The extreme value theorem says that if the domain of a 'continuous' function is compact then both the max and min of the function lies in the domain set. My question is: can the 'continuity' be replaced by 'bounded' function keeping the domain compact and the theorem continue to hold? I haven't seen a proof, but if not, I'd be interested in seeing a counterexample.
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I guess I found the counterexample myself. Consider the following function defined on $[0,1]$. $$f(x) = \left \{ \begin{array}{ll} 2x & \mbox{ if } 0 \leq x < 0.5 \\ 0.5 & \mbox{ if } 0.5 \leq x \leq 1 \end{array} \right . $$
The sup of the function is $1$, but there is no point in the domain $[0,1]$ that 'attains' this point.
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You cannot expect much help from compactness if you drop continuity. Take the function $x\mapsto\operatorname{sg}x-x$ on the compact interval $[-1,1]$ where 'sg' is the sign in $\{-1,0,1\}$, for a counterexample. Here the maximum and minimum of the function value do not exist, so the $\sup$ and $\inf$ of the function are not attained for any value in the domain.
In geneneral, for a function for which continuity is not demanded, topological properties of the domains and codomain have no influence on the kind of behaviour that the function can exhibit.
answered Sep 26 '13 at 9:25
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$\begingroup$ Thanks, I actually found similar counterexample (see below) before I could see this answer. $\endgroup$ – Pagol Sep 26 '13 at 9:32
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1$\begingroup$ @Swaprava: Yes I saw that. Your rapidity of finding a counterexample makes me think: did you really try to solve this yourself before asking? $\endgroup$ – Marc van Leeuwen Sep 26 '13 at 9:46
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Consider the following bounded function defined on the closed interval $[0,1]$.
Let $f(x)=x$ for $0\leq x < 1$, and $f(1)=0$.
Then $$ \sup_{x \in [0,1]} f(x)=1, $$ but there is no element $x_0$ in the domain for which $f(x_0)=1$.
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Let $\varphi: \mathbb R \to [0,1]$ be your favourite bijection. Then consider the function $f(x) = \arctan \varphi(x)$. You have $\sup_{x\in [0,1]} |f(x)| = \sup_{y \in \mathbb R} \left\lvert\arctan y\right\rvert = \tfrac\pi2$ yet the supremum is never attained.
