samples/texts/2952501/page_29.md

Indeed, in that case the liminf of the quantities in the left-hand side of (6.2) are greater than or equal to the right-hand side, for any choice of the forest $\mathcal{F}_k$. As the sum of the quantities on the right-hand side of (6.2) over these choices is equal to 1, necessarily the desired convergence holds.

Let us thus show (6.5). Set $\varphi(l) := 4^{-l}C(l)$. We have

$$A_s=\sum _{l=2k+1}^{\mathrm{\infty }}\frac{\phi (l)}{\phi (p)}\sum _{q=0}^p{P}_{l-(2k+1)}(Y_s=q)P_{m_k}(Y_{s-r}=p-q)\mathrm{.}A\_s\; =\; \backslash sum\_\{l=2k+1\}^\{\backslash infty\}\; \backslash frac\{\backslash varphi(l)\}\{\backslash varphi(p)\}\; \backslash sum\_\{q=0\}^\{p\}\; P\_\{l-(2k+1)\}(Y\_s=q)\; P\_\{m\_k\}(Y\_\{s-r\}=p-q).$$

First, as $\theta$ is a critical offspring distribution, we get from [26] that, for any $q \ge 0$,

$$\frac{P_{m_k}(Y_{s-r}=p-q)}{P_{m_k}(Y_s=p-q)}\stackrel{s\to \mathrm{\infty }}{\to }1.\backslash frac\{P\_\{m\_k\}(Y\_\{s-r\}\; =\; p-q)\}\{P\_\{m\_k\}(Y\_s\; =\; p-q)\}\; \backslash xrightarrow\{s\; \backslash to\; \backslash infty\}\; 1.$$

Thus, for any $l \ge 2k+1$, for every $\varepsilon > 0$, for any sufficiently large $s$,

This implies that:

$$A_s\ge (1-\epsilon )\sum _{l=2k+1}^{\mathrm{\infty }}\frac{\phi (l)}{\phi (p)}{P}_{l-(2k+1)+m_k}(Y_s=p)\mathrm{.}A\_s\; \backslash ge\; (1\; -\; \backslash varepsilon)\; \backslash sum\_\{l=2k+1\}^\{\backslash infty\}\; \backslash frac\{\backslash varphi(l)\}\{\backslash varphi(p)\}\; P\_\{l-(2k+1)+m\_k\}\; (Y\_s\; =\; p).$$

Now, from the asymptotics of $C(l)$, we have that, for some $l_0 \ge 0$, for any $l \ge l_0$, we have

$$\phi (l)\ge (1-\epsilon )\phi (l-(2k+1)+m_k),\backslash varphi(l)\; \backslash ge\; (1\; -\; \backslash varepsilon)\backslash varphi(l\; -\; (2k\; +\; 1)\; +\; m\_k),$$

so that,

$$A_s\ge (1-\epsilon {)}^2\sum _{l=m_k\vee l_0}^{\mathrm{\infty }}\frac{\phi (l)}{\phi (p)}{P}_l(Y_s=p)\mathrm{.}A\_s\; \backslash ge\; (1\; -\; \backslash varepsilon)^2\; \backslash sum\_\{l=m\_k\; \backslash lor\; l\_0\}^\{\backslash infty\}\; \backslash frac\{\backslash varphi(l)\}\{\backslash varphi(p)\}\; P\_l(Y\_s\; =\; p).$$

Recall that $\varphi(l) = lh(l)$, which yields:

the last equality stemming from (5.12).

Finally, we use once again the fact that, for any fixed $l$,

$$P_l(Y_s=p)\stackrel{s\to \mathrm{\infty }}{\to }0,P\_l\; (Y\_s\; =\; p)\; \backslash xrightarrow\{s\; \backslash to\; \backslash infty\}\; 0,$$

to get that, for any $\epsilon > 0$,

$$\underset{s}{\mathrm{lim\hspace{0.17em}inf}}{A}_s\ge (1-ϵ{)}^2\mathrm{.}\backslash liminf\_s\; A\_s\; \backslash ge\; (1\; -\; \backslash epsilon)^2.$$

As $\epsilon$ was completely arbitrary in the above chain of arguments, we get that

$$\underset{s}{\mathrm{lim\hspace{0.17em}inf}}{A}_s\ge 1.\backslash liminf\_s\; A\_s\; \backslash ge\; 1.$$

This completes the proof of the proposition. $\square$