We define, for all $x \in \mathbb{R}$, $\lambda^*(x) = \sup_{t \geqslant 0}(tx - \lambda(t))$. We admit that the convergence of the sequence of functions $\left(t \mapsto \frac{\ln\left(E\left(\mathrm{e}^{tS_n}\right)\right)}{n}\right)_{n \in \mathbb{N}^*}$ towards $t \mapsto \ln(\gamma(t))$ is uniform on $\mathbb{R}^+$. Let $\varepsilon > 0$. Show that there exists a rank $n_0 \in \mathbb{N}^*$ such that, for all $t \in \mathbb{R}^+$ and for all $n \in \mathbb{N}^*$, $$n \geqslant n_0 \Longrightarrow \ln\left(E\left(\mathrm{e}^{tS_n}\right)\right) \leqslant n(\lambda(t) + \varepsilon).$$
We define, for all $x \in \mathbb{R}$, $\lambda^*(x) = \sup_{t \geqslant 0}(tx - \lambda(t))$. We admit that the convergence of the sequence of functions $\left(t \mapsto \frac{\ln\left(E\left(\mathrm{e}^{tS_n}\right)\right)}{n}\right)_{n \in \mathbb{N}^*}$ towards $t \mapsto \ln(\gamma(t))$ is uniform on $\mathbb{R}^+$. Let $\varepsilon > 0$. Show that there exists a rank $n_0 \in \mathbb{N}^*$ such that, for all $t \in \mathbb{R}^+$ and for all $n \in \mathbb{N}^*$,
$$n \geqslant n_0 \Longrightarrow \ln\left(E\left(\mathrm{e}^{tS_n}\right)\right) \leqslant n(\lambda(t) + \varepsilon).$$