3
$\begingroup$

Let $C$ be a compact set in $\mathbb{R}^n$.

Let $f \colon v \in C \mapsto k(v) \in \mathbb{R}$ a continuous function. By Weierstrass' theorem, $f$ admits $k_1$ and $k_2$ as maximum and minimum values. Are the functions $k_i(v)$ continuous as functions on their own?


Sorry, I'll try to fix. Let's hope the following makes some sense.

Let $C$ be a compact set in $\mathbb{R}^n$. For each $P$ in $C$, $f_P \colon v \in C \mapsto f_P(v) \in \mathbb{R}$ is a continuous function. By Weierstrass' theorem, $f_P$ admits $k_1(P)$ and $k_2(P)$ as maximum and minimum values. Are the functions $k_i(P)$ continuous as functions on their own?

  • 1
    Maybe a concrete example will help clarify things. Set $C = [0,1]$, and set $f_p$ to be the constant $1$ function if $p$ is rational, and the constant $-1$ function if $p$ is irrational. Certainly the max (and min!) values of $f_p$ are not continuous functions of $p$.2011-09-21

1 Answers 1

7

It seems that we have a continuous function $f:\ C\times C\to{\mathbb R}, \ (x,P)\to f_P(x)$, where $C$ is a compact set in ${\mathbb R}^n$. For a given $P\in C$ one can ask about the value $m(P):=\max\{f_P(x)| x\in C\}$, and the question is whether the new function $m(\cdot):\ P\mapsto m(P)$ is continuous on $C$.

This is indeed the case. Note that $f$ is uniformly continuous on the compact set $C\times C$. This means that given an $\epsilon>0$ there is a $\delta>0$ with $f_{P_0}(x_0) whenever $|x-x_0|<\delta$ and $|P-P_0|<\delta$. For fixed $P_0\in C$ there is a point $x_0\in C$ with $m(P_0)=f_{P_0}(x_0)$, and therefore we have $m(P_0)< f_P(x_0)+\epsilon\leq m(P)+\epsilon$ for all $P$ with $|P-P_0|<\delta$. By symmetry it follows that $|m(P)-m(P_0)|<\epsilon$ as soon as $|P-P_0|<\delta$, which proves that the function $m(\cdot)$ is continuous on $C$.