I have a question during my intership. Given a convergent sequence of continuous et convex functions $\{f_n(x)\}$ defined in $\mathbb{R}^M$. These functions are uniformly Lipschitz continuous which means that there exist a constant C such that

$$|f_n(x)-f_n(y)| \leq C|x-y|, \quad \forall x,y \ \textrm{ in } \ \mathbb{R}^M \ \textrm{ and } \ n \geq 1 . $$

Furthermore, each function $f_n(x)$ has a minimizer.

So the simple convergence + uniformly Lipschitz continuous allow us to prove the convergence is uniform in any compact of $\mathbb{R}^M$.

Now my questionn is that whether we can demonstrate

$$\inf_{x \in \mathbb{R}^M} f_n(x) \to \inf_{x \in \mathbb{R}^M} f(x)$$ whenever $n$ goes to $\infty$?

Here $f(x)$ is the limit of $f_n(x)$ and is supposed that $\inf_{x \in \mathbb{R}^M} f(x)$ is finite.

Thanks a lot!

  • $\begingroup$ What do you mean by "each function $f_n(x)$ has a minimizer"? $\endgroup$ Jul 10, 2012 at 13:09

2 Answers 2


I think there are counterexamples: consider $f_n(x)=\arctan(\frac{x}{n})$, then $f_n(x)$ converge to $0$ and are uniformly Lipschitz continuous since there derivatives are smaller than $1$. However the inf of $\arctan(\frac{x}{n})$ is $-\frac{\pi}{2}$,which is not $0$.

  • 1
    $\begingroup$ Thanks a lot for your reply. I find also a counterexample: $$f_n(x)=\frac{1}{n}|x-n|-1$$ $$f_{\infty}(x)=0$$ then $\{f_n(x)\}_n$ satisfies all the conditions but $$\lim_{n}\inf_{R}f_n(x)=-1\neq 0=\inf_{R}f_{\infty}(x)$$ $\endgroup$
    – Higgs88
    Jul 11, 2012 at 7:57

I assume that by minimizer you mean a point at which function attains its minimum. Then the example of Zhaoting Wei doesn't work but it can be used to construct a counterexample; the idea is that minimizers go to infinity. Consider, for instance, $f_{n}$ equal $-1$ on $(-\infty,-n-1] \cup [n+1,\infty)$, $0$ on $[-n,n]$ and linear otherwise.


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