= Solution
The <continuity> of f makes its prescribed interval preimage B open, and $\hat x\in B$. On B we have $f(x)>f(\hat x)-\varepsilon$, so
$$
\mathbb E e^{nf(X_n)}\geq e^{n(f(\hat x)-\varepsilon)}\mathbb P(X_n\in B).
$$
The open-set lower bound of the <large deviation principle> therefore yields
$$
\liminf_n n^{-1}\log\mathbb E e^{nf(X_n)}\geq f(\hat x)-\varepsilon-\inf_BI\geq f(\hat x)-I(\hat x)-\varepsilon.
$$
Thus the claimed \b[lower bound holds at every chosen point]. When $I(\hat x)=\infty$ its right side is minus infinity and the assertion is immediate.
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