Continuity of Legendre transform

Question

Let $I \subset \mathbb{R}$ be an interval, and $f_n, f: I \rightarrow \mathbb{R}$ a convex function; then its Legendre transform is the function $f^{\ast}: I^{\ast} \rightarrow \mathbb{R}$ defined by $$ f^{\ast}\left(x^{\ast}\right)=\sup _{x \in I}\left(x^{\ast} x-f(x)\right), \quad x^{\ast} \in I^{} $$ where sup denotes the supremum, and the domain $I^{\ast}$ is $$ I^{\ast}=\{ x^{\ast} \in R: \sup _{x \in I}\left(x^{\ast} x-f(x)\right)<\infty \}.$$ The transform is always well-defined when $f(x)$ is convex.

$$Question$$ Assume that $\frac{1}{n}f_{n}(x) \to f(x)$. Is it true that $\sup_{x \in I}\left(x^{\ast} x-\frac{1}{n}f_{n}(x)\right) \to f^{\ast}(x^{\ast})?$

Answer 1

Conditions in terms of the function $f$ that are simultaneously necessary and sufficient for such convergence are given in this paper and in its arXiv version. The form of those conditions depends on whether either or both endpoints of the interval $I$ are finite or not.

In particular, by Corollary 2a of that paper, the desired convergence will always hold in your setting if both endpoints of the interval $I$ are finite.