Does the minima of a sequence of convex convergent functions converge? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T09:21:02Z http://mathoverflow.net/feeds/question/96711 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/96711/does-the-minima-of-a-sequence-of-convex-convergent-functions-converge Does the minima of a sequence of convex convergent functions converge? gmravi2003 2012-05-11T21:06:43Z 2012-05-12T04:46:12Z <p>Suppose $f_1,f_2,\ldots$ is a sequence of convex functions that converges to a continuous convex $f$. Let $x_1^*,x_2^*$ be their respective (not necessarily unique) minima, and let y be a minima of $f$ (once again need not be unique). Can we prove that there exists a version of $x_1^*,x_2^*,\ldots$ such that $x_n^*\rightarrow y$ ? </p> http://mathoverflow.net/questions/96711/does-the-minima-of-a-sequence-of-convex-convergent-functions-converge/96716#96716 Answer by Alex Gittens for Does the minima of a sequence of convex convergent functions converge? Alex Gittens 2012-05-11T21:47:10Z 2012-05-11T21:47:10Z <p>No; here's a counterexample: let $f = 0$ and consider the minimizer $y = 0.$ Then you can construct convex functions which converge to $0$ pointwise but whose minima are always moving away from $y =0,$ e.g. $f_n(x) = (x - n)^2/n^n.$ </p> http://mathoverflow.net/questions/96711/does-the-minima-of-a-sequence-of-convex-convergent-functions-converge/96717#96717 Answer by Will Sawin for Does the minima of a sequence of convex convergent functions converge? Will Sawin 2012-05-11T21:48:22Z 2012-05-12T04:46:12Z <p>No. Let $f_n=x^2/n$ for $n$ odd and $(x_1)^2/n$ for $n$ even. Then $x_n^*$ is an alternating sequence of $1$s and $0$s, which does not converge to anything. But $f_n$ converges pointwise to $f=0$.</p> <p>We can modify this to make the convergence uniform, by using an absolute value instead of a square, or to make $f$ nonconstant, by adding $\max(|x-1/2|,1)$ to $f_n$.</p> <p>If $f$ has a unique minimum, the statement is true. Let $a$ be the $\lim\inf$ of $x_n^*$ and $b$ be the $\lim\sup$. Let $y$ be the unique minimum of $f$. Assume $a&lt; y$. Then $f_n(a)$ converges to $f(a)$, and $f_n(y)$ converges to $f(y)$, and since $f(a)>f(y)$, $f_n(a)\leq f_n(y)$ only finitely many times. But every time $x_n^*\leq a$ we have $f_n(a)\leq f_n(y)$ since $a$ is closer to the minimum $x_n^*$ then $y$. Since that occurs infinitely many times, this is a contradiction.</p> <p>Therefore $a\geq y$. Similarly $y\geq b$, and by properties of $\lim\sup$ and $\lim\inf$ we have $b\geq a$, so $a=b=y$ and the limit is $y$.</p>