## Musings of a single variable, pt. 1

In this post I’m going to explore several theorems that are generally agreed to be the most important elements of single-variable differential calculus. These are the Mean Value Theorem (and its generalized counterpart, employed in the proof of L’Hopital’s Rule), the Intermediate Value Theorem, the Extreme Value Theorem, and lastly, L’Hopital’s Rule itself. I consider the latter a significant result simply due to its practical utility in the evaluation of limits.

One who is not familiar with the reasoning behind the Monotone Convergence Theorem and the so-called “Peak Point Lemma” certainly cannot lay claim to understanding the theorem of Bolzano-Weierstrass. In the same way, the understanding of these major single-variable results is crucial to maintain a firm footing in the multivariable setting that seems to be so full of “reductio ad single-variable” proof techniques.

The first thing I’d like to touch on is the Intermediate Value Theorem. I’ve seen this proved using the nested interval theorem and a binary-search-esque method, but the proof I will give is slightly different. The basic idea of the proof involves considering the set of points in the domain where the function is negative, invoking the completeness axiom to obtain its least upper bound, and arguing that this least upper bound (by the continuity of $f$) is indeed the intermediate point we are looking for, that is, the point where the function is zero.

Intermediate Value Theorem. Let $f : [a,b] \to \mathbb{R}$ be a continuous function with $f(a) < 0$ and $f(b) > 0$. Then there exists $c \in (a,b)$ such that $f(c) = 0$.
Proof. Consider the set $S = \{ x \in [a,b] \mid f(x) < 0 \}$. Certainly $S \subseteq [a,b]$ and $a \in S$, so as a nonempty, bounded set, $S$ has a least upper bound $c$ by the completeness axiom. Certainly $a \leq c$ since $c$ is an upper bound for $S$ and $a \in S$. Also $c \leq b$ because $b$ is also an upper bound for $S$, and $c$ is the least upper bound for $S$. So $c \in [a,b]$. Also, $c - \frac{1}{n} < c$ for any $n \in \mathbb{N}$ so it is not an upper bound for $S$. Hence there exists $x_n \in S$ with $c - \frac{1}{n} < x_n \leq c$. Certainly $f(x_n) < 0$ and $|x_n - c| < 1/n \to 0$ as $n \to \infty$, so $(x_n) \to c$. However, $f$ is continuous at $c$, and so $(f(x_n)) \to f(c)$. Since $f(x_n) < 0$, we have that $f(c) \leq 0$. This shows that $c \neq b$, since $f(b) > 0$, and thus $c + \frac{1}{N} < b$ for large enough $N$. Then for all $n \geq N$ we obtain $c + \frac{1}{n} \in [a,b]$. Also, $c + \frac{1}{n} > c$ and hence $c + \frac{1}{n} \notin S$. Therefore, $f(c + \frac{1}{n}) \geq 0$, which is true for all $n \geq N$. Again we note that $(c+\frac{1}{n}) \to c$, so by continuity of $f$ at $c$ we obtain $(f(c+\frac{1}{n})) \to f(c)$. Hence $f(c) \geq 0$. Combining this with our previous inequality we obtain $f(c) = 0$ as required.

Note that by defining a simple auxiliary function, this is easily generalized:

Corollary. Let $f : [a,b] \to \mathbb{R}$ be a continuous function with $f(a) < z$ and $f(b) > z$. Then there exists $c \in [a,b]$ such that $f(c) = z$.

Something to think about: how can we define a continuous function on a closed subset of an ordered field which does not possess the least-upper-bound property (for example $\mathbb{Q}$) such that the conclusion of the Intermediate Value Theorem fails to hold?

Before attacking the likes of the Extreme Value Theorem, we have to prove that the image of any function on a closed interval is bounded.

Boundedness theorem. Let $f : [a, b] \to \mathbb{R}$ be a continuous function. Then $f$ is a bounded function on $[a,b]$, that is, there exists $m, M \in \mathbb{R}$ such that $m \leq f(x) \leq M$ for all $x \in [a,b]$.
﻿Proof. We will show $f$ is bounded above. Assume this is not the case. Then for every natural $n$ we obtain some $x_n \in [a,b]$ such that $f(x_n) > n$. Since the sequence $(x_n)$ is bounded (as a subset of a closed interval), the Bolzano-Weierstrass theorem allows us to select a subsequence $(x_{k(p)})$ converging to some $x_0 \in [a,b]$. However, $f$ is continuous and so the sequence $(f(x_{k(p)}))$ converges to $f(x_0) \in \mathbb{R}$. However by construction $f(x_{k(p)}) > k(p) \geq p$ for all $p \in \mathbb{N}$ and hence the sequence $(f(x_{k(p)})) \to \infty$, a contradiction. So $f$ is bounded above. By the same argument $-f$ is bounded above, and hence $f$ is bounded both above and below, as stated.

Extreme Value Theorem. Let $f : [a,b] \to \mathbb{R}$ be continuous. Then $f$ attains its maximum and minimum value on $[a,b]$. That is, there exist $a_1, a_2 \in [a,b]$ with $f(a_1) \leq f(x) \leq f(a_2)$ for all $x \in [a,b]$.
Proof
. From the boundedness theorem, the image set $f([a,b])$ is bounded. It is also (trivially) nonempty, so by the completeness axiom it has a supremum and infimum, denote these $\alpha$ and $\beta$ respectively. Since $\alpha$ is the supremum, we can proceed as follows. For each natural $n \in \mathbb{N}$ there is some $x_n \in [a,b]$ such that $\alpha - \frac{1}{n} < f(x_n)$, for otherwise $\alpha$ would not be the supremum. Now, we have a sequence $(x_n)_{n=1}^\infty$, with $(f(x_n))_{n=1}^\infty$ converging to $\alpha$. Invoke the Bolzano-Weierstrass theorem, which gives a subsequence $(x_{k(p)})_{k=1}^\infty$ converging to some $c \in [a,b]$ (the limit lies within the interval, because $[a,b]$ is a closed interval). But now we note $(f(x_{k(p)}))_{k=1}^\infty$ converges to $f(c) \in [a,b]$ by the continuity of $f$, however as a subsequence, we see that $(f(x_{k(p)}))_{k=1}^\infty$ must also be converging to $\alpha$. Hence by the uniqueness of the limit, $f(c) = \alpha$. This demonstrates the existence of a point where $f$ attains its maximum value. The proof concerning $\beta$ is done similarly.
Alternate proof
. A theorem from elementary topology tells us that the continuous image of a compact set is again compact. This makes our life a lot easier, because if the image set is compact, it is (by the Heine-Borel theorem) closed and bounded. Every compact subset of $\mathbb{R}$ contains its supremum and infimum, so the proof is complete.

To be continued…