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. Choose $\epsilon >0$ with $\epsilon <M-f(c)$. By continuity at $c$, there exists a $\delta >0$ such that $|t-c|<\delta$ implies $|f(t)-f(c)|<\epsilon$. So now pick a point like $c-\delta /2$, and split the interval into $[a,c-\delta /2]$ and $[c-\delta /2,c]$.

• Since $c-\delta /2<c$, there exists $x\in X$ such that $c-\delta /2<x$ (otherwise $c-\delta /2$ would be a smaller upper bound for $X$). So $\sup V_{c-\delta /2}\leq \sup V_{x}<M$. This means that $f(t)\leq \sup V_{c-\delta /2}<M$ for all $t\in [a,c-\delta /2]$.
• But now if $t\in [c-\delta /2,c]$, then $|t-c|<\delta$, so $|f(t)-f(c)|<\epsilon$. This means $f(t)<f(c)+\epsilon <M$.

Now we can choose $M'=\max\{\sup V_{c-\delta /2},f(c)+\epsilon \}$. Then whatever $t\in [a,c]$ happens to be, we can say $f(t)\leq M'$.

If $c<b$ then by continuity we can find points $t$ to the right of $c$ where $\sup V_{t}<M$, which contradicts the fact that $c$ is an upper bound of such points.

Therefore, $c=b$, which implies that $M=\sup V_{b}=\sup V_{c}<M$, a contradiction. So the assumption that $f(c)<M$ was false, and we conclude $f(c)=M$.