$\lim_{k\rightarrow \infty}\frac{2^k}{\gamma}\log\mathbb{E}[e^{-\gamma \frac{X}{2^k}}]$

Question

I am trying to find a limit for this expression $$\lim_{k\rightarrow \infty}\frac{2^k}{\gamma}\log\mathbb{E}[e^{-\gamma \frac{X}{2^k}}]$$ I have so far found these bounds: $$\frac{1}{\gamma}\log\left(\mathbb{E}[e^X]\right)^{-\gamma}\leq\lim_{k\rightarrow \infty}\frac{2^k}{\gamma}\log\mathbb{E}[e^{-\gamma \frac{X}{2^k}}]\leq \frac{1}{\gamma}\log \mathbb{E}[e^{-\gamma X}]$$

I was wondering if there are better bounds in particular a better lower bound estimate. $\gamma \in (0,\infty)$ and $X \in L^\infty(\Omega,\mathcal{F},\mathbb{P})$

Answer 1

$$\lim_{s\to0+}\frac1s\log E(\mathrm e^{-sX})=-E(X)$$

Answer 2

Since $log$ is concave

$\frac{2^k}{\gamma}\log\mathbb{E}[e^{-\gamma \frac{X}{2^k}}]\ge\frac{2^k}{\gamma}\mathbb{E}[\log e^{-\gamma \frac{X}{2^k}}]=\frac{2^k}{\gamma}\mathbb{E}[-\gamma \frac{X}{2^k}]=-\mathbb{E}[X]$