convergence of infimum - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T01:38:50Z http://mathoverflow.net/feeds/question/101829 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101829/convergence-of-infimum convergence of infimum Higgs88 2012-07-10T07:03:03Z 2012-07-10T14:42:30Z <p>Hello everyone,</p> <p>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</p> <p>|f_n(x)-f_n(y)|&lt;=C|x-y|, for all x,y in R^M and n>=1</p> <p>Furthermore, each function f_n(x) has a minimizer.</p> <p>So the simple convergence + uniformly Lipschitz continuous allow us to prove the convergence is uniform in any compact of R^M.</p> <p>Now my questionn is that whether we can demonstrate </p> <p>inf_{R^M}f_n(x) converges to inf_{R^M}f(x), n goes to infty?</p> <p>here f(x) is the limit of f_n(x) and is supposed that inf_{R^M}f(x) fini.</p> <p>Thanks a lot! </p> http://mathoverflow.net/questions/101829/convergence-of-infimum/101851#101851 Answer by Zhaoting Wei for convergence of infimum Zhaoting Wei 2012-07-10T14:14:36Z 2012-07-10T14:14:36Z <p>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 $-\frac{\pi}{2}$ï¼Œwhich is not $0$.</p> http://mathoverflow.net/questions/101829/convergence-of-infimum/101856#101856 Answer by Mateusz Wasilewski for convergence of infimum Mateusz Wasilewski 2012-07-10T14:42:30Z 2012-07-10T14:42:30Z <p>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.</p>