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 $\sup\_{x \in I}\left(x^{\ast} x-\frac{1}{n}f\_{n}(x)\right) \to f^{\ast}(x^{\ast})?$



Continuity of Legendre transform
Conditions in terms of the function $f$ that are simultaneously necessary and sufficient for such convergence are given in [this paper](https://www.heldermann.de/JCA/JCA26/JCA261/jca26005.htm) and in its [arXiv version](https://arxiv.org/abs/1307.3806). 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.


