Proof. For given $\varepsilon>0$ , there exists $k\in {\mathbb N}$ such that $\unicode{x3bb} ^k\operatorname {diam}(Y)<\varepsilon $ . Let $\delta =\gamma ^{-k}$ , where $\gamma $ is the one in (4.1). For $\sigma ,\sigma ^\prime \in \Sigma ^m$ , $d_{\gamma }(\sigma , \sigma ^{\prime })<\delta =\gamma ^{-k}$ implies that, for all $i<k$ , $\sigma _i\!=\sigma ^{\prime }_i$ , and hence, for any $x,y\in Y$ ,

$$ \begin{align*} & d_0(\Theta_{\! f}(\sigma), \Theta_{\! f}(\sigma^\prime)) \\ &\quad \leq\sup_{x,y\in Y} d_0(\phi_{\sigma_1}\circ\cdots\circ\phi_{\sigma_k}\circ\phi_{\sigma_{k+1}}\cdots\circ\phi_{\sigma_n}(x), \\ &\qquad\qquad\qquad \phi_{\sigma_1}\circ\cdots\circ\phi_{\sigma_k}\circ\phi_{\sigma^{\prime}_{k+1}}\cdots\circ \phi_{\sigma^{\prime}_n}(y))\\ &\quad <\unicode{x3bb}^k \sup_{x,y\in Y} d_0(\phi_{\sigma_{k+1}}\circ\cdots\circ\phi_{\sigma_n}(x),\phi_{\sigma^{\prime}_{k+1}}\cdots\circ \phi_{\sigma^{\prime}_n}(y))\\ &\quad <\unicode{x3bb}^k \operatorname{diam}(Y)<\varepsilon, \end{align*} $$

and hence the continuity with respect to $\sigma $ . Since $\Sigma ^m$ is compact, the attractor $\Lambda (\Phi _{\! f})$ is compact as it is the continuous image of a compact set.

Proof. For $p\in {\mathbb N}$ , let ${\boldsymbol P}(f^p)=\{P_1^p,P_2^p,\ldots ,P_{m_p}^p\}$ and let $m_1=m$ . We start by proving the first inclusion, that is $\Lambda (f)\subset \Lambda (f^p)$ . Let $x\in \Lambda (f)= \bigcap _{n\in {\mathbb N}}\overline {{f}^n(X)}$ , that is, for every $n\in {\mathbb N}$ , $x\in \overline {f^n(X)}$ . Accordingly,

$$ \begin{align*} x\in \bigcap_{n\in{\mathbb N}}\overline{(f^p)^n(X)}= \Lambda(f^p). \end{align*} $$

Thus, $\Lambda (f)\subset \Lambda (f^p)$ . For the third inclusion, that is, $\Lambda (\Phi _{f^p})\subset \Lambda (\Phi _{\! f})$ , let $\Phi _{f^p}=\{\varphi _1,\varphi _2,\ldots ,\varphi _{m_p}\}$ . Then, for every $\omega =(\omega _1,\omega _2,\ldots )\in \Sigma ^{m_p}$ there exists $\sigma ^{\omega }=(\sigma ^{\omega _1},\sigma ^{\omega _2}, \ldots )$ , where $\sigma ^{\omega _i}=(\sigma ^{\omega _i}_1,\ldots ,\sigma ^{\omega _i}_p)\in \Sigma ^{m}_p$ , such that

$$ \begin{align*} \Theta_{\! f^p}(\omega)=\bigcap_{n\in{\mathbb N}}\varphi_{\omega_1}\circ\varphi_{\omega_2}\circ\cdots\circ\varphi_{\omega_n}(Y)=\bigcap_{n\in{\mathbb N}} \phi_{\sigma^{\omega_1}_1}\circ\cdots\circ\phi_{\sigma^{\omega_1}_p}\circ\cdots\circ\phi_{\sigma^{\omega_n}_p}(Y). \end{align*} $$

Thus, $\Lambda (\Phi _{f^p})\!=\! \bigcup _{\omega \in \Sigma ^{m_p}}\Theta _{\! f^p}(\omega )\!=\!\bigcup _{\{\sigma ^\omega \;:\;\omega \in \Sigma ^{m_p}\}}\Theta _{\! f}(\sigma ^\omega )\subset \bigcup _{\sigma \in \Sigma ^{m}}\Theta _{\! f}(\sigma )=\Lambda (\Phi _{\! f}) $ . Finally, for the second inclusion, let $x\in \Lambda (f^p)=\bigcap _{n\in {\mathbb N}}\overline {f^{pn}(X)}$ . Then, for all $n\in {\mathbb N}$ , there exists $y_n\in X$ such that $d_0(x,f^{pn}(y_n))<1/n$ . By definition of $\Phi _{f^p}$ , there exists $\omega ^n=(\omega _1^n,\omega _2^n,\ldots ,\omega _n^n,\ldots )\in \Sigma ^{m_p}$ such that $f^{pn}(y_n)=\varphi _{\omega _1^n}\circ \cdots \circ \varphi _{\omega _n^n}(y_n)$ , where $\varphi _{\omega _i^n}\in \Phi _{f^p}$ . This implies that, for all $n\in {\mathbb N}$ , $d_0(x,\varphi _{\omega _1^n}\circ \cdots \circ \varphi _{\omega _n^n}(y_n))<1/n$ . By compactness of $\Sigma ^{m_p}$ , there exists a subsequence $\{n_k\}$ and $\omega \in \Sigma ^{m_p}$ such that $\omega ^{n_k}\to \omega $ . Since, by Lemma 4.6, $\Theta _{\! f^p}$ is continuous,

$$ \begin{align*} d_0(x, \Theta_{\! f^p}(\omega))= \lim_{k\to\infty} d_0(x, \Theta_{\! f^p}(\omega^{n_k}))=\lim_{k\to\infty} d_0(x,f^{pn_k}(y_{n_k}))=0. \end{align*} $$

Hence, by definition of the attractor, $x\in \Lambda (\Phi _{f^p})$ .

Proof. Let $\delta ,\delta ^\prime>0$ , $n\in {\mathbb N}$ , $x\in Y$ and $\sigma =(\sigma _1,\sigma _2,\ldots ,\sigma _n,\ldots )$ . Then

$$ \begin{align*} d_0(\Theta_{\! f^{\delta}}(\sigma), \Theta_{\! f^{\delta^\prime}}(\sigma))&\leq d_0(\Theta_{\! f^{\delta}}(\sigma),\phi_{\sigma_1}^{\delta}\circ \cdots \circ \phi_{\sigma_n}^{\delta}(x) )\\ &\quad +d_0(\Theta_{\! f^{\delta^\prime}}(\sigma),\phi_{\sigma_1}^{\delta^{\prime}}\circ \cdots \circ \phi_{\sigma_n}^{\delta^{\prime}}(x))\\ &\quad +d_0(\phi_{\sigma_1}^{\delta}\circ \cdots \circ \phi_{\sigma_n}^{\delta}(x),\phi_{\sigma_1}^{\delta^{\prime}}\circ \cdots \circ \phi_{\sigma_n}^{\delta^{\prime}}(x))\\ & \leq \unicode{x3bb}^nd_0(\Theta_{\! f^{\delta}}(\tau^n\sigma),x)+ \unicode{x3bb}^n d_0(\Theta_{\! f^{\delta^\prime}}(\tau^n\sigma),x)\\ &\quad +d_0(\delta,\delta^{\prime}){\kern-1pt} +{\kern-1pt}d_0(\phi_{\sigma_1}{\kern-1pt}\circ{\kern-1pt} \phi_{\sigma_2}^{\delta}{\kern-1pt}\circ \cdots \circ \phi_{\sigma_n}^{\delta}(x),\phi_{\sigma_1}{\kern-1pt}\circ \phi_{\sigma_2}^{\delta^\prime}{\kern-1pt}\circ \cdots \circ \phi_{\sigma_n}^{\delta^{\prime}}(x))\\ & \leq 2\unicode{x3bb}^n\operatorname{diam}(Y) +d_0(\delta,\delta^{\prime})\\&\quad +\unicode{x3bb} d_0(\phi_{\sigma_2}^{\delta}\circ \cdots \circ \phi_{\sigma_n}^{\delta}(x),\phi_{\sigma_2}^{\delta^\prime}\circ \cdots \circ \phi_{\sigma_n}^{\delta^{\prime}}(x)). \end{align*} $$

Iterating the above argument yields

$$ \begin{align*} \begin{split} d_0(\Theta_{\! f^{\delta}}(\sigma), \Theta_{\! f^{\delta^\prime}}(\sigma)) &\leq \lim_{n\to\infty}\{2\unicode{x3bb}^n\operatorname{diam}(Y)+ d_0(\delta,\delta^{\prime})(1+\unicode{x3bb}+\unicode{x3bb}^2+\cdots+ \unicode{x3bb}^n)\}\\ &=(1-\unicode{x3bb})^{-1}d_0(\delta,\delta^{\prime}), \end{split} \end{align*} $$

and letting $a=1/(1-\unicode{x3bb} )$ concludes the proof.

Proof of Proposition 4.2

Let $p\in {\mathbb N}$ be such that $m_p\unicode{x3bb} ^p\leq \tfrac 12$ . Lemma 4.8 asserts that $\Lambda (f^\delta )\subset \Lambda (\Phi _{(f^\delta )^p})$ . Hence, by Theorem 2.8, it suffices to prove that, for every $\varepsilon>0$ small enough, there exists $\delta \in B_{\varepsilon }(0)$ such that the attractor $\Lambda (\Phi _{(f^\delta )^p})$ is disjoint from $\partial {\boldsymbol P}(f^\delta )=\partial {\boldsymbol P}(f)$ .

Suppose, to the contrary, that, for every $\delta \in B_{\varepsilon }(0)$ , $\Lambda (\Phi _{(f^\delta )^p})\cap \partial {\boldsymbol P}(f)\neq \emptyset $ . Accordingly, there exists $\omega (\delta )\in \Sigma ^{m_p}$ for which $\Theta _{\! (f^\delta )^{p}}(\omega (\delta ))\in \partial {\boldsymbol P}(f)$ . By definition, $\partial {\boldsymbol P}(f)=\bigcup _{P\in {\boldsymbol P}(f)}\partial P$ . Therefore, there exists $P_i\in {\boldsymbol P}(f)$ and $A\subset B_{\varepsilon }(0)$ with $\mu _d(A)\geq \mu _d(B_{\varepsilon }(0))/m = C_d\varepsilon ^d/m$ ( $\mu _d$ is the d-dimensional Lebesgue measure) such that, for all $\delta \in A,\ \Theta _{\! (f^\delta )^p}(\omega (\delta ))\in \partial P_i $ .

Moreover, for each $k\in {\mathbb N}$ , there exist $\omega ^*= (\omega ^*_1, \omega ^*_2,\ldots ,\omega ^*_k)\in \Sigma ^{m_p}_k$ such that the set defined as

$$ \begin{align*} A_k(\omega^*)= \{\delta\in A: \omega(\delta)_j=\omega^*_j,j\leq k\} \end{align*} $$

is non-empty and $\mu _d(A_k(\omega ^*))\geq \mu _d(A)/m_p^k\geq C_d\varepsilon ^dm^{-1}{m_p}^{-k}$ . Accordingly, for $\omega (\delta )\in A_k(\omega ^*)$ and the IFS associated to $\Phi _{(f^\delta )^p}=\{\varphi _1^\delta ,\varphi _2^\delta ,\ldots ,\varphi _{m_p}^\delta \}$ ,

$$ \begin{align*} \partial P_i\ni \Theta_{\! (f^\delta)^p}(\omega(\delta))&=\varphi_{\omega^*_1}^{\delta}\circ\Theta_{\! (f^\delta)^p}(\tau \omega(\delta))= \varphi_{\omega^*_1}^{\delta}\circ\varphi_{\omega^*_2}^{\delta}\circ\cdots \circ\varphi_{\omega^*_k}^{\delta}\circ\Theta_{\! (f^\delta)^p}(\tau^k\omega(\delta)), \end{align*} $$

where $\tau $ is the left shift as defined above and $\varphi _{\omega _j^*}^\delta = \phi ^\delta _{\sigma ^p_{j,1}}\circ \phi _{\sigma ^p_{j,2}}^\delta \circ \cdots \circ \phi _{\sigma ^p_{j,p}}^\delta $ , for $\omega ^*_j=(\sigma ^p_{j,1},\sigma ^p_{j,2},\ldots ,\sigma ^p_{j,p})\in \Sigma ^m_p$ with $\phi ^\delta _{\sigma ^p_{j,s}}\in \Phi _{f^\delta }$ .

Next, for some $\bar {x}\in X$ , define $\theta :A_k(\omega ^*)\to X$ as $\theta (\delta )=\varphi _{\omega ^*_1}^{\delta }\circ \varphi _{\omega ^*_2}^{\delta }\circ \cdots \circ \varphi _{\omega ^*_k}^{\delta }(\bar {x})$ . Then

$$ \begin{align*} d_0(\Theta_{\! (f^\delta)^p}(\omega(\delta)),\theta(\delta))\leq\unicode{x3bb}^{pk}d_0(\Theta_{\! (f^\delta)^p}(\tau^k\omega(\delta)),\bar{x})\leq \operatorname{diam}(Y)\unicode{x3bb}^{pk}=:B_*\unicode{x3bb}^{pk}. \end{align*} $$

Thus, $\theta (\delta )$ belongs to a $B_*\unicode{x3bb} ^{pk}$ neighbourhood of $\partial P_i$ . Since $\partial P_i$ is contained in the union of finitely many ${\mathcal C}^2$ manifolds, the Lebesgue measure of a $B_*\unicode{x3bb} ^{pk}$ neighbourhood of $\partial P_i$ is bounded above by $C\unicode{x3bb} ^{pk} \mu _{d-1}(\partial P_i)$ for a fixed constant $C>0$ . Accordingly,

(4.6) $$ \begin{align} \mu_d(\theta(A_k(\omega^*)))\leq C\unicode{x3bb}^{pk} \mu_{d-1}(\partial P_i). \end{align} $$

On the other hand,

$$ \begin{align*} \mu_d(\theta(A_k(\omega^*)))=\int_{A_k(\omega^*)}|\kern-1pt \det(D\theta(\delta))|\,d\delta, \end{align*} $$

where, by definition of $\theta (\delta )$ , . Note that

$$ \begin{align*} \|D\varphi_{\sigma_1^*}+D\varphi_{\sigma_1^*}D\varphi_{\sigma_2^*}+\cdots+D\varphi_{\sigma_1^*}\cdots D\varphi_{\sigma_{k-1}^*}\hspace{-1pt}\|\leq \frac{\unicode{x3bb}^p}{1-\unicode{x3bb}^p}, \end{align*} $$

where $\|\cdot \|$ is the standard operator norm defined as $\|L\|=\sup _{\|v\|=1}\|Lv\|$ for any linear operator $L:{\mathbb R}^d\to {\mathbb R}^d$ . It follows that, for all $v\in {\mathbb R}^d$ ,

$$ \begin{align*} \|D\theta(\delta)v\|\geq \|v\|-\frac{\unicode{x3bb}^p}{1-\unicode{x3bb}^p}\|v\|\geq\frac{1}{2}\|v\|, \end{align*} $$

since $\unicode{x3bb} ^p\leq {1}/{2m^p}$ and $m^p\geq 2$ . Hence, the eigenvalues of $D\theta (\delta )$ are larger, in modulus, than $\tfrac 12$ . Accordingly, $|\kern-2.5pt \det (D\theta (\delta ))|\geq 2^{-d}$ and

(4.7) $$ \begin{align} \mu_d(\theta(A_k(\omega^*)))\geq 2^{-d} \mu_d(A_k(\omega^*))\geq C_d 2^{-d}\varepsilon^dm^{-1}{m_p}^{-k}, \end{align} $$

which, for k large enough, is in contradiction with (4.6), and this concludes the proof.

Proof. The fact that $\|\phi _{\sigma _1}\circ \cdots \circ \phi _{\sigma _p}(y)-y\|\leq 2\delta $ implies that, for each $x\in B_{c_*\delta }(y)$ ,

$$ \begin{align*} \begin{split} \|\phi_{\sigma_1}\circ\cdots\circ \phi_{\sigma_p}(x)-y\|&\leq \|\phi_{\sigma_1}\circ\cdots\circ \phi_{\sigma_p}(x)-\phi_{\sigma_1}\circ\cdots\circ \phi_{\sigma_p}(y)\|+2\delta\\ &\leq (\unicode{x3bb}^p c_*+2)\delta\leq c_*\delta. \end{split} \end{align*} $$

Hence, $\phi _{\sigma _1}\circ \cdots \circ \phi _{\sigma _p}(B_{c_*\delta }(y))\subset B_{c_*\delta }(y)$ . The lemma follows by the contraction mapping theorem.

Proof. We proceed by induction on p. If $\sigma _1,\omega _1\in \{1,\ldots ,m\}$ and $x_{\sigma _1}(\Phi )=x_{\omega _1}(\Phi )$ , then we can simply make the perturbation $\tilde \phi _{\sigma _1}(x)=\phi _{\sigma _1}(x)+\eta $ for some $\|\eta \|<\varepsilon /2$ . This proves the statement for $p=1$ . To simplify the notation, we keep calling $\Phi $ also the perturbed IFS. We suppose that the statement is true for p, after a perturbation of size at most $(1-2^{-p})\varepsilon $ , and we prove it for $p+1$ .

Since $A(p+1)=\{x_\sigma (\Phi )\;:\; \sigma \in \Sigma ^m_{p+1,*}(\Phi )\}$ is a finite discrete set,

$$ \begin{align*} \delta^*_{p+1}=\min\Big\{1,\inf_{\substack{x,y\in A(p+1)\\ x\neq y}}\|x-y\|\Big\}>0. \end{align*} $$

Let $\delta \leq c_*^{-2}\delta ^*_{p+1}/2$ (recall that $c_*= {2}/({1-\unicode{x3bb} })$ ). Suppose that $z:=x_\sigma (\Phi )=x_\omega (\Phi )$ and $\omega _{q}=\sigma _{p+1}$ , where $\sigma \in \Sigma ^m_{p+1}(\Phi )$ , $\omega \in \Sigma ^m_{q}(\Phi )$ , $q\leq p+1$ and $\sigma \neq \omega $ . Let ${j\in \{0,\ldots , q-1\}}$ be the largest integer such that $\sigma _{p+1-j}=\omega _{q-j}$ . If $j=q-1$ , then it must be that $q\leq p$ ; otherwise, we would have $\omega =\sigma $ . Hence,

$$ \begin{align*} \phi_{\omega_{1}}\circ \cdots\circ\phi_{\omega_q}(z)\hspace{-0.5pt}=\hspace{-0.5pt}z\hspace{-0.5pt}=\hspace{-0.5pt}\phi_{\sigma_1}\circ \cdots \circ\phi_{\sigma_{p+1}}(z)\hspace{-0.5pt}=\hspace{-0.5pt}\phi_{\sigma_1}\circ \cdots \circ\phi_{\sigma_{p+1-q}}\circ \phi_{\omega_{1}}\circ \cdots\circ\phi_{\omega_q}(z). \end{align*} $$

That is, $\phi _{\sigma _1}\circ \cdots \circ \phi _{\sigma _{p+1-q}}(z)=z$ . It follows, by the inductive hypothesis, that there exist $k\in {\mathbb N}$ and $\Theta \in {\mathcal A}_s$ , $s\leq q$ , such that

$$ \begin{align*} \begin{split} & \phi_{\sigma_{1}}\circ \cdots \circ\phi_{\sigma_{p+1-q}}=\Theta^k,\\ &\phi_{\omega_{1}}\circ \cdots\circ\phi_{\omega_q}=\Theta^j, \end{split} \end{align*} $$

so $\phi _{\sigma _1}\circ \cdots \circ \phi _{\sigma _{p+1}}=\Theta ^{k+j}$ , as claimed. It remains to consider the case $j<q-1$ . Let $\psi := \phi _{\omega _{q-j}}\circ \cdots \circ \phi _{\omega _q}$ . Then $\psi =\phi _{\sigma _{p+1-j}}\circ \cdots \circ \phi _{\sigma _{p+1}}$ and

$$ \begin{align*} [\psi\circ \phi_{\omega_1}\circ \cdots \circ \phi_{\omega_{q-j-1}}]( \psi(z))=\psi(z)= [\psi\circ \phi_{\sigma_1}\circ \cdots \circ\phi_{\sigma_{p-j}}](\psi(z)), \end{align*} $$

where, by construction, $\omega _{q-j-1}\neq \sigma _{p-j}$ . By renaming the indices, we are thus reduced to the case $\omega _{q}\neq \sigma _{p+1}$ . The following lemma is useful for analysing this case.

Sub-lemma 4.12. If, for $j\in \{1,\ldots , p\}$ ,

$$ \begin{align*} \phi_{\sigma_{j+1}}\circ\cdots \circ \phi_{\sigma_{p+1}}(B_\delta(z))\cap B_\delta(z)\neq \emptyset, \end{align*} $$

then $\phi _{\sigma _{j+1}}\circ \cdots \circ \phi _{\sigma _{p+1}}(z)=z$ and $\sigma _j=\sigma _{p+1}$ , and the same for $\omega $ .

Proof. Lemma 4.10 implies that there exists $z_1\in B_{c_*\delta }(z)$ such that $\phi _{\sigma _{j+1}}\circ \cdots \circ \phi _{\sigma _{p+1}}(z_1)=z_1$ . In addition,

$$ \begin{align*} \|z-\phi_{\sigma_{1}}\circ\cdots \circ \phi_{\sigma_{j}}(z_1)\|=\|\phi_{\sigma_{1}}\circ\cdots \circ \phi_{\sigma_{p+1}}(z)-\phi_{\sigma_{1}}\circ\cdots \circ \phi_{\sigma_{p+1}}(z_1)\|\leq\unicode{x3bb} c_*\delta. \end{align*} $$

Thus, $\phi _{\sigma _{1}}\circ \cdots \circ \phi _{\sigma _{j}}(B_{c_*\delta }(z))\cap B_{c_*\delta }(z)\neq \emptyset $ , and hence Lemma 4.10 implies that there exists $z_2\in B_{c_*^2\delta }(z)$ such that $\phi _{\sigma _{1}}\circ \cdots \circ \phi _{\sigma _{j}}(z_2)=z_2$ . The definition of $\delta $ implies that $z_1=z_2$ , which, in turn, implies that $z_1=z$ . Hence,

$$ \begin{align*} &\phi_{\sigma_{1}}\circ\cdots \circ \phi_{\sigma_{j}}(z)=z,\\ &\phi_{\sigma_{j+1}}\circ\cdots \circ \phi_{\sigma_{p+1}}(z)=z, \end{align*} $$

and, by the inductive hypothesis, this is possible only if $\sigma _{j}=\sigma _{p+1}$ . The argument for $\omega $ is identical.

If there exists $j\in \{1,\ldots , p\}$ and $k\in \{1,\ldots , q\}$ such that

(4.9) $$ \begin{equation}\begin{aligned} &\phi_{\sigma_{j+1}}\circ\cdots \circ \phi_{\sigma_{p+1}}(B_\delta(z))\cap B_\delta(z)\neq \emptyset,\\ &\phi_{\omega_{k+1}}\circ\cdots \circ \phi_{\omega_{q}}(B_\delta(z))\cap B_\delta(z)\neq \emptyset, \end{aligned}\end{equation} $$

then, by Sub-Lemma 4.12, $\phi _{\sigma _{j+1}}\circ \cdots \circ \phi _{\sigma _{p+1}}(z)=z=\phi _{\omega _{k+1}}\circ \cdots \circ \phi _{\omega _{q}}(z)$ , which contradicts our inductive hypothesis. Thus, if the first inequality is satisfied for some j, the second cannot be satisfied for any k, and vice versa. Suppose that there does not exist k for which the second inequality of (4.9) is satisfied (the other possibility being completely analogous).

Define the perturbation

$$ \begin{align*} \tilde\phi_k=\begin{cases} \phi_k\circ h_{z,\delta} &\mathrm{if}\ k=\omega_{q},\\ \phi_k&\mathrm{ otherwise,} \end{cases} \end{align*} $$

where, by analogy with (4.18), for $v\in {\mathbb R}^d$ , $\|v\|=1$ ,

$$ \begin{align*} h_{z,\delta}(x)=\begin{cases} x\quad&\text{for all } x\not\in B_{c_*^{-1}\delta}(z),\\ x+c_*^{-3}\delta^3g(1-c_*\delta^{-1}\|x-z\|) v &\mathrm{otherwise}. \end{cases} \end{align*} $$

Note that $\|h_{z,\delta }-\mathrm {id}\|_{{\mathcal C}^2}\leq 2^{-p-1}\varepsilon _0$ . Since $h_{z,\delta }(B_{c_*^{-1}\delta }(z))\subset B_{c_*^{-1}\delta }(z)$ , it follows that the effect of the perturbation is always confined to $B_{\delta }(z)$ and its images. Moreover, by Sub-Lemma 4.12, if there exists j such that $\phi _{\sigma _{j+1}}\circ \cdots \circ \phi _{\sigma _{p+1}}(B_\delta (z))\cap B_\delta (z)\neq \emptyset $ , we have seen that $\sigma _{j}=\sigma _{p+1}\neq \omega _q$ , so next we apply a map, $\phi _{\sigma _j}$ , that has not been modified. On the other hand, if $x\in B_\delta (z)$ and for some k we have $\omega _k=\omega _{q}$ , we have $\phi _{\omega _{k+1}}\circ \cdots \circ \phi _{\omega _q}(x)\not \in B_\delta (z)$ . Next, we apply $\phi _{\omega _k}$ outside the region where it has been modified. It follows that, for each $x\in B_\delta (z)$ ,

$$ \begin{align*} \begin{split} & \tilde\phi_{\omega_{1}}\circ\cdots \circ \tilde\phi_{\omega_{q}}(x)=\phi_{\omega_{1}}\circ\cdots \circ \phi_{\omega_{q}}\circ h_{z,\delta}(x),\\ &\tilde \phi_{\sigma_{1}}\circ\cdots \circ \tilde \phi_{\sigma_{p+1}}(x)=\phi_{\sigma_{1}}\circ\cdots \circ \phi_{\sigma_{p+1}}(x). \end{split} \end{align*} $$

Calling $z(\delta )$ the unique fixed point of $\phi _{\omega _{1}}\circ \cdots \circ \phi _{\omega _{q}}\circ h_{z,\delta }$ ,

Since $z(0)=z$ and

$$ \begin{align*} \partial_\delta h_{z,\delta}=3c_*^{-3}\delta^2g(1-c_*\delta^{-1}\|z(\delta)-z\|) v +c_*^{-2}\delta \|z(\delta)-z\|g'(1-c_*\delta^{-1}\|z(\delta)-z\|) v, \end{align*} $$

we have that it is not possible that $z(\delta )=z$ for all $\delta \leq c_*^{-2}\delta ^*_{p+1}/2$ . Thus, we can make a perturbation for which the two fixed points are different, and, for $\delta $ small, they cannot be equal to the other fixed points.

For any other couple of elements $\sigma \in \Sigma ^m_{p+1}(\Phi ),\omega \in \Sigma ^m_{q}(\Phi )$ , we can repeat the same process and obtain the perturbation with two different fixed points, as above. Note that, as the size of the perturbation is $\delta <c_*^{-1}\delta ^*_{p+1}/2$ , the distance between the newly obtained fixed points in $\bigcup _{q\leq p}{\mathcal A}_q$ stays positive as the perturbation does not move the fixed points more than $\delta ^*_{p+1}/2$ .

Proof. The last statement of the proposition is an immediate consequence of the first part and Lemma 4.10. As for the first part, note that if, for each $\sigma \in \Sigma ^m_{p, *}$ , $p\leq N$ , the fixed points of $\phi _\sigma :=\phi _{\sigma _1}\circ \cdots \circ \phi _{\sigma _p}$ do not belong to $D^{N}(\tilde \Phi )$ , then the proposition holds with $\delta _\varepsilon $ small enough. Thus, it suffices to prove the latter fact.

By Remark 4.13, it follows that there exists $\varepsilon _0>0$ such that, for each $\varepsilon \in (0,\varepsilon _0)$ and $\varepsilon $ -perturbation $\tilde \Phi $ of $\Phi $ , for all $\sigma \in \Sigma ^m_{p,*}(\Phi )$ , $p\leq N$ , the inverse map of $\phi _\sigma $ is well defined in $D_\sigma (\tilde \Phi )$ . In addition, by Lemma C.3, $\Sigma ^m_{p,*}(\Phi )=\sigma ^m_{p,*}(\tilde \Phi )$ . From now on we assume that $\varepsilon \leq \varepsilon _0$ .

We can then apply Proposition 4.11 to obtain the IFS $\Phi ^0$ , a $\varepsilon /4$ perturbation of $\phi $ , where the fixed points differ unless they are associated with sequences composed by the repetition of the same word. Next, we want to proceed by induction on the sequences in $\Sigma ^m_{N,*}(\Phi )$ .

To this end, it is necessary to have an order structure on $\Sigma ^m_{N,*}(\Phi )$ . We introduce the following order: $0\prec \sigma $ for all $\sigma \neq 0$ , and if $p>q$ and $\sigma \in \Sigma ^m_{p,*}(\Phi )$ , $\sigma '\in \Sigma ^m_{q,*}(\Phi )$ , then $\sigma '\preceq \sigma $ . If $p=q$ , then the $\sigma $ are ordered lexicographically. This is a total ordering, and hence we can arrange them as sequences $\{\sigma ^i\}_{i\in {\mathbb N}}$ with $\sigma ^j\prec \sigma ^i$ if and only if $i>j$ . Next, define $\ell (j)$ to be the length of the word $\sigma ^j$ : that is $\sigma ^j\in \Sigma ^{m}_{\ell (j),*}$ . Recall the definition of fixed points (4.8). It is convenient to set $\sigma ^0=0$ and $x_0(\phi )=\emptyset $ . Also, let $\Lambda _0=\max \{\|(D_x\phi _i)^{-1}\|_{{\mathcal C}^0(V_i)}\;:\; \phi _i\in \Phi ^0\}$ .

The idea is to define a sequence of perturbations $\Phi ^k$ , $\|\Phi ^{k+1}-\Phi ^{k}\|_{{\mathcal C}^2}=\varepsilon _k\leq \varepsilon 2^{-k-1}$ , such that, for all $j\leq k$ ,

(4.16) $$ \begin{align} x_{\sigma^j}(\Phi^{k})\not \in D^{N}(\Phi^{k}). \end{align} $$

Note that the above implies that there exists $\Lambda>2$ such that

$$ \begin{align*} \Lambda\geq \max\{\|(D_x\phi_i)^{-1}\|_{{\mathcal C}^0(U_i)}\;:\; \phi_i\in\Phi^k\} \end{align*} $$

for all $k\in {\mathbb N}$ . In particular, the above implies that

(4.17) $$ \begin{align} D^{N}(\Phi^{k+1})\subset D^{N}_{2\Lambda^{N}\varepsilon_k}(\Phi^k). \end{align} $$

Using the notation introduced just before Proposition 4.11, let

$$ \begin{align*} \bar \delta_k=\min\Big\{1,\frac1{2c_*}\inf_{\substack{x,y\in A_{N}(\Phi^k)\\ x\neq y}}\|x-y\|\Big\}>0. \end{align*} $$

We proceed by induction on $\sigma $ . For $\sigma ^0$ , the statement in the induction is trivially true. We assume that it is true for $\sigma ^k$ and we prove the statement for $\sigma ^{k+1}$ . Let

$$ \begin{align*} \delta_k=\min \{\min\{d_0(x_{\sigma^j}(\Phi^k), D^{N}(\Phi^k))\;:\; j\leq k\}, \bar \delta_k/4\}. \end{align*} $$

We consider $\varepsilon _k$ -perturbations of $\Phi ^k$ with $\varepsilon _k\leq \Lambda ^{-N}\delta _k/4$ . For $\sigma \in \Sigma ^m_p$ , we use the notation $\phi _{\sigma }=\phi _{\sigma _1}\circ \cdots \circ \phi _{\sigma _p}$ . If $x_{\sigma ^{k+1}}(\Phi ^k)\not \in D^{N}(\Phi ^k)$ , then we set $\Phi ^{k+1}=\Phi ^k$ and the induction step is satisfied. Otherwise, as before, for some $a>2$ , and any $\delta \in (0,1/ \sqrt a), v\in V$ and $\bar x\in {\mathbb R}^d$ , we define $h_{\bar x,\delta ,v}:{\mathbb R}^d\to {\mathbb R}^d$ as

(4.18) $$ \begin{align} h_{\bar x,\delta,v}(x)=\begin{cases} x\quad&\text{for all } x\not\in B_\delta(\bar x),\\ x+\delta^3g(1-\delta^{-1}\|x-\bar x\|) v &\mathrm{otherwise}, \end{cases} \end{align} $$

where $g\in {\mathcal C}^\infty ({\mathbb R},{\mathbb R}_+)$ is a monotone function such that $g(y)=0$ for all $y\leq 0$ , $g(y)=1$ for all $y\geq 1/2$ and $\|g'\|_\infty < a$ . Let $p=\ell (\sigma ^{k+1})$ . For each $\delta>0$ and $v\in {\mathbb R}^d$ , $\|v\|\leq 1$ , we consider the perturbations $\Phi _{\delta , v}=\{\phi _{i,\delta , v}\}$ defined by

$$ \begin{align*} \phi_{i,\delta,v}(x)=\begin{cases} \phi_{i}(x) &\mathrm{if}\ i\neq \sigma^{k+1}_p,\\ \phi_{\sigma^{k+1}_p}\circ h_{x_{\sigma^{k+1}}(\Phi^k),\delta,v}(x) &\mathrm{if}\ i= \sigma^{k+1}_p, \end{cases} \end{align*} $$

where $\Phi ^k=\{\phi _i\}$ . Note that $\|\phi _i-\phi _i\circ h_{\bar x,\delta ,v}\|_{{\mathcal C}^2}\leq C_g\delta $ and $\|\phi _i\circ h_{\bar x,\delta ,v}\|_{{\mathcal C}^3}\leq C_g$ for some constant $C_g>1$ . Thus, these are $\varepsilon _k$ perturbations provided $\delta \leq C_g^{-1}\varepsilon _k\leq C_g^{-1}\Lambda ^{-N}\delta _k/4$ .

Lemma 4.18. There exist $ C_*>0$ and $\delta _*\in (0, \min \{C_g^{-1}\varepsilon _k, \Lambda ^{-2N}\})$ such that, for all $\delta \leq \delta _*$ , $\|v\|\leq 1$ and each $j\leq k+1$ ,

$$ \begin{align*} \frac 32\delta^2\|v\|\Lambda^{-N}\leq \|x_{\sigma^{j}}(\Phi^k)-x_{\sigma^j}(\Phi_{\delta, v})\|\leq \frac{\delta}2. \end{align*} $$

Moreover, $\partial _vx_{\sigma ^j}(\Phi _{\delta , v})$ is invertible and

$$ \begin{align*} \|(\partial_vx_{\sigma^j}(\Phi_{\delta, v}))^{-1}\|\leq C_*\delta^{-3}\Lambda^{N}. \end{align*} $$

Proof. Let $q=\ell (\sigma ^j)$ . If $\sigma ^j_s\neq \sigma ^{k+1}_p$ , for all $s\leq q$ , then $x_{\sigma ^{j}}(\Phi ^k)=x_{\sigma ^j}(\Phi _{\delta , v})$ . Otherwise, let $\bar s$ be the largest such that, for all $s\leq \bar s$ , $\sigma ^j_s\neq \sigma ^{k+1}_p$ ,

$$ \begin{align*} \begin{split} &\phi_{\sigma^j_{\bar s},\delta,v}\circ \cdots \circ\phi_{\sigma^j_q,\delta,v}(x_{\sigma^j}(\Phi_{\delta, v}))\\ &\quad =\phi_{\sigma^j_{\bar s},\delta,v}\circ \cdots \circ\phi_{\sigma^j_q,\delta,v}\circ \phi_{\sigma^j_1,\delta,v}\circ \cdots\circ \phi_{\sigma^j_{\bar s-1},\delta,v}\\ &\qquad \times ( \phi_{\sigma^j_{\bar s},\delta,v}\circ \cdots \circ\phi_{\sigma^j_q,\delta,v}(x_{\sigma^j}(\Phi_{\delta, v}))). \end{split} \end{align*} $$

Note that, $(\sigma ^j_{\bar s},\ldots ,\sigma ^j_{q},\sigma ^j_1,\ldots , \sigma ^j_{\bar s-1},)=\sigma ^{j_1}$ for some $j_1<k+1$ . Moreover, ${\sigma ^{j}_q=\sigma ^{k+1}_p}$ and

$$ \begin{align*} x_{\sigma^{j_1}}(\Phi_{\delta,v})=\phi_{\sigma^j_{\bar s},\delta,v}\circ \cdots \circ\phi_{\sigma^j_q,\delta,v}(x_{\sigma^j}(\Phi_{\delta, v})). \end{align*} $$

By hypothesis,

$$ \begin{align*} \|x_{\sigma^{j}}(\Phi_{\delta,v})-x_{\sigma^{j}}(\Phi^k)\|\leq\Lambda^p\|x_{\sigma^{j_1}}(\Phi_{\delta,v})-x_{\sigma^{j_1}}(\Phi^k)\|. \end{align*} $$

We can thus consider only the case in which $\sigma ^{j}_q=\sigma ^{k+1}_p$ . Let $1\leq \bar s<q$ be the largest integer, if it exists, such that $\sigma ^j_s= \sigma ^{k+1}_p$ for all $s\leq \bar s$ . Then for $y\in B_\delta (x_{\sigma ^{k+1}}(\Phi ^k))$ ,

$$ \begin{align*} x_{\sigma^{j}}(\Phi_{\delta,v})=\phi_{\sigma^j_{1},\delta,v}\circ \cdots \circ\phi_{\sigma^j_{\bar s},\delta,v} \circ \phi_{\sigma^j_{\bar s+1}}\circ \cdots \circ\phi_{\sigma^j_{q}}(h_{x_{\sigma^{k+1}}(\Phi^k),\delta,v}(y)). \end{align*} $$

Since, by construction, $h_{x_{\sigma ^{k+1}}} (B_\delta (x_{\sigma ^{k+1}}(\Phi ^k)))\subset B_\delta (x_{\sigma ^{k+1}}(\Phi ^k))$ , we have that applying $\phi _{\sigma ^j_{\bar s},\delta ,v}$ differs from applying $\phi _{\sigma ^j_{\bar s}}$ only if

$$ \begin{align*} \phi_{\sigma^j_{\bar s+1}}\circ \cdots \circ\phi_{\sigma^j_{q}}(B_\delta(x_{\sigma^{k+1}}(\Phi^k))) \bigcap B_\delta(x_{\sigma^{k+1}}(\Phi^k))\neq \emptyset. \end{align*} $$

But then Lemma 4.10 implies that

$$ \begin{align*} \|x_{(\sigma^j_{s+1}, \ldots, \sigma^j_q)}(\Phi^k)-x_{\sigma^{k+1}}(\Phi^k)\|\leq c_*\delta. \end{align*} $$

Then our choice of $\delta $ and the induction hypothesis implies that $x_{\sigma ^{k+1}}(\Phi ^k)=x_{(\sigma ^j_{s+1}, \ldots , \sigma ^j_q)}(\Phi ^k)\not \in D^{N}(\Phi ^k)$ , which is contrary to our current assumption. It follows that, provided $x_{\sigma ^{j}}(\Phi _{\delta ,v})\in B_\delta (x_{\sigma ^{k+1}}(\Phi ^k))$ ,

$$ \begin{align*} x_{\sigma^{j}}(\Phi_{\delta,v})=\phi_{\sigma^j}(h_{x_{\sigma^{k+1}}(\Phi^k),\delta,v}(x_{\sigma^{j}}(\Phi_{\delta,v}))). \end{align*} $$

To simplify notation, let $z(\delta ,v)=x_{\sigma ^{j}}(\Phi _{\delta ,v})$ , $h_{\delta ,v}=h_{x_{\sigma ^{k+1}}(\Phi ^k),\delta ,v}$ and $\phi _{\delta ,v}=\phi _{\sigma ^j}\circ h_{\delta ,v}$ . We can study $z(\delta ,v)$ by applying the implicit function theory, which yields

If $z(\delta ,v)\in B_{\delta /2}(x_{\sigma ^{k+1}}(\Phi ^k))$ , then $h_{\delta ,v}(z(\delta ,v))=z(\delta ,v)+\delta ^3 v$ , and $\partial _\delta h_{\delta ,v} (z(\delta ,v))=3\delta ^2v$ . Thus, setting $A:=D_{z(\delta ,v)+\delta ^3v}\phi _{\delta ,v}$ we obtain

Since the maximal eigenvalue of is bounded by $(1-\unicode{x3bb} )^{-1}\unicode{x3bb} $ , there exists a $\delta _*$ such that, for all $\delta \leq \delta _*$ , $z(\delta ,v)\in B_{\delta /2}(x_{\sigma ^{k+1}}(\Phi ^k))$ . Moreover,

and thus $z(\delta ,v)\not \in B_{(3/2)\delta ^2\|v\|\Lambda ^{-N}}(x_{\sigma ^{k+1}}(\Phi ^k))$ . Finally, for each $\delta \leq \delta _0$ ,

from which the last statement of the lemma follows.

By Lemma 4.18, equation (4.17) and our choice of $\varepsilon _{k}$ , $x_{\sigma ^j}(\Phi _{\delta ,v})\not \in D^{N}_{3\delta _k/4}(\Phi ^{k})$ and $D^{N}(\Phi ^{k+1})\subset D^{N}_{\delta _k/4}(\Phi ^{k})$ for all $j\leq k$ . Thus, $x_{\sigma ^j}(\Phi _{\delta ,v})\not \in D^{N}(\Phi ^{k+1})$ for all $j\leq k$ . We are left with $x_{\sigma ^{k+1}}(\Phi ^{k+1})$ , recalling that $x_{\sigma ^{k+1}}(\Phi ^k) \in D^{N}(\Phi ^k)$ . Let $\omega \in \Sigma ^m_{q,i}$ , $q\leq N$ and $M \in M_{\omega ,i}^{N}(\Phi ^k)$ such that $x_{\sigma ^{k+1}}(\Phi ^k)\in M$ . Then

$$ \begin{align*} \phi_{\omega_1}\circ\cdots\circ\phi_{\omega_q}(x_{\sigma^{k+1}}(\Phi^k))\in \overline{\psi_i(D_i)}. \end{align*} $$

First, suppose that $\omega _q\neq \sigma ^{k+1}_p$ . It follows that, for $y\in B_{\delta /2}(x_{\sigma ^{k+1}}(\Phi ^k))$ ,

$$ \begin{align*} \phi_{\omega_1, \delta,v}\circ\cdots\circ\phi_{\omega_q, \delta,v}(y)\neq \phi_{\omega_1}\circ\cdots\circ\phi_{\omega_q}(y) \end{align*} $$

only if, for some $s< q$ , $\omega _s=\sigma ^{k+1}_p$ and

(4.19) $$ \begin{align} \phi_{\omega_{s+1}}\circ\cdots\circ\phi_{\omega_q}(B_{\delta}(x_{\sigma^{k+1}}(\Phi^k)))\cap B_{\delta}(x_{\sigma^{k+1}}(\Phi^k))\neq \emptyset, \end{align} $$

but this is ruled out by our choice of $\delta _k$ and Proposition 4.11. The above discussion shows that $M\cap B_{\delta /2}(x_{\sigma ^{k+1}}(\Phi ^k))$ is an element of $M_{\omega , i}^{N}(\Phi _{\delta ,v})$ as well. Hence, it suffices to ensure that $x_{\sigma ^{k+1}}(\Phi _{\delta ,v})\not \in M$ . Since Lemma 4.18 shows that varying v the fixed point visits an open ball, and since M has zero measure, it follows that there exists an open set of v which yields the wanted property.

It remains to analyse the case $\omega _q= \sigma ^{k+1}_p$ . In this case, for $y\in B_{\delta /4}(x_{\sigma ^{k+1}}(\Phi ^k))$ ,

$$ \begin{align*} \phi_{\omega_1, \delta,v}\circ\cdots\circ\phi_{\omega_q, \delta,v}(y)=\phi_{\omega_1, \delta,v}\circ\cdots\circ\phi_{\omega_q}(y+\delta^3 v)\neq \phi_{\omega_1}\circ\cdots\circ\phi_{\omega_q}(y+\delta^3 v) \end{align*} $$

only if equation (4.19) is satisfied, which, by our choice of $\delta _k$ , is possible only if $x_\omega (\Phi ^k)=x_\sigma (\Phi ^k)$ . But then Proposition 4.11 implies that there exist $\sigma ^r\preceq \sigma ^{k+1}$ such that $\phi _\omega =\phi _{\sigma ^r}^{m_1}$ and $\phi _\sigma =\phi _{\sigma ^r}^{m_2}$ , which would mean that

$$ \begin{align*} \overline{\psi_i(D_i)}\ni \phi_{\sigma^r}^{m_1}(x_{\sigma^{k+1}}(\Phi^k))=x_{\sigma^{k+1}}(\Phi^k). \end{align*} $$

Again, Lemma 4.18 allows to find an open set of v for which $x_{\sigma ^{k+1}}(\Phi ^k)\not \in \overline {\psi _i(D_i)}$ . The last possibility is that

$$ \begin{align*} \phi_{\omega_1, \delta,v}\circ\cdots\circ\phi_{\omega_q, \delta,v}(y)=\phi_{\omega_1, \delta,v}\circ\cdots\circ\phi_{\omega_q}(y+\delta^3 v). \end{align*} $$

This implies that the perturbed manifold M is displaced by, at most, $2\Lambda ^{N}\delta ^3\|v\|$ whereas Lemma 4.18 implies that the fixed point moves by at least $(3/2)\delta ^2\|v\|\Lambda ^{-N}\geq 2\Lambda ^{N}\delta ^3\|v\|$ . Hence, again we have an open set of v, which produces perturbations with the wanted property. As a last observation, note that if there are other manifolds $M\in D^{N}(\Phi ^k)$ such that $x_{\sigma ^{k+1}}(\Phi ^k)\in M$ , then we can repeat the same argument and we have just a smaller open set of v that does the job. This concludes the overall induction and hence the proof of Proposition 4.17.

Proof. Before starting the proof, we need to introduce some language.

Consider an IFS $\Phi _{\! f}=\{\phi _1,\phi _2,\ldots ,\phi _m\}$ associated to f with contraction coefficient $\unicode{x3bb} $ . Let $\varepsilon _0\in (0,\varepsilon /4)$ be small enough. Then, by Proposition 4.17, there exist $\delta _*\leq \delta _\varepsilon \in (0,\varepsilon _1)$ and an $\varepsilon _0$ -perturbation (here, and in the following, by perturbation we mean a function that is ${\mathcal C}^2$ close and with a uniformly bounded ${\mathcal C}^3$ norm) of $ \Phi _{\! f}$ (which, abusing the notation, we still call $\Phi _{\! f}$ ) such that, for every $p\leq N$ , $\sigma =(\sigma _1,\ldots ,\sigma _p)\in \Sigma ^{m}_{p,*}$ and $\xi \in D^{N}_{\delta _*}(\Phi _{\! f}\!)\cap V_p$ ,

(4.20) $$ \begin{align} \phi_{\sigma_1}\circ\phi_{\sigma_{2}}\circ\cdots\circ\phi_{\sigma_p} (B_{\delta_*}(\xi))\cap B_{\delta_*}(\xi)=\emptyset. \end{align} $$

Note that there exists $\epsilon _0\leq \varepsilon _0$ such that (4.20) persists for $\epsilon _0$ -perturbations of $\Phi $ .

By compactness, for $\delta \in (0,\min \{\delta _{*}/2,\delta _N\})$ , where $\delta _N>0$ is such that it satisfies the condition of Remark 4.13 for each $\epsilon _0$ -perturbation, there exists a finite open cover $\{B_{\delta /2}(z_i)\}_{i=1}^t$ of $\overline {D_{\delta }^N(\Phi _{\! f}\!)}$ (which, by definition, contains $\partial {\boldsymbol P}(f^N) $ ) such that, for each i, $z_i\in P_j$ and $B_{\delta }(z_i)\subset V_j$ , for some j. (See the discussion at the beginning of §4.4 for the definition of $V_j$ .)

Let $Y=\overline Y\subset {\mathbb R}^d$ be compact, such that $\overline {\phi (Y)}\subset Y$ for all $\phi \in \Phi $ .

By convention, we set ${\mathcal Z}^1_0(\Phi )=\{Y\}$ and ${\mathcal Z}^1_1(\Phi ):=\{\psi _{\omega }\;;\;\omega \in \{1,\ldots ,l_{*}\}\}$ to be the collection of the manifolds $\psi _i(A)$ for $A\in {\mathcal W}^N_i$ , as defined in (4.13). Also, we call ${\mathcal Z}^1_k(\Phi ):=\{\psi ^k_{\omega }\;:\;\omega \in \{1,\ldots ,l_{*}\}^k\}$ the manifolds consisting of the intersection of the k manifolds $\{\psi _{\omega _i}\;:\;i\in \{1,\ldots ,k\}; i\neq j\!\implies \!\psi _{\omega _i}\not\hskip-3pt\curlywedge\hskip2pt\psi_{\omega_j}\} $ . (These are indeed manifolds; see Definition 2.16. To simplify notation, we use $\psi _\omega $ both for the manifold and for the map that defines it.) By construction, ${\mathcal Z}^1_k(\Phi )=\emptyset $ for $k>l_0$ . In addition, the maximal dimension of the manifolds in ${\mathcal Z}^1_k(\Phi )$ , for $k>1$ , is $d-2$ (since the boundary manifolds are pairwise transversal; see Definition 2.16). Note that ${\mathcal Z}^1_1(\Phi _{\! f}\!)$ is a collection that covers the boundary manifolds; for simplicity, we call the elements of ${\mathcal Z}^1_1(\Phi _{\! f}\!)$ , from now on, original boundary manifolds. For each $s\in {\mathbb N}$ and $k_1,\ldots , k_n\in {\mathbb N}\cup \{0\}$ , let ${{\mathcal Z}^s_0=\{Y\}, U_0=Y}$ and (to alleviate notation, from now on we write $\phi _{\sigma _i}^{-1}$ to mean the inverse of $\phi _{\sigma _i}|_{U_{\sigma _i}}\!$ , while the domain of $\phi _{\sigma _i}^{-1}\circ \psi $ consists of the points where the composition is well defined)

$$ \begin{align*} &{\mathcal Z}_{k}^{s+1}(\Phi)=\bigg\{\phi_{\sigma_1}^{-1}\circ\psi_{\omega_1}\cap \cdots\cap \phi_{\sigma_n}^{-1}\circ\psi_{\omega_n}\;:\; n\in{\mathbb N}, k_1,\ldots, k_n\in{\mathbb N}\cup\{0\}, \sum_{i=1}^nk_i=k, \\ &\phantom{{\mathcal Z}_{k}^{s+1}(\Phi)=\{\ \ } \psi_{\omega_i}\in{\mathcal Z}_{k_i}^s(\Phi), \sigma_i\in\Sigma^m_{1,*}(\Phi); i\neq j,\sigma_i=\sigma_j\!\implies\! \psi_{\omega_i}\not\hskip-3pt\curlywedge\hskip2pt\psi_{\omega_j}\bigg\},\\ &{\mathcal Z}_{*}^{s}(\Phi)=\bigcup_{k\in{\mathbb N}\cup\{0\}}{\mathcal Z}_{k}^{s}(\Phi). \end{align*} $$

Note that ${\mathcal Z}^s_*$ contains the admissible pre-images of the boundary manifolds under composition of at most s maps in $\Phi $ and all their intersections. In particular, the sets in ${\mathcal Z}^s_1$ cover $\partial {\mathcal P}(f^s)$ , and we call them boundary manifolds. In addition, if a set belongs to ${\mathcal Z}^s_k(\Phi )$ , then, by definition, it is determined by the intersection of the pre-images of k original boundary manifolds. Also, we remark that ${\mathcal Z}^{s+1}_k\supset {\mathcal Z}^s_k$ since $\phi _0=\mathrm {id}$ (see Remark 4.14 and Definition (4.12)). Next, let $N_s$ be the maximal number of manifolds that can intersect in ${\mathcal Z}^s_*$ : that is, ${\mathcal Z}^s_k=\emptyset $ for $k>N_s$ . We have seen that $N_1\leq l_1$ . Moreover, each original boundary manifold can have at most $m^s$ different pre-images obtained by the compositions of s maps. This implies that, at each point, we have at our disposal at most $\sum _{s'=0}^{s-1} m^{s'}l_1$ different manifolds from $\bigcup _i M^N_{0,i}$ to intersect. If $m=1$ , then $N_s\leq sl_1$ ; if $m\geq 2$ , then $N_s\leq ({m^{s}-1})/({m-1}) l_1< 2m^{s-1}l_1$ . Accordingly, $N_s\leq s m^{s-1} l_1$ . (We remark that, by definition, the pre-images are taken via invertible maps, and hence the manifolds cannot self-intersect.)

Our goal is to produce a sequence of perturbations $\Phi ^s$ of $\Phi _{\! f}=:\Phi ^0$ such that $\Phi ^{s}$ is a $2^{-s}\varepsilon $ perturbation of $\Phi ^{s-1}$ with the following property.

1. (⋆) The set ${\mathcal Z}^{s}_k(\Phi ^s)$ consists of manifolds of dimension strictly smaller than $d-j$ for all $k> jm^{j-1}l_1$ , whereas ${\mathcal Z}^{s}_1(\Phi ^s)$ consists of $d-1$ dimensional manifolds. This property persists for small perturbations of $\Phi ^s$ .

Note that the above implies that ${\mathcal Z}^{s}_k(\Phi ^{s})=\emptyset $ for each $s\in {\mathbb N}$ and $k> dm^{d-1}l_1$ . Accordingly, at most $dm^{d-1} l_1$ pre-images of the original boundary manifolds under composition of at most s elements of $\Phi ^s$ can have non-empty intersections. In turn, defining $f_s(x)=\phi _i(x)$ for $x\in P_i$ and $\phi _i\in \Phi ^s$ , we obtain a perturbation of f smaller than $\sum _{j=1}^{s}2^{-j}\varepsilon \leq \varepsilon $ such that ${\mathcal P}(f_s^s)$ has at most $2^{dm^{d-1} l_1}$ elements meeting at point.

Indeed, suppose p elements of ${\mathcal P}(f_s^s)$ meet at a point x. The boundaries of such elements in a neighbourhood small enough of x consist of codimension one manifolds belonging to ${\mathcal Z}^s_1(\Phi ^s)$ , and they have to intersect at x. Suppose that the total number of such boundary manifolds is q. Then x must belong to a manifold in ${\mathcal Z}^s_q(\Phi ^s)$ , and hence it must be ${q\leq dm^{d-1} l_1}$ . Note that we can uniquely define a partition element by specifying on which side it lies with respect to all its boundary manifolds. Since there are at most $2^q$ possibilities, it must be $ p\leq 2^q$ . It follows that $p\leq 2^{dm^{d-1} l_1}$ . The lemma then follows by choosing $s=N$ .

It remains to prove property $(\star )$ . We proceed by induction. If $s=1$ and $k\in \{1,\ldots ,l_1\}$ , then the manifolds in ${\mathcal Z}^1_k$ are indeed of codimension at least one, and the manifolds in ${\mathcal Z}^1_1$ are of codimension one, whereas, if $k>l_1$ , then ${\mathcal Z}^1_k=\emptyset $ , so $\Phi _{\! f}=\Phi ^0$ satisfies our hypothesis. We assume that the hypothesis is verified for some s and prove it for $s+1$ .

Let $\epsilon _s\leq \min \{2^{-s-1}\varepsilon ,\epsilon _0\}$ be such that all the $\epsilon _s$ -perturbations of $\Phi ^s$ still satisfy $(\star )$ . This implies that, provided $\Phi ^{s+1}$ is a $\epsilon _s$ perturbation of $\Phi ^s$ , ${\mathcal Z}_*^{s'}(\Phi ^{s+1})$ has the wanted property for all $s'\leq s$ .

Accordingly, we must analyse only sets of the type $\phi _{\sigma _1}^{-1}\circ \psi _{1}\cap \cdots \cap \phi _{\sigma _n}^{-1}\circ \psi _n$ , where $\psi _i\in {\mathcal Z}_{k_i}^s(\Phi ^s)$ and $\phi _{\sigma _i}\in \Phi ^{s}\cup \{\mathrm {id}\}$ , $\sigma _i\in \{0,\ldots ,m\}$ . (By an innocuous abuse of notation here we use $\psi _i$ to refer to generic elements.) By definition, such sets are elements of ${\mathcal Z}^{s+1}_k(\Phi ^s)$ , with $k=\sum _{i=1}^n k_i$ . Note that the $\psi _i\in {\mathcal Z}^s_0(\Phi ^s)$ do not contribute to the intersection. We can thus assume, without loss of generality, that $k_i>0$ . Note that if $n=1$ , then the manifolds belong to $\bigcup _{i=0}^m\phi _i^{-1}({\mathcal Z}_*^s(\Phi ^s))\subset {\mathcal Z}_*^{s+1}(\Phi ^{s})$ which have automatically the wanted property, and so has any $\epsilon _s$ -perturbation. We consider thus only the case $n\geq 2$ . In addition, if $\phi _{\sigma _i}=\phi _{\sigma _j}$ , $i\neq j$ , then $\phi _{\sigma _i}^{-1}\circ \psi _i\cap \phi _{\sigma _j}^{-1}\circ \psi _j=\phi _{\sigma _i}^{-1}\circ (\psi _j\cap \psi _j)$ , and since, by definition, $\psi _i\not\hskip-3pt\curlywedge\hskip2pt\psi_j $ , $\psi _j\cap \psi _j\in {\mathcal Z}^s_{k_i+k_j}(\Phi ^s)$ . Hence, we can substitute to the intersection of the manifolds $\phi _{\sigma _i}^{-1}\circ \psi _i\cap \phi _{\sigma _j}^{-1}\circ \psi _j$ the manifold $\phi _{\sigma _i}^{-1}\circ (\psi _j\cap \psi _j)$ . We can thus assume, without loss of generality, that $i\neq j$ implies that $\phi _{\sigma _i}\neq \phi _{\sigma _j}$ .

We define the map $F:{\mathbb R}^d\to {\mathbb R}^{nd}$ by $F(x):=(\phi _{\sigma _1}(x),\ldots , \phi _{\sigma _n}(x))$ and the stratified sub-variety $C=\{(\psi _{1}(x_0), \ldots \psi _{n}(x_n))\;:\; x_i\in \overline D_{i}\}$ , where $D_i\subset {\mathbb R}^{d_i}$ is the domain of the map $\psi _i$ . By Lemma B.4, for a constant c to be chosen later, there exists a $\tfrac 12 c\epsilon _s$ -perturbation $\hat F=(\hat \phi _{\sigma _1}(x),\ldots , \hat \phi _{\sigma _n}(x))$ of F, transversal to C. If $\phi _{\sigma _i}\neq \mathrm {id}$ for all i, then we set $\tilde F=\hat F$ . If, for some i, $\phi _{\sigma _i}=\mathrm {id}$ , then $\hat \phi _{\sigma _i}$ is a small perturbation of identity, and hence it is invertible with ${\mathcal C}^3$ inverse. (Indeed, if $\|h-\mathrm {id}\|_{{\mathcal C}^1}=\alpha <1$ , then h is a diffeomorphism. In fact, if $h(x)=h(y)$ ,

which implies that $\|x-y\|\leq \alpha \|x-y\|$ : that is $x=y$ . Thus, h is globally invertible, and the claim follows by the inverse function theorem.) By possibly relabelling, we can assume that $i=1$ . Then we define

$$ \begin{align*} \tilde F(x)=(\tilde \phi_{\sigma_1}, \ldots, \tilde \phi_{\sigma_n})=(x,\hat\phi_{\sigma_2}\circ\hat \phi_{\sigma_1}^{-1},\ldots,\hat\phi_{\sigma_n}\circ\hat \phi_{\sigma_1}^{-1}). \end{align*} $$

$\tilde F$ is still transversal to C and, if c is small enough, by Lemmata C.1 and C.2, it is a $c\epsilon _s$ -perturbation of F. Let $d_i=d-j_i$ be the dimension of the manifold $\psi _{i}$ . Then ${k_i\leq j_im^{j_i-1}l_1}$ . Lemma B.4 implies that the sets $\bigcap _{i=1}^n\phi _{\sigma _i}^{-1}\circ \psi _i$ are manifolds with dimension (actually, they are stratified sub-varieties, but we can restrict them to manifolds without loss of generality)

$$ \begin{align*} \sum_{i=1}^n (d-j_i)-(n-1)d=d-\sum_{i=1}^nj_i=:d-\bar j\leq d-\max\{j_i\}-1. \end{align*} $$

Note that

$$ \begin{align*} k:=\sum_{i=1}^nk_i\leq \sum_{i=1}^n j_im^{j_i-1}l_1\leq \sum_{i=1}^n j_im^{\bar j-1}l_1=\bar j m^{\bar j-1}l_1. \end{align*} $$

Accordingly, if $k>j m^{j-1}l_1$ , then $\bar j\geq j+1$ and the manifold has a dimension strictly smaller than $d-j$ , as required. (Of course, if $k>dm^{d-1}l_1$ , then the intersection is empty.)

We would then like to define a perturbed IFS $\tilde \Phi ^{s}$ as

$$ \begin{align*} \tilde \phi_{k}=\begin{cases} \tilde \phi_{\sigma_i}&\mathrm{if}\ k=\sigma_i,\\ \phi_k&\mathrm{ otherwise}. \end{cases} \end{align*} $$

Unfortunately, this may perturb the new manifolds $\psi _i$ as well, since they are now defined via pre-images of $\tilde {\Phi }^{s}$ . This is the last problem we need to take care of. To this end, we make the perturbation only locally starting from the ball $B_{\delta }(z_1)$ . Once we check that the perturbation is as required in this ball, we will consider the other balls, making the new perturbations small enough not to upset the property obtained in $B_{\delta /2}(z_1)$ .

Let $g\in {\mathcal C}^\infty ({\mathbb R}^d,[0,1])$ be such that

$$ \begin{align*} g(z)= \begin{cases}1,\quad z\in B_{\delta/2}(z_1),\\ 0, \quad z\notin B_{3\delta/4}(z_1), \end{cases} \end{align*} $$

and $\|g\|_{{\mathcal C}^r}\leq C\delta ^{-r}$ for $r\in \{0,1,2,3\}$ for some $C>0$ . Define, for each $i\in \{1,\ldots ,m\}$ ,

$$ \begin{align*} \phi_{i,1}(x)=\phi_{i}(x)+g(x)(\widetilde \phi_i(x)-\phi_i(x)). \end{align*} $$

Provided we choose c small enough,

$$ \begin{align*} \| \phi_{i,1}-\phi_{i}\|_{{\mathcal C}^2}=\|g\|_{{\mathcal C}^2}\|\widetilde \phi_i-\phi_i\|_{{\mathcal C}^2}<C\delta^{-2}C_\#c\epsilon_s <\epsilon_s/4. \end{align*} $$

Then we define the perturbation $\Phi ^{s,1}=\{\phi _{i,1}\}$ . Note that the $\Phi ^{s,1}$ equals $\Phi ^{s}$ outside the ball $B_\delta (z_1)$ and agrees with $\tilde \Phi ^s$ inside the ball $B_{\delta /2}(z_1)$ .

Recall that we perturbed the system in order to control the intersection of the manifolds $\phi _{\sigma _1}^{-1}\circ \psi _1\cap \cdots \cap \phi _{\sigma _n}^{-1}\circ \psi _n$ . By construction, each $\psi _i$ is the intersection of manifolds $\phi _{\sigma ^{i,j}_{s}}^{-1}\circ \cdots \phi _{\sigma ^{i,j}_1}^{-1}\circ \bar \psi _{i,j}$ with $\bar \psi _{i,j}\in {\mathcal Z}_1^1(\Phi ^s)$ and $\sigma ^{i,j}\in \{0,\ldots ,m\}^s$ . We are thus interested in $A:=\phi _{\sigma _{i},1}^{-1}\circ \phi _{\sigma ^{i,j}_{s},1}^{-1}\circ \cdots \circ \phi _{\sigma ^{i,j}_1,1}^{-1}\circ \bar \psi _{i,j}\cap B_{\delta }(z_1) $ , which are the perturbation of ${\phi _{\sigma _i}^{-1}\circ \psi _i}$ . Let

$$ \begin{align*} h_i(x)=\begin{cases} x&\mathrm{if}\ x\not\in B_\delta(z_1),\\ \phi_{\sigma^{i}}^{-1}\circ \phi_{\sigma^{i},1}&\mathrm{ otherwise}. \end{cases} \end{align*} $$

By choosing $\epsilon _s$ small enough, we have $\|h_i- \mathrm {id}\|_{{\mathcal C}^2}\leq \delta /4$ ; in particular, $h_i$ is a diffeomorphism. It follows that $h_i(B_\delta (z_1))\subset B_\delta (z_1)$ . Let $x\in A$ . Then Proposition 4.17 implies that

$$ \begin{align*} \begin{split} \phi_{\sigma^{i,j}_{1},1}\circ\cdots\phi_{\sigma^{i,j}_s,1}\circ \phi_{\sigma_i,1}(x)&=\phi_{\sigma^{i,j}_{1},1}\circ\cdots\phi_{\sigma^{i,j}_s,1}\circ \phi_{\sigma_i}\circ h_i(x)\\ &=\phi_{\sigma^{i,j}_1}\circ\cdots\circ \phi_{\sigma^{i,j}_s}\circ \phi_{\sigma_i}\circ h_i(x)\\ &=\phi_{\sigma^{i,j}_1}\circ\cdots\circ \phi_{\sigma^{i,j}_s}\circ\phi_{\sigma^{i},1}(x), \end{split} \end{align*} $$

and hence the part of $\psi _i$ contained in $\phi _{\sigma ^{i},1}(B_\delta (z_1))$ is unchanged. This implies that, inside the ball $B_{\delta /2}(z_1)$ , the IFS $\Phi ^{s,1}$ has the wanted property for the manifold $\phi _{\sigma _{1},1}^{-1}\circ \psi _1\cap \cdots \cap \phi _{\sigma _n,1}^{-1}\circ \psi _n$ . Moreover, by the openness of the transversality property, there exists $\epsilon _{s,1}\leq \epsilon _s/4$ such that the wanted property persists in $B_{\delta /2}(z_1)$ for each $\epsilon _{s,1}$ perturbation. We can now consider all the other pre-images and do the same procedure with ${\epsilon _{s,j}\leq 4^{-j-1}\epsilon _s}$ for the jth intersection manifold. In this way, we can construct an IFS $\Phi ^{s,q}=:\tilde \Phi ^{s,1}$ for some $q\in {\mathbb N}$ that is a $\epsilon _s/3$ perturbation of $\Phi ^s$ and has the wanted property in $B_{\delta /2}(z_1)$ . We can then repeat the same procedure in the ball $B_\delta (z_2)$ to obtain an IFS $\tilde \Phi ^{s,2}$ that is a $ \epsilon _s/9$ perturbation of $\tilde \Phi ^{s,1}$ , small enough not to upset what we have achieved in $B_{\delta /2}(z_1)$ . Iterating such a construction, we finally obtain $\Phi ^{s+1}=\tilde \Phi ^{s,t}$ , which has the wanted property on all the space since $\{B_{\delta /2}(z_i)\}_{i=1}^t$ is a covering of $ D^N(\Phi ^s_{\! f}\!)$ and is a $\sum _{i=1}^{t}\epsilon _s 3^{-k}\leq \epsilon _s\leq 2^{-s-1}\varepsilon $ perturbation of $\Phi ^s$ . This concludes the induction argument.

Proof of Proposition 4.3

Let f be a piecewise smooth contraction with contraction coefficient $\unicode{x3bb} <1$ and maximal partition ${\boldsymbol P}(f)=\{P_1, P_2, \ldots , P_m\}$ . If $\unicode{x3bb} <1/2m$ , then the proof is complete; otherwise, we have the following.

Let the IFS associated to f be $\Phi _{\! f}=\{\phi _1,\phi _2,\ldots ,\phi _m\}$ and let $l_1=\max \{c_d l_0,d\}$ , where $l_0$ is the number of boundary manifolds in $\partial {\boldsymbol P}(f)$ , d is the dimension of the space and $c_d$ is the maximum number of original boundary manifold overlaps (as defined in (4.13)). Let $N\in {\mathbb N}$ be the least number such that $\unicode{x3bb} ^N 2^{ dm^{d-1} l_1} <1/4$ . By Lemma 4.19, for $\varepsilon>0$ small enough, there exists a piecewise contraction $\tilde {f}$ such that $d_2(f,\tilde {f})<\varepsilon $ and no more than $2^{dm^{d-1} l_1}$ elements of the partition $\partial {\boldsymbol P}(\tilde f^N)$ have a non-empty intersection of their closure.

Accordingly, for each $x\in X$ , there is a $\delta (x)$ such that $B_{\delta (x)}(x)$ intersects at most $2^{dm^{d-1} l_1}$ elements of ${\boldsymbol P}(\tilde f^N)$ . Since X is compact, we can extract a finite cover $\{B_{\delta (x_j)/2}(x_j)\}$ . Set $\delta =\tfrac 12\min \{\delta (x_j)\}$ and let $k\in {\mathbb N}$ be such that, for any partition element $P\in {\boldsymbol P}(\tilde {f}^{kN})$ , $\operatorname {diam}(\tilde {f}^{kN}(P))<\delta /2$ ; hence $\tilde {f}^{kN}(P)\subset B_{\delta (x_j)}(x_j)$ for some j. Therefore, it can intersect at most $2^{dm^{d-1} l_1}$ elements of ${\boldsymbol P}(\tilde f^N)$ .

To conclude, let L be the number of elements of ${\boldsymbol P}(\tilde {f}^{kN})$ . Then $\#{\boldsymbol P}(\tilde {f}^{2kN})\leq L2^{dm^{d-1} l_1}$ and $\#{\boldsymbol P}(\tilde {f}^{jkN})\leq L(2^{dm^{d-1} l_1})^{j}$ for $j\in {\mathbb N}$ . Since $\unicode{x3bb} ^{kN}(2^{dm^{d-1} l_1})<1/4$ , there exists $j_*\in {\mathbb N}$ such that $L(2^{dm^{d-1} l_1})^{j_*}\unicode{x3bb} ^{j_*kN}< 1/2$ . Hence $\tilde {f}^{j_*kN}$ is strongly contracting.