Proof. Given $L>0$ , $0<\varepsilon <\varepsilon _1$ , choose $\delta ^{\prime }_1\in (0,1)$ such that $|f'x|<\delta ^{\prime }_1$ implies there exists a unique $c\in C(f)$ , with $d(x,c)<\varepsilon $ . Then, define $\delta _1:=\min \{ \delta ^{\prime }_, \Lambda (f)^{-L}\}$ .

Assume $a\in \mathbb {N}$ satisfies $|f'x_a|<\delta _1$ . We check that $[a,a+l(a)[$ is $(L,\varepsilon )$ -shadowing.

First, note that $l(a)\geq L$ . Indeed,

$$ \begin{align*} l(a)=\frac{-\!\log |f'x_a|}{\unicode{x3bb}(f)}>\frac{-\!\log \delta_1}{\unicode{x3bb}(f)}> \frac{ L\log \Lambda(f)}{\unicode{x3bb}(f)}=L. \end{align*} $$

Second, observe that $[a,a+l(a)[$ is $\varepsilon $ -shadowing.

Since $\delta _1<\delta ^{\prime }_1$ , there exists a unique $c_a\in C(f)$ such that $d(x_a,c_a)<\varepsilon $ . Now, for all $0<h<l(a)$ ,

$$ \begin{align*} \begin{aligned} d(x_{a+h},f^h c_a)&\leq \Lambda(f)^{h-1}d(x_{a+1},f c_a)\\ &\leq \Lambda(f)^{h-1} \cdot |f'x_a|d(x_a,c_a)\leq \Lambda(f)^{l(a)-1}\cdot |f'x_a| d(x_a,c_a)\\ &\leq \Lambda(f)^{-\!\log|f'x_a| / \unicode{x3bb}(f)}|f'x_a| d(x_a,c_a)\leq d(x_a,c_a) <\varepsilon. \end{aligned} \end{align*} $$

The second inequality follows from the fact that $d(x_a,c_a)<\varepsilon <\varepsilon _1$ and $|f'|$ is monotone decreasing from $x_a$ to $c_a$ ; the last inequality follows from our choice of $\delta <\delta ^{\prime }_1$ , which guarantees $\Lambda (f)^{-\!\log |f'x_a| / \unicode{x3bb} (f)}|f'x_a|\leq 1$ .

Proof. Since I is a PSI, $ G^I(a_{n+1}-1)<1$ . If $a_{n+1}\in A$ , then $F(a_{n+1})=2^{K+1}\cdot \, |f'x_{a_{n+1}}|<1$ and

$$ \begin{align*} G^I(a_{n+1})=G^I(a_{n+1}-1)\cdot F(a_{n+1})<G^I(a_{n+1})=G^I(a_{n+1}-1)<1, \end{align*} $$

which contradicts our assumption that I is right maximal.

Proof. Up to changing notation, we assume $a_1\leq b_1$ .

Note that for $j\leq l$ , $s_j=a_i$ with $i\leq n$ or $b_j$ with $j\leq m$ . Thus:

1. (1) all $s_j\in A$ ;

2. (2) $[s_j,s_{j+1}[$ is a segment of some $[a_i,a_{i+1}[$ or $[b_i,b_{i+1}[$ , with $s_i=a_i$ or $b_i$ ; hence, a shadowing interval.

It remains to check that $G^J(k)<1$ for all $k\in [s_1,s_{l+1}[$ . Now, if $k\leq a_{n+1}$ , then $G^J(k)=G^{I_1}(k)<1$ . If $k>a_{n+1}$ , then $b_1<k<b_{m+1}$ and

$$ \begin{align*} G^J(k)=G^{I_1}(b_1)\cdot G^{I_2}(k)<1, \end{align*} $$

since $\operatorname {I_1}\cap \operatorname {I_2}\neq \emptyset $ implies $b_1<a_{n+1}$ .

Proof. We proceed by contradiction. Let $I_1, I_2$ be as above. We can assume that $a_1<b_1$ , $\operatorname {supp}I_1\cap \operatorname {supp}I_2\neq \emptyset $ , and $I_2$ is not contained in $I_1$ . Thus, $b_1<a_{n+1}$ and $a_{n+1}<b_{m+1}$ .

By Lemma 2.4, $I_1\cup I_2$ should be a piecewise shadowing interval. However, this implies $G^{I_1}(a_{n+1})=G^{I_1}(b_1)G^{I_2}(a_{n+1})<1$ , which contradicts $I_1$ being maximal.

Proof. By definition of right maximality,

$$ \begin{align*} \begin{aligned} 1\leq G(a_{n+1})&=\prod_{a_1\leq k<a_{n+1}}F( k)\\ &=\prod_{a\in A\cap [a_1,a_{n+1}[}^n (|f'x_{a}|2^{K+1}) \cdot \prod_{k\notin A} \Lambda\\ &\leq \bigg(\!\prod_{a\in A\cap [a_1,a_{n+1}[}^n (|f'x_{a}|2^{K+1}) \bigg) \cdot \Lambda^{a_{n+1}-a_1}. \end{aligned} \end{align*} $$

Taking the logarithm, we get

$$ \begin{align*} 0\leq \sum_{a\in A\cap [a_1,a_{n+1}[}^n (-\unicode{x3bb}\cdot l(a)+(K+1)\log 2)+(a_{n+1}-a_1)\cdot \unicode{x3bb}. \end{align*} $$

Thus,

$$ \begin{align*} \begin{aligned} a_{n+1}-a_1\geq \sum_{a\in A\cap [a_1,a_{n+1}[} \bigg( l(a)-\frac{K+1}{\unicode{x3bb}}\cdot \log2 \bigg). \end{aligned}\\[-40pt] \end{align*} $$

Proof. First, note that as f has only non-flat critical points and $\varepsilon $ is small, $|f'|$ is monotone from $x_b$ to $c_b$ and from $f^{b-a}c_a$ to $c_b$ . Thus, by the mean value theorem,

$$ \begin{align*} d(x_{b+1},f^{b-a+1}c_a)\leq \max\{|f'x_b|, |f' f^{b-a} c_a| \}\cdot d(x_b,f^{b-a}c_a). \end{align*} $$

We consider three cases.

Case A.1: When I is non-switching and $|f'x_b|\geq |f'f^{b-a}c_a|$ . By the previous discussion,

$$ \begin{align*} d(x_{b+1},f^{b-a+1}c_a)\leq |f'x_b| \cdot d(x_b,f^{b-a}c_a)\leq 2^{K+1}|f'x_b| \cdot d(x_b,f^{b-a}c_a). \end{align*} $$

Case A.2: When I is non-switching and $|f'x_b|\leq |f'f^{b-a}c_a|$ . Then,

$$ \begin{align*} d(x_{b+1},f^{b-a+1}c_a)\leq |f' f^{b-a} c_a|\cdot d(x_b,f^{b-a}c_a). \end{align*} $$

Then, we compare $|f'f^{b-a}c_a|$ with $|f'x_b|$ . Since I is non-switching, $d(f^{b-a}c_a,c_b)<d(f^{b-a}c_a,x_b)+d(x_b,c_b)\leq 2d(x_b,c_b)$ . Thus,

$$ \begin{align*} \begin{aligned} \frac{|f' f^{b-a}c_a|}{|f'x_b|}&=\frac{|f^{(d_b)}(c_b)(f^{b-a}c_a-c_b)^{d_b}\cdot g_{c_b}(f^{b-a}c_a)|}{|f^{(d_b)}(c_b)(x_b-c_b)^{d_b}\cdot g_{c_b}(x_b)|}\\ & \leq \frac{|(2(x_b-c_b))^{d_b}|}{|(x_b-c_b)^{d_b}|}\cdot \frac{| g_{c_b}(f^{b-a}c_a)|}{|g_{c_b}(x_b)|}\\ &\leq 2^{d_b}\cdot 2\leq 2^{K+1}, \end{aligned} \end{align*} $$

where the last inequality follows from (1.1) and (1.2).

Case B: When I is switching. Since $|f'|$ is decreasing from $x_b$ to $c_b$ , we have

$$ \begin{align*}d(x_{b+1}, f c_b)\leq |f'x_b|\,d(x_b,c_b).\end{align*} $$

We conclude

$$ \begin{align*} \begin{aligned} d(x_{b+1}, f c_b)&<|f'x_b|\,d(x_b,c_b)\\ &<|f'x_b|\,d(x_b,f^{b-a}c_b)< 2^{K+1}|f'x_b|\,d(x_b,f^{b-a}c_b). \end{aligned}\\[-34pt] \end{align*} $$

Proof. Fix $\varepsilon>0$ small, then Lemma 2.2 gives $\delta _1>0$ . Then, choose a $0<\delta <\delta _1$ . Let $x\in X$ and $a_1\in A$ .

Claim 1. There is a finite sequence of pairs $(I_s,a^{\ast }_s)$ , $s=1,\ldots S$ , with $I_s$ a finite collection of integer intervals and $a^{\ast }_s\in \mathbb {N}$ such that:

1. (1) $(I_s,a^{\ast }_s)$ is an IPSI, $s=1,\ldots , S-1$ ;

2. (2) $\min I_s=a_1$ ;

3. (3) for $1<s\leq S$ , $\operatorname {supp} I_{s-1}\subseteq {\operatorname {supp }} I_{s}$ , $a^{\ast }_{s-1}<a^{\ast }_s$ ;

4. (4) $S>1$ is the minimal $s\geq 1$ such that $a^{\ast }_s=\max I_s+1$ ;

5. (5) $I_S$ is a PSI and it is right maximal.

Proof. We build by induction a sequence of IPSIs satisfying conditions $(1)$ , $(2)$ , and $(3)$ until condition $(4)$ holds. Then, we check condition $(5)$ .

For $s=1$ , we define $(I_1,a^{\ast }_1)$ by setting $I_1:=\{ [a_1,b_1[ \}$ , with $b_1:=a_1+l(a_1)-(({K+1})/{\unicode{x3bb} })\cdot \log 2$ and $a^{\ast }_1=a_1$ .

Note that $[a_1,b_1[$ is a shadowing interval by Lemma 2.2, so $(I_1,a^{\ast }_1)$ is an IPSI.

Now, assume by induction that for some $s\geq 1$ , there are some IPSIs $(I_1,a^{\ast }_1),\ldots $ , $(I_s,a^{\ast }_s)$ satisfying conditions $(2),(3)$ , and $a^{\ast }_t\leq \max I_t$ for $t\leq s$ . The induction ends when $s+1=\max I_{s+1}+1$ .

Now, we build a new IPSI according to the following two cases.

Case A: $A\ \cap\ ]a^{\ast }_s, \max I_s]\neq \emptyset $ . Write $I_s:=\{ [a_1,a_2[,\ldots , [a_r,b[ \}$ . Notice that since $I_s$ is an IPSI, $[a_r,b[$ is a shadowing interval with $a^{\ast }_s\in [a_r,b[$ .

Fix $a^{\ast }_{s+1}:=\min \{ a\in A\ \cap\ ]a^{\ast }_s,\max I_s] \}\in [a_r,b[$ . Note that $[a_r,a^{\ast }_{s+1}[$ is either switching or non-switching.

Case A.1: $[a_r,a^{\ast }_{s+1}[$ is non-switching. Define

$$ \begin{align*} \begin{aligned} &b':=\min\{ i>b, i\in \mathbb{N}: G^{I_s}_{a^{\ast}_{s+1}}(i)\geq 1 \};\\ &I_{s+1}:=\{ [a_1,a_2[,\ldots,[a_r,b'[ \}. \end{aligned} \end{align*} $$

Note that these two are well defined, since $G^{I_s}_{a^{\ast }_{s+1}}$ is a map defined on $[a_1,\infty [$ . We now verify that $(I_{s+1},a^{\ast }_{s+1})$ is indeed an IPSI.

First, note that by our choice of $b'$ , automatically, $G^{I_{s+1}}_{a^{\ast }_{s+1}}(k)< 1$ for all $k\in \operatorname {supp}I_{s}$ .

Note that by our choice, $a^{\ast }_{s+1}>a^{\ast }_s$ , so $A\ \cap\ ]a^{\ast }_{s},a^{\ast }_{s+1}[\ =\emptyset $ . Now, since $G^J_{-}(k)$ depends on $A\cap [a_1,-[$ , we see that for $a_1\leq k\leq a^{\ast }_{s+1}$ ,

$$ \begin{align*} \begin{aligned} G^{I_{s+1}}_{a^{\ast}_{s+1}}(k)&=G^{I_{s+1}}_{a^{\ast}_s}(k)\\ &=G^{I_s}_{a^{\ast}_s}(k); \end{aligned} \end{align*} $$

and for $a^{\ast }_{s+1}<k<b'$ ,

$$ \begin{align*} \begin{aligned} G^{I_{s+1}}_{a^{\ast}_{s+1}}(k)&=\prod_{a_1\leq a_j<k, a_j\in A_{a^{\ast}_{s+1}}} (|f'x_{a_j}|2^{K+1})\prod_{a\leq j<k, j\notin A_{a^{\ast}_{s+1}}} \Lambda\\ &=G^{I_s}_{a^{\ast}_s}(a^{\ast}_{s+1})(|f'x_{a^{\ast}_{s+1}}|2^{K+1})\Lambda^{k-a^{\ast}_{s+1}-1} \end{aligned} \end{align*} $$

for all $1\leq k\leq b'$ . So, for $a_1\leq a_i \leq k \leq a_{i+1} \leq a^{\ast }_{s+1}$ , by inductive hypothesis, $d(x_k,f^{k-a_i}c_{a_i})<G^{I_{s+1}}_{a^{\ast }_{s+1}}(k)\varepsilon $ . For $a_r< a^{\ast }_{s+1}<k<b'$ ,

(2.1) $$ \begin{align} d(x_k,f^{k-a_i}c_{a_r})&\leq \Lambda^{k-a^{\ast}_{s+1}-1}d(x_{a^{\ast}_{s+1}+1},f^{a^{\ast}_{s+1}-a_r+1}c_{a_r}) \nonumber\\ &\leq \Lambda^{k-a^{\ast}_{s+1}-1}(|f'x^{\ast}_{a_{s+1}}|2^{K+1})d(x_{a^{\ast}_{s+1}},f^{a^{\ast}_{s+1}-a_r}c_{a_r}) \nonumber\\ &\leq G^{I_s}_{a^{\ast}_s}(a_{s+1})(|f'x_{a^{\ast}_{s+1}}|2^{K+1})\Lambda^{k-a^{\ast}_{s+1}-1}\varepsilon \\ &=G^{I_s}_{a^{\ast}_{s+1}}(k)\varepsilon,\nonumber \end{align} $$

where the second inequality follows from $a^{\ast }_{s+1}\in [a_r,b'[$ being non-switching and Proposition 2.7; the third inequality follows from $a^{\ast }_{s+1}\in [a_r,b[$ and inductive assumption; the fourth equality follows from our definition of $G^{I_{s+1}}_{a^{\ast }_{s+1}}$ . From (2.1) and the fact that $G^{I_{s+1}}_{a^{\ast }_{s+1}}(k)<1$ , it is easy to check that $[a_r, b'[$ is shadowing.

Case A.2: $[a_r,b[$ is switching. Define

$$ \begin{align*} \begin{aligned} &b':=\min\{ i>b, i\in \mathbb{N}: G^{I_s}_{a^{\ast}_{s+1}}(i)\geq 1 \};\\ &a_{r'}:=a^{\ast}_{s+1};\\ &I_{s+1}:=\{ [a_1,a_2[,\ldots,[a_r,a^{\ast}_{s+1}[,[a^{\ast}_{s+1},b'[ \};\\ &a_{r+1}:=a^{\ast}_{s+1}. \end{aligned} \end{align*} $$

We verify that $I_{s+1}$ is an IPSI.

For $a_1\leq k\leq a^{\ast }_{s+1}$ , since by our choice $a^{\ast }_{s+1}>a^{\ast }_s$ ,

$$ \begin{align*} G^{I_{s+1}}_{a^{\ast}_{s+1}}(k)=G^{I_s}_{a^{\ast}_s}(k); \end{align*} $$

and for $a^{\ast }_{s+1}<k<b'$ ,

$$ \begin{align*} G^{I_{s+1}}_{a^{\ast}_{s+1}}(k)=G^{I_s}_{a^{\ast}_s}(a^{\ast}_{s+1})(|f'x_{a^{\ast}_{s+1}}|2^{K+1})\Lambda^{k-a^{\ast}_{s+1}-1}. \end{align*} $$

Thus, for $a_1\leq k\leq a^{\ast }_{s+1}$ , naturally $d(x_k,f^{k-a_i}c_{a_i})<G^{I_{s+1}}_{a^{\ast }_{s+1}}(k)\varepsilon $ . For $a^{\ast }_{s+1}<k<b'$ , as $k\in [a_{r'},b'[$ and $a_{r'}=a^{\ast }_{s+1}$ by our construction,

$$ \begin{align*} \begin{aligned} d(x_k,f^{k-a_{s+1}}c_{a_{s+1}})&\leq \Lambda^{k-a_{s+1}-1}\,d(x_{a_{s+1}+1},f c_{a_{s+1}})\\ &\leq \Lambda^{k-a_{s+1}-1}(|f'x_{a_{s+1}}|2^{K+1})\,d(x_{a_{s+1}},f^{a_{s+1}-a_{l_s}}c_{a_{l_s}})\\ &\leq G^{I_s}_{a^{\ast}_s}(a_{s+1})(|f'x_{a_{s+1}}|2^{K+1})\Lambda^{k-a_{s+1}-1}\varepsilon\\ &=G^{J_{n+1}}_{a^{\ast}_{s+1}}(k)\varepsilon\leq \varepsilon, \end{aligned} \end{align*} $$

where the second inequality follows from Lemma 2.7, applied to $a^{\ast }_{s+1}\in [a_r,b[$ , the switching interval.

The integer intervals $[a_1,a_2[,\ldots [a_{r-1},a_r[$ are all shadowing by inductive assumption. The interval $[a_r,a^{\ast }_{s+1}[$ is also shadowing, since $[a_r,a^{\ast }_{s+1}[ \ \subseteq [a_r,b[$ , and the latter is shadowing by inductive assumption. Additionally, $[a^{\ast }_{s+1},b'[$ is shadowing by the arguments above as well as the definition of $b'$ .

Case B: $A\ \cap\ ]a^\ast _s, \max I_s]=\emptyset $ . In this case, we define

$$ \begin{align*} \begin{aligned} &I_{s+1}:=I_s:=\{ [a_1,a_2[,\ldots, [a_r,b[ \};\\ &a^{\ast}_{s+1}:=\max I_s +1;\\ &S:=s+1. \end{aligned} \end{align*} $$

By our assumption of this case, $a_{s+1}>b$ , for $I_s$ ,

$$ \begin{align*} A_{a^{\ast}}=A\cap [a_1,a_s]=A\cap[a_1,b[. \end{align*} $$

In this case, for all $a_1<k<b$ ,

$$ \begin{align*} G^{I_{S}}(k)=G^{I_{S}}_{a^{\ast}_s}(k)< 1, \end{align*} $$

so $I_S$ is a PSI by the definition of b.

Note that in Case B, the constructed sequence $(I_s,a^{\ast }_s)$ satisfies the first three conditions $(1)$ , $(2)$ , and $(3)$ by construction. It remains to check that $(I_s,a^{\ast }_s)$ also satisfies conditions $(4)$ and $(5)$ .

Condition $(4)$ follows by induction hypothesis.

Then, we check condition $(5)$ . By construction, I is a PSI, so it remains to check that it is maximal. Note that by our construction in Case B, $I_S=I_{S-1}$ . By our definition of $b'$ for $I_{S-1}$ , we know that $I_S$ is right maximal.

Now, it remains to show that for all $x\in X$ with $|(f^n)'x\rightarrow \infty |$ , the inductions end up with Case B within a finite amount of steps.

We prove by contradiction. Note that for each $x\in X$ , the collection of visits near the critical set A is a monotone non-decreasing sequence in integers. Thus, by induction, when s goes to infinity, $a^{\ast }_s$ is also monotonely non-decreasing. Thus, $a^{\ast }_{s}\geq s$ . Hence, for all $k\geq a_1$ , $k\leq a^{\ast }_k<b$ , and $k\in [a_1,b[$ .

On the one had, note that by definition of IPSI, for this k,

$$ \begin{align*} 1> G^{I_k}_{a^{\ast}_k}(k)\geq |(f^{k-a_1})'x_{a_1}|. \end{align*} $$

On the other hand, we have assumed $|(f^{k-a_1})' x_{a_1}|\rightarrow \infty $ . For k large enough, this yields a contradiction.

This finishes the proof of the claim.

Proof. Since $b_n\geq n$ , $b_n \rightarrow \infty $ for all $x\in X$ .

Since $\nu \in \mathbb {P}_{\mathrm {erg}}(f)$ , by Birkhoff’s ergodic theorem, for $\nu $ -a.e. $x\in X$ ,

$$ \begin{align*} \sum_{i=0}^{b_n-1}\Psi (x_i)&=b_n\cdot \int \Psi \,d\nu+o(b_n),\\\sum_{i=0}^{a_n-1}\Psi (x_i)&=a_n\cdot \int \Psi \,d\nu+o(a_n). \end{align*} $$

Subtracting the second equation from the first, we have

(2.2) $$ \begin{align} \sum_{i=a_n}^{b_n-1}\Psi (x_i)=(b_n-a_n)\cdot \int \Psi \,d\nu +o(b_n)+o(a_n). \end{align} $$

By the first assumption, $a_n\leq b_n$ . Hence, an $o(a_n)$ is also an $o(b_n)$ . Moreover, by the second assumption, $ 0\leq \sum _{i=a_n}^{b_n-1}\Psi x_i\leq \eta $ , $ \sum _{i=a_n}^{b_n-1}\Psi x_i=o(b_n)$ . Replacing corresponding terms in (2.2), we have

$$ \begin{align*} 0=(b_n-a_n)\cdot \int \Psi \,d\nu +o(b_n). \end{align*} $$

Now that $\int \Psi \,d\nu \neq 0$ as assumed, we have $b_n-a_n=o(b_n)$ .

Since $b_n-a_n=|b_n-n|+|a_n-n|$ ,

$$ \begin{align*} \begin{aligned} b_n-n=o(n)=a_n-n. \end{aligned}\\[-36pt] \end{align*} $$

Proof. Fix $\varepsilon>0$ , $L\in \mathbb {N}^+$ .

Fix $\delta _1$ as in Proposition 2.8. Then, take $0<\delta <\delta _1$ such that

$$ \begin{align*} -\!\log \delta> L\cdot \bigg(L+\frac{K+1}{\unicode{x3bb}}\bigg). \end{align*} $$

Let $\nu \in \mathbb {P}_{\mathrm {{erg}}}(f)$ with $\unicode{x3bb} (\nu )>0$ .

Step 1: The construction of $\mathcal {S}(x)$ . This implies $\nu (X_0)=1$ , where $X_0:=\{ x\in X: |(f^n)'x|\rightarrow \infty \}$ . By Proposition 2.8, there exists $X_1\subseteq X_0$ , $\nu (X_1)=1$ , such that for all $x\hspace{-1pt}\in\hspace{-1pt} X_1$ , there exists a right maximal PSI $J_1(x)$ , with $ \min J_1(x)\hspace{-1pt}=\hspace{-1pt}\min A(x)$ . Consecutively, given $l\in \mathbb {N}^+$ , applying Proposition 2.8 again to $X_l$ with $a_{l+1}:=\min \{a\in A(x): a>\max J_l(x) \}$ , we get $X_{l+1}$ with $\nu (X_{l+1})=1$ and $J_{l+1}(x)$ , PSI with $\min J_{l+1}(x)=a_{l+1}$ . Take $X':=\bigcap _{l\geq 0}X_l$ . Then, $\nu (X')=1$ and for all $x\in X'$ , there exists a collection of finite right maximal PSIs $\mathcal {E}(x):=\{ J_l=J_l(x) \}_{l\in \mathbb {N^+}}$ .

Fix $x\in X'$ . We choose the initial position $a_{l+1}$ to be $ \min \{ a\in A: a>\max J_l \}$ . So, by Lemma 2.5, each $J, J'\in \mathcal {E}(x)$ are disjoint. Thus, we may assume $\max J_l<\min J_{l+1}$ for all $l\in \mathbb {N}$ . In particular, given $x\in X'$ , and $J, J'\in \mathcal {E}(x)$ , J and $J'$ cannot be nested. Moreover, for all $J\in \mathcal {E}(x)$ , by Lemma 2.6,

$$ \begin{align*} \begin{aligned} |\!\operatorname{supp} J | &\geq \sum_{a_i\in A\cap J}\bigg(l(a_i)-\frac{K+1}{\unicode{x3bb}} \log 2\bigg)\\ &\geq \sum_{a_i\in A\cap J}\bigg(l(a_i)-\frac{K+1}{\unicode{x3bb}}\bigg). \end{aligned} \end{align*} $$

Remember that each $J\in \mathcal {E}(x)$ is a PSI, which is a collection of shadowing intervals. For each $x\in X'$ , define

$$ \begin{align*} \mathcal{S}(x):=\{ I=[a,b[\subseteq \mathbb{N}: \text{ there exists } J(x)\in \mathcal{E}(x) \ \text{ such that } \ I\in J(x), |I|\geq L \}, \end{align*} $$

that is, $\mathcal {S}(x)$ consists of shadowing intervals in some $J\in \mathcal {E}(x)$ with length greater than L.

Step 2: The ‘tail’ interval. For the sake of simplicity, we first introduce some notation. Abbreviate $\overline {J}:=\operatorname {supp} J$ for $J\in \mathcal {E}(x)$ . For all $n\in \mathbb {N}^+$ , let $l_n$ be the maximal integer such that $\min \overline {J}_{l_n}<n$ .

Claim that if we define

(1) $$ \begin{align} a_n= \begin{cases} \min \overline{J}_{l_n} & \text{if } n\in \overline{J}_{l_n}, \\ n & \text{otherwise}, \end{cases} \end{align} $$

(2) $$ \begin{align} b_n= \begin{cases} \max \overline{J}_{l_n} & \text{if } n\in \overline{J}_{l_n},\\ n & \text{otherwise}, \end{cases} \end{align} $$

then $b_n-a_n=o(n)$ . To prove the claim, we use Lemma 2.9.

Define

(3) $$ \begin{align} \Psi(x):= \begin{cases} -\unicode{x3bb}(f) & \text{if } |f'x|\geq \delta, \\ -(\log|f'x|+(K+1)\log 2) & \text{if } |f'x|< \delta. \end{cases} \end{align} $$

We check that $a_n, b_n$ , and $\Psi $ satisfy all the hypotheses of Lemma 2.9. First, we check that $\int \Psi \,d\nu \neq 0$ . By definition,

$$ \begin{align*} \begin{aligned} \int \Psi \,d\nu&= \int_{\{ |f'|<\delta \}} -(\log |f'x|+(K+1)\log2)\,d\nu +\int_{\{|f'|\geq \delta\}}-\unicode{x3bb}(f)\,d\nu\\ &\leq -\int \log |f'x| \,d\nu - \int_{\{ |f'|<\delta \}} (K+1)\log2 \,d\nu<0. \end{aligned} \end{align*} $$

The first inequality follows from the fact that $\unicode{x3bb} (f)\geq \log |f'x|$ for all $x\in X$ . The last inequality follows from our assumption that $\unicode{x3bb} (\nu )=\int \log |f'|\,d\nu>0$ and our choice of $\delta $ being small.

When $n\notin \overline {J}_{l_n}$ , then $a_n=b_n=n$ and $\sum _{i=a_n}^{b_n-1}\Psi (x_i)=0$ . Assume $n\in \overline {J}_{l_n}$ . Remember that by definition, $J_{l_n}$ is a right maximal PSI. Thus, $G^{J_{l_n}}(\max \overline {J}_{l_n}+1)\geq 1$ , which implies

$$ \begin{align*} \begin{aligned} 0&\leq \log G^{J_{l_n}}(\max \overline{J}_{l_n}+1)\\ &=\sum_{j\in \overline{J}_{l_n}\cap A^c}\unicode{x3bb}(f)+\sum_{a\in \overline{J}_{l_n}\cap A}(\log |f'|+(K+1)\log2)\\ &=-\sum_{i=a_n}^{b_n} \Psi(x_i)\\ &=-\sum_{i=a_n}^{b_n-1}\Psi(x_i)+\unicode{x3bb}(f), \end{aligned} \end{align*} $$

that is, $\sum _{j=a_n}^{b_n-1}\Psi (x_j)\leq \unicode{x3bb} (f)$ . The last equality follows from Lemma 2.3: $b_n=\max \overline {J}_{l_n}+1\notin A$ , so $\log \Psi (x_{b_n})=\unicode{x3bb} (f)$ .

Meanwhile, since $J_{l_n}$ is a PSI, $G^{J_{l_n}}(\max \overline {J}_{l_n})\leq 1$ . So, $\sum _{j=a_n}^{b_n-1}\Psi (x_j)\geq 0$ . So, $a_n$ , $b_n$ , and $\Psi $ satisfy all the requirements of Lemma 2.9.

Step 3: Verifying the properties in Theorem 2.1.

(1) Since elements in $\mathcal {S}(x)$ are constructed in Proposition 2.8 by visits to critical points, the map $x\mapsto \mathcal {S}(x)$ is naturally measurable.

(2) Note that since elements in $\mathcal {E}(x)$ are finite, pairwise disjoint, so will elements be in $\mathcal {S}(x)$ . Moreover, by construction, elements in $\mathcal {S}(x)$ are naturally shadowing intervals.

(3) Let

$$ \begin{align*} \psi(x):=1_{\{ |f'|<\delta \}}\left(\frac{-\!\log|f'|+K+1}{\unicode{x3bb}(f)}\right). \end{align*} $$

Note that $\psi (x)\geq 0$ and $\psi (x_j)>0$ if and only if $j\in \bigcup \mathcal {E}(x)$ . In particular,

$$ \begin{align*} \sum_{i=1}^{l_n-1} |\overline{J}_i|\leq \sum_{i=0}^{n-1} \psi(x_i)\leq \sum_{i=1}^{l_n} |\overline{J}_i|. \end{align*} $$

Thus, $\sum _{i=1}^{l_n-1} |\overline {J}_i|=\sum _{i=0}^{n-1} \psi (x_i)+o(n)$ , by Step 2.

Now,

$$ \begin{align*} \sum_{I\in \mathcal{S}(x), I^{-}<n} |I|\leq \sum_{i=0}^{n-1}(L\cdot 1_{\{ |f'|<\delta \}}). \end{align*} $$

Thus, by our choice of $\delta $ ,

$$ \begin{align*} \begin{aligned} |\mathcal{S}(x)\cap [0,n[|&=\sum_{I\in \mathcal{S}(x), I^{-}<n} |I|\geq \sum_{i=1}^{l_n-1} |\overline{J}_{i}|-\sum_{i=0}^{n} (L\cdot 1_{\{ |f'|<\delta \}})\\ &\geq \sum_{i=0}^{n-1}\psi(x_i) -\sum_{i=0}^{n} (L\cdot 1_{\{ |f'|<\delta \}})+o(n). \end{aligned} \end{align*} $$

The proof is concluded.

Proof. Note that $H(t)$ is a function defined on $(0,1)$ , increasing from $0$ to $\tfrac 12$ and decreasing from $\tfrac 12$ to $1$ .

By definition, every finite collection in $\mathcal {T}_L(n)$ consists of disjoint intervals in $[0,n[$ of lengths at least L. Thus, there can be at most $\llcorner {n}/{L} \lrcorner $ such intervals maximal in the union. These give

$$ \begin{align*} \sum_{i=0}^{\llcorner {n}/{L} \lrcorner}\binom{n}{2i} \end{align*} $$

many possible types. By assumption, $ ({1}/{L})<\tfrac 14$ . Thus, each $\binom {n}{2i}\leq \binom {n}{2\llcorner {n}/{L} \lrcorner }$ and

$$ \begin{align*} \begin{aligned} \sum_{i=0}^{\llcorner {n}/{L} \lrcorner} \binom{n}{i} &\leq \dfrac{1}{2}\cdot 2\llcorner \dfrac{n}{L} \lrcorner\cdot \binom{n}{2\llcorner \dfrac{n}{L} \lrcorner}\\ &\leq \dfrac{1}{2}\exp\left(n\cdot H\left(\dfrac{2}{L}\right)+o(n)\right) \end{aligned} \end{align*} $$

using de Moivre’s approximation.

Proof. Simply note that given a $(({\varepsilon }/{2}),n)$ -Bowen ball $B(z,n,({\varepsilon }/{2}))$ of $f^m X$ and a covering $\mathcal {B}$ of X by $(({\varepsilon }/{2}),m)$ -Bowen balls, each intersection $f^{-m} B(z,n,({\varepsilon }/{2}))\cap B$ with $B\in \mathcal {B}$ is contained in a $(\varepsilon ,m+n)$ -Bowen ball.

Proof. Given $\varepsilon , \gamma $ , let $(X',\varepsilon ,L,\varphi , \{ X_I \}, \mathcal {E}(x))$ be a partial shadowing on $(X,f)$ . Take $C:=C_{\gamma , \varepsilon }$ as in Lemma 3.2 such that

$$ \begin{align*} r\left(f,n,\dfrac{\varepsilon}{2},X\right)\leq C \cdot \exp (n\cdot (h_{\mathrm{top}}(f)+\gamma)). \end{align*} $$

Since we have defined $h(\{ X_I \}):=\limsup _{n\rightarrow \infty }\max _{|I|=n}({1}/{n})\log |X^{\varepsilon }_I|$ , for $\gamma>0$ given, there exists $D=D_{\gamma }$ such that for all $n\in \mathbb {N}^+$ ,

$$ \begin{align*} \max_{|I|=n}|X_I|\leq D \cdot \exp ( n\cdot (h(\{ X_I\})+\gamma) ). \end{align*} $$

Take one $\nu \in \mathbb {P}_{\mathrm {erg}}(f)$ .

Step 1: Finding a suitable Borel subset. By definition of a partial shadowing, the function $\varphi $ is strictly positive on X. So, there exists a Borel subset $Z\subseteq X'$ such that $\nu (Z)=\nu (X')$ and for all $x\in Z$ ,

$$ \begin{align*} \lim_{n\rightarrow \infty}\dfrac{1}{n}\sum_{i=0}^{n-1}\varphi(x_i)=\int \varphi \,d\nu. \end{align*} $$

We then subdivide Z into smaller pieces. Define the n-combinatorical types of $\mathcal {E}(x)$ to be

$$ \begin{align*} T(\mathcal{E}(x),n):=\bigcup_{I\in \mathcal{E}(x)} I\cap [0,n[. \end{align*} $$

Note that $T(\mathcal {E}(x),n)\in \mathcal {T}_L(n)$ by the zeroth assumption on partial shadowing. For a given type $T\in \mathcal {T}_L(n)$ , let

$$ \begin{align*} Z^T:=\{ x\in Z: T(\mathcal{E}(x),n)=T \}. \end{align*} $$

By the second assumption on partial shadowing, $Z^T$ is a measurable subset of X. By Lemma 3.1,

$$ \begin{align*} |\mathcal{T}_L(n)|\leq \exp\bigg(n\cdot H\bigg(\dfrac{2}{L}\bigg)+o(n)\bigg). \end{align*} $$

Thus, for all $n\in \mathbb {N}^+$ ,

$$ \begin{align*} r(f,n,\varepsilon,Z)\leq \exp\bigg(n\cdot H\bigg(\dfrac{2}{L}\bigg)+o(n)\bigg) \cdot \max_{T\in \mathcal{T}(x,n)} \{ r(f,n,\varepsilon,Z^T) \}. \end{align*} $$

Step 2: Entropy estimate. By construction, given each $Z^T$ , the combinatorical type for $x\in Z^T$ is a fixed union of disjoint intervals $\mathcal {E}_n(x)=T$ . Denote by $T^c$ the collection of integer intervals which are complements of those in T.

Note that

$$ \begin{align*} \begin{aligned} h(f,\nu,\varepsilon)&=\limsup_{n\rightarrow \infty} \dfrac{1}{n} \log r(f,n,\varepsilon,Z)\\ &\leq \limsup_{n\rightarrow \infty}\dfrac{1}{n}\log\left(\exp\left(n\cdot H\left(\dfrac{2}{L}\right)+o(n)\right)\cdot \max_{T\in \mathcal{T}} \{ r(f,n,\varepsilon_0,Z^T) \}\right)\!. \end{aligned} \end{align*} $$

Meanwhile, by our definition of Z and Lemma 3.3, for all $n\in \mathbb {N}^+$ ,

$$ \begin{align*} \begin{aligned} &\hspace{-12pt} \max_{T\in \mathcal{T}} \{ r(f,n,\varepsilon_0,Z^T) \}\\ &\leq \prod_{[a,a'[\in T}r\bigg(f,a'-a,\dfrac{\varepsilon}{2},f^a Z^T\bigg)\prod_{[b,b'[\in T^c} r\bigg(f,b'-b,\frac{\varepsilon}{2},f^b Z^T\bigg). \end{aligned} \end{align*} $$

Since each $[a,a'[$ is part of the system of covering, $r(f,a-a',{\varepsilon _0}/{2},f^a Z^T_k)\leq \max _{|I|=n}|X_I|$ , as $f^a Z^T$ could be covered by a union of $(a'-a,\varepsilon )$ -Bowen balls. Thus, we may apply Lemma 3.2 to see that

$$ \begin{align*} \begin{aligned} & \max_{T\in \mathcal{T}} \{ r(f,n,\varepsilon_0,Z^T) \}\\ &\leq \bigg(\!\prod_{[a,a'[\in T} \max_{|I|=n}|X_I| \bigg)\cdot \bigg(\!\prod_{[b,b'[\in T^c} r\bigg(f,b'-b,\frac{\varepsilon}{2},f^b Z^T\bigg) \bigg), \end{aligned} \end{align*} $$

where

$$ \begin{align*} \begin{aligned} & \prod_{[a,a'[\in T} \max_{|I|=n}|X_I| \leq \exp\bigg((h(\{ X_I \})+\gamma)\cdot \bigg(\!\sum_{I\in T} |I\cap [0,n[| \bigg)+|T|\log D\bigg),\end{aligned} \end{align*} $$

$$ \begin{align*} \begin{aligned} & \prod_{[b,b'[\in T^c} r\left(f,b'-b,\frac{\varepsilon}{2},f^b Z^T\right) \leq \exp\bigg((h_{\mathrm{top}}(f)+\gamma)\cdot \bigg(\!\sum_{I\in T^c}|I\cap [0,n[| \bigg)+|T^c|\log C\bigg) \end{aligned} \end{align*} $$

for all $x\in Z^T$ .

Note that, by our assumption (4) and assumption (1), for all $x\in Z^T$ , $T=\mathcal {E}_n(x)$ and

$$ \begin{align*} \begin{aligned} \liminf_{n\rightarrow \infty}\dfrac{1}{n} \bigg(\!\sum_{I\in \mathcal{E}_n(x)}|I|-\bigg(1-\frac{1}{L}\bigg)\sum_{i=0}^{n-1}\varphi(x_i)\bigg)&\geq 0,\\ \limsup_{n\rightarrow \infty} \frac{|\mathcal{E}_n(x)|}{n}&\leq \dfrac{1}{L}. \end{aligned} \end{align*} $$

Thus,

$$ \begin{align*} h(f,\nu,\varepsilon) &\leq H\left(\dfrac{2}{L}\right)+ \frac{\log D}{L}+\frac{\log{C}}{L}+h_{\mathrm{{top}}}(f)\\ & \quad-\bigg( \bigg(1-\dfrac{1}{L}\bigg)\int \varphi \,d\nu \bigg)(h_{\mathrm{{top}}}(f)-h(\{ X_I \}))+2\gamma. \end{align*} $$

Just let

$$ \begin{align*} K(\varepsilon,\gamma,L):=\lim_{n\rightarrow \infty} H\left(\dfrac{2}{L}\right)+\frac{\log C}{L}+\frac{\log D}{L}. \end{align*} $$

With Lemma 3.1, one verifies easily that $K(\varepsilon ,\gamma ,L)\rightarrow 0$ when $\varepsilon $ and $\gamma $ are fixed, and $L\rightarrow \infty $ .

Proof. First, fix $\gamma _0>0$ .

Then, take $\varepsilon _0$ small, so that $h(f,\nu ,\varepsilon _0)<h(f,\nu )+\gamma _0$ for all $\nu \in \mathbb {P}_{\mathrm {{erg}}}(f)$ . This is possible since $C(f)$ is a finite set and f is $C^{\infty }$ . Third, note that by Theorem 3.4, with $\varepsilon _0, \gamma _0$ fixed,

$$ \begin{align*} \lim_{L\rightarrow \infty} K(\varepsilon_0,\gamma_0,L)=0. \end{align*} $$

Thus, we may take $L_0$ large so that $K(\varepsilon _0,\gamma _0,L_0)<\gamma _0$ and $({1}/{L_0})<\gamma _0$ .

By assumption, $\alpha :=\lim _{\delta \to 0}\limsup _{k\to \infty }\int _{\{ |f'|<\delta \}} -\!\log |f'|\,d\mu _k>0$ . So, we may take $\delta _0>0$ and take a subsequence if necessary, so that for all $k\in \mathbb {N}_{+}$ ,

$$ \begin{align*} \frac{1}{\unicode{x3bb}}\int_{\{ |f'|<\delta_0 \}} -\!\log|f'|\,d\mu_k>\frac{\alpha}{\unicode{x3bb}}-\gamma_0. \end{align*} $$

Finally, shrink $\delta _0$ , so that we get the sequence $\mathcal {S}(x)$ as in Theorem 2.1 for $([0,1],f,\varepsilon _0,L_0)$ . Then, define

$$ \begin{align*} \varphi_0(x):=-1_{\{ x:|f'x|<\delta_0 \}}\frac{\log |f'(x)|}{\unicode{x3bb}(f)}. \end{align*} $$

Take one $\mu _k$ . We check that the $\mathcal {S}(x)$ satisfies all the conditions for being partially shadowing. Take $X'$ to be the subset such that:

1. (1) $\mu _k(X')>\tfrac 12$ ;

2. (2) $|\mathcal {S}(x)\cap [0,n[|\geq (1-({1}/{L_0}))\sum _{i=0}^{n-1}\varphi _0(x_i)-g_{n}(x)$ .

The existence of $X'$ is guaranteed by Theorem 2.1.

For each I, integer interval, define $ X_I $ to be $C(f)$ . In this case, $h(\{X_I\})=0$ . We now verify that $(X',\varepsilon _0,L_0,\varphi _0,C(f),\mathcal {S}(x))$ really gives a partial shadowing for $(f,\mu _k)$ .

1. (1) By construction, for all $I\in \mathcal {S}(x)$ , $|I|\geq L_0$ .

2. (2) By construction, $I=[a,b[\ \in \mathcal {S}(x)$ is $\varepsilon $ -shadowing, so $x_a\in B(c,\varepsilon ,|I|)$ for some $c\in C(f)$ .

3. (3) This is obvious, as I is characterized by visiting to $\{ |f'x|<\delta _0 \}$ .

4. (4) This is property (3) of Theorem 2.1.

Now, apply Theorem 3.4 to $(X',f,\varepsilon _0.L_0,\varphi _0,\mu _k)$ . We get

$$ \begin{align*} h(\mu_k,\varepsilon_0)\leq \left(1-(1-\gamma_0)\left(\frac{\alpha}{\unicode{x3bb}}-\gamma_0\right)\right) h_{\mathrm{{top}}}(f)+2\gamma_0. \end{align*} $$

So,

$$ \begin{align*} \begin{aligned} h_{\mathrm{{top}}}(f)-h(\mu_k)&\geq h_{\mathrm{{top}}}(f)-h(\mu_k,\varepsilon_0)-\gamma_0\\ &\geq h_{\mathrm{{top}}}(f)-\left(1-(1-\gamma_0)\left(\frac{\alpha}{\unicode{x3bb}}-\gamma_0\right)\right)h_{\mathrm{{top}}}(f)-3\gamma_0\\ &\geq \frac{\alpha}{\unicode{x3bb}} h_{\mathrm{{top}}}(f)-4\gamma_0-\gamma_0\left(\frac{\alpha}{\unicode{x3bb}}-\gamma_0\right)h_{\mathrm{{top}}}(f). \end{aligned} \end{align*} $$

As the choice of $\gamma _0$ is arbitrary, we get the main theorem.