# extreme value theorem

Let $a$ and $b$ be real numbers with $a, and let $f$ be a continuous   , real valued function on $[a,b]$. Then there exists $c,d\in[a,b]$ such that $f(c)\leq f(x)\leq f(d)$ for all $x\in[a,b]$.

Proof. We show only the existence of $d$. By the boundedness theorem $f([a,b])$ is bounded above; let $l$ be the least upper bound of $f([a,b])$. Suppose, for a contradiction   , that there is no $d\in[a,b]$ such that $f(d)=l$. Then the function

 $g(x)=\frac{1}{l-f(x)}$

is well defined and continuous on $[a,b]$. Since $l$ is the least upper bound of $f([a,b])$, for any positive real number $M$ we can find $\alpha\in[a,b]$ such that $f(\alpha)>l-\frac{1}{M}$, then

 $M<\frac{1}{l-f(\alpha)}\textrm{.}$

So $g$ is unbounded  on $[a,b]$. But by the boundedness theorem $g$ is bounded   on $[a,b]$. This contradiction finishes the proof.

Title extreme value theorem ExtremeValueTheorem 2013-03-22 14:29:21 2013-03-22 14:29:21 classicleft (5752) classicleft (5752) 7 classicleft (5752) Theorem msc 26A06 Weierstrass extreme value theorem