Let $X$ be a random variable, following the distribution $P$. Let $\delta>0$ and $$ g(\alpha):=-\alpha\log(E_{P}[e^{-\frac{X}{\alpha}}])-\alpha\delta. $$
How can I tell that $\lim_{\alpha\to 0}g(\alpha)=ess\,inf X$?
This was part of a proof in P.13 from the paper http://proceedings.mlr.press/v130/zhou21d/zhou21d-supp.pdf. The explanation was omitted.
You can verify that the limit holds if $X$ is a constant random variable, so we really just need to go through the definition of the essential infimum. I will consider the limit for when $\alpha > 0$.
If $c$ is the aforementioned essential infimum, then $X \geq c$ almost everywhere. Then since $-\alpha \log(x)$ is a decreasing function in $x$, and $e^{-x/\alpha}$ is also decreasing in $x$, and $c \leq X$ $$-\alpha \log(E_P[e^{-X/\alpha}]) - a\delta \geq -\alpha \log(E_P[e^{-c/\alpha}]) - \alpha \delta.$$ If we take limits as $\alpha \to 0^+$, the right hand side simplifies to $c$, and we obtain $$\lim_{\alpha \to 0^+} g(\alpha) \geq c.$$
The other side seems worse, but hopefully it follows from the definition. Let $\epsilon > 0$: by definition of the essential infimum, the set $A_\epsilon := \{\omega : X(\omega) < c + \epsilon\}$ has positive measure. Then assuming $\alpha > 0$, $$-\alpha \log(E_P[e^{-X/\alpha}]) = -\alpha \log(E_P[e^{-X/\alpha}(1_{A_\epsilon} + 1_{A_\epsilon^C})]) \leq -\alpha \log(E_P[e^{-X/\alpha}1_{A_\epsilon}]).$$ We have again used that $-\alpha \log(x)$ is decreasing in $x$, and $$E_P[e^{-X/\alpha}] \geq E_P[e^{-X/\alpha}1_{A_\epsilon}]$$ Now, again use the fact that $e^{-x/\alpha}$ is decreasing in $x$ and $-\alpha \log(x)$ is decreasing in $x$, with $X < c + \epsilon$ to get the inequality $$-\alpha \log(E_P[e^{-X/\alpha}1_{A_\epsilon}]) \leq -\alpha \log(E_P[e^{-(c+\epsilon)/\alpha} 1_{A_\epsilon}]) = -\alpha \log(e^{-(c+\epsilon)/\alpha}P(A_\epsilon)) = c+\epsilon - \alpha \log(P(A_\epsilon)).$$ Take limits of the above as $\alpha \to 0$ to conclude for arbitrary $\epsilon > 0$, $$\lim_{\alpha \to 0^+} -\alpha \log(E_P[e^{-X/\alpha}]) - \delta \alpha \leq \lim_{\alpha \to 0^+}c+\epsilon - \alpha \log(P(A_\epsilon)) - \delta \alpha = c + \epsilon.$$ This concludes the proof, assuming that I haven't made a sign error.