If the derivative is positive, then the function is increasing

We all know Rolle’s theorem. It says that if $f: [a,b] \to \mathbb R$ is continuous on $[a,b]$ , differentiable on $(a,b)$ and satisfies $f(a) = f(b)$ , then there must be a $c \in (a,b)$ such that $f'(c) = 0$ . Few of us, though, know the proof of Rolle’s theorem. Unfortunately, it is given without proof in high school, and it is usually not proven in university courses.

In this post, I intend to give a proof of Rolle’s theorem as well as the mean value theorem, and we will see how this will allow us to prove the famous fact that $f$ is increasing if and only if $f'(x) \ge 0$ .

Theorem. Every real sequence has a monotone subsequence.
Proof. Let $(x_n)$ be a sequence of real numbers. Define: $A = \{n \in \mathbb N: \forall m\ge n, x_m \ge x_n\}$ .
We have two possibilities: either $A$ is finite, or $A$ is infinite. If $A$ is finite, then it is either empty or non-empty. We will show that in all the three cases, we can construct a monotone subsequence of $(x_n)$ .

Case 1. $A = \varnothing$ . Then, by the definition of $A$ , for each $n \in \mathbb N$ , there exists an $m \ge n$ such that $x_m < x_n$ . If $m=n$ , then it’s not true that $x_m < x_n$ , so we can say that there exists an $m > n$ such that $x_m < x_n$ . For $n=1$ , we have an $n_1 > 1$ such that $x_{n_1} < x_1$ . For $n=n_1$ , we have an $n_2 > n_1$ such that $x_{n_2} < x_{n_1}$ . Proceeding inductively, for $n=n_{k}$ , we have an $n_{k+1} > n_k$ such that $x_{n_{k+1}} < x_{n_k}$ . We have therefore constructed a decreasing subsequence $(x_{n_k})$ of $(x_n)$ .

Case 2. $A$ is non-empty but finite, say $A = \{m_1, \ldots, m_t\}$ where $m_1 < \cdots < m_t$ . Let $N = m_t + 1$ . Then, for all $n \ge N$ , we have $n \notin A$ , so $\exists m > n$ such that $x_m < x_n$ . We then proceed as in case 1 to obtain a decreasing subsequence $(x_{n_k})$ of $(x_n)$ .

Case 3. $A$ is infinite.
By the well-ordering theorem, $A$ has a minimal element, say $n_1$ . Also, by the well-ordering theorem, $A - \{n_1\}$ has a minimal element, say $n_2$ . We have that $n_1 < n_2$ . Again, by the well-ordering theorem, $A - \{n_1, n_2\}$ has a minimal element, say $n_3$ , and we have $n_1 < n_2 < n_3$ . Proceeding as such, we may construct a strictly increasing sequence $(n_i) \subset A$ ; in other words, $A \supset \{n_1, n_2, n_3, \ldots\}$ with $n_1 < n_2 < n_3 < \cdots$
For a given $i$ , we have $n_i \in A$ , so for all $m\ge n_i$ , $x_m \ge x_{n_i}$ . In particular, the subsequence $(x_{n_k})$ of $(x_n)$ is increasing. QED.

Proposition. If $(x_n)$ is increasing (resp. decreasing) and bounded above (resp. below), then $(x_n)$ is convergent.
Proof. We will just show that if $(x_n)$ is increasing and bounded above then it’s convergent. Suppose that $(x_n)$ is increasing and that it is bounded above by $A$ . Let $\ell = \sup\{x_n: n\in \mathbb N\}$ . Then, $\ell \le A < \infty$ . Let $\epsilon > 0$ . Then, by definition, there is an $n_0$ such that $\ell - x_{n_0} < \epsilon$ . As $(x_n)$ is increasing, we get $\ell - x_n \le \ell - x_{n_0} < \epsilon$ for all $n\ge n_0$ . Therefore, $(x_n)$ is convergent to $\ell$ . QED.

Corollary. (Bolzano-Weiertrass) Every bounded real sequence has a convergent subsequence.
Proof. Let $(x_n) \subset \mathbb R$ be a bounded sequence. Then, $(x_n)$ has a monotone subsequence, and this subsequence is bounded and is therefore convergent. QED.

Proposition. Let $f: I \subset \mathbb R \to \mathbb R$ be a function defined on an open interval $I$ around a given point $x$ , and suppose that $f$ is continuous at $x$ . Let $(x_n)$ be a any sequence converging to $x$ , and suppose WLOG that $(x_n) \subset I$ . Then, the sequence $(f(x_n))$ converges to $f(x)$ .
Proof. Let $\epsilon > 0$ . As $f$ is continuous at $x$ , there exists a $\delta > 0$ such that for all $y \in I$ , $|y - x| < \delta \implies |f(y) - f(x)| < \epsilon$ . As $x_n \to x$ , there exists an $n_0$ such that for all $n \ge n_0$ , $|x_n - x| < \delta.$ Hence for all $n\ge n_0$ , we have $|f(x_n) - f(x)| < \epsilon$ . QED.

Theorem. (Extreme Value Theorem) Let $f: [a,b] \subset \mathbb R \to \mathbb R$ be a continuous function. Then, $f$ attains a maximum and a minimum, i.e. there are points $c,c' \in [a,b]$ such that $f(c') \le f(x) \le f(c)$ for all $x \in [a,b]$ .
Proof. We will show that $f$ attains a maximum. That $f$ attains a minimum follows from $g = -f$ having a maximum. Let $S = f([a,b]) = \{f(x): x\in [a,b]\}$ . We have that $S$ is bounded above. Indeed, suppose that it is not bounded above. Then, there is a sequence $(y_n)$ in $S$ such that $\lim_{n \to \infty} y_n = \infty$ . Now for each $n$ , $y_n = f(x_n)$ for some $x_n \in [a,b]$ . The sequence $(x_n)$ is contained in $[a,b]$ , so it is bounded, and Bolzano-Weierstrass implies that it has a convergent subsequence $(x_{n_k})$ . We have $x_{n_k} \to x$ for some $x\in [a,b]$ . Hence, as $f$ is continuous at $x$ , we get $f(x_{n_k}) \to f(x)$ . Now for each $k$ , $y_{n_k} = f(x_{n_k})$ . This implies that $(y_{n_k})$ converges to $f(x)$ , contradicting the fact that $y_n \to \infty$ .
Let $z = \sup S$ and let us show that $z \in S$ . As $z= \sup S$ , we can find a sequence $(z_n) \subset S$ converging to $z$ . Now for each $n$ , $z_n = f(t_n)$ for some $t_n \in [a,b]$ . The sequence $(t_n)$ is bounded, and so it has a convergent subsequence $(t_{n_k})$ converging to some $t \in [a,b]$ . As $f$ is continuous at $t$ , we have that $f(t_{n_k}) \to f(t)$ . But, $f(t_{n_k}) = z_{n_k}$ , so $z_{n_k} \to f(t)$ . As $(z_{n_k})$ is a subsequence of $(z_n)$ , we have that $z_{n_k} \to z$ . By uniqueness of limit, we have $z = f(t)$ . Therefore, $z \in S$ .
Write $z = f(c), c \in [a,b]$ . Let $x \in [a,b]$ . We have that $f(x) \in S$ , and as $z = \sup S$ , we get $f(x) \le z = f(c)$ . This shows that $f$ attains a maximum. QED.

Proposition. Let $f: I \subset \mathbb R \to \mathbb R$ , where $I$ is an open interval. Let $c$ be an interior point of $I$ and suppose that $f$ has a local extremum at $c$ . If $f$ is differentiable at $c$ , then $f'(c) = 0$ .
Proof. WLOG suppose that $c$ is a local maximum. We have:

$f'(c) = \lim_{x \to c^{-}} \frac{\overbrace{f(x) - f(c)}^{\le 0}}{\underbrace{x-c}_{\le 0}} \ge 0$

and

$f'(c) = \lim_{x \to c^{+}} \frac{\overbrace{f(x) - f(c)}^{\le 0}}{\underbrace{x-c}_{\ge 0}} \le 0$

Hence, $f'(c) = 0$ . QED.

Theorem. (Rolle’s Theorem) Let $f: [a,b] \to \mathbb R$ be a continuous function. Suppose that $f(a) = f(b)$ and that $f$ is differentiable on $(a,b)$ . Then, there exists a $c \in (a,b)$ such that $f'(c) =0$ .
Proof. We know that $f$ attains a maximum and a minimum on $[a,b]$ , say at $c$ and $c'$ , respectively. If $c,c' \in \{a,b\}$ , then $f$ is constant on $[a,b]$ and thus $f'(x) = 0$ for all $x \in [a,b]$ and there’s nothing to prove. Hence suppose that one of $c$ and $c'$ is in $(a,b)$ , say $c$ . By the above proposition, $f'(c) = 0$ . QED.

Theorem. (Mean Value Theorem) Let $f: [a,b] \to \mathbb R$ be a continuous function. Suppose that $f$ is differentiable on $(a,b)$ . Then, there exists a $c \in (a,b)$ such that

$f'(c) = \frac{f(a) - f(b)}{a-b}$

Proof. Let

$g(x) = \frac{f(b) - f(a)}{b-a} x - f(x)$

$g$ satisfies the hypothesis of Rolle’s theorem, so there exists a $c \in (a,b)$ such that $g'(c) = 0$ . But

$g'(c) = 0 \iff f'(c) = \frac{f(a) - f(b)}{a-b}$

so we are done. QED.

Corollary. Let $f: I \subset \mathbb R \to \mathbb R$ , where $I$ is an open interval, and suppose that $f$ is differentiable. We have that $f$ is increasing if and only if $f'(x) \ge 0$ .
Proof. Suppose that $f$ is increasing. We have for all $x \in I$ :

$f'(x) = \lim_{y \to x^{+}} \frac{\overbrace{f(y) - f(x)}^{\ge 0}}{\underbrace{y - x}_{\ge 0}} \ge 0$

Conversely, suppose that $f'(x) \ge 0$ for all $x \in I$ . Let $x,y \in I$ such that $x<y$ . $f$ is continuous on $[x,y]$ and differentiable on $(x,y)$ , so by the mean value theorem, there exists a $c \in (x,y)$ such that

$f'(c) = \frac{f(y) - f(x)}{y-x}$