The Extreme Value Theorem: a bootstrap approach

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At the very end of last week I was teaching the first-year mathematics students about the Extreme Value Theorem.

Here is a reminder of what the theorem says.

Theorem (Extreme Value Theorem)
Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exist points $c$ and $d$ in $[a,b]$ such that, for all $x \in [a,b]$, we have $f(c)\leq f(x) \leq f(d)$.

Here we have

\[f(c)=\min\{f(x):x\in[a,b]\}=\inf\{f(x):x\in[a,b]\}\]

and 

\[f(d)=\max\{f(x):x\in[a,b]\}=\sup\{f(x):x\in[a,b]\}\,.\]

Unfortunately I only had a few minutes left for this, ran out of time, and didn't finish the proof. As I am on a very tight schedule this year, I had to refer the students to the the typed notes for the rest of the proof. On the other hand, I did finish the first part of the proof, showing that a continuous real-valued function $f$ on a closed and bounded interval $[a,b]$ is always bounded above, i.e., the set $f([a,b])$ is bounded above.

In this post I want to see how you can use just this fact to prove the rest of the Extreme Value Theorem without having to repeat aspects of the proof.

I will take the following lemma as my starting point.

Lemma
Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exists $M>0$ such that, for all $x \in [a,b]$, we have $f(x)\leq M$. $\quad\square$

Our first step is to see that we can use this lemma to show that $f$ is bounded below.

Corollary
Let $a,b$ be real numbers with $a<b$, and let $f:[a,b]\to\R$ be continuous. Then there exists $M>0$ such that, for all $x \in [a,b]$, we have $f(x)\geq {-M}$.

We can prove this Corollary by applying the Lemma to either of the functions $-f$ or $|f|$, where the latter function is the function $x \mapsto |f(x)|$. $\quad\square$

Proof of the Extreme Value Theorem

We already know that the set $f([a,b])=\{f(x):x\in[a,b]\}$ is bounded (i.e., both bounded below and bounded above). We now prove the existence of a suitable point $d$. (The proof for $c$ is the same, but flipped as usual.)

Set $S=\sup f([a,b]) =\sup \{f(x):x\in[a,b]\} \in \R\,.$ We need to show that there exists a point $d \in[a,b]$ such that $f(d)=S$. Assume, towards a contradiction, that there is no such point $d$. In that case, since $S$ is an upper bound for $f([a,b])$, we must actually have $f(x)<S$ for all $x \in [a,b]$. 

Consider the function $g:[a,b]\to \R$ defined by $g(x)=1/(S-f(x))$. Then, since $f(x)<S$ for all $x\in[a,b]$, $g$ is a well-defined, continuous, real-valued function on $[a,b]$, and moreover $g(x)>0$ for all $x\in[a,b]$.

By our Lemma, $g$ is bounded above, i.e., there exists $M>0$ such that, for all $x \in [a,b]$, we have $g(x)\leq M$.

Since these real numbers are strictly positive, we can take reciprocals and obtain, for all $x \in [a,b]$, that $1/g(x) \geq 1/M$, i.e., $S-f(x) \geq 1/M$. Note that $1/M>0$. Rearranging we have, for all $x \in [a,b]$, that

\[f(x) \leq S - \frac1M\,.\]

But then $S-1/M$ is an upper bound for $f([a,b])$, and $S-1/M < S$, which contradicts the fact that $S$ is the least upper bound for $f([a,b])$. (That is the definition of supremum!)

We have obtained a contradiction, and so there must be a point $d\in[a,b]$ with $f(d)=S$, as required.

The proof for $c$ is similar, or you can apply the result about maximum to the function $-f$ to find a point $c$ where $-f$ attains its maximum value, and then $f$ attains its minimum value there. $\quad\square$

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