Using Lemma \[reclem1\](b) for each bound segment and Lemma \[outside\] for each free segment in between two consecutive free returns, we have $$J^{n+1}(c_0) \geq \delta L^{2 \lambda n(1-\frac{\alpha}{\lambda})}\geq  L^{ \lambda n}.$$ This completes the proof of Lemma \[derive0\].

The next expansion estimate at deep return times will be used in a crucial way in the construction of the parameter set $\Delta$.

\[exp\] If $c \in C$ and $\nu\leq n+1$ is a deep return time of $c_0=fc$, then $$J^{\nu}(c_0) \cdot D_{\nu}(c_0) \geq \sqrt{d_C(c_{\nu})}.$$

Let $0<n_1<\cdots<n_t<\nu$ denote all free returns in the first $\nu$ iterates of $c_0$, with $p_1,\cdots,p_t$ the corresponding bound periods. Let $$\Theta_{n_k}=\sum_{i=n_k}^{n_k+p_k-1}d_i^{-1}(c_0) \ \ \ \text{
and } \ \ \ \Theta_{0}=\sum_{i=0}^{\nu-1}d_i^{-1}(c_0)
-\sum_{k=1}^{t} \Theta_{n_k}.$$

[*Step 1 (Estimate for bound segments):*]{} Observe that $$\begin{aligned}
& \frac{\Theta_{n_k}}{J^{n_k+p_k}(c_0)} =\frac{1}{J^{p_k}(c_{n_k}) d_C(c_{n_k})
d_S(c_{n_k})} + \sum_{i = n_k+1}^{n_k+p_k-1}
\frac{1}{J^{n_k+p_k-i}(c_i) d_C(c_i) d_S(c_i)}.\end{aligned}$$ To estimate the first term we use Lemma \[reclem1\](b) to obtain $$\label{f1-deep}
\frac{1}{J^{p_k}(c_{n_k}) d_C(c_{n_k})
d_S(c_{n_k})}\leq \sigma^{-1}
(d_C(c_{n_k}))^{- \frac{16 \alpha}{\lambda}}.$$ To estimate the second term we let $\tilde c$ be the critical point to which $c_{n_k}$ is bound. By using Lemma \[dist\] and (\[derivative6\]) in the proof of Lemma \[dist\], which implies $d_C(c_i)>\frac{1}{2} d_C(\tilde c_{i-n_k-1})$ and $d_S(c_i)> \frac{1}{2} d_S(\tilde c_{i-n_k-1})$ for $i \in [n_k+1,
n_k+p_k-1]$, 