The extreme value theorem requires that a function be continuous on a closed interval $[a,b]$ for it to necessarily take on a max and min, but I've been thinking and it seems to me that as long as it is defined for all numbers in a closed interval it will take on a max and min on that interval.
Is this correct?
Best Answer
Consider $f:[-1,1] \to \mathbb{R},$
$$f(x)=\begin{cases} |x| &, x\in [-1,1]\setminus \{ 0\}\\ 1&, x=0 \end{cases}$$
then this function does not have a global minimum.
Another function would be $g: [-1,1] \to \mathbb{R}$,
$$g(x)=\begin{cases} \frac{1}{x+1}+\frac{1}{x-1} &, |x| \leq 1\\ 0&, |x|=1 \end{cases}$$