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 values that $f$ takes on as the input ranges from $a$ to $x$ (inclusive).

Let $M=\sup\{f(x):a\leq x\leq b\}=\sup V_{b}$ (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 $f(x)=M$. The idea now is to locate the leftmost point where $f$ attains $M$ by taking the supremum of $X$. But we have a small problem, which is that $X$ might be empty (it is however always bounded, so we don't need to worry about that part). This can happen if $f(a)=M$, in which case $\sup V_{a}=M$. But if that's the case, we have already found a point where $f$ equals $M$, so we're actually done!

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 for sake of contradiction that $f(c)<M$. Choose $\epsilon >0$ with $\epsilon <M-f(c)$. Continuity of $f$ at $c$ implies that there exists $\delta >0$ such that $x\in (c-\delta ,c+\delta )\cap [a,b]$ implies $|f(x)-f(c)|<\epsilon$. Now, $c-\delta$ cannot be an upper bound of $X$, so there exists some $x_{0}\in X$ such that $c-\delta <x_{0}\leq c$. Thus for $x\in [a,x_{0}]$ we have $f(x)\leq \sup V_{x_{0}}<M$. Also, for $x\in [x_{0},c]\subseteq (c-\delta ,c+\delta )\cap [a,b]$ we have $f(x)<f(c)+\epsilon <M$. So $f$ is bounded above by $\max\{\sup V_{x_{0}},f(c)+\epsilon \}<M$ on $[a,c]$, which means $\sup V_{c}<M$. If $c<b$, then there are points to the right of $c$ where $f$ continues to stay below $f(c)+\epsilon <M$, which contradicts the fact that $c$ is an upper bound of $X$. So $c=b$, which means $M=\sup V_{b}=\sup V_{c}<M$, a contradiction. The assumption that $f(c)<M$ was false, and we conclude $f(c)=M$.

Takeaways

• "less than" vs "bounded away from". If we have a sequence like $0.1,0.01,0.001,\ldots$, then we can say this is greater than 0. but we cannot say it is "bounded away from zero", because it gets closer and closer to it in the limit. similarly, in the proof above, we wanted the values of f to be less than M even after taking the supremum of values of f, which meant that we had to bound f from above, not by M, but by something less than M (say m). That way, we could say that $\sup\{{\text{some set}}\}\leq m<M$, so that even after taking sup, we are strictly less than M.
• there are "easier" ways to prove this theorem, for instance by using the machinery of sequences. but the nice thing about this proof is that it uses only the least upper bound property of real numbers. so you get to see how facts about continuity are derived "by hand".