convergence of infimum - MathOverflow most recent 30 from http://mathoverflow.net2013-05-23T01:38:50Zhttp://mathoverflow.net/feeds/question/101829http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/101829/convergence-of-infimumconvergence of infimumHiggs882012-07-10T07:03:03Z2012-07-10T14:42:30Z
<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)|<=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#101851Answer by Zhaoting Wei for convergence of infimumZhaoting Wei2012-07-10T14:14:36Z2012-07-10T14:14:36Z<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 $-\frac{\pi}{2}$,which is not $0$.</p>
http://mathoverflow.net/questions/101829/convergence-of-infimum/101856#101856Answer by Mateusz Wasilewski for convergence of infimumMateusz Wasilewski2012-07-10T14:42:30Z2012-07-10T14:42:30Z<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>