Does the minima of a sequence of convex convergent functions converge?

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 $x_n^*\rightarrow y$ ?

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. $f_n(x) = (x - n)^2/n^n.$

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 $f=0$.

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 $f_n$.

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 $a< y$. Clearly $f( (a+y)/2) > f(y)$. By convexity, $f(a)> f((a+y)/2)$. So for $n$ sufficiently large, $f_n(a)> f_n((a+y)/2)> f_n(y)$. But this implies that the minimum of $f_n$ is greater than $(a+y)/2$. So the $\lim \inf$ of $x_n^*$ is greater than or equal to $(a+y)/2$, which is greater than $a$. This is a contradiction so $a\geq y$. By symmetry $b\leq y$. Hence because $a \leq b$, $a=b=y$ and $y$ is the limit.

• Thanks Will. The last paragraph is unclear to me. To start did you mean if f has a unique minima? – gmravi2003 May 11 '12 at 22:06
• Yes, I did. What else is confusing? – Will Sawin May 12 '12 at 4:46
• It seems that there should be $(x-1)^2/n$ instead of $(x_1)^2/n$. – Alex Ravsky Jan 19 '15 at 5:27
• Why $x_n^*\leq a$ occurs infinitely many times? – Alex Ravsky Jan 19 '15 at 5:40
• @AlexRavsky fixed. – Will Sawin Jan 19 '15 at 14:23