free segment out of $C_{\delta}$. For the new free segments, we can now replace $\delta$ by $\delta^{\frac{1}{20}}$ in (\[deep\]); for the bound segments, we use (\[sublem2\]) with $d_C
> \delta$. We then obtain $$\label{quatro}
\frac{\Theta_0}{J^{\nu}(c_0)}\leq \frac{1}{\sigma
\delta^{\frac{1}{20}}} +  \sum_{i  \in [0, \nu-1] \setminus (\cup
[n_k, n_k + p-1])} K L^{-\frac{1}{3} \lambda (\nu - i)} \delta^{-
\frac{18 \alpha}{\lambda}} < \frac{1}{\delta^{\frac{1}{3}}}.$$

[*Step 3 (Proof of the Lemma):*]{}  From the assumption that $\nu$ is a deep free return, we have $$|d_C(c_{n_k})|^{-1}\leq
|d_C(c_{\nu})|^{-2}\prod_{j\colon n_j\in(n_k,\nu)}
|d_C(c_{n_j})|^{-2}.$$ Substituting this into (\[sublem2\]) gives $$\label{plu}
J^{n_k+p_k}(c_0)^{-1}\Theta_{n_k}\leq
|d_C(c_{{\nu}})|^{-\frac{36\alpha}{\lambda}}\prod_{j\colon
n_j\in(n_k,\nu)}|d_C(c_{n_j})|^{-\frac{36\alpha}{\lambda}}.$$ Meanwhile, splitting the orbit from time $n_k+p_k+1$ to $\nu$ into bound and free segments and we have $$\label{plu2}
J^{\nu-n_k-p_k}(c_{n_k+p_k})^{-1}\leq \left(\prod_{j\colon
n_j\in(n_k,\nu)} J^{p_j}(c_{n_j})\right)^{-1}.$$ Multiplying (\[plu\]) with (\[plu2\]) gives $$\begin{aligned}
J^{\nu}(c_0)^{-1}  \Theta_{n_k} \leq
|d_C(c_{\nu})|^{-\frac{36\alpha}{\lambda}}\prod_{j: \ n_j \in
(n_k, \nu)} \left(J^{p_j}(c_{n_j}) \cdot
|d_C(c_{n_j})|^{\frac{36\alpha}{\lambda}}\right)^{-1}\end{aligned}$$ $$\begin{aligned}
\leq |d_C(c_{\nu})|^{-\frac{36\alpha}{\lambda}} \prod_{j: \ n_j
\in (n_k, \nu)} (d_C(c_{n_j}))^{\frac{1}{2}}  \ \leq \
\delta^{(t-k)/2}|d_C(c_{\nu})|^{-\frac{36\alpha}{\lambda}}.\end{aligned}$$ where for the second inequality we use Lemma \[reclem1\](b) for $J^{p_j}(c_{n_j})$, and for the last we use $d_C(c_{n_j}) < \delta$. Thus $$\begin{aligned}
 \sum_{n_k \in [0, \nu-1]} J^{\nu}(c_{0}) ^{-1}\Theta_{n_k}&\leq
|d_C(c_{\nu})|^{-\frac{36\alpha}{\lambda}}
\sum_{k=1}^{t}\delta^{(t-k)/2}\leq
2|d_C(c_{\nu})|^{-\frac{36\alpha}{\lambda}}.\end{aligned}$$ Combining this with (\[quatro\]) we obtain $$\begin{aligned}
J^{\nu}(c_{0})^{-1}D_{\nu}^{-1}=
\sqrt {L}\left( \sum_{1 \leq k \leq t} J^{\nu}(c_{0}) ^{-1}
\Theta_{n_k}+ J^{\nu}(c_{0}) ^{-1} \Theta_0\right)\leq
\frac{1}{\sqrt{d_C(c_{\nu})}}.\end{aligned}$$ This completes the 