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 $V_x=f([a,x])=\{f(t):a\le t\le x\}$ to be the values that $f$ takes on as the input ranges from $a$ to $x$ (inclusive).

Let $M=sup\{f(x):a\le x\le b\}=sup{V}_b$ (this number exists by the boundedness theorem) 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 immediate.

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 $c=supX$. 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 want to find $M^{\prime }<M$ such that $f(t)<M^{\prime }$ for all $t\in [a,c]$. That would mean that $sup{V}_c\le M^{\prime }<M$. To do this, we split the interval into two parts. Choose $ϵ>0$ with $ϵ<M-f(c)$. By continuity at $c$, there exists a $\delta >0$ such that $|t-c|<\delta$ implies $|f(t)-f(c)|<ϵ$. 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 $sup{V}_{c-\delta /2}\le sup{V}_x<M$. This means that $f(t)\le sup{V}_{c-\delta /2}<M$ for all $t\in [a,c-\delta /2]$.
• But now if $t\in [c-\delta /2,c]$, then $|t-c|<\delta$, so $|f(t)-f(c)|<ϵ$. This means $f(t)<f(c)+ϵ<M$.

Now we can choose $M^{\prime }=max\{sup{V}_{c-\delta /2},f(c)+ϵ\}$. Then whatever $t\in [a,c]$ happens to be, we can say $f(t)\le M^{\prime }$.

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 $f(c)=M$.