I am reading Pugh's Real Mathematical Analysis, and in chapter 1, section 6, ``The Skeleton of Calculus,'' Pugh supplies a proof of the Extreme Value Theorem. I am having trouble understanding one particular point in the proof. Note that he proves the existence of maximums, and leaves minimums for the reader.

Theorem 23. A continuous function $f$ defined on an interval $[a, b]$ takes on absolute minimum and absolute maximum values: for some $x_0, x_1 \in [a,b]$ and for all $x \in [a,b]$,

$f(x_0) \leq f(x) \leq f(x_1)$.

Proof. Let $M = \sup f(t)$ as $t$ varies in $[a,b]$. This exists since the values of a continuous function defined on an interval $[a,b]$ form a bounded subset of $\mathbb R$ (his Theorem 22). Consider the set $X = \{x \in [a,b] : \sup V_x < M\}$ where $V_x$ is the set of values of $f(t)$ as $t$ varies on $[a,x]$.

Case 1: $f(a) = M$. Then $f$ takes on a maximum at $a$ and the theorem is proved.

Case 2: $f(a) < M$. Then $X$ is nonempty and we can consider the supremum of $X$, say $c$. Here's where I lose him---I don't seem to understand what's going on here. If $f(c) < M$, we choose $\epsilon > 0$ with $\epsilon < M - f(c)$. By continuity, there exists a $\delta > 0$ such that $|t - c| < \delta \implies |f(t) - f(c)| < \epsilon$. I understand the use of continuity and the application of the definition of a continuous function, but I don't understand the motivation for considering the continuity of $f$ at $c$. Thus, $\sup V_c < M$. I don't understand how we can deduce that $\sup V_c < M$ by considering the continuity of $f$ at $c$.

If you could help me understand this proof, and specifically how considering the continuity of $f$ at $c$ helps us determine that $\sup V_c < M$, I would greatly appreciate it.

  • $\begingroup$ You know nothing else about $f$, other than it is continuous. On the other hand, continuity is necessary for proving the existence of a maximum, because it's easy to find an example of a function not continuous at a single point of its domain that lacks a maximum: just consider $f(x)=-x^2$ for $x\in[-1,1]$, but $x\ne0$, and $f(0)=-1$. $\endgroup$ – egreg Jun 28 '16 at 0:05

The idea is simple. If $f(c) < M$ then by continuity we have an interval around $c$ where the values of $f$ are less than $M$. In fact a stronger statement is true and that is what we need. We can find an interval of $[c - h, c + h]$ such that all values of $f$ are less than a specific number $M'$ which is itself less than $M$.

Next it should be easy to see that $c - h < c$ and $c - h \in X$ (why??). and therefore supremum of $f$ on $[a, c - h]$ is less than $M$, say equal to $M''$ and $M'' < M$. Now the values of $f$ in $[c - h, c + h]$ are less than $M'$ so it follows that supremum of $f$ on $[a, c + h]$ does not exceed $\max(M', M'')$. And hence the supremum of $f$ on $[a, c + h]$ is also less than $M$ and $c + h \in X$. This is not possible as $c = \sup X$.

A much better proof is to use theorem $22$ of your book. Assume $f(x) \neq M$ for all $x \in [a, b]$. Then $g(x) = 1/(M - f(x))$ is continuous on $[a, b]$ and hence by theorem $22$ it is bounded on $[a, b]$. Now $M = \sup f(x)$ and hence given any $\epsilon > 0$ there is an $x$ such that $M - f(x) < \epsilon$ so that $g(x) > 1/\epsilon$. This shows that $g(x)$ is unbounded. This contradiction shows that we must have $f(x) = M$ for some $x \in [a, b]$.

BTW the proof in your question uses "supremum principle" (every non-empty set with upper bound has a supremum). You should try to devise other proofs based on principles like Nested Interval Principle and Heine Borel Theorem.

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My guess: the idea is to prove that $f(c)=M$. He proves it by contradiction. Assume that $f(c)<M$ then by continuity we can go a bit further to $c+\delta$ and still have all functional values being under $M$. It contradicts the choice of $c$ as the supremum of $X$.

P.S. It is easy, but you have to mention also that $c\in[a,b]$.

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  • $\begingroup$ Right. How exactly does your answer address the question as to how we deduce that $\sup V_c < M$? Reading what you've written here, I see how we contradicted that $c$ was $\sup X$, but my principle question remains. Thanks. $\endgroup$ – Alex Ortiz Jun 27 '16 at 23:08
  • $\begingroup$ Sorry, I did not understand what the principle question was. If $f(c)<M$ then by continuity the functional values in $[c-\delta,c]$ is under $M$ with some margin. However, $c-\delta\in X$ (by definition of supremum), so the functional values in $[a,c-\delta]$ is under $M$ with some margin too. Those margins ensures that $\sup V_c<M$. $\endgroup$ – A.Γ. Jun 27 '16 at 23:22
  • $\begingroup$ I think I see! If we can show that $c \in X$, it will follow that $\sup V_c < M$. Thus, we want to show that $c \in X$, so what we do is observe that by continuity, we can find some open interval $I = (c - \delta,c+\delta)$ such that $t \in I \implies ft < M$ as well. Then, since $c = \sup X$, $c - \delta \in X$. That means that $V_c = V_{[a,c]} = V_{[a,c - \delta]} \cup V_{(c-\delta,c]}$. Further, $\sup V_{[a,c-\delta]} < M$ since $c - \delta \in X$. Also, since $t \in I \implies ft \leq \sup V_I < M$. Is that right? A bit long-winded, but I wanted to make it explicit. $\endgroup$ – Alex Ortiz Jun 28 '16 at 0:10
  • $\begingroup$ That's right. The only trouble is that $f(t)<M$ gives only $\sup f(t)\le M$ (with equality). To ensure that $\sup f(t)<M$ (strictly) you need to say $f(t)<M-\epsilon$. That's why "some margin" $\epsilon$ comes to play in the proof. $\endgroup$ – A.Γ. Jun 28 '16 at 0:22

As usual from what I have seen of Pugh's book, his proof is not very clear.

I assume Case 1 is understood, so assume Case 2 holds.

Note that if $x \in X$, then every number $y \in [a,x]$ is also in $X$ (in other words, $x \in X \iff [a,x] \subset X$). This is because the supremum cannot increase if we compute it over a smaller set: $\sup\{f(t) : t \in [a,y]\} \leq \sup\{f(t) : t \in [a,x]\} < M$. We will use this handy fact below.

Recall that $c$ has been defined as $\sup X$, which exists because $X$ is nonempty in Case 2, and because $X$ is bounded above by $b$.

Choose an arbitrary $\gamma \in (0,c-a]$. By definition of the supremum of $X$, there is an element $x \in X$ with $x > c - \gamma$. By the handy fact, this means that $[a,c - \gamma]\subset [a,x] \subset X$. Since this is true for arbitrarily small positive $\gamma$, it follows that $[a,c) \subset X$.

The goal is to show that $f(c) = M$. Note that since $f(x) < M$ for all $x \in (a,c)$, and $f$ is continuous, the only way that this can be false is if $f(c) < M$.

So, suppose for a contradiction that $f(c) < M$. Then there is some $\epsilon > 0$ such that $f(c) < f(c) + \epsilon < M$. So, $f(c) < M - \epsilon$.

Now, since $f$ is continuous, it can't instantly "jump" above $M$ as soon as $x > c$. In other words, we will also have $f(x) < M$ in some neighborhood around $c$.

Expressing this more formally, there is some $\delta > 0$ such that $f(t) < M - \epsilon/2 < M$ whenever $t \in (c-\delta, c + \delta)$. Therefore, the supremum of $f$ on $(c-\delta , c + \delta)$ does not exceed $M - \epsilon/2$. But the supremum of $f$ on $[a,c-\delta]$ is also strictly smaller than $M$. Therefore, the supremum of $f$ on the entire interval $[a,c+\delta) = [a, c-\delta] \cup (c-\delta, c+\delta)$ is strictly smaller than $M$. Consequently, by the handy fact above, every number less than $c + \delta$ is in $X$. But this contradicts $\sup X = c$.

So, our assumption that $f(c) < M$ is false, and consequently we must have $f(c)= M$.

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  • $\begingroup$ Can you elaborate a little more on how $\forall \epsilon \in (0,c-a]$, $[a, c - \epsilon] \subset [a, x] \subset X$ implies $[a,c) \subset X$? I am struggling to see how you get the semi-open interval. $\endgroup$ – Alex Ortiz Jun 28 '16 at 1:37
  • 1
    $\begingroup$ @AOrtiz I'm just using the fact that if $[a,c-\epsilon] \subset X$ for all sufficiently small values of $\epsilon$, then $[a,c) \subset X$. Proof: suppose that $y \in [a,c)$. Then $a \leq y < c$, so there is some $\epsilon > 0$ such that $a \leq y \leq c - \epsilon < c$, so $y \in [a, c-\epsilon]$, hence $y \in X$. $\endgroup$ – Bungo Jun 28 '16 at 1:44
  • $\begingroup$ Thank you. That made it very clear. Minor nitpick: if possible, would you rephrase your answer in the language used in my question? I notice you have used $\epsilon$ more than a couple times in different contexts. Thanks again. $\endgroup$ – Alex Ortiz Jun 28 '16 at 1:56
  • $\begingroup$ OK, I'm now using $\gamma$, $\epsilon$, and $\delta$ without reusing any of them, and I am using $\epsilon$ and $\delta$ consistently with Pugh. $\endgroup$ – Bungo Jun 28 '16 at 2:25
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    $\begingroup$ @AOrtiz: This is a subtle but important point. Note that if $f(t) < M$ whenever $t \in (c-\delta, c+\delta)$, this does not guarantee that $\sup \{f(t) : t \in (c - \delta, c+\delta)\} < M$, only that $\sup \{f(t) : t \in (c - \delta, c+\delta)\} \leq M$. We need a little extra margin in order to ensure that the supremum is strictly less than $M$. See the definition of $X$ to recall why we need the supremum to be strictly less than $M$. $\endgroup$ – Bungo Jun 28 '16 at 3:00

There are two things going on: compactness/Bolzano-Weierstrass, and continuity.

If $X=[a,b]$ is closed and bounded, it is compact by the Heine-Borel theorem.

Take $M = \sup_{x \in X=[a,b]}f(x)$. Now consider any sequence $x_n$ in $X=[a,b]$ satisfying $$ M- \dfrac{1}{n} \le f(x_n) \le M. $$ (Note that because $M$ is the supremum of $f(x)$ over $X$, there cannot be any $f(x_n)>M$.)

Now because $X=[a,b]$ is compact, the Bolzano-Weierstrass theorem implies it has a convergent subsequence $x_{n_k} \rightarrow x^*$. Consider $$ M - \dfrac{1}{n_k} \le f(x_{n_k}) \le M. $$ If we take the limit, $$ \lim_{n_k \rightarrow \infty} M - \dfrac{1}{n_k} \le \lim_{n_k \rightarrow \infty} f(x_{n_k}) \le M, $$ by the sequential criterion for continuity $\lim_{x_n \rightarrow x^*}f(x_n)=f(x)$, so that $$ M \le f(\lim_{n_k \rightarrow \infty} x_{n_k}) \le M, $$ or $$ M \le f(x^*) \le M, $$ This implies $x^* \in X = [a,b]$ is a maximizer of $f$, since it achieves the supremum.

Note that none of this really has to do with $X=[a,b]$: the real ingredients of the proof are the definition of the sup, compactness, and the sequential criterion for continuity.

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  • $\begingroup$ While I like this approach in general, if I recall correctly, at this point in the book, we only have the least upper bound property of $\mathbb R$, so the game is to prove the claim with that alone. $\endgroup$ – Alex Ortiz Apr 2 at 21:33

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