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# extreme value theorem

Extreme Value Theorem. Let $a$ and $b$ be real numbers with $a<b$, 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.

## Mathematics Subject Classification

26A06*no label found*

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