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 $V_{x}=f([a,x])=\{f(t):a\leq t\leq x\}$ to be the image of $f$ up to and including $x$.

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

Our goal now is to find some $x$ such that $f(x)=M$. If $f(a)=M$ this is easy.

So 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 want to find $M'<M$ such that $f(t)<M'$ for all $t\in [a,c]$. That would mean that $\sup V_{c}\leq M'<M$. To do this, we split the interval into two parts. By continuity of $f$ at $c$, we can choose $\epsilon >0$ with $\epsilon <M-f(c)$. So now pick a point like $c-\delta /2$, and split the interval into $[a,c-\delta /2]$ and $[c-\delta /2,c]$.

• Consider $t=c-\delta /2$. Then since $t<c$, there exists $x\in X$ such that $t<x$. So $\sup V_{t}\leq \sup V_{c}<M$ so $f(t)\leq \sup V_{t}<M$. This also means that for all $x\in [a,c-\delta /2]$ we have $f(x)\leq \sup V_{t}<M$.
• But now if $c-\delta /2\leq t\leq c$, then By continuity at $c$, there exists a $\delta >0$ such that $|t-c|<\delta$ implies $|f(t)-f(c)|<\epsilon$. This means $f(t)<f(c)+\epsilon <M$.

Now we can choose $M'=\max\{\sup V_{c-\delta /2},f(c)+\epsilon \}$.

If $c<b$ then

If $t\leq c-\delta /2$ 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 $c-\delta /2\leq t\leq c$ then $|t-c|<\delta$ so $f(t)<f(c)+\epsilon <M$. 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.

Therefore, $c=b$, which implies that . So the assumption that $f(c)<M$ was false, and we conclude $f(c)=M$.

If $c=b$ then $M=\sup V_{b}=\sup V_{c}<M$, a contradiction.