User:IssaRice/Extreme value theorem

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

For $x\in [a,b]$, define $V_x=f([a,x])=\{f(t):t\in [a,x]\}$ to be the image of $f$ up to and including $x$.

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 $c=supX$. We already know $f(c)\le M$, 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 $ϵ<M-f(c)$. By continuity at $c$, there exists a $\delta >0$ such that $|t-c|<\delta$ implies $|f(t)-f(c)|<ϵ$. This means $f(t)<f(c)+ϵ<M$. If $t\le c-\delta /2$ then there exists some $x\in X$ such that $t<x$. This means $sup{V}_t\le sup{V}_x<M$ so $t\in X$. Otherwise if $c-\delta /2\le t\le c$ then $|t-c|<\delta$ 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 $M^{\prime }<M$ such that $f(t)<M^{\prime }$ for all $t\in [a,c]$. That way, $sup{V}_c\le M^{\prime }<M$. But $M^{\prime }=sup{V}_c$ works.

$sup{V}_{c-\delta /2}<M$ so let $M^{\prime }=(M+sup{V}_{c-\delta /2})/2$. Then $sup{V}_{c-\delta /2}<M^{\prime }<M$ and if $v\in V_{c-\delta /2}$ we have $v\le sup{V}_{c-\delta /2}<M^{\prime }$.

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