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Hello everyone,

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

|f_n(x)-f_n(y)|<=C|x-y|, for all x,y in R^M and n>=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 R^M.

Now my questionn is that whether we can demonstrate

inf_{R^M}f_n(x) converges to inf_{R^M}f(x), n goes to infty?

here f(x) is the limit of f_n(x) and is supposed that inf_{R^M}f(x) fini.

Thanks a lot!

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What do you mean by "each function $f_n(x)$ has a minimizer"? –  Zhaoting Wei Jul 10 '12 at 13:09
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2 Answers

up vote 0 down vote accepted

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 $0$.

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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 $$\lim_{n}\inf_{R}f_n(x)=-1\neq 0=\inf_{R}f_{\infty}(x)$$ –  Higgs88 Jul 11 '12 at 7:57
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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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