have that for any $\delta \in (0, \lambda)$ $$\limsup_{n \to \infty} \frac{1}{n} \log E |\hat{N}^{\lambda n}_n| 
= \limsup_{n \to \infty} \frac{1}{n} \log(I_1 + I_2) \leq \Delta_{(\lambda-\delta)}.$$ Letting $\delta \to 0$ and using continuity and monotonicity of $\Delta_{\lambda}$ as a function of $\lambda$ we obtain the sought result.

Proposition \[lambda\_sup\] can now be applied to prove the upper bounds for Lemma \[pop\_lambda\_as\].

\[upper\_pop\_lambda\]$ $

If $\lambda < \lambda_{crit}$ ($\Delta_\lambda > 0$) then $$\label{subcrit_upper}
\limsup_{t \to \infty} \frac{1}{t} \log |N_t^{\lambda t}| \leq \Delta_\lambda 
\qquad P \text{-a.s.}$$

If $\lambda > \lambda_{crit}$ ($\Delta_\lambda < 0$) then $$\label{supercrit1_upper}
\lim_{t \to \infty} |N_t^{\lambda t}| = 0 \qquad P \text{-a.s.}$$ and $$\label{supercrit2_upper}
\limsup_{t \to \infty} \frac{1}{t} \log P (|N_t^{\lambda t}| > 0) \leq \Delta_{\lambda}  \text{.}$$

For any $\lambda >0$ let $\hat{N}_n^{\lambda n}$ be as in the previous proposition and fix $\epsilon > 0$. Then the Markov inequality gives $$P \Big( |\hat{N}_n^{\lambda n}| > \mathrm{e}^{(\Delta_\lambda + \epsilon)n} \Big) \leq 
\mathrm{e}^{-(\Delta_\lambda + \epsilon)n} E |\hat{N}_n^{\lambda n}|$$ and from Proposition \[lambda\_sup\] the right hand side decays exponentially fast. Therefore by Borel-Cantelli Lemma $$P \Big( \Big\{|\hat{N}_n^{\lambda n}| > \mathrm{e}^{(\Delta_\lambda + \epsilon)n} \Big\} \text{ i.o.}\Big) = 0.$$ This is equivalent to saying that $$|\hat{N}_n^{\lambda n}| \leq \mathrm{e}^{(\Delta_\lambda + \epsilon)n} \text{ eventually } 