TITLE: convergence of infimum
QUESTION [5 upvotes]: I have a question that I encountered during my internship:
Consider a convergent sequence of continuous, convex functions $\{f_n(x)\}_n$ defined in $\mathbb{R}^M$. These functions are uniformly Lipschitz continuous, that is, $\exists C\in\mathbb{R}$ such that:
$$\forall x,y \in \mathbb{R^M},\forall n\ge1\quad |f_n(x)-f_n(y)|\le C|x-y|.$$
Furthermore, each function $f_n(x)$ has a minimizer. The properties of simple convergence and uniform Lipschitz continuity allow us to prove that the convergence is uniform in any compact of $\Bbb R^M$.
My question is:
Can we demonstrate  that $\inf_{\Bbb R^M}f_n(x)$ converges to $\inf_{\Bbb R^M}f_{\infty}(x)$ as $n\rightarrow\infty$,
where $f_{\infty}(x)$ is the limit of $f_n(x)$ and it is supposed that $\inf_{\mathbb{R^M}}f(x)$ is finite?
Thanks a lot!

REPLY [5 votes]: No. Take the sequence
$$
f_n : x\in \mathbb R \to \frac 1 n \lvert x - n \rvert -1 \in \mathbb R
$$
We have
$$
\lim_n f_n(x) = 0 \quad\forall x\in \mathbb R\\
\inf_{\mathbb R} f_n(x) = f_n(n) = -1\\
\lvert f_n(x) - f_n(y) \rvert = \frac 1 n \big\lvert \lvert x - n \rvert - \lvert y - n\rvert \big\rvert \leq \frac 1 n \lvert x - y\rvert \leq \lvert x -y \rvert
$$
the functions $f_n$ converge punctually to $f_\infty = 0$, they are convex and uniformly Lipschitz continuous but
$$
\lim_n \left(\inf_{\mathbb R} f_n(x) \right) = -1 \neq \inf_{\mathbb R} f_\infty(x) = 0
$$