9

Suppose we have a sequence of functions $f_n(x)$ that converge to a limiting function $f(x) = \lim_{n \to \infty} f_n(x)$ for $\forall x \in [a,b]$. I was wondering under what conditions the following holds? \begin{equation*} \lim_{n \to \infty} \inf_{x \in [a,b]} f_n(x) = \inf_{x \in [a,b]} f(x). \end{equation*}

MikeL
  • 627
  • 2
    You need to use $\inf$ instead of $\min$, as, the minimum may not exist. – James Mar 20 '15 at 13:15
  • @Moya I disagree; there they are assuming uniform convergence, here the OP is asking for conditions. – Ian Mar 20 '15 at 13:22
  • MBP: can you come up with an example where this doesn't hold? I am having trouble, even with some simple discontinuous examples (like $x^n$ or the approximation of $1_{\mathbb{Q}}$). – Ian Mar 20 '15 at 13:23
  • There is a concept called $\Gamma$-convergence, that guarantees that limit and minimization can be interchanged. For details, see http://en.wikipedia.org/wiki/%CE%93-convergence – sranthrop Mar 20 '15 at 13:25
  • @Ian I couldn't come up with an example that this doesn't hold. That's why I was wondering if this holds under quite weak conditions. – MikeL Mar 20 '15 at 13:25

1 Answers1

6

As already mentioned by sranthop, $\Gamma$-convergence is the precise notion to capture minima of function sequences.

Here is a sequence of functions, that does not have the desired property (here $[a,b]=[0,1]$): $$ f_n(x) : = \begin{cases} 0 & x \in (0,\frac1n)\\ 1 & x\not\in (0,\frac1n) \end{cases}. $$ Then $f_n(x) \to 1 =:f(x)$ for all $x$. Hence $$ \lim_{n\to\infty} \inf_{x\in[0,1]} f_n(x) = 0 \ne \inf_{x\in[0,1]} f(x) =1. $$
However, you can deduce the inequality $$ \operatorname*{lim sup}_{n\to\infty} \inf_{x\in[a,b]} f_n(x) \le \inf_{x\in[a,b]} f(x)$$ for pointwise converging $f_n$ towards $f$: Take $\epsilon>0$. Then there is $x_\epsilon$ such that $f(x_\epsilon)\le \inf_x f(x) +\epsilon$. Then we have $$ \inf_x f_n(x) \le f_n(x_\epsilon), $$ passing to the limit yields $$ \operatorname*{lim sup}_{n\to\infty} f_n(x) \le f(x_\epsilon) \le \inf_x f(x) +\epsilon. $$ Now, $\epsilon>0$ was arbitrary, the inequality is proven.
daw
  • 49,113
  • 2
  • 38
  • 76