Given a nice real valued functional $C$ on some probability space $(\Omega, \mathcal F, P_0)$ we have the following Donsker-Varadhan variational representation
$$\log E_{P_0}\left[e^C\right]=\sup_{P\sim P_0}\{E_P[C]-D_{KL}(P||P_0)\},$$
where the supremum is achieved at the measure $P^\ast=\frac{e^C}{E_{P_0}[e^C]}P_0$. This can be verified by looking at $D_{KL}(P||P^\ast)$.
We say that a sequence of measures $\mu_0^\epsilon$ on $(\Omega, \mathcal F)$ satisfies the Laplace principle with rate $I$ if for all continuous and bounded real valued functions $f$ we have
$$\lim_{\epsilon\to 0}\epsilon \log E_{\mu_0^\epsilon}\left[e^{f/\epsilon}\right]=\sup_{\omega\in\Omega}\{f(\omega)-I(\omega)\}.$$
Then using Bryc's lemma if we have exponential tightness we can say that $\mu^\epsilon$ satisfies a large deviations principle (LDP) with rate function $I$. Similarly, if $\mu^\epsilon$ satisfy a LDP with rate function $I$ and the measures satisfy a moment condition, then by Varadhan's lemma we can say that the Laplace principle holds. The relationship between large deviations and Laplace principle is clear to me.
However, applying the Donsker-Varadhan representation to the Laplace principle yields
$$\lim_{\epsilon\to 0}\epsilon \log E_{\mu_0^\epsilon}\left[e^{f/\epsilon}\right]=\lim_{\epsilon\to 0} \sup_{\mu^\epsilon\sim \mu_0^\epsilon}\{E_{\mu^\epsilon}[f]-\epsilon D_{KL}(\mu^\epsilon||\mu_0^\epsilon)\}=\sup_{\omega\in\Omega}\{f(\omega)-I(\omega)\}.$$
It seems like there is some message here but I am unable to find it.
What is the relationship between $\lim_{\epsilon\to 0} \sup_{\mu^\epsilon\sim \mu_0^\epsilon}\{E_{\mu^\epsilon}[f]-\epsilon D_{KL}(\mu^\epsilon||\mu_0^\epsilon)\}$ and $\sup_{\omega\in\Omega}\{f(\omega)-I(\omega)\}$?
For an example of a precise question, suppose that there is a sequence of optimizers of the measure supremum, call them $\mu^{\epsilon, \ast}$ and suppose that there is an optimizer $\omega^\ast$ in the other supremum. Then is it true that $\lim_{\epsilon\to 0} E_{\mu^{\epsilon,\ast}}[f]=f(\omega^\ast)$ and $\lim_{\epsilon\to 0} \epsilon D_{KL}(\mu^{\epsilon,\ast}||\mu_0^\epsilon)=I(\omega^\ast)$? This doesn't seem at all obvious why this should happen.