac{1}{C_vN(v)^2}e^{pnh_{\mathrm{top}}(f,O)}.\\[-37pt] \end{align*} $$

Proof. Our first constraint on $\varepsilon $ is that it has to be small so that

$$ \begin{align*} 0<\varepsilon\leq\min \bigg\{\frac{C_0^\kappa-1}{Cr_0^\beta},1 \bigg\}. \end{align*} $$

Recall that the diameter of M is smaller than $1$ . There are the following two cases.

Case 1. $\varepsilon (x)\hspace{-0.5pt}=\hspace{-0.5pt}\varepsilon r_0^\beta $ . In this case, we note that $x\hspace{-1pt}\in\hspace{-1pt} O(r_0)$ and $1\hspace{-1pt}<\hspace{-1pt}C_0\hspace{-1pt}\leq\hspace{-1pt} \min \{m(D f|_{E^{u}(x)}), m(D f^{-1}|_{E^{s}(x)})\}$ . Therefore,

$$ \begin{align*} \begin{aligned} C\varepsilon(x)\leq C_0^\kappa-1\leq\min\{m(D f|_{E^{u}(x)})^{\kappa}-1,m(D f^{-1}|_{E^{s}(x)})^{\kappa}-1\}. \end{aligned} \end{align*} $$

Case 2. $ \varepsilon (x)=\varepsilon d(x,\partial O)^\beta $ . Now we have $d(x,\partial O)\leq r_0$ . Since $\beta =\beta ^u\vee \beta ^s$ , $C=C^u\wedge C^s$ , and $\kappa =\kappa ^u\vee \kappa ^s$ , the required inequality is an immediate result of Assumption US(ii) and (iii).

Proof. We give the proofs for equations (3.5) and (3.6). The proofs for $f^{-1}$ are almost identical, the major change being the substitution of $f^{-1}$ for f.

For the companion inequality in equation (3.5), there are four cases. It is worth pointing out that Case 2 is the most interesting, since it is the case where both x and $f(x)$ are close to $\partial O$ , and hence there is little hyperbolicity.

Case 1. $Q(x)=Q(f(x))=\varepsilon ^{{2}/({\alpha -\delta })}r_0^\gamma $ . In this case, equation (3.5) is straightforward since for any $x\in O$ ,

$$ \begin{align*} \begin{aligned} \frac{C}{\|Df|_{E^s(x)}\|^{-\kappa}+(C-1)}\leq1=\frac{Q(f(x))}{Q(x)}=1\leq\frac{1}{C}(m(D f|_{E^{u}(x)})^{\kappa}-1)+1. \end{aligned} \end{align*} $$

Case 2. $Q(x)=\varepsilon ^{{2}/({\alpha -\delta })}d(x, \partial O)^\gamma $ and $Q(f(x))=\varepsilon ^{{2}/({\alpha -\delta })}d(f(x), \partial O)^\gamma $ . Equivalently, $d(x, \partial O) \leq r_0\leq r_0^u$ and $d(f(x), \partial O) \leq r_0\leq r_0^s$ . Taking Assumption US(ii) and (iii) into consideration, we find in this case that

$$ \begin{align*} \bigg(\frac{d(f(x), \partial O)}{d(x, \partial O)}\bigg)^{\gamma^u}\leq\frac{1}{C^u}(m(D f|_{E^{u}(x)})^{\kappa^u}-1)+1 \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \bigg(\frac{d(x, \partial O)}{d(f(x), \partial O)}\bigg)^{\gamma^s}&\leq\frac{1}{C^s}(m(D f^{-1}|_{E^{s}(f(x))})^{\kappa^s}-1)+1\notag\\ &=\frac{1}{C^s}(\|Df_{E^{s}(x)}\|^{-\kappa^s}-1)+1, \end{aligned} \end{align*} $$

where in the last equality, we used that $m(D f^{-1}|_{E^{s}(f(x))})^{\kappa ^s}=\|D f|_{E^{s}(x)}\|^{-\kappa ^s}$ .

If $({d(f(x),\partial O)}/{d(x,\partial O)})\geq 1$ , on the one hand,

$$ \begin{align*} \frac{Q(f(x))}{Q(x)}=\bigg(\frac{d(f(x),\partial O)}{d(x,\partial O)}\bigg)^\gamma>1\geq\frac{C}{\|Df|_{E^s(x)}\|^{-\kappa}+(C-1)}. \end{align*} $$

On the other hand, since $\gamma \leq \gamma ^u$ , $C\leq C^u$ , $\kappa ^u\leq \kappa $ , we have

$$ \begin{align*} \begin{aligned} \frac{Q(f(x))}{Q(x)}&=\bigg(\frac{d(f(x),\partial O)}{d(x,\partial O)}\bigg)^\gamma\notag\\ &\leq\bigg(\frac{d(f(x), \partial O)}{d(x, \partial O)}\bigg)^{\gamma^u}\notag\\ &\leq\frac{1}{C^u}(m(D f|_{E^{u}(x)})^{\kappa^u}-1)+1\notag\\ &\leq\frac{1}{C}(m(D f|_{E^{u}(x)})^{\kappa}-1)+1. \end{aligned} \end{align*} $$

Therefore, the companion inequality in equation (3.5) holds when $({d(f(x),\partial O)}/ {d(x,\partial O)})\geq 1$ in Case 2.

Now if $0<({d(f(x),\partial O)}/{d(x,\partial O)})\leq 1$ , then

$$ \begin{align*} \frac{Q(f(x))}{Q(x)}\leq1\leq\frac{1}{C}(m(D f|_{E^{u}(x)})^{\kappa}-1)+1. \end{align*} $$

Using $\gamma \leq \gamma ^s$ , $C\leq C^s$ , $\kappa ^s\leq \kappa $ , we obtain

$$ \begin{align*} \begin{aligned} \frac{Q(f(x))}{Q(x)}&=\bigg(\frac{d(f(x),\partial O)}{d(x,\partial O)}\bigg)^\gamma\notag\\ &\geq\bigg(\frac{d(f(x), \partial O)}{d(x, \partial O)}\bigg)^{\gamma^s}\notag\\ &\geq\bigg[\frac{1}{C^s}(\|Df_{E^{s}(x)}\|^{-\kappa^s}-1)+1\bigg]^{-1}\notag\\ &\geq\bigg[\frac{1}{C}(\|Df_{E^{s}(x)}\|^{-\kappa}-1)+1\bigg]^{-1}\notag\\ &=\frac{C}{\|Df|_{E^s(x)}\|^{-\kappa}+(C-1)}. \end{aligned} \end{align*} $$

This completes the proof of equation (3.5) in Case 2.

Case 3. $Q(x)=\varepsilon ^{{2}/({\alpha -\delta })}d(x, \partial O)^\gamma $ and $Q(f(x))=\varepsilon ^{{2}/({\alpha -\delta })}r_0^\gamma $ . Now $d(x, \partial O)^\gamma \leq r_0^\gamma \leq d(f(x), \partial O)^\gamma $ , so it follows that

$$ \begin{align*} \frac{Q(f(x))}{Q(x)}=\frac{r_0^\gamma}{d(x, \partial O)^\gamma}\geq1\geq\frac{C}{\|Df|_{E^s(x)}\|^{-\kappa}+(C-1)}. \end{align*} $$

In this case, since $d(x, \partial O)\leq r_0^u$ and $({d(f(x),\partial O)}/{d(x,\partial O)})\geq 1$ , we have

$$ \begin{align*} \begin{aligned} \frac{Q(f(x))}{Q(x)}&\leq\bigg(\frac{d(f(x),\partial O)}{d(x,\partial O)}\bigg)^\gamma\notag\\ &\leq\frac{1}{C}(m(D f|_{E^{u}(x)})^{\kappa}-1)+1 \end{aligned} \end{align*} $$

as in Case 2 when $({d(f(x),\partial O)}{d(x,\partial O)})\geq 1$ .

Case 4. $Q(x)=\varepsilon ^{{2}/({\alpha -\delta })}r_0^\gamma $ and $Q(f(x))=\varepsilon ^{{2}/({\alpha -\delta })}d(f(x), \partial O)^\gamma $ . In this case, $d(f(x), \partial O)^\gamma \leq r_0^\gamma \leq d(x, \partial O)^\gamma $ , then

$$ \begin{align*} \frac{Q(f(x))}{Q(x)}=\frac{d(f(x), \partial O)^\gamma}{r_0^\gamma}\leq1\leq\frac{1}{C}(m(D f|_{E^{u}(x)})^{\kappa}-1)+1. \end{align*} $$

Notice that $d(f(x), \partial O)\leq r_0^s$ and $0<({d(f(x),\partial O)}/{d(x,\partial O)})\leq 1$ , so we obtain

$$ \begin{align*} \begin{aligned} \frac{Q(f(x))}{Q(x)}&\geq\bigg(\frac{d(f(x),\partial O)}{d(x,\partial O)}\bigg)^\gamma\notag\\ &\geq\frac{C}{\|Df|_{E^s(x)}\|^{-\kappa}+(C-1)} \end{aligned} \end{align*} $$

as in Case 2 when $0<({d(f(x),\partial O)}/{d(x,\partial O)})\leq 1$ .

Hence, the companion inequality in equation (3.5) is proved. From Lemma 3.1, it remains to show that $C\varepsilon (x)\leq 1-\|Df|_{E^s(x)}\|^{\kappa }$ for small $\varepsilon>0$ such that

$$ \begin{align*} 0<\varepsilon\leq\min\bigg\{1,\frac{C_0^\kappa-1}{Cr_0^\beta},\frac{1-C_0^{-\kappa}}{Cr_0^\beta}, \frac{1}{A_s^\beta(C^f)^{\kappa^s}}\bigg\}, \end{align*} $$

where $A_s:=[({1}/{C^s})((C^f)^{\kappa ^s}-1)+1]^{1/\gamma ^s}>1$ and $C^f:={\max }_{x\in M}\{{\max }\{\|D_xf\|, \|D_x f^{-1}\|\}\}$ .

For any $x\in O$ , $f(x)\in O$ , there are the following two cases.

Case 1. $\varepsilon (f(x))=\varepsilon r_0^\beta $ . In this case, observe that $f(x)\in O(r_0)$ , so using equation (3.1), we get

$$ \begin{align*} C_0\leq m(Df^{-1}|_{E^s(f(x))})=\|Df|_{E^s(x)}\|^{-1}, \end{align*} $$

and hence

$$ \begin{align*} C\varepsilon(x)\leq C\varepsilon r_0^\beta\leq1-C_0^{-\kappa}\leq1-\|Df|_{E^s(x)}\|^{\kappa}. \end{align*} $$

Case 2. $\varepsilon (f(x))=\varepsilon d(f(x),\partial O)^\beta $ . Now $d(f(x),\partial O)\leq r_0^s$ , and from Assumption US(iii), we find

$$ \begin{align*} \bigg(\frac{d(x, \partial O)}{d(f(x), \partial O)}\bigg)^{\beta}\leq A_s^\beta \end{align*} $$

and

$$ \begin{align*} \begin{aligned} C^s d(f(x),\partial O)^{\beta^s}&\leq m(Df^{-1}|_{E^s(f(x))})^{\kappa^s}(1- m(Df^{-1}|_{E^s(f(x))})^{-\kappa^s})\notag\\ &=m(Df^{-1}|_{E^s(f(x))})^{\kappa^s}(1- \|D f|_{E^{s}(x)}\|^{\kappa^s})\notag\\ &\leq(C^f)^{\kappa^s}(1-\|D f|_{E^{s}(x)}\|^{\kappa^s}). \end{aligned} \end{align*} $$

Therefore, we obtain that

$$ \begin{align*} \begin{aligned} C\varepsilon(x)&\leq \varepsilon C d(x,\partial O)^\beta\notag\\ &\leq\varepsilon A_s^\beta C^sd(f(x), \partial O)^\beta\notag\\ &\leq \varepsilon A_s^\beta(C^f)^{\kappa^s} (1-\|D f|_{E^{s}(x)}\|^{\kappa^s})\notag\\ &\leq 1-\|D f|_{E^{s}(x)}\|^{\kappa}. \end{aligned}\\[-35pt] \end{align*} $$

Proof. Since $F_{x,y}$ is $C^{1+\alpha }$ , there is $ \hat {C}>0$ such that for any $v,w\in S_{x}$ ,

$$ \begin{align*} \|D_vF_{x,y}-D_wF_{x,y}\|\leq \hat{C}\|v-w\|^\alpha. \end{align*} $$

Fix $\varepsilon>0$ small enough so that

$$ \begin{align*} \varepsilon^{{\delta}/({\alpha-\delta})}\leq\min\bigg\{1,\frac{1}{(2\sqrt{2})^{\alpha-{\delta}/{2}}\hat{C}}\bigg\}. \end{align*} $$

Part (1) is free and part (2) is a special case of part (3) when $w=0$ since $D_0\phi _{x,y}=0$ . Then what we need to check is property (3).

In fact, we have

$$ \begin{align*} \begin{aligned} \|D_v\phi_{x,y}-D_w\phi_{x,y}\|&=\|D_vF_{x,y}-D_wF_{x,y}\|\notag\\ &\leq \hat{C}\|v-w\|^{\alpha-{\delta}/{2}}\|v-w\|^{\delta/2}\notag\\ &\leq (2\sqrt{2})^{\alpha-{\delta}/{2}}\hat{C}Q(x)^{\delta/2} Q(x)^{\alpha-\delta}\|v-w\|^{\delta/2}\notag\\ &\leq (2\sqrt{2})^{\alpha-{\delta}/{2}}\hat{C}\varepsilon^{{\delta}/({\alpha-\delta})}\cdot\varepsilon\varepsilon(x)\|v-w\|^{\delta/2}\notag\\ &\leq \varepsilon\varepsilon(x)\|v-w\|^{\delta/2}. \end{aligned} \end{align*} $$

By considering $f^{-1}$ in the charts $\exp _y$ of point y and $\exp _x^{-1}$ of $f^{-1}(y)$ when x is near $f^{-1}(y)$ , we can apply a similar treatment to $F_{x,y}^{-1}$ . The details are left to the reader.

Proof. We follow the same strategy of [Reference Sarig22, Proposition 4.11].

(1) To prove the non-emptiness of $V_x^{u}\cap V_x^{s}$ , we fix $V_x^{t}$ with representing function $\sigma _x^t$ . From a pedagogical perspective, the ideas of the proof are very simple. If we suppose the intersections of $V_x^{u}$ and $V_x^{s}$ are non-empty, it corresponds to points $(v^u,v^s)$ that satisfy $\exp _x(v^u,\sigma _x^u(v^u))=\exp _x(\sigma _x^s(v^s),v^s)$ , so it is the solution of the equation:

$$ \begin{align*} \begin{cases} v^u=\sigma_x^s(v^s), \\ v^s=\sigma_x^u(v^u). \end{cases} \end{align*} $$

Hence, the trick of the proof is to find the fixed points of $\sigma _x^s \circ \sigma _x^u$ on $B_x^u(Q(x))$ .

We first note that $\sigma _x^s \circ \sigma _x^u$ maps the ball $B_x^u(10^{-2}Q(x))$ into itself from equation (4.1) if $\varepsilon>0$ is small enough. It follows that $\sigma _x^s \circ \sigma _x^u$ is an $\varepsilon $ -contraction of $B_x^u(10^{-2}Q(x))$ into itself. Indeed, from condition (AM2), we have $\|D(\sigma _x^s\hspace{-0.5pt} \circ\hspace{-0.5pt} \sigma _x^u)\|_{C^0}\hspace{-0.5pt}\leq\hspace{-0.5pt} \|D\sigma _x^s\|_{C^0}\|D\sigma _x^u\|_{C^0}\hspace{-0.5pt}\leq\hspace{-0.5pt} \varepsilon $ . By the Banach fixed point theorem, $\sigma _x^s \circ \sigma _x^u$ has a unique fixed point: $(\sigma _x^s \circ \sigma _x^u)(v_0^u)=v_0^u,$ where $v_0^u\in B_x^u(10^{-2}Q(x))$ .

Let $v_0^s:=\sigma _x^u(v_0^u)$ , $w:=(v_0^u,v_0^s)$ , then $V_x^{u}$ intersects $V_x^{s}$ at $P:=\exp _{x}(w)$ . Indeed, $P \in V_x^{u}$ is evident because $v_0^s=\sigma _x^u(v_0^u)$ and $\|v_0^u\| \leq 10^{-2} Q(x)<Q(x).$ Since $v_0^u=(\sigma _x^s \circ \sigma _x^u)(v_0^u)=\sigma _x^s (v_0^s)$ and

$$ \begin{align*} \|v_0^s\|\leq\|\sigma_x^u(0)\|+\operatorname{Lip}(\sigma_x^u)\|v_0^u\| \leq10^{-3}Q(x)+\sqrt{\varepsilon}\cdot10^{-2}Q(x)\leq10^{-2}Q(x)<Q(x), \end{align*} $$

we have $P \in V_x^{s}$ . Moreover, we also see that $\|w\|\leq \|v_0^u\|+\|v_0^s\| \leq 50^{-1} Q(x)<d(x,\partial O)$ , and thus $P\in O$ .

To prove the uniqueness in the bigger ball $B_x^u(Q(x))$ , we note that $\sigma _x^s \circ \sigma _x^u$ maps $B_x^u(Q(x))$ into itself if $\varepsilon>0$ is small enough, and the same calculation as above shows that such a map has a unique fixed point even in $B_x^u(Q(x))$ .

(2) Now we verify that P is a Lipschitz function of $V_x^{u}, V_x^{s}$ . For every $x\in O$ , suppose $V_{i}^{t}\in \mathscr {V}_x^{t}$ are represented by $\sigma ^t_{i}$ , $i=1,2$ . Let $P_{i}=\exp _{x}(w_i)=\exp _{x}(v^u_{i}, v^s_{i})$ denote the intersection points of $V_{i}^{u} \cap V_{i}^{s}$ . We saw above that $v^s_{i}=\sigma ^u_{i}(v^u_{i})$ and that $v^u_{i}=\sigma ^s_{i}(v^s_{i})$ because $v^u_{i}$ is a fixed point of $\sigma ^s_{i}\circ \sigma ^u_{i}$ .

Since $\exp _x$ is 2-Lipschitz, this gives that $d(P_1,P_2)\leq 2\|w_1-w_2\|$ . We have

$$ \begin{align*} \|w_1-w_2\|&\leq\|v^u_1-v^u_2\|+\|v^s_1-v^s_2\|\notag\\ &=\|\sigma_1^s(v^s_1)-\sigma_2^s(v^s_2)\|+\|\sigma_1^u(v^u_1)-\sigma_2^u(v^u_2)\|\notag\\ &\leq\|\sigma_1^s(v^s_1)-\sigma_1^s(v^s_2)\|+\|\sigma_1^s(v^s_2)-\sigma_2^s(v^s_2)\|\\&\quad+\|\sigma_1^u(v^u_1)-\sigma_1^u(v^u_2)\|+\|\sigma_1^u(v^u_2)-\sigma_2^u(v^u_2)\|\notag\\ &\leq\sqrt{\varepsilon}(\|v^s_1-v^s_2\|+\|v^u_1-v^u_2\|)+\|\sigma_1^s-\sigma_2^s\|_{C^0}+\|\sigma_1^u-\sigma_2^u\|_{C^0}. \end{align*} $$

Therefore, we conclude that

$$ \begin{align*} \begin{aligned} d(P_1,P_2)&\leq2( \|v^u_1-v^u_2\|+\|v^s_1-v^s_2\|)\notag\\ &\leq\frac{2}{1-\sqrt{\varepsilon}}(d_{C^0}(V^u_1,V^u_2)+d_{C^0}(V^s_1,V^s_2)). \end{aligned}\\[-43pt] \end{align*} $$

Proof. We give the proof for part (1), the proof of part (2) follows in a similar manner and is left to the reader.

First, we fix

$$ \begin{align*} 0<\varepsilon\leq\min\bigg\{1,\frac{1}{\gamma-1},(2\gamma C_1)^{-({\alpha-\delta})/{2}}, \bigg(\frac{C_1}{2}\bigg)^{({\alpha-\delta})/{4}}, \frac{C-1}{1+C_2}\bigg\}. \end{align*} $$

Since $d(f(x),y)<\delta ^u(\varepsilon (x))Q(x)^2Q(f(x))$ , we have $({d(f(x),y)}/{d(f(x),\partial O)})<\delta ^u(\varepsilon (x))Q(x)^2<\varepsilon $ . By the fact that $(1+a)^\gamma \leq 1+2\gamma a$ if $0<a\leq ({1}/({\gamma -1}))$ , we obtain

$$ \begin{align*} \begin{aligned} Q(y)&\leq\varepsilon^{{2}/({\alpha-\delta})}\min\{r_0^\gamma,(d(f(x),\partial O)+d(y,f(x)))^\gamma\}\notag\\ &\leq\varepsilon^{{2}/({\alpha-\delta})}\min\bigg\{r_0^\gamma,d(f(x),\partial O)^\gamma\bigg(1+2\gamma\frac{d(y,f(x))}{d(f(x),\partial O)}\bigg)\bigg\}\notag\\ &\leq Q(f(x))+2\gamma\varepsilon^{{2}/({\alpha-\delta})}d(y,f(x))\notag\\ &\leq Q(f(x))+ \frac{d(y,f(x))}{C_1}. \end{aligned} \end{align*} $$

Recall that $0<\delta ^u(\varepsilon (x))<\varepsilon \varepsilon (x)$ in Proposition 3.1 and using Lemma 3.2, we have

$$ \begin{align*} \begin{aligned} &\frac{C_1Q(y)+d(f(x),y)+C_1C_2\varepsilon\varepsilon(x)Q(x)}{C_1Q(x)}\\ &\quad\leq\frac{Q(f(x))}{Q(x)}+\frac{2d(y,f(x))}{C_1Q(x)}+C_2\varepsilon\varepsilon(x)\notag\\ &\quad\leq\frac{Q(f(x))}{Q(x)}+\frac{2}{C_1}\varepsilon^{{4}/({\alpha-\delta})}\delta^u(\varepsilon(x))+C_2\varepsilon\varepsilon(x)\notag\\ &\quad\leq \frac{1}{C}(m(Df|_{E^u(x)})^\kappa-1)+1+(1+C_2)\varepsilon\varepsilon(x)\notag\\ &\quad\leq \frac{1}{C}(m(Df|_{E^u(x)})^\kappa-1)+1+\frac{C-1}{C}(m(Df|_{E^u(x)})^\kappa-1)\notag\\ &\quad=m(Df|_{E^u(x)})^\kappa, \end{aligned} \end{align*} $$

proving part (1).

Proof. The proof is a straightforward adaption of arguments in [Reference Sarig22, Proposition 4.12]. We give the proof in the case of $F_{x,y}$ . The case of $F_{x,y}^{-1}$ is symmetric.

(1) The well-defined nature of $(G_{x,y}^{\sigma _x^u})^{-1}$ is a consequence of the expansion property of $Df|_{E^u(x)}$ . For every $v_x^u\in B_x^u(Q(x))$ , by property (AM2), we know that $\|D_{v_x^u}{\sigma ^u_x}\|\leq 1$ if $0<\varepsilon \leq \min \{1,{C}/{4}\}$ . Therefore, from Proposition 3.1(2) and Lemma 3.1, we have

$$ \begin{align*} \begin{aligned} \|D_{v_x^u}G_{x,y}^{\sigma^u_x}\|&\geq m(D_{x,y}^{uu})-\|D_{x,y}^{us}\|\ \|D_{v_x^u}{\sigma^u_x}\|-\|D_{(v_x^u,\sigma^u_x(v_x^u))}\phi^u_{x,y}\|(1+\|D_{v_x^u}{\sigma^u_x}\|)\notag\\ &\geq m(Df|_{E^{u}(x)})-\varepsilon\varepsilon(x)-\varepsilon\varepsilon(x)-2\varepsilon\varepsilon(x)\notag\\ &\geq m(Df|_{E^{u}(x)})^\kappa-4\varepsilon\varepsilon(x)>1. \end{aligned} \end{align*} $$

It follows that $G_{x,y}^{\sigma ^u_x}$ is $[m(Df|_{E^{u}(x)})-4\varepsilon \varepsilon (x)]$ -expanding and hence one-to-one, and $(G_{x,y}^{\sigma _x^u})^{-1}$ is well defined on $G_{x,y}^{\sigma _x^u}(B_x^u(Q(x)))$ .

By invariance of the domain, to prove $G_{x,y}^{\sigma _x^u}(B_x^u(Q(x)))\supset B_y^u(Q(y))$ , the problem reduces to show that

$$ \begin{align*} \|G_{x,y}^{\sigma^u_x}(v_x^u)\|\geq Q(y) \end{align*} $$

for any $v_x^u\in \partial B_x^u(Q(x))$ .

According to

$$ \begin{align*} \begin{aligned} \|G_{x,y}^{\sigma^u_x}(0)\|&=\|D_{x,y}^{us}\circ\sigma^u_x(0)+\phi^u_{x,y}(0,\sigma^u_x(0))\|\notag\\ &\leq\|D_{x,y}^{us}\|\ \|\sigma^u_x(0)\|+\|\phi_{x,y}(0,0)\|+\operatorname{Lip}(\phi_{x,y})\|\sigma^u_x(0)\|,\\ \end{aligned} \end{align*} $$

by property (AM1) and Proposition 3.1(2), we have

(4.4) $$ \begin{align} \|G_{x,y}^{\sigma^u_x}(0)\|\leq2\varepsilon\varepsilon(x)\times 10^{-3}Q(x)+d(f(x),y). \end{align} $$

In particular, using Lemma 4.1 for $C_1=1, C_2=5$ , and the fact that $G_{x,y}^{\sigma ^u_x}$ is $[m(Df|_{E^{u}(x)})-4\varepsilon \varepsilon (x)]$ -expanding, we have that for any $v_x^u\in \partial B_x^u(Q(x))$ ,

$$ \begin{align*} \begin{aligned} \|G_{x,y}^{\sigma^u_x}(v_x^u)\|&\geq \|G_{x,y}^{\sigma^u_x}(v_x^u)-G_{x,y}^{\sigma^u_x}(0)\|-\|G_{x,y}^{\sigma^u_x}(0)\|\notag\\ &\geq (m(Df|_{E^{u}(x)})-4\varepsilon\varepsilon(x))Q(x)-2\varepsilon\varepsilon(x)\times 10^{-3}Q(x)-d(f(x),y)\notag\\ &\geq (m(Df|_{E^{u}(x)})^\kappa-5\varepsilon\varepsilon(x))Q(x)-d(f(x),y)\notag\\ &\geq Q(y). \end{aligned} \end{align*} $$

(2) We first prove $\sigma ^u_y$ satisfies property (AM1). Denote $v_{0x}^u:=(G_{x,y}^{\sigma ^u_x})^{-1}(0)$ , since $G_{x,y}^{\sigma ^u_x}$ is $[m(Df|_{E^{u}(x)})-4\varepsilon \varepsilon (x)]$ -expanding on $B_x^u(Q(x))$ , in particular,

$$ \begin{align*} \|D_{v_y^u}(G_{x,y}^{\sigma^u_x})^{-1})\|<1\quad \text{for all } v_y^u\in G_{x,y}^{\sigma_x^u}(B_x^u(Q(x))). \end{align*} $$

By equation (4.4), we have

$$ \begin{align*} \begin{aligned} \|v_{0x}^u\|&=\|(G_{x,y}^{\sigma^u_x})^{-1}(0)-(G_{x,y}^{\sigma^u_x})^{-1}\circ G_{x,y}^{\sigma^u_x}(0)\|\notag\\ &\leq\|D(G_{x,y}^{\sigma^u_x})^{-1})\|_{C^0}\|G_{x,y}^{\sigma^u_x}(0)\|\notag\\ &\leq2\varepsilon\varepsilon(x)\times 10^{-3}Q(x)+d(f(x),y)\notag\\ &\leq3\varepsilon\varepsilon(x)\times 10^{-3}Q(x). \end{aligned} \end{align*} $$

Note that from equation (3.6) of Lemma 3.2, it follows that $\|D_{x,y}^{ss}\|\leq \|Df|_{E^s(x)}\|^\kappa +\varepsilon \varepsilon (x)\leq 1$ if $0<\varepsilon \leq \min \{1,C\}$ . According to properties (AM1), (AM2), and Lemma 4.1 for $C_3=10^{-3}, C_4=13$ , we obtain

$$ \begin{align*} \begin{aligned} \|\sigma^u_y(0)\|&=\|L_{x,y}^{\sigma_x^u}(v_{0x}^u)\|\notag\\ &=\|D_{x,y}^{ss}\circ\sigma_x^u(v_{0x}^u)+D_{x,y}^{su}(v_{0x}^u)+ \phi_{x,y}^s(v_{0x}^u,\sigma_x^u(v_{0x}^u))\|\notag\\ &\leq\|D_{x,y}^{ss}\|(\|\sigma_x^u(0)\|+\operatorname{Lip}(\sigma_x^u)\|v_{0x}^u\|)+\|D_{x,y}^{su}\|\ \|v_{0x}^u\|+ \|\phi_{x,y}^s(0,0)\|\\&\quad+\operatorname{Lip}(\phi^s_{x,y})(1+\operatorname{Lip}(\sigma_x^u))\|v_{0x}^u\|\notag\\ &\leq(\|Df|_{E^s(x)}\|^\kappa+\varepsilon\varepsilon(x))10^{-3}Q(x)+12\varepsilon\varepsilon(x)\times 10^{-3}Q(x)+d(f(x),y)\notag\\ &\leq 10^{-3}Q(y), \end{aligned} \end{align*} $$

proving property (AM1).

Next we prove $\|D\sigma ^u_y\|_{C^0}\leq \sqrt {\varepsilon }$ , where the norm is taken in $G_{x,y}^{\sigma _x^u}(B_x^u(Q(x)))$ .

For all $v_y^u\in G_{x,y}^{\sigma _x^u}(B_x^u(Q(x)))$ , $v_x^u=(G_{x,y}^{\sigma _x^u})^{-1}(v_y^u)\in B_x^u(Q(x))$ . Thanks to $\|D_{v_y^u} (G_{x,y}^{\sigma ^u_x})^{-1})\|<1$ , property (AM2) is easy to be verified from

$$ \begin{align*} \begin{aligned} \|D_{v_y^u}\sigma^u_y\|&=\|D_{x,y}^{ss}\cdot D_{v_{x}^u}\sigma_x^u\cdot D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}+D_{x,y}^{su}\cdot D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\\&\quad+ D_{v_{x}^u}\phi_{x,y}^s\cdot D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\|\notag\\ &\leq\|D_{x,y}^{ss}\|\ \|D\sigma_x^u\|_{C^0}+3\varepsilon\varepsilon(x)\notag\\ &\leq\|Df|_{E^s(x)}\|\sqrt{\varepsilon}+4\varepsilon\varepsilon(x)\notag\\ &\leq\sqrt{\varepsilon} \end{aligned} \end{align*} $$

if $0<\varepsilon \leq \min \{1,({C}/{4})^2\}$ .

Finally, we have to show that $\sigma ^u_y$ satisfies property (AM3) also. From

$$ \begin{align*} \sigma_y^u=L_{x,y}^{\sigma_x^u}\circ(G_{x,y}^{\sigma_x^u})^{-1} =[D^{ss}_{x,y}\circ\sigma_x^u(\cdot)+D^{su}_{x,y}(\cdot)+\phi^s_{x,y}(\cdot,\sigma_x^u(\cdot))]\circ(G_{x,y}^{\sigma_x^u})^{-1}, \end{align*} $$

we have

$$ \begin{align*} \begin{aligned} \|D\sigma_y^u\|_{C^{\delta/2}}&=\|(D^{ss}_{x,y}\cdot D\sigma_x^u+D^{su}_{x,y} +D\phi^s_{x,y})\cdot D(G_{x,y}^{\sigma_x^u})^{-1}\|_{C^{\delta/2}}\notag\\ &\leq(\|D^{ss}_{x,y}\|\cdot \|D\sigma_x^u\|_{C^{\delta/2}}+\|D^{su}_{x,y}\|+2\|D\phi_{x,y}\|_{C^{\delta/2}(S_x)}) \|D(G_{x,y}^{\sigma_x^u})^{-1}\|_{C^{\delta/2}}.\notag\\ \end{aligned} \end{align*} $$

Proof of Claim

For $v_y^u,w_y^u\in G_{x,y}^{\sigma _x^u}(B_x^u(Q(x)))$ , notice that $D_{v_y^u}(G_{x,y}^{\sigma _x^u})^{-1}\cdot D_{v_x^u}G_{x,y}^{\sigma _x^u}$ and $D_{w_x^u}G_{x,y}^{\sigma _x^u}\cdot D_{w_y^u}(G_{x,y}^{\sigma _x^u})^{-1}$ are identity maps, where $v_y^u=G_{x,y}^{\sigma _x^u}(v_x^u)$ , $w_x^u=(G_{x,y}^{\sigma _x^u})^{-1}(w_y^u)$ . Thus, we have

$$ \begin{align*} \begin{aligned} &\|D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}-D_{w_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\|\notag\\ &\quad=\|D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\cdot(D_{w_x^u}G_{x,y}^{\sigma_x^u}-D_{v_x^u}G_{x,y}^{\sigma_x^u})\cdot D_{w_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\|\notag\\ &\quad=\|D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\cdot(D_{x,y}^{us}D_{w_x^u}\sigma_x^u+D_{w_x^u}\phi_{x,y}^u\\&\qquad-D_{x,y}^{us}D_{v_x^u}\sigma_x^u-D_{v_x^u}\phi_{x,y}^u)\cdot D_{w_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\|. \end{aligned} \end{align*} $$

First, to estimate $\|D_{w_x^u}\phi _{x,y}^u-D_{v_x^u}\phi _{x,y}^u\|$ in parentheses, a little manipulation is needed. If $0<\varepsilon \leq 1$ , by property (AM3) and Proposition 3.1(3), we can get

$$ \begin{align*} \begin{aligned} &\|D_{w_x^u}\phi^u_{x,y}(\cdot,\sigma_x^u(\cdot))-D_{v_x^u}\phi^u_{x,y}(\cdot,\sigma_x^u(\cdot))\|\notag\\ &\quad=\left\|D_{(w_x^u,\sigma_x^u(w_x^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{w_x^u}\sigma_x^u \end{array}\right]-D_{(v_x^u,\sigma_x^u(v_x^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_x^u}\sigma_x^u \end{array}\right]\right\|\notag\\ &\quad\leq\left\|D_{(w_x^u,\sigma_x^u(w_x^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{w_x^u}\sigma_x^u \end{array}\right]-D_{(w_x^u,\sigma_x^u(w_x^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_x^u}\sigma_x^u \end{array}\right]\right\|\notag\\ &\qquad+\left\|D_{(w_x^u,\sigma_x^u(w_x^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_x^u}\sigma_x^u \end{array}\right]-D_{(v_x^u,\sigma_x^u(v_x^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_x^u}\sigma_x^u \end{array}\right]\right\|\notag\\ &\quad\leq\frac{\varepsilon\varepsilon(x)}{2}\|w_x^u-v_x^u\|^{\delta/2}+4\varepsilon\varepsilon(x)\|w_x^u-v_x^u\|^{\delta/2}\notag\\ &\quad=\frac{9\varepsilon\varepsilon(x)}{2}\|w_x^u-v_x^u\|^{\delta/2}. \end{aligned} \end{align*} $$

For the rest in the parentheses, from property (AM3), we find

$$ \begin{align*} \|D_{x,y}^{us}D_{w_x^u}\sigma_x^u-D_{x,y}^{us}D_{v_x^u}\sigma_x^u\|\leq\tfrac{1}{2}\varepsilon\varepsilon(x)\|w_x^u-v_x^u\|^{\delta/2}; \end{align*} $$

therefore, we obtain that

$$ \begin{align*} \begin{aligned} &\|D_{v_y^u}(G_{x,y}^{\sigma_x^u})^{-1}-D_{w_y^u}(G_{x,y}^{\sigma_x^u})^{-1}\|\notag\\ &\quad\leq \|D(G_{x,y}^{\sigma_x^u})^{-1}\|^2_{C^0}\ 5\varepsilon\varepsilon(x)\|w_x^u-v_x^u\|^{\delta/2}\notag\\ &\quad\leq 5\varepsilon\varepsilon(x)\|D(G_{x,y}^{\sigma_x^u})^{-1}\|^{2+{\delta}/{2}}_{C^0}\|w_y^u-v_y^u\|^{\delta/2} \end{aligned} \end{align*} $$

or

$$ \begin{align*} \operatorname{Hol}_{\delta/2}(D(G^{\sigma_x^u}_{x,y})^{-1})\leq5\varepsilon\varepsilon(x)\|D(G_{x,y}^{\sigma_x^u})^{-1}\|^{2+{\delta}/{2}}_{C^0}\leq5\varepsilon\varepsilon(x). \end{align*} $$

Using $G_{x,y}^{\sigma ^u_x}$ is $[m(Df|_{E^{u}(x)})-4\varepsilon \varepsilon (x)]$ -expanding again, that is,

$$ \begin{align*} \|D(G_{x,y}^{\sigma^u_x})^{-1}\|_{C^0}\leq\frac{1}{m(Df|_{E^{u}(x)})-4\varepsilon\varepsilon(x)}, \end{align*} $$

we get finally

$$ \begin{align*} \begin{aligned} \|D(G^{\sigma_x^u}_{x,y})^{-1}\|_{C^{\delta/2}}&=\|D(G^{\sigma_x^u}_{x,y})^{-1}\|_{C^0}+ \operatorname{Hol}_{\delta/2}(D(G^{\sigma_x^u}_{x,y})^{-1})\notag\\ &\leq\frac{1}{m(Df|_{E^{u}(x)})-4\varepsilon\varepsilon(x)}+5\varepsilon\varepsilon(x)\notag\\ &<1 \end{aligned} \end{align*} $$

if $0<\varepsilon <\min \{1,{C}/{5(C^f+4)}\}$ . This completes the proof of the claim.

We are now turning to the proof of $\|D\sigma _y^u\|_{C^{\delta /2}}\leq \tfrac 12$ . From Proposition 3.1(2) and (3), we have $\|D\phi _{x,y}\|_{C^{\delta /2}(S_x)}\leq 2\varepsilon \varepsilon (x)$ . By using equation (3.6) of Lemma 3.2 and combining with the claim above, we obtain that

$$ \begin{align*} \begin{aligned} \|D\sigma_y^u\|_{C^{\delta/2}}&\leq(\|D^{ss}_{x,y}\|\cdot \|D\sigma_x^u\|_{C^{\delta/2}}+\|D^{su}_{x,y}\|+2\|D\phi_{x,y}\|_{C^{\delta/2}(S_x)}) \|D(G_{x,y}^{\sigma_x^u})^{-1}\|_{C^{\delta/2}}\notag\\ &\leq\|D^{ss}_{x,y}\|\cdot \|D\sigma_x^u\|_{C^{\delta/2}}+\|D^{su}_{x,y}\|+2\|D\phi_{x,y}\|_{C^{\delta/2}(S_x)}\notag\\ &\leq\tfrac{1}{2}(\|Df|_{E^s(x)}\|+\varepsilon\varepsilon(x))+5\varepsilon\varepsilon(x)\notag\\ &\leq \tfrac{1}{2} \end{aligned} \end{align*} $$

when $0<\varepsilon \leq \min \{1, {C}/{11}\}$ , and the proof of this proposition is completed.

Proof.

1. (1) Since $\mathscr {F}_{x,y}^{u}[V_x^u]=\exp _{y}\{(v_y^u,\sigma _y^u(v_y^u)):\|v_y^u\|\leq Q(y)\}$ , it suffices to check that

   $$ \begin{align*} F_{x,y}(0,\sigma^u_x(0))=(G_{x,y}^{\sigma_x^u}(0),L_{x,y}^{\sigma_x^u}(0))\in\{(v_y^u,\sigma_y^u(v_y^u)):\|v_y^u\|\leq Q(y)\}, \end{align*} $$
   or check $\|G_{x,y}^{\sigma _x^u}(0)\|\leq Q(y)$ , which is straightforward.

2. (2) For any $V_y^s\in \mathscr {V}_{y}^s$ with the representing function $\sigma _y^s$ , by Proposition 4.1(1), $\mathscr {F}_{x,y}^{u}[V_x^u]$ and $V_y^s$ intersect at a single point. According to Definition 4.4, $f(V_x^u)$ contains a u-manifold $\mathscr {F}_{x,y}^{u}[V_x^u]\in \mathscr {V}_{y}^u$ ; therefore, $f(V_x^u)$ and $V_y^s$ intersect at least at one point.

Since

$$ \begin{align*} f(V_x^u)=\exp_{y}\{(v_y^u,\sigma_y^u(v_y^u)):v_y^u\in G_{x,y}^{\sigma_x^u}(B_x^u(Q(x)))\} \end{align*} $$

with $\operatorname {Lip}(\sigma _y^u)\leq \sqrt {\varepsilon }$ , by the Kirszbraun theorem, we can extend $\sigma _y^s$ to a Lipschitz function $\tilde {\sigma }_y^s$ on $G_{x,y}^{\sigma _x^u}(B_x^u(Q(x)))$ with $\operatorname {Lip}(\tilde {\sigma }_y^s)=\operatorname {Lip}(\sigma _y^s)\leq \sqrt {\varepsilon }$ . The extension represents a Lipschitz manifold $\tilde {V}_y^s\supset V_y^s$ . Hence, the proof is quite similar to that given earlier in Proposition 4.1(1) to conclude that $f(V_x^u)\cap \tilde {V}_y^s$ consists of a single point and so is omitted.

Proof. The proof is similar to the proof of [Reference Sarig22, Proposition 4.14]. We prove the proposition for $\mathscr {F}^u_{x,y}$ and leave the case of $\mathscr {F}^s_{x,y}$ to the reader.

Let $V_{1}, V_{2}$ with representing functions $\sigma _{1}, \sigma _{2}$ , and let $\tilde {\sigma }_{1}, \tilde {\sigma }_{2}$ be the representing functions of $\mathscr {F}^u_{x,y}(V_{1}), \mathscr {F}^u_{x,y}(V_{2}),$ that is, $\tilde {\sigma }_{i}=L^{\sigma _i}_{x,y}\circ (G^{\sigma _i}_{x,y})^{-1}|_{B_y^u(Q(y))}$ , $i=1,2$ .

(1) For any $v_y^u\in B_y^u(Q(y))$ , denote $v_{1x}^u=(G^{\sigma _1}_{x,y})^{-1}(v_y^u)$ , $v_{2x}^u=(G^{\sigma _2}_{x,y})^{-1}(v_y^u)$ . Some tedious manipulation yields

$$ \begin{align*} \begin{aligned} &\|\tilde{\sigma}_{1}(v_y^u)-\tilde{\sigma}_{2}(v_y^u)\|=\|\tilde{\sigma}_{1}\circ G^{\sigma_1}_{x,y}(v_{1x}^u)-\tilde{\sigma}_{2}\circ G^{\sigma_1}_{x,y}(v_{1x}^u)\|\notag\\ &\quad\leq\|\tilde{\sigma}_{1}\circ G^{\sigma_1}_{x,y}(v_{1x}^u)-\tilde{\sigma}_{2}\circ G^{\sigma_2}_{x,y}(v_{1x}^u)\|+\|\tilde{\sigma}_{2}\circ G^{\sigma_2}_{x,y}(v_{1x}^u)-\tilde{\sigma}_{2}\circ G^{\sigma_1}_{x,y}(v_{1x}^u)\|\notag\\ &\quad\leq\|D_{x,y}^{ss}(\sigma_1(v_{1x}^u)-\sigma_2(v_{1x}^u))\|+\|\phi_{x,y}^s(v_{1x}^u,\sigma_1(v_{1x}^u))-\phi_{x,y}^s(v_{1x}^u,\sigma_2(v_{1x}^u))\|\\&\qquad+\operatorname{Lip}(\tilde{\sigma}_{2}) \|G^{\sigma_2}_{x,y}(v_{1x}^u)-G^{\sigma_1}_{x,y}(v_{1x}^u)\|\notag \end{aligned} \end{align*} $$

$$ \begin{align*} \begin{aligned} &\qquad+\operatorname{Lip}(\phi_{x,y})) \|\sigma_2(v_{1x}^u)-\sigma_1(v_{1x}^u)\|\notag\\ &\quad\leq(\|D_{x,y}^{ss}\|+\|D_{x,y}^{us}\|+ 2\operatorname{Lip}(\phi_{x,y}))\|\sigma_2(v_{1x}^u)-\sigma_1(v_{1x}^u)\|\notag\\ &\quad\leq(\|Df|_{E^s(x)}\|+6\varepsilon\varepsilon(x))\|\sigma_2(v_{1x}^u)-\sigma_1(v_{1x}^u)\|. \end{aligned} \end{align*} $$

We remind the reader that $C\varepsilon (x)\leq 1-\|Df|_{E^s(x)}\|^\kappa \leq 1-\|Df|_{E^s(x)}\|$ and introduce a new constant

(4.5) $$ \begin{align} a:=\inf_{x\in O}\frac{\|Df|_{E^s(x)}\|^{1/2}-\|Df|_{E^s(x)}\|}{1-\|Df|_{E^s(x)}\|}>0, \end{align} $$

then if $0<\varepsilon \leq ({aC}/{6})$ , we get finally

$$ \begin{align*} \begin{aligned} \|\tilde{\sigma}_{1}-\tilde{\sigma}_{2}\|_{C^0}&\leq(\|Df|_{E^s(x)}\|+aC\varepsilon(x))\|\sigma_{1}-\sigma_{2}\|_{C^0}\notag\\ &\leq\|Df|_{E^s(x)}\|^{1/2}\|\sigma_{1}-\sigma_{2}\|_{C^0}. \end{aligned} \end{align*} $$

(2) In view of part (1), our task now is to estimate $\|D\tilde {\sigma }_{1}- D\tilde {\sigma }_{2}\|_{C^{0}}$ in terms of $\|D\sigma _{1}-D\sigma _{2}\|_{C^{0}}$ and $\|\sigma _{1}-\sigma _{2}\|_{C^{0}}^{\delta /2}$ . We will show that

$$ \begin{align*} \begin{aligned} &\|D\tilde{\sigma}_{1}-D\tilde{\sigma}_{2}\|_{C^0}\notag\\ &\quad\leq(\|Df|_{E^s(x)}\|+3\varepsilon\varepsilon(x))\|D\sigma_1-D\sigma_2\|_{C^0} +\tfrac{1}{2}(\|Df|_{E^s(x)}\|+25\varepsilon\varepsilon(x))\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}. \end{aligned} \end{align*} $$

For any $v_y^u\in B_y^u(Q(y))$ , let $v_{ix}^u=(G^{\sigma _i}_{x,y})^{-1}(v_y^u)$ as part (1), where $i=1,2$ . We have

$$ \begin{align*} \begin{aligned} &\|D_{v_y^u}\tilde{\sigma}_{1}-D_{v_y^u}\tilde{\sigma}_{2}\|\notag\\ &\quad=\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_1}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_2}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_1}_{x,y})^{-1} -D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\\ &\qquad+\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\qquad+\|D_{v_{2x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_2}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|, \end{aligned} \end{align*} $$

so we need to estimate each term above. We divide our proof of part (2) into four steps.

First, note that the last term is easy to estimate because $\|D(G_{x,y}^{\sigma _2})^{-1}\|<1$ and $L_{x,y}^{\sigma _1}(\cdot )=D^{ss}_{x,y}\circ \sigma _1(\cdot )+D^{su}_{x,y}(\cdot )+\phi ^s_{x,y}(\cdot ,\sigma _1(\cdot ))$ ,

$$ \begin{align*} \begin{aligned} &\|D_{v_{2x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_2}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq \|D_{v_{2x}^u}L^{\sigma_1}_{x,y} -D_{v_{2x}^u}L^{\sigma_2}_{x,y}\|\notag\\ &\quad\leq\|D^{ss}_{x,y}\|\cdot\|D_{v_{2x}^u}\sigma_1-D_{v_{2x}^u}\sigma_2\| +\|D_{v_{2x}^u}\phi^s_{x,y}(\cdot,\sigma_1(\cdot))-D_{v_{2x}^u}\phi^s_{x,y}(\cdot,\sigma_2(\cdot))\|. \end{aligned} \end{align*} $$

However, if $0<\varepsilon \leq 1$ , using Proposition 3.1(2), (3), and property (AM3), we get

$$ \begin{align*} \begin{aligned} &\|D_{v_{2x}^u}\phi^s_{x,y}(\cdot,\sigma_1(\cdot))-D_{v_{2x}^u}\phi^s_{x,y}(\cdot,\sigma_2(\cdot))\|\notag\\ &\quad=\left\|D_{(v_{2x}^u,\sigma_1(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_1 \end{array}\right]-D_{(v_{2x}^u,\sigma_2(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_2 \end{array}\right]\right\|\notag \end{aligned} \end{align*} $$

$$ \begin{align*} \begin{aligned} &\quad\leq\left\|D_{(v_{2x}^u,\sigma_1(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_1 \end{array}\right]-D_{(v_{2x}^u,\sigma_1(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_2 \end{array}\right]\right\|\notag\\ &\qquad+\left\|D_{(v_{2x}^u,\sigma_1(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_2 \end{array}\right]-D_{(v_{2x}^u,\sigma_2(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_2 \end{array}\right]\right\|\notag\\ &\quad\leq\varepsilon\varepsilon(x)\|D\sigma_1-D\sigma_2\|_{C^0}+4\varepsilon\varepsilon(x)\|\sigma_1-\sigma_2\|^{\delta/2}_{C^0}. \end{aligned} \end{align*} $$

Consequently, we infer that the last term satisfies

(4.6) $$ \begin{align} &\|D_{v_{2x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_2}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq(\|Df|_{E^s(x)}\|+2\varepsilon\varepsilon(x))\|D\sigma_1-D\sigma_2\|_{C^0}+4\varepsilon\varepsilon(x)\|\sigma_1-\sigma_2\|^{\delta/2}_{C^0}. \end{align} $$

The next thing to do is to estimate the second term:

$$ \begin{align*} \begin{aligned} &\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}-D_{v_{2x}^u}L^{\sigma_1}_{x,y}\|\notag\\ &\quad\leq\|D^{ss}_{x,y}\|\cdot\|D_{v_{1x}^u}\sigma_1-D_{v_{2x}^u}\sigma_1\| +\|D_{v_{1x}^u}\phi^s_{x,y}(\cdot,\sigma_1(\cdot))-D_{v_{2x}^u}\phi^s_{x,y}(\cdot,\sigma_1(\cdot))\|. \end{aligned} \end{align*} $$

Similarly, if $0<\varepsilon \leq 1$ , using Proposition 3.1(2), (3), and property (AM3), we have

$$ \begin{align*} \begin{aligned} &\|D_{v_{1x}^u}\phi^s_{x,y}(\cdot,\sigma_1(\cdot))-D_{v_{2x}^u}\phi^s_{x,y}(\cdot,\sigma_1(\cdot))\|\notag\\ &\quad=\left\|D_{(v_{1x}^u,\sigma_1(v_{1x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_1 \end{array}\right]-D_{(v_{2x}^u,\sigma_1(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_1 \end{array}\right]\right\|\notag\\ &\quad\leq\left\|D_{(v_{1x}^u,\sigma_1(v_{1x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_1 \end{array}\right]-D_{(v_{1x}^u,\sigma_1(v_{1x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_1 \end{array}\right]\right\|\notag\\ &\qquad+\left\|D_{(v_{1x}^u,\sigma_1(v_{1x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_1 \end{array}\right]-D_{(v_{2x}^u,\sigma_1(v_{2x}^u))}\phi^s_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{2x}^u}\sigma_1 \end{array}\right]\right\|\notag\\ &\quad\leq\frac{\varepsilon\varepsilon(x)}{2}\|v_{1x}^u-v_{2x}^u\|^{\delta/2}+2\varepsilon\varepsilon(x)\|v_{1x}^u-v_{2x}^u\|^{\delta/2}\notag\\ &\quad=\frac{5\varepsilon\varepsilon(x)}{2}\|v_{1x}^u-v_{2x}^u\|^{\delta/2}. \end{aligned} \end{align*} $$

Due to $G_{x,y}^{\sigma _i}(\cdot )=D_{x,y}^{uu}(\cdot )+D_{x,y}^{us}\sigma _i(\cdot )+\phi _{x,y}^u(\cdot ,\sigma _i(\cdot ))$ , we obtain

(4.7) $$ \begin{align} \|v_{1x}^u-v_{2x}^u\|&=\|(G^{\sigma_1}_{x,y})^{-1}(v_y^u)-(G^{\sigma_2}_{x,y})^{-1}(v_y^u)\|\notag\\ &=\|(G^{\sigma_1}_{x,y})^{-1}\circ (G^{\sigma_2}_{x,y}\circ(G^{\sigma_2}_{x,y})^{-1}(v_y^u))-(G^{\sigma_1}_{x,y})^{-1}\circ(G^{\sigma_1}_{x,y}\circ(G^{\sigma_2}_{x,y})^{-1}(v_y^u))\|\notag\\ &\leq\|D(G^{\sigma_1}_{x,y})^{-1}\|_{C^0}\|G^{\sigma_2}_{x,y}(v_{2x}^u)-G^{\sigma_1}_{x,y}(v_{2x}^u)\|\notag\\ &\leq\|D_{x,y}^{us}\|\ \|\sigma_2(v_{2x}^u)-\sigma_1(v_{2x}^u)\|+\operatorname{Lip}(\phi_{x,y})\|\sigma_2(v_{2x}^u)-\sigma_1(v_{2x}^u)\|\notag\\ &\leq2\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}\notag\\ &\leq\|\sigma_2-\sigma_1\|_{C^0}.\end{align} $$

Hence, the second term is estimated as

(4.8) $$ \begin{align} &\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\circ D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1} -D_{v_{2x}^u}L^{\sigma_1}_{x,y}\circ D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq\|D^{ss}_{x,y}\|\cdot\frac{1}{2}\|v_{1x}^u-v_{2x}^u\|^{\delta/2}+\frac{5\varepsilon\varepsilon(x)}{2}\|v_{1x}^u-v_{2x}^u\|^{\delta/2}\notag\\ &\quad\leq(\tfrac{1}{2}\|Df|_{E^s(x)}\|+3\varepsilon\varepsilon(x))\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}. \end{align} $$

Another step in the proof of part (2) is to establish an estimate of the first term. Since $\|D_{v_{1x}^u}L^{\sigma _1}_{x,y}\|<1$ and $\|D_{v_y^u}(G^{\sigma _i}_{x,y})^{-1}\|<1$ , $i=1,2$ , we have

$$ \begin{align*} \begin{aligned} &\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_1}_{x,y})^{-1} -D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq\|D_{v_y^u}(G^{\sigma_1}_{x,y})^{-1}-D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad=\|D_{v_y^u}(G^{\sigma_1}_{x,y})^{-1}(D_{v_{2x}^u}G^{\sigma_2}_{x,y}-D_{v_{1x}^u}G^{\sigma_1}_{x,y})D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq\|D_{v_{2x}^u}G^{\sigma_2}_{x,y}-D_{v_{1x}^u}G^{\sigma_2}_{x,y}\|+\|D_{v_{1x}^u}G^{\sigma_2}_{x,y}-D_{v_{1x}^u}G^{\sigma_1}_{x,y}\|. \end{aligned} \end{align*} $$

From the proof in property (AM3) of Proposition 4.2 and equation (4.7) above, we obtain that

$$ \begin{align*} \|D_{v_{2x}^u}G^{\sigma_2}_{x,y}-D_{v_{1x}^u}G^{\sigma_2}_{x,y}\|\leq5\varepsilon\varepsilon(x)\|v_{2x}^u-v_{1x}^u\|^{\delta/2}\leq5\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}. \end{align*} $$

At the same time, if $0<\varepsilon \leq 1$ , using Proposition 3.1(2), (3), and property (AM3), we get

$$ \begin{align*} \begin{aligned} &\|D_{v_{1x}^u}\phi^u_{x,y}(\cdot,\sigma_2(\cdot))-D_{v_{1x}^u}\phi^u_{x,y}(\cdot,\sigma_1(\cdot))\|\notag\\ &\quad=\left\|D_{(v_{1x}^u,\sigma_2(v_{1x}^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_2 \end{array}\right]-D_{(v_{1x}^u,\sigma_1(v_{1x}^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_1 \end{array}\right]\right\|\notag\\ &\quad\leq\left\|D_{(v_{1x}^u,\sigma_2(v_{1x}^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_2 \end{array}\right]-D_{(v_{1x}^u,\sigma_2(v_{1x}^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_1 \end{array}\right]\right\|\notag\\ &\qquad+\left\|D_{(v_{1x}^u,\sigma_2(v_{1x}^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_1 \end{array}\right]-D_{(v_{1x}^u,\sigma_1(v_{1x}^u))}\phi^u_{x,y}\left[\begin{array}{@{}c@{}} I \\ D_{v_{1x}^u}\sigma_1 \end{array}\right]\right\|\notag\\ &\quad\leq\varepsilon\varepsilon(x)\|D\sigma_2-D\sigma_1\|_{C^0}+2\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}. \end{aligned} \end{align*} $$

Hence, by using property (AM3) and equation (4.7) again, we get for this term that

(4.9) $$ \begin{align} &\|D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_1}_{x,y})^{-1} -D_{v_{1x}^u}L^{\sigma_1}_{x,y}\cdot D_{v_y^u}(G^{\sigma_2}_{x,y})^{-1}\|\notag\\ &\quad\leq\|D_{v_{2x}^u}G^{\sigma_2}_{x,y}-D_{v_{1x}^u}G^{\sigma_2}_{x,y}\|+\|D_{x,y}^{us}D_{v_{1x}^u}\sigma_2-D_{x,y}^{us}D_{v_{1x}^u}\sigma_2\|\notag\\ &\qquad+\|D_{v_{1x}^u}\phi^u_{x,y}(\cdot,\sigma_2(\cdot))-D_{v_{1x}^u}\phi^u_{x,y}(\cdot,\sigma_1(\cdot))\|\notag\\ &\quad\leq5\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|^{\delta/2}+\tfrac{1}{2}\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}\notag\\ &\qquad+\varepsilon\varepsilon(x)\|D\sigma_2-D\sigma_1\|_{C^0}+2\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}\notag\\ &\quad=\varepsilon\varepsilon(x)\|D\sigma_2-D\sigma_1\|_{C^0}+\frac{15}{2}\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}. \end{align} $$

Finally, plugging all estimates in equations (4.6), (4.8), and (4.9) together, it holds that

$$ \begin{align*} \begin{aligned} &\|D\tilde{\sigma}_{1}-D\tilde{\sigma}_{2}\|_{C^0}\notag\\ &\quad\leq\varepsilon\varepsilon(x)\|D\sigma_2-D\sigma_1\|_{C^0}+\frac{15}{2}\varepsilon\varepsilon(x)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}\notag\\ &\qquad+\big(\tfrac{1}{2}\|Df|_{E^s(x)}\|+3\varepsilon\varepsilon(x)\big)\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}\notag\\ &\qquad+(\|Df|_{E^s(x)}\|+2\varepsilon\varepsilon(x))\|D\sigma_1-D\sigma_2\|_{C^0}+4\varepsilon\varepsilon(x)\|\sigma_1-\sigma_2\|^{\delta/2}_{C^0}\notag\\ &\quad\leq(\|Df|_{E^s(x)}\|+3\varepsilon\varepsilon(x))\|D\sigma_1-D\sigma_2\|_{C^0} +\frac{1}{2}(\|Df|_{E^s(x)}\|+29\varepsilon\varepsilon(x))\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}\notag\\ &\quad\leq\|Df|_{E^s(x)}\|^{1/2}(\|D\sigma_1-D\sigma_2\|_{C^0}+\|\sigma_2-\sigma_1\|_{C^0}^{\delta/2}) \end{aligned} \end{align*} $$

if $0<\varepsilon \leq \min \{1,{aC}/{29}\}$ , where a is defined in equation (4.5). With the estimates of part (1), we conclude finally that

$$ \begin{align*} \|\tilde{\sigma}_{1}-\tilde{\sigma}_{2}\|_{C^1}\leq\|Df|_{E^s(x)}\|^{1/2}(\|\sigma_{1}-\sigma_{2}\|_{C^{1}} +\|\sigma_{1}-\sigma_{2}\|_{C^{0}}^{\delta/2}), \end{align*} $$

proving part (2).

Proof. Fix a countable subset $\mathscr {A}$ of O such that

$$ \begin{align*} O=\bigcup_{z\in\mathscr{A}}B(z,r(z)), \end{align*} $$

where $r(z)=\varepsilon \min \{\delta ^u(\varepsilon \cdot \varepsilon (z)),\delta ^s(\varepsilon \cdot \varepsilon (z))\}\cdot Q(z)^2\cdot \min \{Q(f(z)),Q(f^{-1}(z))\}$ .

(1) For all $r>0$ , since $O(r)=\{x\in O: d(x,\partial O)\geq r\}$ is compact, if necessary, we can remove some elements of $\mathscr {A}$ in $O(r)$ to make the set $\{x \in \mathscr {A}: d(x,\partial O)\geq r\}$ finite and maintain the above denseness assumption.

(2) As mentioned above, we can fix a countable subset $\mathscr {A}$ of O such that $O=\bigcup _{z\in \mathscr {A}}B(z,r(z))$ and satisfying discreteness. For every $x\in O$ and any $n\in \mathbb {Z}$ , since $f^n(x)\in O$ , there exist $\{x_n\}_{n\in \mathbb {Z}}\in \mathscr {A}^{\mathbb {Z}}$ such that $d(f^n(x),x_{n})\leq r(x_n)$ for all $n\in \mathbb {Z}$ . For $\varepsilon>0$ sufficiently small, since $r(x_n)\leq ({\varepsilon }/{2}) Q(x_n)\leq ({\varepsilon }/{2}) d(x_n,\partial O)$ , it follows that

$$ \begin{align*} \begin{aligned} d(f^n(x),\partial O)&\leq d(f^n(x),x_n)+d(x_n,\partial O) \notag\\ &\leq e^\varepsilon d(x_n,\partial O), \end{aligned} \end{align*} $$

and hence

$$ \begin{align*} \varepsilon(f^n(x))\leq e^{\beta\varepsilon}\varepsilon(x_n)\quad\text{and}\quad Q(f^n(x))\leq e^{\gamma\varepsilon}Q(x_n). \end{align*} $$

Analogously, from $d(f^n(x),\partial O)\geq d(x_n,\partial O)-d(f^n(x),x_n)\geq e^{-\varepsilon } d(x_n,\partial O)$ , we have

$$ \begin{align*} \varepsilon(f^n(x))\geq e^{-\beta\varepsilon}\varepsilon(x_n),\ Q(f^n(x))\geq e^{-\gamma\varepsilon}Q(x_n). \end{align*} $$

Therefore, we obtain $e^{-\beta \varepsilon }\leq {\varepsilon (f^n(x))}/{\varepsilon (x_n)}\leq e^{\beta \varepsilon }$ and $e^{-\gamma \varepsilon }\leq {Q(f^n(x))}/{Q(x_n)}\leq e^{\gamma \varepsilon }$ if $\varepsilon>0$ is small enough. It remains to prove part (b).

First we claim that

(5.1) $$ \begin{align} \varepsilon(x_{n+1})\leq A_u^\beta e^{2\beta\varepsilon}\varepsilon(x_n), \end{align} $$

where $A_u:=[({1}/{C^u})((C^f)^{\kappa ^u}-1)+1]^{1/\gamma ^u}>1$ .

In practice, by using $e^{-\varepsilon }\leq ({d(f^n(x),\partial O)}/{d(x_n,\partial O)})\leq e^{\varepsilon }$ , equation (5.1) holds in the case $\varepsilon (f^n(x))=\varepsilon r_0^\beta $ since

$$ \begin{align*} \begin{aligned} \varepsilon(x_{n+1})&\leq\varepsilon r_0^\beta\notag\\ &\leq\varepsilon d(f^n(x),\partial O)^\beta\notag\\ &\leq\varepsilon e^{\beta\varepsilon} d(x_n,\partial O)^\beta, \end{aligned} \end{align*} $$

which implies that $\varepsilon (x_{n+1})\leq A_u^\beta e^{2\beta \varepsilon }\varepsilon (x_n)$ when $0<\varepsilon \leq A_u^\beta $ .

However, if $\varepsilon (f^n(x))=\varepsilon d(f^n(x),\partial O)^\beta $ , from Assumption US(ii) and $e^{-\varepsilon }\leq ({d(f^n(x),\partial O)}/{d(x_n,\partial O)})\leq e^{\varepsilon }$ again, we obtain for all $n\in \mathbb {Z}$ ,

$$ \begin{align*} \begin{aligned} d(x_{n+1},\partial O)&\leq e^{\varepsilon} d(f^{n+1}(x),\partial O)\notag\\ &\leq A_ue^\varepsilon d(f^n(x),\partial O)\notag\\ &\leq A_ue^{2\varepsilon} d(x_n,\partial O). \end{aligned} \end{align*} $$

Then $\varepsilon (x_{n+1})\leq A_u^\beta e^{2\beta \varepsilon }\varepsilon (x_n)$ , which completes the proof of the claim.

Denote $A:=({1}/{C})((C^f)^{\kappa }-1)+1>1$ , and according to property (a) and equation (3.5) of Lemma 3.2, we get for all $n\in \mathbb {Z}$ ,

(5.2) $$ \begin{align} Q(x_{n+1})&\leq e^{\gamma\varepsilon} Q(f^{n+1}(x))\notag\\ &\leq Ae^{\gamma\varepsilon} Q(f^n(x))\notag\\ &\leq Ae^{2\gamma\varepsilon} Q(x_n), \end{align} $$

and

(5.3) $$ \begin{align} Q(f(x_{n+1}))&\leq A Q(x_{n+1})\notag\\ &\leq A^2e^{2\gamma\varepsilon} Q(x_n)\notag\\ &\leq A^3e^{2\gamma\varepsilon} Q(f(x_n)). \end{align} $$

Combining equations (5.1), (5.2), and (5.3), we have

$$ \begin{align*} \begin{aligned} d(f^{n+1}(x),x_{n+1})&\leq r(x_{n+1})\notag\\ &\leq\varepsilon\delta^u(\varepsilon\varepsilon(x_{n+1})) Q(x_{n+1})^2Q(f(x_{n+1}))\notag\\ \end{aligned}\end{align*} $$

$$ \begin{align*} \begin{aligned} &\leq\varepsilon\delta^u(\varepsilon A_u^\beta e^{2\beta\varepsilon}\varepsilon(x_n))\cdot A^2e^{4\gamma\varepsilon}Q(x_n)^2\cdot A^3e^{2\gamma\varepsilon}Q(f(x_n))\notag\\ &=A^5 \cdot \varepsilon e^{6\gamma\varepsilon}\delta^u(A_u^\beta\cdot \varepsilon e^{2\beta\varepsilon}\varepsilon(x_n)) Q(x_n)^2Q(f(x_n)). \end{aligned} \end{align*} $$

Now we can verify the half of $x_n\rightarrow x_{n+1}$ for all $n\in \mathbb {Z}$ from the local monotonicity of $\delta ^u(\cdot )$ if $\varepsilon>0$ is small enough. In fact,

$$ \begin{align*} \begin{aligned} d(f(x_n),x_{n+1})&\leq d(f(x_n),f^{n+1}(x))+d(f^{n+1}(x),x_{n+1}) \notag\\ &\leq C^fd(x_n,f^n(x))+d(f^{n+1}(x),x_{n+1})\notag\\ &\leq\varepsilon[C^f\delta^u(\varepsilon\varepsilon(x_n))+A^5\cdot e^{6\gamma\varepsilon}\delta^u(A_u^\beta\cdot \varepsilon e^{2\beta\varepsilon}\varepsilon(x_n))]Q(x_n)^2Q(f(x_n))\notag\\ &\leq\delta^u(\varepsilon(x_n))Q(x_n)^2Q(f(x_n)), \end{aligned} \end{align*} $$

where the last inequality holds if $0<\varepsilon \leq \min \{1,{1}/{A_u^\beta e^{2\beta }},{1}/({C^f+A^5 e^{6\gamma }})\}$ .

Similarly, we can perform as above to conclude that

$$ \begin{align*} d(f^{-1}(x_{n+1}),x_n)\leq\delta^s(\varepsilon(x_{n+1}))Q(x_{n+1})^2Q(f^{-1}(x_{n+1})) \end{align*} $$

if $\varepsilon>0$ is small enough.

Therefore, $x_n\rightarrow x_{n+1}$ for all $n\in \mathbb {Z}$ when $\varepsilon>0$ is small enough. The proof is completed.

Proof. The proof is quite similar to the proof of [Reference Sarig22, Proposition 4.15]. We treat the case of u-admissible manifolds and s-admissible manifolds in the proof of items (1,2,4,5) and (3), respectively. By suitable modification to the proof, all the other cases are similar in its opposite direction.

(1) We divide our proof into three steps. First, we verify that if the limit exists, then it is independent of the choice of $V_{-n}^{u}$ . Since $\mathscr {F}_{-1,0}^{u} \circ \cdots \circ \mathscr {F}_{-n, -n+1}^{u}[V^u_{-n}]$ is a u-admissible manifold at $x_{0}$ , by using Proposition 4.3, for any other choice of u-admissible manifolds $W_{-n}^{u}$ at $ x_{-n}$ , we have

$$ \begin{align*} \begin{aligned} &d_{C^{0}}(\mathscr{F}_{-1,0}^{u} \circ \cdots \circ \mathscr{F}_{-n, -n+1}^{u}[V^u_{-n}], \mathscr{F}_{-1,0}^{u} \circ \cdots \circ \mathscr{F}_{-n, -n+1}^{u}[W^u_{-n}])\notag\\ &\quad\leq\prod_{i=1}^n \|Df|_{E^s(x_{-i})}\|^{1/2}d_{C^{0}}(V^u_{-n},W^u_{-n})\notag\\ &\quad\leq\prod_{i=1}^n \|Df|_{E^s(x_{-i})}\|^{1/2}\underset{n \rightarrow \infty}{\longrightarrow}0, \end{aligned} \end{align*} $$

since $d_{C^{0}}(V^u_{-n},W^u_{-n})\leq 2Q(x_{-n})<1$ if $\varepsilon>0$ is small enough and $\{x_{-n}\}_{n\geq 0}$ have constant subsequence. Thus, if the limit exists, then it is independent of $V_{-n}^{u}$ .

The next thing to do is to prove that the limit exists. For every $m>n $ , we have

$$ \begin{align*} \begin{aligned} &d_{C^{0}}(\mathscr{F}_{-1,0}^{u} \circ \cdots \circ \mathscr{F}_{-n, -n+1}^{u}[V^u_{-n}], \mathscr{F}_{-1,0}^{u} \circ \cdots \circ \mathscr{F}_{-m, -m+1}^{u}[V^u_{-m}])\notag\\ &\quad\leq\prod_{i=1}^n \|Df|_{E^s(x_{-i})}\|^{1/2}d_{C^{0}}(V^u_{-n},\mathscr{F}_{-n-1,-n}^{u} \circ \cdots \circ \mathscr{F}_{-m, -m+1}^{u}[V^u_{-m}])\notag\\ &\quad\leq\prod_{i=1}^n \|Df|_{E^s(x_{-i})}\|^{1/2}\underset{n \rightarrow \infty}{\longrightarrow}0. \end{aligned} \end{align*} $$

Hence, $\{\mathscr {F}_{-1,0}^{u} \circ \cdots \circ \mathscr {F}_{-n, -n+1}^{u}[V^u_{-n}]\}_{n\geq 1}$ is a Cauchy sequence and therefore converges.

Finally, we have to show the admissibility of the limit. Since $\mathscr {F}_{-1,0}^{u} \circ \cdots \circ \mathscr {F}_{-n, -n+1}^{u}[V^u_{-n}] \underset {n \rightarrow \infty }{\longrightarrow } V^{u}[\{x_{-n}\}_{n\geq 0}]$ , we have $\sigma ^{(n)} \underset {n \rightarrow \infty }{\longrightarrow } \sigma $ uniformly on $B_{x_0}^u(Q(x_0))$ , where $\sigma $ and $\sigma ^{(n)}$ represent $V^{u}[\{x_{-n}\}_{n\geq 0}]$ and $\mathscr {F}_{-1,0}^{u} \circ \cdots \circ \mathscr {F}_{-n, -n+1}^{u}[V^u_{-n}]$ , respectively.

Due to property (AM3), that is, $\operatorname {Hol}_{\delta /2}(D\sigma ^{(n)})\leq \tfrac 12$ and $\|D\sigma ^{(n)}\|_{C^0}\leq \tfrac 12$ , by the Arzelà–Ascoli theorem, $\text {there exists } n_{k} \uparrow \infty $ such that $D\sigma ^{(n_k)}\underset {k \rightarrow \infty }{\longrightarrow } \omega $ uniformly where $\|\omega \|_{C^{\delta /2}} \leq \tfrac 12$ . Thus, for any $v\in B_{x_0}^u(Q(x_0))$ ,

$$ \begin{align*}\sigma^{(n_k)}(v)=\sigma^{(n_k)}(0)+\int_0^1 D_{\unicode{x3bb} v}\sigma^{(n_k)}\cdot vd\unicode{x3bb}\underset{k \rightarrow \infty}{\longrightarrow} \sigma(0)+\int_0^1 \omega(\unicode{x3bb} v)\cdot vd\unicode{x3bb}\end{align*} $$

and hence $\sigma $ is differentiable, and $D\sigma =\omega $ . We also see that $\{D\sigma ^{(n)}\}$ can only have one limit point. Consequently, $D\sigma ^{(n)}\underset {n \rightarrow \infty }{\longrightarrow } D\sigma $ uniformly. It follows that $\|\sigma (0)\| \leq 10^{-3}Q(x_0)$ , $\|D\sigma \|_{C^0} \leq \sqrt {\varepsilon }$ , and $\|D\sigma \|_{C^{\delta /2}}\leq \tfrac 12$ ; hence the u-admissibility of $V^{u}(\underline {x}^-)$ .

(2) By the continuity of $\mathscr {F}_{-1,0}^{u}$ and the definition of $\mathscr {F}_{-1,0}^{u}$ , we have

$$ \begin{align*} \begin{aligned} V^{u}(\{x_{-n}\}_{n\geq0})&=\lim _{n \rightarrow\infty}(\mathscr{F}_{-1,0}^{u} \circ \cdots \circ \mathscr{F}_{-n, -n+1}^{u})[V^u_{-n}]\notag\\ &=\mathscr{F}_{-1,0}^{u}\Big[\lim _{n \rightarrow\infty}(\mathscr{F}_{-2,-1}^{u} \circ \cdots \circ \mathscr{F}_{-n, -n+1}^{u})[V^u_{-n}]\Big]\notag\\ &=\mathscr{F}_{-1,0}^{u}[ V^{u}(\{x_{-n-1}\}_{n\geq0})]\notag\\ &\subset f[ V^{u}(\{x_{-n-1}\}_{n\geq0})]. \end{aligned} \end{align*} $$

(3) We write

$$ \begin{align*} V^{s}(\{x_{n}\}_{n\geq0})=\exp_{x_0}\{(\sigma^s_{0}(v^s_0), v^s_0):\|v^s_0\| \leq Q(x_0)\} \end{align*} $$

and

$$ \begin{align*} V^{s}(\{x_{n+k}\}_{n\geq0})=\exp_{x_k}\{(\sigma^s_k(v^s_k), v^s_k):\|v^s_k\| \leq Q(x_{k})\}\quad \text{for all } k\geq0. \end{align*} $$

Admissibility implies that $\|\sigma ^s_k(0)\| \leq 10^{-3} Q(x_k)$ and $Lip(\sigma ^s_k)\leq \sqrt {\varepsilon }$ for all $k\geq 0$ .

By part (2), $f^{k}[V^s(\{x_{n}\}_{n\geq 0})]\subset V^{s}(\{x_{n+k}\}_{n\geq 0}) \subset \exp _{x_k}(S_{x_k})$ for all $k \geq 0$ . Therefore, for any $y, z \in V^s(\{x_{n}\}_{n\geq 0})$ , one can write $f^{k}(y)=\exp _{x_{k}}(\sigma ^s_k(v^s_k), v^s_k)$ and $f^{k}(z)=\exp _{x_k}(\sigma ^s_k(w^s_k), w^s_k)$ .

For all $k \geq 0$ , since the exponential map has Lipschitz constant less than two, we get

$$ \begin{align*} \begin{aligned} d(f^{k}(y),f^{k}(z))&\leq2(\|v^s_{k}-w^s_{k}\|+\mathrm{Lip}(\sigma^s_{k})\|v^s_{k}-w^s_{k}\|)\notag\\ &\leq4\|v^s_{k}-w^s_{k}\| \end{aligned} \end{align*} $$

if $0<\varepsilon \leq 1$ .

Notice that

$$ \begin{align*} (\sigma^s_{k}(v^s_{k}), v^s_{k})&=F_{x_{k-1},x_{k}}(\sigma^s_{k-1}(v^s_{k-1}), v^s_{k-1})\quad \text{and}\quad (\sigma^s_{k}(w^s_{k}), w^s_{k})\\&=F_{x_{k-1},x_{k}}(\sigma^s_{k-1}(w^s_{k-1}), w^s_{k-1}), \end{align*} $$

or in the s-direction,

$$ \begin{align*} v^s_{k}=D^{ss}_{x_{k-1},x_{k}}(v^s_{k-1})+D^{su}_{x_{k-1},x_{k}}\circ\sigma^s_{k-1}(v^s_{k-1})+\phi^s_{x_{k-1},x_{k}}(\sigma^s_{k-1}(v^s_{k-1}), v^s_{k-1}) \end{align*} $$

and

$$ \begin{align*} w^s_{k}=D^{ss}_{x_{k-1},x_{k}}(w^s_{k-1})+D^{su}_{x_{k-1},x_{k}}\circ\sigma^s_{k-1}(w^s_{k-1})+\phi^s_{x_{k-1},x_{k}}(\sigma^s_{k-1}(w^s_{k-1}), w^s_{k-1}). \end{align*} $$

So we have

$$ \begin{align*} \begin{aligned} \tfrac{1}{4}d(f^{k}(y),f^{k}(z))&\leq\|v^s_{k}-w^s_{k}\|\notag\\ &\leq(\|D^{ss}_{x_{k-1},x_{k}}\|+\|D^{su}_{x_{k-1},x_{k}}\|+2\operatorname{Lip}(\phi^s_{x_{k-1},x_{k}}) )\|v^s_{k-1}-w^s_{k-1}\|\notag\\ &\leq\|Df|_{E^s(x_{k-1})}\|^{1/2}\|v^s_{k-1}-w^s_{k-1}\| \end{aligned} \end{align*} $$

as at the end of the proof of Proposition 4.3(1). Thus,

$$ \begin{align*} \begin{aligned} d(f^{k}(y),f^{k}(z))&\leq4\prod_{i=0}^{k-1}\|Df|_{E^s(x_i)}\|^{1/2}\|v^s_0-w^s_0\|\notag\\ &\leq8Q(x_0)\prod_{i=0}^{k-1}\|Df|_{E^s(x_i)}\|^{1/2}. \end{aligned} \end{align*} $$

(4) The inclusion $\subseteq $ is trivial. In fact, notice that $V^{u}(\underline {x}^-)$ is a u-admissible manifold at $x_0$ , it is contained in $\exp _{x_0}(S_{x_0})$ , that is, for every $x\in V^{u}(\underline {x}^-)$ , $x\in \exp _{x_0}(S_{x_0})$ . By part (2), for every $k \geq 0$ ,

$$ \begin{align*} f^{-k}(x) \in f^{-k}[V^{u}(\{x_{-n}\}_{n\geq0})] \subseteq V^{u}(\{x_{-n-k}\}_{n \geq 0}) \subset \exp_{x_{-k}}(S_{x_{-k}}) \end{align*} $$

because $V^{u}(\{x_{-n-k}\}_{n \geq 0})$ is a u-admissible manifold at $x_{-k}$ . We have $\subseteq $ .

Now we prove $\supseteq $ . An outline of the proof of this inclusion is as follows. For every point x satisfying $f^{-k}(x)\in \exp _{x_{-k}}(S_{x_{-k}})$ for all $k \geq 0$ , write $x=\exp _{x_{0}}(v^u_{0}, v^s_{0})$ . We show that $x \in V^{u}(\{x_{-n}\}_{n\geq 0})$ by proving that $v^s_{0}=\sigma ^u(v^u_{0})$ , where $\sigma ^u$ is the function which represents $V^{u}(\{x_{-n}\}_{n\geq 0})$ .

Introduce for this purpose the point $\bar {x}=\exp _{x_{0}}(\bar {v}^u_{0}, \bar {v}^s_{0})\in V^{u}(\{x_{-n}\}_{n\geq 0})$ , where $\bar {v}^u_{0}=v^u_{0}$ and $\bar {v}^s_{0}=\sigma ^u(\bar {v}^u_{0})$ . Our task now is to prove $\bar {v}^s_{0}=v^s_{0}$ .

For every $k \geq 0$ , write

$$ \begin{align*} f^{-k}(x)=\exp_{x_{-k}}(v^u_{-k}, v^s_{-k})\quad \text{and}\quad f^{-k}(\bar{x})=\exp_{x_{-k}}(\bar{v}^u_{-k}, \bar{v}^s_{-k})\in V^{u}(\{x_{-n-k}\}_{n \geq 0}), \end{align*} $$

where $\|v^u_{-k}\|,\|v^s_{-k}\|,\|\bar {v}^u_{-k}\|,\|\bar {v}^u_{-k}\| \leq Q(x_{-k})$ for all $k \geq 0$ .

By Proposition 3.1 in its version for $f^{-1}$ , for every $k \geq 0, F_{x_{-k-1}x_{-k}}^{-1}=\exp _{x_{-k-1}}^{-1} \circ f^{-1} \circ \exp _{x_{-k}}$ can be put in the form

$$ \begin{align*} \begin{aligned} &F^{-1}_{x_{-k-1},x_{-k}}(v_{-k}^u,v_{-k}^s)\notag\\ &\quad=(\bar{D}^{uu}_{x_{-k},x_{-k-1}}(v_{-k}^u)+\bar{D}^{us}_{x_{-k},x_{-k-1}}(v_{-k}^s) +\bar{\phi}^u_{x_{-k},x_{-k-1}}(v_{-k}^u,v_{-k}^s), \bar{D}^{ss}_{x_{-k},x_{-k-1}}(v_{-k}^s)\\&\qquad+\bar{D}^{su}_{x_{-k},x_{-k-1}}(v_{-k}^u)+\bar{\phi}^s_{x_{-k},x_{-k-1}}(v_{-k}^u,v_{-k}^s)) \end{aligned} \end{align*} $$

under the basis $(E^u,E^s)$ for all $v_{-k}=(v_{-k}^u,v_{-k}^s)\in S_{x_{-k}}$ , where

$$ \begin{align*}\bar{D}^{\tau t}_{x_{-k},x_{-k-1}}&=P_{x_{-k-1}}^{\tau}\circ D_{f^{-1}(x_{-k})}\exp^{-1}_{x_{-k-1}}\circ D_{x_{-k}}f^{-1}|_{E^t(x_{-k})}:B_{x_{-k}}^t(Q(x_{-k}))\\&\quad\rightarrow E^\tau(x_{-k-1}), \tau,t=u,s.\end{align*} $$

Since $d(f^{-1}(x_{-k}), x_{-k-1})<\delta ^s(\varepsilon (x_{-k}))$ , in the s-direction, we have

$$ \begin{align*}\|\bar{D}^{su}_{x_{-k},x_{-k-1}}\|\leq\varepsilon\varepsilon(x_{-k}),\end{align*} $$

$$ \begin{align*} m(D f^{-1}|_{E^{s}(x_{-k})})\leq m(\bar{D}^{ss}_{x_{-k},x_{-k-1}})+\varepsilon\varepsilon(x_{-k}).\end{align*} $$

While in the u-direction, we have

$$ \begin{align*}\|\bar{D}^{us}_{x_{-k},x_{-k-1}}\|\leq\varepsilon\varepsilon(x_{-k}),\end{align*} $$

$$ \begin{align*}\|\bar{D}^{uu}_{x_{-k},x_{-k-1}}\|\leq\|D f^{-1}|_{E^{u}(x_{-k})}\|+\varepsilon\varepsilon(x_{-k}),\end{align*} $$

and moreover,

$$ \begin{align*} \operatorname{Lip}(\bar{\phi}_{x_{-k},x_{-k-1}})\leq\varepsilon\varepsilon(x_{-k}), \end{align*} $$

where $\bar {\phi }=(\bar {\phi }^u,\bar {\phi }^s)$ .

Let $\Delta v^u_{-k}:=v^u_{-k}-\bar {v}^u_{-k}$ and $\Delta v^s_{-k}:=v^s_{-k}-\bar {v}^s_{-k}$ . Observe that for every $k \geq 0$ , $(v^u_{-k-1}, v^s_{-k-1})=F_{x_{-k-1}x_{-k}}^{-1}(v^u_{-k}, v^s_{-k})$ , and $(\bar {v}^u_{-k-1}, \bar {v}^s_{-k-1})=F_{x_{-k-1}x_{-k}}^{-1}{ }(\bar {v}^u_{-k}, \bar {v}^s_{-k})$ , we have

$$ \begin{align*} \begin{aligned} \|\Delta v^u_{-k-1}\|&\leq\|\bar{D}^{uu}_{x_{-k},x_{-k-1}}\|\cdot \|\Delta v_{-k}^u\|+\|\bar{D}^{us}_{x_{-k},x_{-k-1}}\|\cdot \|\Delta v_{-k}^s\|\\ &\quad+\operatorname{Lip}(\bar{\phi}^u_{x_{-k},x_{-k-1}})(\|\Delta v_{-k}^u\|+\|\Delta v_{-k}^s\|)\notag\\ &\leq(\|Df^{-1}|_{E^u(x_{-k})}\|+2\varepsilon\varepsilon(x_{-k}))\|\Delta v^u_{-k}\|+2\varepsilon\varepsilon(x_{-k})\|\Delta v^s_{-k}\| \notag \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \|\Delta v^s_{-k-1}\|&\geq m(\bar{D}^{ss}_{x_{-k},x_{-k-1}})\|\Delta v_{-k}^s\|-\|\bar{D}^{su}_{x_{-k},x_{-k-1}}\|\cdot \|\Delta v_{-k}^u\|\\ &\quad-\operatorname{Lip}(\bar{\phi}^s_{x_{-k},x_{-k-1}})(\|\Delta v_{-k}^u\|+\|\Delta v_{-k}^s\|)\notag\\ &\geq (m(Df^{-1}|_{E^s(x_{-k})})-2\varepsilon\varepsilon(x_{-k}))\|\Delta v^s_{-k}\|-2\varepsilon\varepsilon(x_{-k})\|\Delta v^u_{-k}\|. \end{aligned} \end{align*} $$

We claim that an easy induction gives $\|\Delta v^u_{-k}\|\leq \|\Delta v^s_{-k}\| \leq \|\Delta v^s_{-k-1}\|$ for all $k\geq 0$ . In fact, for $k=0$ , this is because $\|\Delta v^u_{0}\|=0$ by definition and $C\varepsilon \varepsilon (x_{-k})\leq m(Df^{-1}|_{E^s(x_{-k})})-1$ if $\varepsilon>0$ is small enough from Lemma 3.1. Assume by induction that $\|\Delta v^u_{-k}\|\leq \|\Delta v^s_{-k}\| \leq \|\Delta v^s_{-k-1}\|$ . Notice that $C\varepsilon (x_{-k})\leq 1-\|Df^{-1}|_{E^u(x_{-k})}\|$ from the case $f^{-1}$ of Lemma 3.2, and choosing a smaller $\varepsilon $ if necessary such that $\varepsilon \leq \min \{1,{C}/{4}\}$ , then we have

$$ \begin{align*} \begin{aligned} \|\Delta v^u_{-k-1}\| &\leq (\|Df^{-1}|_{E^u(x_{-k})}\|+2\varepsilon\varepsilon(x_{-k}))\|\Delta v^u_{-k}\|+2\varepsilon\varepsilon(x_{-k})\|\Delta v^s_{-k}\|\notag\\ & \leq(\|Df^{-1}|_{E^u(x_{-k})}\|+4\varepsilon\varepsilon(x_{-k}))\|\Delta v^s_{-k}\|\leq \|\Delta v^s_{-k}\| \leq \|\Delta v^s_{-k-1}\| \\ \end{aligned} \end{align*} $$

and

$$ \begin{align*} \begin{aligned} \|\Delta v^s_{-k-2}\| &\geq (m(Df^{-1}|_{E^s(x_{-k-1})})-2\varepsilon\varepsilon(x_{-k-1}))\|\Delta v^s_{-k-1}\|-2\varepsilon\varepsilon(x_{-k-1})\|\Delta v^u_{-k-1}\|\notag\\ &\geq(m(Df^{-1}|_{E^s(x_{-k-1})})-4\varepsilon\varepsilon(x_{-k-1})) \|\Delta v^s_{-k-1}\|\geq \|\Delta v^s_{-k-1}\|. \end{aligned} \end{align*} $$

Hence, the claim is verified.

We now return to the proof of this part. If we introduce a new constant

$$ \begin{align*} b:=\inf_{x\in O}\frac{m(Df^{-1}|_{E^s(x)})-m(Df^{-1}|_{E^s(x)})^{1/2}}{m(Df^{-1}|_{E^s(x)})-1}>0, \end{align*} $$

and choosing a smaller $\varepsilon $ if necessary such that $0<\varepsilon \leq ({bC}/{4})$ , then by Assumption US(iii), we see that

$$ \begin{align*} \begin{aligned} \|\Delta v^s_{-k-2}\|&\geq(m(Df^{-1}|_{E^s(x_{-k-1})})-bC\varepsilon(x_{-k-1}))\|\Delta v^s_{-k-1}\|\notag\\ &\geq m(Df^{-1}|_{E^s(x_{-k-1})})^{1/2}\|\Delta v^s_{-k-1}\|\notag \end{aligned} \end{align*} $$

for all $k\geq 0$ . Therefore, we get $\|\Delta v^s_{-k}\| \geq \prod _{i=0}^{k-1} m(Df^{-1}|_{E^s(x_{-i})})^{1/2} \|\Delta v^s_{0}\| $ and hence either $\|\Delta v^s_{0}\|=0$ or $\|\Delta v^s_{-k}\| \underset {k \rightarrow \infty }{\longrightarrow } \infty .$ However, $\|\Delta v^s_{-k}\|=\|v^s_{-k}-\bar {v}^s_{-k}\| \leq 2Q(x_{-k})<2$ , so $\|\Delta v^s_{0}\|=0 .$ Recall that $\Delta v^s_{0}=v^s_{0}-\bar {v}^s_{0}$ , we have $v^s_{0}=\bar {v}^s_{0}$ and then $v^s_0=\bar {v}^s_0=\sigma ^u(\bar {v}^u_0)=\sigma ^u(v^u_0)$ by definition. Thus, $x=\exp _{x_0}(v^u_0, \sigma ^u(v^u_0)) \in V^{u}(\{x_{-n}\}_{n\geq 0})$ .

(5) Given $n>N$ , let $V_{-n}^{u}$ be a u-admissible manifold at $x_{-n}$ and let $W_{-n}^u$ be a u-admissible manifold at $y_{-n}$ . For the sake of simplicity in notation, let $\mathscr {F}_{u}^{\ell }(V_{-n}^{u})$ (respectively $\mathscr {F}_{u}^{\ell }(W_{-n}^{u}))$ denote the result of applying $\mathscr {F}^{u}_{\cdot ,\cdot } \ell $ times to $V_{-n}^u$ using the path $x_{-n} \rightarrow \cdots \rightarrow x_{-n+\ell }$ (respectively using $y_{-n} \rightarrow \cdots \rightarrow y_{-n+\ell }$ ).

Here, $\mathscr {F}_{u}^{n-N}(V_{-n}^{u})$ and $\mathscr {F}_{u}^{n-N}(W_{-n}^{u})$ are u-admissible manifolds at $x_{-N}$ and $y_{-N}$ , respectively. Let $\sigma _{-N}, \widetilde {\sigma }_{-N}$ be their representing functions. Admissibility implies that

$$ \begin{align*} \begin{aligned} &\|\sigma_{-N}-\widetilde{\sigma}_{-N}\|_{C^0} \leq\|\sigma_{-N}\|_{C^0}+\|\widetilde{\sigma}_{-N}\|_{C^0}\leq2 Q(x_{-N})<1, \\ &\|D\sigma_{-N}-D\widetilde{\sigma}_{-N}\|_{C^0} \leq\|D\sigma_{-N}\|_{C^0}+\|D\widetilde{\sigma}_{-N}\|_{C^0}\leq2 \sqrt{\varepsilon}<1 \end{aligned} \end{align*} $$

if $\varepsilon $ is small enough.

For every $0\leq k\leq N$ , represent $\mathscr {F}_{u}^{n-k}[V_{-n}^{u}]$ and $\mathscr {F}_{u}^{n-k}[W_{-n}^{u}]$ by functions $\sigma _{-k}$ and $\widetilde {\sigma }_{-k}$ , respectively. By Proposition 4.3, we have

(5.4) $$ \begin{align} \|\sigma_{-k+1}-\widetilde{\sigma}_{-k+1}\|_{C^0} \leq \|Df|_{E^s(x_{-k})}\|^{1/2}\|\sigma_{-k}-\widetilde{\sigma}_{-k}\|_{C^0} \end{align} $$

and

(5.5) $$ \begin{align} \|D\sigma_{-k+1}-D\widetilde{\sigma}_{-k+1}\|_{C^0} \leq \|Df|_{E^s(x_{-k})}\|^{1/2}(\|D\sigma_{-k}-D\widetilde{\sigma}_{-k}\|_{C^0}+2\| \sigma_{-k}-\widetilde{\sigma}_{-k}\|_{C^0}^{\delta / 2}). \end{align} $$

Iterating equation (5.4), from $k=N$ and going down, we get

$$ \begin{align*} \begin{aligned} \|\sigma_{-k}-\widetilde{\sigma}_{-k}\|_{C^0} &\leq\|Df|_{E^s(x_{-k-1})}\|^{1/2}\|\sigma_{-k-1}-\widetilde{\sigma}_{-k-1}\|_{C^0}\notag\\ &\leq\prod^{N}_{i=k+1}\|Df|_{E^s(x_{-i})}\|^{1/2}\|\sigma_{-N}-\widetilde{\sigma}_{-N}\|_{C^0}\notag\\ &\leq\prod^{N}_{i=k+1}\|Df|_{E^s(x_{-i})}\|^{1/2}, \end{aligned} \end{align*} $$

and hence $d_{C^0}(\mathscr {F}_{u}^{n}[V_{-n}^{u}], \mathscr {F}_{u}^{n}[W_{-n}^{u}]) \leq \prod ^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{1/2}$ . Passing to the limit ${n \rightarrow \infty }$ , we obtain

$$ \begin{align*} d_{C^0}(V^{u}(\{x_{-n}\}_{n\geq0}), V^{u}(\{y_{-n}\}_{n\geq0})) \leq \prod^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{1/2}. \end{align*} $$

Apply $\|\sigma _{-k}-\widetilde {\sigma }_{-k}\|_{C^0} \leq \prod ^{N}_{i=k+1}\|Df|_{E^s(x_{-i})}\|^{1/2}$ to equation (5.5), and set $c_{-k}:=\|D\sigma _{-k}-D\widetilde {\sigma }_{-k}\|_{C^0}$ for every $0\leq k\leq N$ . Then $c_{-k+1} \leq \|Df|_{E^s(x_{-k})}\|^{1/2}(c_{-k}+2\prod ^{N}_{i=k+1} \|Df|_{E^s(x_{-i})}\|^{\delta /4} )$ for every $0\leq k\leq N$ . Now an easy induction gives that for every $0 \leq k \leq N$ ,

$$ \begin{align*} \begin{aligned} c_0 &\leq \prod^{k}_{i=1}\|Df|_{E^s(x_{-i})}\|^{1/2} c_{-k}+2\times\bigg[\prod^{k}_{i=1}\|Df|_{E^s(x_{-i})}\|^{1/2}\cdot\prod^{N}_{i=k+1}\|Df|_{E^s(x_{-i})}\|^{\delta/4}\notag\\ &\quad+\prod^{k-1}_{i=1}\|Df|_{E^s(x_{-i})}\|^{1/2}\cdot\prod^{N}_{i=k}\|Df|_{E^s(x_{-i})}\|^{\delta/4} +\cdots+\|Df|_{E^s(x_{-1})}\|^{1/2}\\ &\quad\times\prod^{N}_{i=2}\|Df|_{E^s(x_{-i})}\|^{\delta/4}\bigg]. \end{aligned} \end{align*} $$

We now take $k=N$ , paying attention to the inequalities $\|Df|_{E^s(x_{-i})}\|^{1/2} \leq \|Df|_{E^s(x_{-i})}\|^{\delta /4}$ for all $i=1,\ldots ,N$ since $0<\delta <\min \{1,\alpha -{\beta }/{\gamma }\}$ and noting that $c_{-N} \leq 1$ , we have

$$ \begin{align*} c_{0} \leq \prod^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{1/2}+2 N \prod^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{\delta/4}<(2 N+1) \prod^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{\delta/4}. \end{align*} $$

Immediately, it follows that $d_{C^1}(\mathscr {F}^n_{u}[V_{-n}^{u}], \mathscr {F}^n_{u}[W^u_{-n}]) \kern1pt{\leq}\kern1pt 2(N+1) \prod ^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{\delta /4}$ . In part (1), we saw that $\mathscr {F}^n_{u}[V_{-n}^{u}]$ and $\mathscr {F}_{u}^{n}[W_{-n}^{u}]$ converge to $V^{u}(\{x_{-n}\}_{n \geq 0})$ and $V^{u}(\{y_{-n}\}_{n \geq 0})$ in $C^{1}$ , respectively. If we pass to the limit as $n \rightarrow \infty $ , finally we get

$$ \begin{align*}d_{C^1}(V^{u}(\{x_{-n}\}_{n \geq 0}), V^{u}(\{y_{-n}\}_{n \geq 0})) \leq 2(N+1) \prod^{N}_{i=1}\|Df|_{E^s(x_{-i})}\|^{\delta/4}.\\[-50pt] \end{align*} $$

Proof. Let $\underline {x}\in \Sigma ^{\#}$ be an $\varepsilon $ -rppo. From Proposition 4.1(1), there is a singleton set ${\{x\}:=V^u(\underline {x}^-)\cap V^s(\underline {x}^+)\subset O}$ , we claim that $\underline {x}$ shadows x and $x\in O$ . By Proposition 5.2(2), for all $k\geq 0$ , we have

$$ \begin{align*} f^{-k}(x)\in f^{-k}[V^u(\{x_{-n}\}_{n\geq0})]\subset V^u(\{x_{-n-k}\}_{n\geq0})\subset\exp_{x_{-k}}(S_{x_{-k}}) \end{align*} $$

and

$$ \begin{align*} f^k(x)\in f^k[V^s(\{x_n\}_{n\geq0})]\subset V^s(\{x_{n+k}\}_{n\geq0})\subset\exp_{x_k}(S_{x_k}), \end{align*} $$

that is, $f^{k}(x)\in \exp _{x_k}(S_{x_k})$ for all $k \in \mathbb {Z}$ , and hence $\underline {x}$ shadows x.

If $y\in O$ is any other point shadowed by $\underline {x}$ , according to Definition 5.2 and Proposition 5.2(4), it must lie in $V^u(\underline {x}^-)\cap V^s(\underline {x}^+)=\{x\}$ .

Proof. (1) Notice that for all $\underline {x}\in \Sigma ^{\#}$ , from the shadowing lemma in Lemma 5.1 and Definition 5.2, we can denote $x=\pi (\underline {x})$ which satisfies $f^{n}(f\circ \pi (\underline {x}))=f^{n+1}(x)\in \exp _{x_{n+1}}(S_{x_{n+1}})$ for all $n \in \mathbb {Z}$ . However, since $\sigma (\underline {x})=\{x_{n+1}\}_{n\in \mathbb {Z}}\in \Sigma ^{\#}$ , using the shadowing lemma in Lemma 5.1 again, we can denote $y=\pi \circ \sigma (\underline {x})$ which satisfies $f^{n}(\pi \circ \sigma (\underline {x}))=f^{n}(y)\in \exp _{x_{n+1}}(S_{x_{n+1}})$ for all $n \in \mathbb {Z}$ by Definition 5.2 again. From the uniqueness of the shadowing point, $y=f(x)$ , that is, $\pi \circ \sigma (\underline {x})=f\circ \pi (\underline {x})$ for all $\underline {x}\in \Sigma ^{\#}$ .

(2) First we prove $\pi (\Sigma ^\#)\supset O^{\#}$ . For every $x\in O^{\#}\subset O$ , looking closely into the proof of Proposition 5.1(2), we see that there exists a sequence $\{x_n\}_{n\in \mathbb {Z}}\in \Sigma $ that satisfies $d(f^n(x),x_n)<\varepsilon Q(x_n)$ and $d(x_n,\partial O)\geq e^{-\varepsilon }d(f^n(x),\partial O)$ for all $n\in \mathbb {Z}$ . By the definition of $O^{\#}$ , $d(x_n,\partial O)\geq e^{-\varepsilon }d(f^n(x),\partial O)$ means there exist sequences $n_k,m_k\uparrow \infty $ for which $d(x_{n_k},\partial O)$ and $d(x_{-m_k},\partial O)$ are bounded away from zero. By the discreteness of $\mathscr {A}$ (Proposition 5.1(1)), $x_n$ must repeat some symbol infinitely often in the past and (possibly a different symbol) in the future, that is, $\{x_n\}_{n\in \mathbb {Z}}\in \Sigma ^{\#}$ . Since $d(f^n(x),x_n)<\varepsilon Q(x_n)$ , that is, $f^n(x)\in \exp _{x_n}(T_{x_n}M(\varepsilon Q(x_n)))\subset \exp _{x_n}(S_{x_n})$ if $\varepsilon>0$ is small enough, from the uniqueness of the shadowing point, we have $x=\pi (\{x_n\}_{n\in \mathbb {Z}})\in \pi (\Sigma ^{\#})$ for all $n\in \mathbb {Z}$ . Thus, we get $\pi (\Sigma ^\#)\supset O^{\#}$ .

In the opposite direction, for every $\underline {x}=\{x_n\}_{n\in \mathbb {Z}}\in \Sigma ^{\#}$ , the shadowing lemma in Lemma 5.1 claims there is a unique $x\in O$ such that $x=\pi (\underline {x})$ . Since $d(f^n(x),x_n)\leq ({1}/{50})Q(x_n)\leq ({1}/{50})d(x_n,\partial O)$ from Proposition 4.1(1), we have

$$ \begin{align*} d(f^n(x),\partial O)\geq d(x_n,\partial O)-d(f^n(x),x_n)\geq \frac{49}{50}d(x_n,\partial O), \end{align*} $$

and then $x\in O^{\#}$ . Therefore, $\pi (\Sigma ^\#)\subset O^{\#}$ .

(3) Write $\{\pi (\underline {x})\}=V^u(\underline {x}^-)\cap V^s(\underline {x}^+)$ , $\{\pi (\underline {y})\}=V^u(\underline {y}^-)\cap V^s(\underline {y}^+)$ . From Propositions 4.1(2), 4.3(1), and using induction, we have

$$ \begin{align*} \begin{aligned} d(\pi(\underline{x}),\pi(\underline{y})) &\leq \frac{2}{1-\sqrt{\varepsilon}}[d_{C^0}(V^u(\underline{x}^-),V^u(\underline{y}^-))+d_{C^0}(V^s(\underline{x}^+),V^s(\underline{y}^+))]\notag\\ &\leq \frac{2}{1-\sqrt{\varepsilon}} \bigg[\prod_{i=1}^{N}\|Df|_{E^s(x_{-i})}\|^{1/2}+\prod_{i=1}^{N}\|Df^{-1}|_{E^u(x_{i})}\|^{1/2}\bigg]\notag\\ &\leq 6 \bigg[\prod_{i=1}^{N}\|Df|_{E^s(x_{-i})}\|^{1/2}+\prod_{i=1}^{N}\|Df^{-1}|_{E^u(x_{i})}\|^{1/2}\bigg] \end{aligned} \end{align*} $$

if $0<\varepsilon \leq \tfrac 19$ .

Finally, the continuity is obvious since

$$ \begin{align*} \lim_{N\rightarrow\infty}\prod_{i=1}^{N}\|Df|_{E^s(x_{-i})}\|=\lim_{N\rightarrow\infty}\prod_{i=1}^{N}\|Df^{-1}|_{E^u(x_{i})}\|=0.\\[-48pt] \end{align*} $$

Proof. As proven in [Reference Sarig22, Proposition 6.4], we show the proposition for $V^{s}$ by dividing our proof into three claims, and leave the case of $V^{u}$ to the reader. Suppose $\varepsilon>0$ is small enough.

Proof of Claim 1

Recall that by Proposition 5.2(2), for all $k \geq 0$ , we have

$$ \begin{align*} f^{k}( V^{s}(\underline{y}^+)) \subset V^{s}(\{y_{n+k}\}_{n\geq 0})=\exp_{y_k}\{(\sigma^s_k(v^s_k), v^s_k):\|v^s_k\| \leq Q(y_{k})\}\subset\exp_{y_k}(S_{y_k}), \end{align*} $$

where $\sigma ^s_k$ is the function which represents s-admissible manifold $V^{s}(\{y_{n+k}\}_{n\geq 0})$ at $y_k$ for all $k\geq 0$ . Hence, for any $y \in V^{s}(\underline {y}^+)$ , one can write $f^{k}(y)=\exp _{y_{k}}(\sigma ^s_k(v^s_k), v^s_k)$ , $\|v^s_k\|\leq Q(y_k)$ .

Notice that $(\sigma ^s_{k}(v^s_{k}), v^s_{k})=F_{y_{k-1},y_{k}}(\sigma ^s_{k-1}(v^s_{k-1}), v^s_{k-1})$ , or in the s-direction,

$$ \begin{align*} v^s_{k}=D^{ss}_{y_{k-1},y_{k}}(v^s_{k-1})+D^{su}_{y_{k-1},y_{k}}\circ\sigma^s_{k-1}(v^s_{k-1})+\phi^s_{y_{k-1},y_{k}}(\sigma^s_{k-1}(v^s_{k-1}), v^s_{k-1}). \end{align*} $$

Using equations (4.1), (3.12), and $\|\phi ^s_{y_{k-1},y_{k}}(0,0)\|\leq \varepsilon \varepsilon (y_{k-1})Q(y_{k-1})$ , it follows that

$$ \begin{align*} \begin{aligned} \|v^s_{k}\|&\leq\|D^{ss}_{y_{k-1},y_{k}}\|\ \|v^s_{k-1}\|+\|D^{su}_{y_{k-1},y_{k}}\|\ \|\sigma_{k-1}^s(v^s_{k-1})\|+\|\phi^s_{y_{k-1},y_{k}}(0,0)\|\\&\quad+\operatorname{Lip}(\phi^s_{y_{k-1},y_{k}}) (\|\sigma_{k-1}^s(v^s_{k-1})\|+\|v^s_{k-1}\|))\notag\\ &\leq(\|Df|_{E^s(y_{k-1})}\|+4\varepsilon\varepsilon(y_{k-1}))\|v^s_{k-1}\|+2\varepsilon\varepsilon(y_{k-1})Q(y_{k-1}). \end{aligned} \end{align*} $$

We see that $\|v^s_{k}\| \leq a_{k}$ where $a_{k}$ is defined inductively by

$$ \begin{align*} a_{0}:=Q(y_{0}) \quad \text {and} \quad a_{k}=(\|Df|_{E^s(y_{k-1})}\|+4\varepsilon\varepsilon(y_{k-1}))a_{k-1}+2\varepsilon\varepsilon(y_{k-1})Q(y_{k-1}). \end{align*} $$

We claim that $a_{k}\leq \tfrac 18 Q(y_k)$ for some $k\geq 0$ . If the assertion would not hold, then ${Q(y_k) < 8 a_{k}}$ for all $k\geq 0$ , and hence $a_{k} <(\|Df|_{E^s(y_{k-1})}\|+20\varepsilon \varepsilon (y_{k-1})) a_{k-1}$ for all $k\geq 0$ , which implies that

$$ \begin{align*} a_{k}<\prod_{i=0}^{k-1}(\|Df|_{E^s(y_i)}\|+20\varepsilon\varepsilon(y_i)) a_{0}. \end{align*} $$

Fix $\kappa <\kappa _1<1$ , and introduce a new constant $c_1$ associated to $\kappa _1$ given by

$$ \begin{align*} c_1:=\inf_{x\in O}\frac{\|Df|_{E^s(x)}\|^{\kappa_1}-\|Df|_{E^s(x)}\|}{1-\|Df|_{E^s(x)}\|}>0, \end{align*} $$

then if $0<\varepsilon \leq ({c_1C}/{20})$ , we get

$$ \begin{align*} a_{k}<\prod_{i=0}^{k-1}\|Df|_{E^s(y_i)}\|^{\kappa_1} a_{0}. \end{align*} $$

Whereas by assumption, $a_{k}> \tfrac 18 Q(y_k)$ for all $k\geq 0$ . Since $y_{k-1}\rightarrow y_k$ , from Lemma 4.1 in the case $C_3=C_4=1$ , we obtain

$$ \begin{align*} \begin{aligned} Q(y_k)&> Q(y_k)-d(f(y_{k-1}),y_k)\notag\\ &\geq(\|Df|_{E^s(y_{k-1})}\|^\kappa+\varepsilon\varepsilon(y_{k-1}))Q(y_{k-1})\notag\\ &> \|Df|_{E^s(y_{k-1})}\|^\kappa Q(y_{k-1}). \end{aligned} \end{align*} $$

Therefore,

$$ \begin{align*} \begin{aligned} a_{k}> \tfrac{1}{8}Q(y_k)&>\tfrac{1}{8}\|Df|_{E^s(y_{k-1})}\|^\kappa Q(y_{k-1})\notag\\ &>\frac{1}{8}\prod_{i=0}^{k-1}\|Df|_{E^s(y_{i})}\|^\kappa a_0. \end{aligned} \end{align*} $$

Combining with the above, we have

$$ \begin{align*} \frac{1}{8}\leq \prod_{i=0}^{k-1}\|Df|_{E^s(y_{i})}\|^{\kappa_1-\kappa}\underset{k \rightarrow \infty}{\longrightarrow} 0, \end{align*} $$

which is a contradiction. Consequently, we infer that $\text {there exists } k_0\geq 0$ such that ${a_{k_0}\leq \tfrac 18Q(y_{k_0})}$ .

Moreover, we have $a_{k}\leq \tfrac 18Q(y_k)$ for all k large enough by using induction. In fact, suppose that the inequality is true for $k-1$ , we have by the definition of $a_k$ that

$$ \begin{align*} \begin{aligned} a_{k}&\leq(\|Df|_{E^s(y_{k-1})}\|+4\varepsilon\varepsilon(y_{k-1}))\tfrac{1}{8}Q(y_{k-1})+2\varepsilon\varepsilon(y_{k-1})Q(y_{k-1})\notag\\ &\leq(\|Df|_{E^s(y_{k-1})}\|+20\varepsilon\varepsilon(y_{k-1}))\tfrac{1}{8}Q(y_{k-1})\notag\\ &\leq\|Df|_{E^s(y_{k-1})}\|^{\kappa_1}\tfrac{1}{8}Q(y_{k-1})\notag\\ &\leq\|Df|_{E^s(y_{k-1})}\|^{\kappa}\tfrac{1}{8}Q(y_{k-1})\notag\\ &\leq\tfrac{1}{8}Q(y_k). \end{aligned} \end{align*} $$

Thus, $a_{k}\leq \tfrac 18Q(y_k)$ for all k large enough.

In particular, $\|v^s_k\|\leq \tfrac 18Q(y_k)$ for all k large enough. Since $\|\sigma ^s_{k}(v^s_{k})\| \leq \|\sigma ^s_{k}(0)\|+\operatorname {Lip}(\sigma ^s_{k})\|v^s_{k}\|<(10^{-3}+\sqrt {\varepsilon }) Q(y_k)\leq \tfrac 18Q(y_k)$ , if $\varepsilon $ is small enough, we have

$$ \begin{align*} f^{k}[ V^{s}(\underline{y}^+)] \subseteq \exp_{y_k}\bigg[B_{y_k}^u\bigg(\frac{1}{8}Q(y_k)\bigg)\oplus B_{y_k}^s\bigg(\frac{1}{8}Q(y_k)\bigg)\bigg] \end{align*} $$

for all k large enough.

Proof of Claim 2

For any $y\in V^{s}(\underline {y}^+)$ , since $\pi (\underline {x})=\pi (\underline {y})=x$ , from Proposition 4.1(1), we get

$$ \begin{align*} d(y_k,x_k)\leq d(y_k,f^k(x))+d(f^k(x),x_k)\leq \frac{1}{50}Q(y_k)+\frac{1}{50}Q(x_k), \end{align*} $$

and hence $d(f^k(y),x_k)\leq d(f^k(y),y_k)+d(y_k,x_k)<\rho (M)$ and we have $f^k(y)\in B(x_k, \rho (M))$ . Now for $k>0$ large enough, by Claim 1, we have

$$ \begin{align*} \begin{aligned} \|\!\exp_{x_k}^{-1}f^k(y)\|&\leq\|\!\exp_{y_k}^{-1}f^k(y)\|+d(y_k,x_k)\notag\\ &\leq\frac{1}{4}Q(y_k)+\frac{1}{50}Q(x_k)+\frac{1}{50}Q(y_k)\notag\\ &\leq\tfrac{1}{4}(2Q(y_k)+Q(x_k)). \end{aligned} \end{align*} $$

Note that $Q(x_k)\leq \varepsilon ^{{2}/({\alpha -\delta })}d(x_k,\partial O)$ , then

$$ \begin{align*} \begin{aligned} d(f^k(x),\partial O)&\leq d(x_k,\partial O)+d(x_k,f^k(x))\notag\\ &\leq d(x_k,\partial O)+\frac{1}{50}Q(x_k)\notag\\ &\leq e^{\varepsilon^{{2}/({\alpha-\delta})}}d(x_k,\partial O) \end{aligned} \end{align*} $$

and we have $Q(f^k(x))\leq e^{\gamma \varepsilon ^{{2}/({\alpha -\delta }})} Q(x_k)$ .

Similarly, from

$$ \begin{align*} \begin{aligned} d(f^k(x),\partial O)&\geq d(y_k,\partial O)-d(y_k,f^k(x))\notag\\ &\geq d(y_k,\partial O)-\frac{1}{50}Q(y_k)\notag\\ &\geq e^{-\varepsilon^{{2}/({\alpha-\delta})}}d(y_k,\partial O), \end{aligned} \end{align*} $$

we have $Q(f^k(x))\geq e^{-\gamma \varepsilon ^{{2}/({\alpha -\delta })}} Q(y_k)$ . Thus, we arrive at two useful inequalities:

(6.1) $$ \begin{align} d(y_k,\partial O)\leq e^{\varepsilon^{{2}/({\alpha-\delta})}}d(f^k(x),\partial O)\leq e^{2\varepsilon^{{2}/({\alpha-\delta})}}d(x_k,\partial O) \end{align} $$

and

(6.2) $$ \begin{align} Q(y_k)\leq e^{\gamma\varepsilon^{{2}/({\alpha-\delta})}}Q(f^k(x))\leq e^{2\gamma\varepsilon^{{2}/({\alpha-\delta})}}Q(x_k). \end{align} $$

This leads to

$$ \begin{align*} \begin{aligned} \|\!\exp_{x_k}^{-1}f^k(y)\|&\leq\tfrac{1}{4}(2Q(y_k)+Q(x_k))\notag\\ &\leq\tfrac{1}{4}(2e^{2\gamma\varepsilon^{{2}/({\alpha-\delta})}}+1)Q(x_k)\notag\\ &\leq Q(x_k) \end{aligned} \end{align*} $$

if k is large enough and $0<\varepsilon \leq (({\log 3-\log 2})/{2\gamma })^{({\alpha -\delta })/{2}}$ .

It is now obvious that the claim holds since we have assumed as in [Reference Chen, Hu and Zhou10] that the metric on M is taken in such a way that for any $x\in O$ , $E^u(x)$ and $E^s(x)$ are pairwise orthogonal, we thus prove

$$ \begin{align*} f^{k}[ V^{s}(\underline{y}^+)] \subseteq \exp_{x_k}(S_{x_k}) \end{align*} $$

for all k large enough.