User:IssaRice/Extreme value theorem

Working through the proof in Pugh's book by filling in the parts he doesn't talk about.

Let $M = sup {f (x) : x \in [a, b]}$ and $X = {x \in [a, b] : sup V_{x} < M}$ .

Now suppose $f (a) < M$ . Then $a \in X$ . We already know that $X$ is bounded above, for instance by the number $b$ . We can thus take the least upper bound of $X$ , say $c = sup X$ . We already know $f (c) \leq M$ , so if we can just eliminate the possibility that $f (c) < M$ , we will be done.

So suppose $f (c) < M$ . We can choose $ϵ > 0$ with $ϵ < M - f (c)$ . (Note: it is important here that $ϵ$ does not equal $M - f (c)$ ; choosing this $ϵ$ would be too weak and we would not be able to conclude $sup V_{c} < M$ , rather only that $sup V_{c} \leq M$ .) By continuity at $c$ , there exists a $δ > 0$ such that $| t - c | < δ$ implies $| f (t) - f (c) | < ϵ$ . This means $f (t) < f (c) + ϵ < M$ . If $t \leq c - δ / 2$ then there exists some $x \in X$ such that $t < x$ . This means $sup V_{t} \leq sup V_{x} < M$ so $t \in X$ . Otherwise if $c - δ / 2 \leq t \leq c$ then $| t - c | < δ$ so $f (t) < f (c) + ϵ < M$ . So now what can we say about $sup V_{c}$ ? We want to say $sup V_{c} < M$ . We can do this by showing that there exists a number $M^{'} < M$ such that $f (t) < M^{'}$ for all $t \in [a, c]$ . That way, $sup V_{c} \leq M^{'} < M$ . But $M^{'} = sup V_{c}$ works.

$sup V_{c - δ / 2} < M$ so let $M^{'} = (M + sup V_{c - δ / 2}) / 2$ . Then $sup V_{c - δ / 2} < M^{'} < M$ and if $v \in V_{c - δ / 2}$ we have $v \leq sup V_{c - δ / 2} < M^{'}$ .

Therefore, $c = b$ , which implies that $M = sup V_{b} = sup V_{c} < M$ , a contradiction.