8.2 Monotone Functions and Inverse Functions

Section 8.1 made real functions into objects with a domain and a graph, and combined them. This section asks how a function treats the order of $\mathbb{R}$ : does a larger input give a larger output? We give the property its definition, a method for proving it, and rules for building monotone functions from others. Then we prove the main result: a strictly monotone function is a bijection onto its image, and its inverse is monotone in the same direction. That gives the $n$ -th root as a function, for every real input when $n$ is odd.

8.2.1. Increasing and decreasing functions

Ask a reader whether $x \mapsto 1/x$ is decreasing, and the answer is usually yes. On $\mathbb{R} \setminus \{0\}$ it is not: $-1 < 1$ , but $1/(-1) = -1 < 1 = 1/1$ , so a larger input gave a larger output. The function decreases on $(-\infty, 0)$ and on $(0, \infty)$ , and the jump across $0$ spoils the union. So “decreasing” is never a property of a formula alone. It is a property of a function on a set, and the definition must name the set.

A small case first. On the set $\{1, 2, 3\}$ , the function $x \mapsto x^2$ takes the values $1, 4, 9$ . Each larger input gives a larger output, so the function keeps the order there. On the set $\{-2, 1\}$ it does not: $-2 < 1$ , but $(-2)^2 = 4 > 1 = 1^2$ .

A function is increasing on a set when moving right inside the set never moves its value down.

Definition 8.2.1 (Monotone Function)

Let $f \colon D \to \mathbb{R}$ be a real function (Definition 8.1.1) and let $A \subseteq D$ . The function $f$ is

increasing on $A$ if, for all $x, y \in A$ , $x < y$ implies $f(x) \le f(y)$ ;
strictly increasing on $A$ if, for all $x, y \in A$ , $x < y$ implies $f(x) < f(y)$ ;
decreasing on $A$ if, for all $x, y \in A$ , $x < y$ implies $f(x) \ge f(y)$ ;
strictly decreasing on $A$ if, for all $x, y \in A$ , $x < y$ implies $f(x) > f(y)$ .

It is monotone on $A$ if it is increasing on $A$ or decreasing on $A$ , and strictly monotone on $A$ if it is strictly increasing or strictly decreasing on $A$ . Without “on $A$ ”, the set is the whole domain $D$ . In symbols, “increasing on $A$ ” is $\forall x \in A \ \ \forall y \in A \ \ \bigl( x < y \to f(x) \le f(y) \bigr) .$

In words: we pick two points of $A$ , name the smaller one $x$ , and ask that the values come in the same order ( $\le$ or $<$ ) or in the reverse order ( $\ge$ or $>$ ). Since $x \le y$ means $x < y$ or $x = y$ , an increasing function also satisfies $x \le y \Rightarrow f(x) \le f(y)$ , and a decreasing one satisfies $x \le y \Rightarrow f(x) \ge f(y)$ . A strictly monotone function satisfies the same non-strict forms; the strict inequality fails at $x = y$ .

Negating quantifier by quantifier (Logic and Proof, Proposition 6.2.1), “ $f$ is not increasing on $A$ ” is $\exists x \in A \ \ \exists y \in A \ \ \bigl( x < y \wedge f(x) > f(y) \bigr) ,$ so a failure is shown by two witnesses, as $-1$ and $1$ were for $1/x$ above.

The identity $x \mapsto x$ on $\mathbb{R}$ is strictly increasing, and so is $x \mapsto x^3$ . Sets and Functions used the same word, and proved there that $x \mapsto x^3$ is strictly increasing (Sets and Functions, Example (Maps from calculus) (a)). It also proved that every strictly increasing map between subsets of $\mathbb{R}$ is injective (Sets and Functions, Proposition 2.2.3). We use both facts without proving them again.

Second readingThe two definitions agree; sets with one point

Sets and Functions called a map $f \colon A \to B$ between subsets of $\mathbb{R}$ strictly increasing when $x < x'$ implies $f(x) < f(x')$ . Our $f \colon D \to \mathbb{R}$ is strictly increasing on $A$ exactly when its restriction $f|_A \colon A \to \mathbb{R}$ (Sets and Functions, Definition 2.3.1) is strictly increasing in that sense: both say the same thing about the same pairs of points of $A$ . The codomain plays no part.

Sets with one point: on a set $A$ with at most one point there is no pair $x < y$ , so every function is strictly increasing and strictly decreasing on $A$ , vacuously. Monotonicity says something only about pairs.

Warning. Monotone on two sets does not give monotone on their union. The function $1/x$ is strictly decreasing on $(-\infty, 0)$ and on $(0, \infty)$ , yet not decreasing on $\mathbb{R} \setminus \{0\}$ , by the witnesses $-1 < 1$ found at the start of this section. A proof that splits $A$ into pieces must also compare a point of one piece with a point of the other.

Quick check. Is a constant function on $\mathbb{R}$ increasing? Is $1/x$ decreasing on $\mathbb{R} \setminus \{0\}$ ? Is the indicator $\mathbf{1}_{[0, \infty)}$ (Sets and Functions, Definition 2.3.2) increasing on $\mathbb{R}$ ? For each “yes”, say whether “strictly” holds too.

Solution. A constant is increasing, since $f(x) = f(y)$ satisfies $\le$ ; it is decreasing too, and strictly neither: $0 < 1$ and the values are equal. The function $1/x$ is not decreasing on $\mathbb{R} \setminus \{0\}$ : $-1 < 1$ and $1/(-1) < 1/1$ . The indicator is increasing: for $x < y$ , if its value at $x$ is $1$ , then $0 \le x < y$ and its value at $y$ is $1$ too. It is not strictly increasing: $-2 < -1$ with both values $0$ .

8.2.2. The method: factor out the difference of the inputs

To prove $f$ strictly increasing on $A$ , we fix $x < y$ in $A$ and must show $f(y) - f(x) > 0$ . The hypothesis is about the difference $y - x$ , which is positive. So the natural plan is to make $y - x$ appear as a factor of $f(y) - f(x)$ , and then to sign what is left.

Example (Method: factor out y − x)

How it goes. “Let $x, y \in A$ with $x < y$ .” Write $f(y) - f(x) = (y - x) \, Q(x, y)$ by algebra on the formula for $f$ . Since $y - x > 0$ , the sign of $f(y) - f(x)$ is the sign of $Q(x, y)$ : show $Q(x, y) > 0$ for strictly increasing, $Q(x, y) \ge 0$ for increasing, and $< 0$ or $\le 0$ for the decreasing kinds (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c), (d)). Differences of powers factor by Lemma 1.1.8: $y^n - x^n = (y - x) \sum_{k=0}^{n-1} y^k x^{n-1-k} .$ Quotients are brought over a common denominator first.

When to use it. $f$ is given by a formula built from $+$ , $\cdot$ and $/$ , and the claim is that $f$ is monotone on a set.

Common mistakes. One mistake is to conclude “strictly” from $Q \ge 0$ : a positive number times a non-negative one may be $0$ . Another is to use a sign of $Q$ that holds on part of $A$ only. In that case, split $A$ first, and also compare a point of one piece with a point of another.

We know	We want
$x, y \in A$ , $x < y$ ; the formula for $f$	$f(x) < f(y)$
$y - x > 0$	$f(y) - f(x) > 0$
$f(y) - f(x) = (y - x) \, Q(x, y)$	$Q(x, y) > 0$ : what is left to show

For a quotient, the obvious argument fails.

Example (A quotient of two increasing functions)

Let $g \colon (-2, \infty) \to \mathbb{R}$ , $g(x) = \dfrac{x - 1}{x + 2}$ . Prove that $g$ is strictly increasing.

Idea (Scratch work). The naive attempt. “The numerator $x - 1$ increases with $x$ , so the quotient does.” The denominator increases too, and a quotient of two increasing positive things can do anything: $x/x$ is constant. So the naive attempt proves nothing.

Factor out $y - x$ . For $-2 < x < y$ , bring the difference over the common denominator $(x + 2)(y + 2)$ , which is positive. The numerator is $\begin{aligned} &(y - 1)(x + 2) - (x - 1)(y + 2) \\ &\quad = (xy + 2y - x - 2) - (xy + 2x - y - 2) = 3(y - x) . \end{aligned}$ So $Q(x, y) = 3/\bigl((x + 2)(y + 2)\bigr)$ .

We know	We want
$-2 < x < y$	$g(y) - g(x) > 0$
$g(y) - g(x) = 3(y - x)/\bigl((x + 2)(y + 2)\bigr)$	$Q(x, y) = 3/\bigl((x + 2)(y + 2)\bigr) > 0$
$x + 2 > 0$ , $y + 2 > 0$	done: a quotient of positive elements is positive

Solution. Let $x, y \in (-2, \infty)$ with $x < y$ . Then $x + 2 > 0$ and $y + 2 > 0$ , so the denominators are non-zero. Over the common denominator, $g(y) - g(x) = \frac{(y - 1)(x + 2) - (x - 1)(y + 2)}{(x + 2)(y + 2)} = \frac{3(y - x)}{(x + 2)(y + 2)} .$ The numerator is positive since $y - x > 0$ . The denominator is positive, since a product of positive elements is positive, (O4). Its reciprocal is positive too. Hence $g(y) - g(x) > 0$ , and $g$ is strictly increasing.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Lemma 4.1.1 (common denominator); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 2.4.1 (positives closed under products); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (a) (positive reciprocal); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (a) (positive difference).

When the formula changes at $0$ , the domain splits there into two pieces, and a point of one piece must also be compared with a point of the other.

Example (The squashing function)

Let $\sigma \colon \mathbb{R} \to \mathbb{R}$ , $\sigma(x) = \dfrac{x}{1 + \left| x \right|}$ . Prove that $\sigma$ is strictly increasing on $\mathbb{R}$ , and find its image.

Idea. The absolute value has one formula for $x \ge 0$ and another for $x \le 0$ , so the factoring works on each half separately. Two points in the same half are handled by the method; for one point in each half, the signs of the values decide at once, since $\sigma(x)$ has the sign of $x$ .

Solution. Since $1 + \left| x \right| \ge 1 > 0$ , $\sigma$ is defined on $\mathbb{R}$ and $\sigma(x)$ has the sign of $x$ . Let $x < y$ .

Case $0 \le x < y$ . Here $\sigma(t) = t/(1 + t)$ at $t = x, y$ , and $\sigma(y) - \sigma(x) = \frac{y(1 + x) - x(1 + y)}{(1 + x)(1 + y)} = \frac{y - x}{(1 + x)(1 + y)} > 0 .$

Case $x < y \le 0$ . Here $\sigma(t) = t/(1 - t)$ , and $\sigma(y) - \sigma(x) = \frac{y(1 - x) - x(1 - y)}{(1 - x)(1 - y)} = \frac{y - x}{(1 - x)(1 - y)} > 0 ,$ since $1 - x \ge 1$ and $1 - y \ge 1$ .

Case $x < 0 \le y$ . Then $\sigma(x) < 0 \le \sigma(y)$ .

By trichotomy the three cases cover every pair $x < y$ , so $\sigma$ is strictly increasing.

Its image is $(-1, 1)$ . Every value satisfies $\left| \sigma(x) \right| = \left| x \right|/(1 + \left| x \right|) < 1$ . Every $v \in (-1, 1)$ is a value, at $x = v/(1 - \left| v \right|)$ , which is $v/(1 - v)$ for $v \ge 0$ and $v/(1 + v)$ for $v < 0$ . That is the surjectivity proved in Sets and Functions by solving backwards (Sets and Functions, Example (Solving backwards, proving forwards)).

8.2.3. Building monotone functions from others

The method handles one formula at a time. Most functions are built from simpler ones by the operations of Definition 8.1.2, and monotonicity passes through some of them.

A small case first. On $\mathbb{R}$ , $x \mapsto x$ and $x \mapsto x^3$ are strictly increasing. For $x < y$ we have $x < y$ and $x^3 < y^3$ , and adding the two inequalities gives $x + x^3 < y + y^3$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (b)). So $x \mapsto x + x^3$ is strictly increasing, with no factoring at all. Rule (a) below says this for any two functions.

Proposition 8.2.1 (Rules for Monotone Functions)

Let $f, g \colon D \to \mathbb{R}$ and $A \subseteq D$ .

(a) If $f$ and $g$ are increasing on $A$ , so is $f + g$ ; it is strictly increasing if $f$ or $g$ is.
(b) If $f$ is increasing on $A$ and $c \ge 0$ , then $cf$ is increasing on $A$ , strictly if $f$ is and $c > 0$ ; and $-f$ is decreasing on $A$ , strictly if $f$ is strictly increasing.
(c) If $f$ and $g$ are increasing on $A$ and $f(x) \ge 0$ , $g(x) \ge 0$ for every $x \in A$ , then $fg$ is increasing on $A$ .
(d) If $f$ is increasing on $A$ and $f(x) > 0$ for every $x \in A$ , then $1/f$ is decreasing on $A$ , strictly if $f$ is strictly increasing.
(e) Let $f \colon D \to \mathbb{R}$ , $g \colon E \to \mathbb{R}$ , $A \subseteq D$ and $B \subseteq E$ with $f(A) \subseteq B$ . If $f$ is monotone on $A$ and $g$ on $B$ , then $g \circ f$ is monotone on $A$ . It is increasing if both are increasing or both decreasing, and decreasing if one is increasing and the other decreasing. It is strictly so if both are strictly monotone.

Idea. Each part fixes $x < y$ in $A$ , writes down what the hypotheses say about the values at $x$ and $y$ , and applies one rule for inequalities in an ordered field to them.

Proof. In each part let $x, y \in A$ with $x < y$ .

(a) $f(x) \le f(y)$ and $g(x) \le g(y)$ . Adding the two inequalities gives $f(x) + g(x) \le f(y) + g(y)$ , strictly if one of the two is strict.
(b) If $c = 0$ , $cf$ is constant. Let $c > 0$ . If $f(x) < f(y)$ , multiplying by $c > 0$ keeps the inequality: $cf(x) < cf(y)$ . If $f(x) = f(y)$ , then $cf(x) = cf(y)$ . So $cf(x) \le cf(y)$ , strictly if $f(x) < f(y)$ . Finally, negating reverses an inequality: $f(x) \le f(y)$ gives $-f(y) \le -f(x)$ , strictly if $f(x) < f(y)$ .
(c) $0 \le f(x) \le f(y)$ and $0 \le g(x) \le g(y)$ . Inequalities between non-negative numbers may be multiplied, so $f(x)g(x) \le f(y)g(y)$ .
(d) $0 < f(x) \le f(y)$ . Taking reciprocals of positive numbers reverses the inequality, so $1/f(y) \le 1/f(x)$ . The same holds with $<$ : if $f(x) < f(y)$ , then $1/f(y) < 1/f(x)$ .
(e) The composite is defined on $A$ , since $f(A) \subseteq B \subseteq E$ (Definition 8.1.2). Let $f$ and $g$ be increasing. Then $f(x) \le f(y)$ with both in $B$ . Since $g$ is increasing on $B$ , it also satisfies $u \le v \Rightarrow g(u) \le g(v)$ for $u, v \in B$ , so $g(f(x)) \le g(f(y))$ . If both are strict, $f(x) < f(y)$ and then $g(f(x)) < g(f(y))$ . The other three cases are the same argument with the inequality reversed at each decreasing function: two reversals restore the order, one reverses it.

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (a)–(c) (add, negate, scale inequalities); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c), (d) (reciprocals, products of inequalities); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (b) (strict reciprocal inequality); Definition 8.2.1 (increasing preserves $\le$ ).

Warning. The sign condition in (c) cannot be dropped. The identity $x \mapsto x$ is increasing on $\mathbb{R}$ , but its product with itself, $x \mapsto x^2$ , is not: $-1 < 0$ and $(-1)^2 = 1 > 0 = 0^2$ .

Quick check. On $[0, \infty)$ , the functions $x \mapsto x$ and $x \mapsto x^2 + 1$ are increasing and non-negative. Is $x \mapsto x(x^2 + 1)$ increasing on $[0, \infty)$ , and by which rule?

Solution. Yes, by (c): both factors are increasing on $[0, \infty)$ , and both are $\ge 0$ there. The second factor is increasing because $x \mapsto x^2$ is, by Lemma 1.5.1 (a) with $n = 2$ . Adding the constant $1$ keeps it increasing, by rule (a) with a constant function.

8.2.4. Strict monotonicity reflects the order

A strictly increasing function carries $x < y$ to $f(x) < f(y)$ . For the inverse we need the way back: from $f(x) < f(y)$ to $x < y$ . Because $\le$ on $\mathbb{R}$ is a total order, the way back is free.

Proposition 8.2.2 (Strictly Monotone Functions Reflect the Order)

Let $f \colon D \to \mathbb{R}$ , $A \subseteq D$ , and $x, y \in A$ .

(a) If $f$ is strictly increasing on $A$ , then $x < y \iff f(x) < f(y)$ , and $x \le y \iff f(x) \le f(y)$ .
(b) If $f$ is strictly decreasing on $A$ , then $x < y \iff f(x) > f(y)$ , and $x \le y \iff f(x) \ge f(y)$ .

Idea. The directions $(\Rightarrow)$ are the definition. For the way back, suppose the conclusion fails; totality turns “not $x < y$ ” into $y \le x$ , and the forward direction then contradicts the hypothesis.

Proof.

(a) The direction $(\Rightarrow)$ of the first equivalence is Definition 8.2.1. For $(\Rightarrow)$ of the second: if $x \le y$ , then $x = y$ and $f(x) = f(y)$ , or $x < y$ and $f(x) < f(y)$ ; in both cases $f(x) \le f(y)$ .
$(\Leftarrow)$ of the first: let $f(x) < f(y)$ , and suppose $x < y$ fails. Since $\le$ is a total order with strict part $<$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (a)), $y \le x$ . By $(\Rightarrow)$ of the second equivalence, $f(y) \le f(x)$ . With $f(x) < f(y)$ this gives $f(x) = f(y)$ by antisymmetry, against $f(x) \ne f(y)$ . So $x < y$ .

$(\Leftarrow)$ of the second: let $f(x) \le f(y)$ , and suppose $x \le y$ fails. By totality, $y \le x$ ; and $y \ne x$ , since $y = x$ would give $x \le y$ . So $y < x$ , and then $f(y) < f(x)$ , against $f(x) \le f(y)$ . So $x \le y$ .
(b) If $u < v$ in $A$ , then $f(u) > f(v)$ , so $-f(u) < -f(v)$ , since negating reverses inequalities. Thus $-f$ is strictly increasing on $A$ . By the same rule, $f(x) > f(y) \iff -f(x) < -f(y)$ , and likewise for $\ge$ . Apply (a) to $-f$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (a) (negating reverses inequalities).

Taking $x \le y$ and $y \le x$ together, the second equivalence gives $f(x) = f(y) \Rightarrow x = y$ again: injectivity, which Sets and Functions proved directly (Sets and Functions, Proposition 2.2.3). What is new is the direction $(\Leftarrow)$ , and it rests on totality alone.

Quick check. Which half of Proposition 8.2.2 (a) uses that $\le$ is total, and how?

Solution. The directions $(\Leftarrow)$ . From “not $x < y$ ” totality gives $y \le x$ , hence $f(y) \le f(x)$ , against $f(x) < f(y)$ . In a partial order, “not $x < y$ ” would allow $x$ and $y$ to be incomparable, and the argument would stop.

The converse of Sets and Functions, Proposition 2.2.3 is false: an injective function need not be monotone.

Example (Injective but not monotone)

Let $f \colon \mathbb{R} \to \mathbb{R}$ swap $0$ and $1$ : $f(0) = 1$ , $f(1) = 0$ , and $f(x) = x$ for every other $x$ . Show that $f$ is injective but monotone on no set containing $0$ , $1$ and $2$ . Then do the same for $1/x$ on $\mathbb{R} \setminus \{0\}$ and the set $\{-1, 1, 2\}$ .

Solution. The swap. Checking the three cases of the definition, $f(f(x)) = x$ for every $x$ , so $f$ is its own two-sided inverse and is bijective (Sets and Functions, Theorem 2.4.4). It is not increasing: $0 < 1$ and $f(0) = 1 > 0 = f(1)$ . It is not decreasing: $1 < 2$ and $f(1) = 0 < 2 = f(2)$ .

The reciprocal. If $1/x = 1/y$ , then $x = y$ , taking reciprocals. It is not increasing: $1 < 2$ and $1 > 1/2$ . It is not decreasing: $-1 < 1$ and $1/(-1) = -1 < 1 = 1/1$ , the witnesses found at the start of this section.

8.2.5. The inverse of a strictly monotone function

By Sets and Functions, Theorem 2.4.4, a function has an inverse exactly when it is a bijection. A strictly increasing $f \colon A \to \mathbb{R}$ is injective, but usually not surjective onto $\mathbb{R}$ : $\sigma$ of Example (The squashing function) misses $1$ . Cutting the codomain down to the image repairs that, and the inverse then inherits the order.

A small case first. Let $f(x) = 2x$ on $A = \{1, 2, 3\}$ . Its image is $f(A) = \{2, 4, 6\}$ , and each value comes from exactly one input. So going back is a function from $\{2, 4, 6\}$ to $\{1, 2, 3\}$ : it sends $2 \mapsto 1$ , $4 \mapsto 2$ , $6 \mapsto 3$ . It is again strictly increasing. Its graph $\{(2, 1), (4, 2), (6, 3)\}$ is the graph $\{(1, 2), (2, 4), (3, 6)\}$ of $f$ with each pair swapped.

Theorem 8.2.3 (Inverse of a Strictly Monotone Function)

Let $A \subseteq \mathbb{R}$ and let $f \colon A \to \mathbb{R}$ be strictly increasing. Then:

(a) $f$ , with codomain $f(A)$ , is a bijection $A \to f(A)$ ; write $f^{-1} \colon f(A) \to A$ for its inverse;
(b) $f^{-1}$ is strictly increasing;
(c) $\operatorname{graph} f^{-1} = \{ (y, x) : (x, y) \in \operatorname{graph} f \}$ .

The same holds with “strictly decreasing” in place of “strictly increasing”, in the hypothesis and in (b).

Try it first (ten minutes): take $u < v$ in $f(A)$ . Why should $f^{-1}(u) < f^{-1}(v)$ ?

Idea. The direct attempt applies $f^{-1}$ to $u < v$ , but nothing is known about $f^{-1}$ yet; that is what we are proving. So name the two preimages, $x \mathrel{:=} f^{-1}(u)$ and $y \mathrel{:=} f^{-1}(v)$ . Now $f(x) = u < v = f(y)$ , and the question is about $f$ alone: does $f(x) < f(y)$ force $x < y$ ? That is the direction $(\Leftarrow)$ of Proposition 8.2.2. Part (a) needs no new idea: $f(A)$ is chosen so that every point is reached, and strict monotonicity gives injectivity. Part (c) tests membership of a pair, as in Example (Method: reading a graph as a set of pairs), with each pair $(x, y)$ swapped to $(y, x)$ .

Before reading on: where is the totality of $\le$ used?

Proof.

(a) For $x \in A$ , $f(x) \in f(A)$ by the definition of the image, so $f$ is a function $A \to f(A)$ . It is surjective, since every $v \in f(A)$ is $f(a)$ for some $a \in A$ . It is injective by Sets and Functions, Proposition 2.2.3. So it is a bijection, and by Sets and Functions, Theorem 2.4.4 it has a two-sided inverse $f^{-1} \colon f(A) \to A$ .
(b) Let $u, v \in f(A)$ with $u < v$ , and let $x \mathrel{:=} f^{-1}(u)$ and $y \mathrel{:=} f^{-1}(v)$ , points of $A$ . Since $f \circ f^{-1} = \operatorname{id}_{f(A)}$ , $f(x) = u < v = f(y)$ . By Proposition 8.2.2 (a), $x < y$ , that is, $f^{-1}(u) < f^{-1}(v)$ .
(c) Let $v, x \in \mathbb{R}$ . If $(v, x) \in \operatorname{graph} f^{-1}$ , then $v \in f(A)$ and $x = f^{-1}(v) \in A$ , and $f(x) = f(f^{-1}(v)) = v$ ; so $(x, v) \in \operatorname{graph} f$ . Conversely, if $(x, v) \in \operatorname{graph} f$ , then $x \in A$ and $v = f(x) \in f(A)$ , and $f^{-1}(v) = f^{-1}(f(x)) = x$ , since $f^{-1} \circ f = \operatorname{id}_A$ ; so $(v, x) \in \operatorname{graph} f^{-1}$ . This proves the theorem for strictly increasing $f$ .

Uses: Sets and Functions, Definition 2.5.1 (the image $f(A)$ ); Sets and Functions, Definition 2.4.2 (two-sided inverse).

Second readingThe decreasing case

Let $f$ be strictly decreasing. Part (a) is proved word for word as above, since Sets and Functions, Proposition 2.2.3 covers decreasing maps too. Part (c) used only $f \circ f^{-1} = \operatorname{id}_{f(A)}$ and $f^{-1} \circ f = \operatorname{id}_A$ . For (b), let $u < v$ in $f(A)$ and $x \mathrel{:=} f^{-1}(u)$ , $y \mathrel{:=} f^{-1}(v)$ . Then $f(y) = v > u = f(x)$ , and Proposition 8.2.2 (b), applied to the pair $y, x$ , gives $y < x$ , that is, $f^{-1}(v) < f^{-1}(u)$ . So $f^{-1}$ is strictly decreasing.

Where is totality used? In part (b), inside Proposition 8.2.2, where “not $x < y$ ” became $y \le x$ . The injectivity cited in (a) rests on it too: Sets and Functions, Proposition 2.2.3 is proved by trichotomy. Each hypothesis has a job. Without “strictly”, (a) fails: a constant on $[0, 1]$ is increasing and not injective. Without cutting the codomain to $f(A)$ , (a) fails as well: $\sigma \colon \mathbb{R} \to \mathbb{R}$ has no inverse, because nothing goes back from $1$ .

Part (c) says that the graph of $f^{-1}$ is the graph of $f$ reflected in the line $y = x$ , the inverse relation that Sets and Functions defined (Sets and Functions, Definition 1.4.3). To draw an inverse, draw $f$ and swap the axes.

Quick check. Find the image of $g$ of Example (A quotient of two increasing functions), and its inverse.

Solution. Writing $g(x) = 1 - 3/(x + 2)$ shows that $g$ takes exactly the values in $(-\infty, 1)$ : as $x$ runs over $(-2, \infty)$ , $x + 2$ runs over $(0, \infty)$ , and so does $3/(x + 2)$ . Solving $v = (x - 1)/(x + 2)$ gives $g^{-1}(v) = (1 + 2v)/(1 - v)$ on $(-\infty, 1)$ . Check: with $x = (1 + 2v)/(1 - v)$ , $x + 2 = 3/(1 - v) > 0$ and $x - 1 = 3v/(1 - v)$ , so $g(x) = v$ . By Theorem 8.2.3, $g^{-1}$ is strictly increasing.

Chapter 1 proved directly that roots keep order (Lemma 1.5.2 (d)). The theorem gives a second proof, with the root as an inverse function, and the new part is that roots extend to negative numbers when the exponent is odd.

Example (Roots as inverse functions)

Let $n \in \mathbb{N}$ with $n \ge 1$ , and let $p_n \colon \mathbb{R} \to \mathbb{R}$ , $p_n(x) = x^n$ . Show that the function $a \mapsto \sqrt[n]{a}$ on $[0, \infty)$ is strictly increasing, and that for odd $n$ every real number has exactly one real $n$ -th root.

Solution. On $[0, \infty)$ . By Lemma 1.5.1 (a), $p_n$ is strictly increasing on $[0, \infty)$ . Its image is $[0, \infty)$ : a power of a non-negative number is non-negative ( $0^n = 0$ , and $0 < t^n$ for $t > 0$ by Lemma 1.5.1 (a) with $s = 0$ ), and every $a \ge 0$ is $b^n$ for $b = \sqrt[n]{a} \ge 0$ (Lemma 1.5.1 (b)). By Theorem 8.2.3, applied to the restriction of $p_n$ to $[0, \infty)$ , the inverse $[0, \infty) \to [0, \infty)$ is strictly increasing, and it sends $a$ to the unique $b \ge 0$ with $b^n = a$ , which is $\sqrt[n]{a}$ .

Odd $n$ on $\mathbb{R}$ . Let $n = 2k + 1$ with $k \in \mathbb{N}$ . Then $(-x)^n = \bigl((-x)^2\bigr)^k (-x) = -x^n$ for every real $x$ , by the power laws and $(-x)^2 = x^2$ . Let $x < y$ . If $0 \le x$ , then $x^n < y^n$ by Lemma 1.5.1 (a). If $y \le 0$ , then $0 \le -y < -x$ , so $-y^n = (-y)^n < (-x)^n = -x^n$ . Negating reverses this inequality, so $x^n < y^n$ . If $x < 0 < y$ , then $x^n = -(-x)^n < 0 < y^n$ , since a power of a positive number is positive (Lemma 1.5.1 (a) with $s = 0$ ). So $p_n$ is strictly increasing on $\mathbb{R}$ . Its image is $\mathbb{R}$ : $a \ge 0$ is $(\sqrt[n]{a})^n$ , and $a < 0$ is $\bigl(-\sqrt[n]{-a}\bigr)^n = -(-a)$ . By Theorem 8.2.3, $p_n$ is a bijection $\mathbb{R} \to \mathbb{R}$ with a strictly increasing inverse. Its value at $a$ is the one real $s$ with $s^n = a$ ; we write it $\sqrt[n]{a}$ , for every real $a$ when $n$ is odd. By the formula for the image, $\sqrt[n]{-a} = -\sqrt[n]{a}$ for $a \ge 0$ , and so for every real $a$ .

Uses: Proposition 1.1.1 (a), (b) (power laws); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.3.2 (c) ( $(-x)^2 = x^2$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (a) (negating reverses inequalities); Lemma 1.5.1 (a) with $s = 0$ (powers of positive numbers are positive).

For even $n = 2k$ the second half fails: $(-1)^n = \bigl((-1)^2\bigr)^k = 1 = 1^n$ , so $p_n$ is not injective on $\mathbb{R}$ , and $x^n = (x^2)^k \ge 0$ for every $x$ , so $-1$ is not a value. So $\sqrt[n]{a}$ for $a < 0$ is defined only for odd $n$ .

Diagram. Labels: x, y, y = x, y = x^3, y = sqrt [3]x

The graphs of $x \mapsto x^3$ (solid) and of its inverse $x \mapsto \sqrt[3]{x}$ (dotted) on $\mathbb{R}$ , mirror images in the dashed line $y = x$ , as Theorem 8.2.3 (c) says.

A strictly increasing function on all of $\mathbb{R}$ need not have an interval as its image.

Example (A strictly increasing function with a jump)

Let $H \colon \mathbb{R} \to \mathbb{R}$ , $H(x) = x$ for $x < 0$ and $H(x) = x + 1$ for $x \ge 0$ . Show that $H$ is strictly increasing, find its image, and describe $H^{-1}$ .

Solution. Let $x < y$ . If $y < 0$ , then $H(x) = x < y = H(y)$ . If $0 \le x$ , then $H(x) = x + 1 < y + 1 = H(y)$ . If $x < 0 \le y$ , then $H(x) = x < 0 < 1 \le y + 1 = H(y)$ . So $H$ is strictly increasing.

The image is $(-\infty, 0) \cup [1, \infty)$ : values at $x < 0$ are the negative numbers, and values at $x \ge 0$ are the numbers $\ge 1$ . In particular $1/2$ is not a value, though it lies between the values $H(-1) = -1$ and $H(0) = 1$ ; the image is not an interval. By Theorem 8.2.3, $H^{-1} \colon (-\infty, 0) \cup [1, \infty) \to \mathbb{R}$ , $H^{-1}(v) = v$ for $v < 0$ and $H^{-1}(v) = v - 1$ for $v \ge 1$ , is strictly increasing.

The graph of $H$ : the line $y = x$ up to the open point $(0, 0)$ , then the line $y = x + 1$ from the closed point $(0, 1)$ . The values between $0$ and $1$ are skipped.

Remark (Looking ahead). Analysis on the Real Line, Chapter 7 proves that a continuous injective function on an interval is strictly monotone. Both words are needed: the swap of Example (Injective but not monotone) lives on an interval but jumps, and $1/x$ does not jump but lives on $\mathbb{R} \setminus \{0\}$ , which is not an interval. The function $H$ shows that a strictly monotone function may still jump.

8.2.6. Exercises

A. Check your understanding

Exercise (A1)

These parts are about the proof of Theorem 8.2.3.

(a) Run the proof for $f(x) = x^2$ on $A = [0, 2]$ : find $f(A)$ and $f^{-1}$ , and say what $x$ , $y$ are in part (b) of the theorem for $u = 1$ , $v = 4$ .
(b) Which cited result gives the injectivity in part (a) of the theorem?
(c) Give the idea of the proof of part (b) of the theorem in two sentences.

Solution.

(a) $f$ is strictly increasing on $[0, 2]$ by Lemma 1.5.1 (a). If $0 \le x \le 2$ , then $0 \le x^2 \le 4$ ; and every $v \in [0, 4]$ is $(\sqrt v)^2$ with $0 \le \sqrt v \le 2$ , since $\sqrt v > 2$ would give $v > 4$ . So $f(A) = [0, 4]$ , and $f^{-1}$ is $\sqrt{\phantom{a}}$ on $[0, 4]$ . For $u = 1 < 4 = v$ : $x = \sqrt 1 = 1$ , $y = \sqrt 4 = 2$ , and $f(1) < f(2)$ gives $1 < 2$ .
(b) Sets and Functions, Proposition 2.2.3.
(c) Name the preimages $x = f^{-1}(u)$ , $y = f^{-1}(v)$ , so that $u < v$ becomes $f(x) < f(y)$ , a statement about $f$ . Strict monotonicity reflects the order, by totality of $\le$ , so $x < y$ .

Exercise (A2)

Classify each function as strictly increasing, strictly decreasing, or neither, with a reason.

(a) $x^2$ on $[-1, 1]$ .
(b) $x^2$ on $[0, \infty)$ .
(c) $-x^3$ on $\mathbb{R}$ .
(d) $1/x$ on $(0, \infty)$ .

Solution.

(a) Neither: $-1 < 0$ but $f(-1) = 1 > 0 = f(0)$ , so not increasing; $0 < 1$ and $f(0) = 0 < 1 = f(1)$ , so not decreasing.
(b) Strictly increasing, by Lemma 1.5.1 (a) with $n = 2$ .
(c) Strictly decreasing. The function $x^3$ is strictly increasing on $\mathbb{R}$ , by Sets and Functions, Example (Maps from calculus) (a). Passing to $-f$ reverses it, by Proposition 8.2.1 (b).
(d) Strictly decreasing: for $0 < x < y$ , $1/y - 1/x = -(y - x)/(xy) < 0$ , the method with $Q(x, y) = -1/(xy) < 0$ .

B. Practice

Exercise (B1)

Prove that $f(x) = x + 1/x$ is strictly increasing on $[1, \infty)$ and strictly decreasing on $(0, 1]$ .

Solution. Let $0 < x < y$ . Over the common denominator $xy > 0$ , $f(y) - f(x) = (y - x) - \frac{y - x}{xy} = \frac{(y - x)(xy - 1)}{xy} .$ So $Q(x, y) = (xy - 1)/(xy)$ , and its sign is the sign of $xy - 1$ . If $1 \le x < y$ , then $xy > x \cdot 1 \ge 1$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c)), so $Q > 0$ and $f(x) < f(y)$ . If $0 < x < y \le 1$ , then $0 < xy < 1 \cdot y \le 1$ , so $Q < 0$ and $f(x) > f(y)$ .

Exercise (B2: Prove or disprove)

Prove each true statement and refute each false one with a counterexample.

(a) The composite of two strictly decreasing functions $\mathbb{R} \to \mathbb{R}$ is strictly increasing.
(b) If $f \colon D \to \mathbb{R}$ is strictly increasing and $f(x) > 0$ for all $x \in D$ , then $1/f$ is strictly decreasing.
(c) The sum of a strictly increasing and a strictly decreasing function on $\mathbb{R}$ is monotone.
(d) The function $x \mapsto x^3 - x$ is strictly increasing on $[1/2, \infty)$ .

Solution.

(a) True: Proposition 8.2.1 (e), both strictly decreasing.
(b) True: Proposition 8.2.1 (d).
(c) False. Take $x^3$ , strictly increasing (Sets and Functions, Example (Maps from calculus) (a)), and $-x$ , strictly decreasing. The sum $h(x) = x^3 - x$ has $h(-1) = 0$ , $h(-1/2) = 3/8$ , $h(0) = 0$ : it goes up and then down, so it is neither increasing ( $-1/2 < 0$ , $3/8 > 0$ ) nor decreasing ( $-1 < -1/2$ , $0 < 3/8$ ).
(d) False. Let $f(x) = x^3 - x$ . The method gives $f(y) - f(x) = (y - x)\bigl(x^2 + xy + y^2 - 1\bigr) ,$ and at $x = 1/2$ , $y = 3/5$ the bracket is $0.25 + 0.3 + 0.36 - 1 < 0$ . Indeed $f(1/2) = -3/8 = -375/1000$ and $f(3/5) = -48/125 = -384/1000$ , so $1/2 < 3/5$ and $f(1/2) > f(3/5)$ . Repair: on $[1, \infty)$ the statement is true, since $1 \le x < y$ gives $x^2 \ge 1$ , $xy > 1$ and $y^2 > 1$ , so the bracket is $> 3 - 1 > 0$ .

Exercise (B3)

Let $s \colon \mathbb{R} \to \mathbb{R}$ , $s(x) = x\left| x \right|$ . Prove that $s$ is strictly increasing, find its image, and write down $s^{-1}$ . Is $s^{-1}$ increasing?

Solution. Strictly increasing, split as in Example (The squashing function). Let $x < y$ . If $0 \le x < y$ , then $s(y) - s(x) = y^2 - x^2 > 0$ (Lemma 1.5.1 (a)). If $x < y \le 0$ , then $s(y) - s(x) = -y^2 + x^2 = (x - y)(x + y) > 0 ,$ as both factors are negative (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (d)). If $x < 0 \le y$ , then $s(x) = -x^2 < 0 \le y^2 = s(y)$ .

Image $\mathbb{R}$ : $v \ge 0$ is $s(\sqrt v)$ , and $v < 0$ is $s(-\sqrt{-v}) = -(-v)$ . So $s^{-1}(v) = \sqrt v$ for $v \ge 0$ and $s^{-1}(v) = -\sqrt{-v}$ for $v < 0$ : this is the inverse whose two composites were checked in Sets and Functions (Sets and Functions, Example (Inverses found and checked) (c)). It is strictly increasing by Theorem 8.2.3, with no further computation.

C. Going deeper

Exercise (C1: Find the error)

Here is an argument that $f(x) = x^3 - 3x$ is strictly increasing on $[1, \infty)$ .

(i) Let $1 \le x < y$ .
(ii) By the method of Example (Method: factor out y − x), $f(y) - f(x) = (y - x)(y^2 + xy + x^2 - 3)$ .
(iii) Since $x \ge 1$ and $y \ge 1$ , we get $y^2 + xy + x^2 \ge 3$ .
(iv) Also $y - x > 0$ , so $f(y) - f(x) > 0$ .

Is the conclusion true? Does the argument prove it? Name the line that fails, if any, and repair it.

Solution. The conclusion is true, but the argument does not prove it. Line (iv) fails: (iii) gives only that the bracket is $\ge 0$ , and a positive number times a non-negative one is only $\ge 0$ . A counterexample to the step: $1 \cdot 0 = 0$ . This is the first mistake Example (Method: factor out y − x) warns against: strictness concluded from $Q \ge 0$ .

Repair. Line (i) gives more than (iii) uses: $y > x \ge 1$ , so $y^2 > 1$ (Lemma 1.5.1 (a)), while $x^2 \ge 1$ and $xy \ge x \ge 1$ . Hence $y^2 + xy + x^2 > 3$ , the bracket is $> 0$ , and (iv) holds.

Exercise (C2, used later)

Let $f \colon \mathbb{R} \to \mathbb{R}$ be increasing, with $f(f(x)) = x$ for every $x \in \mathbb{R}$ . Prove that $f = \operatorname{id}_{\mathbb{R}}$ . Then show that “increasing” cannot be dropped.

Solution (Hint 1). Fix $x$ , and split by the sign of $f(x) - x$ .

Solution (Hint 2). Suppose $f(x) > x$ , and apply $f$ to both sides.

Solution. Let $x \in \mathbb{R}$ . By trichotomy, $f(x) > x$ , $f(x) < x$ or $f(x) = x$ .

If $x < f(x)$ , then, $f$ being increasing, $f(x) \le f(f(x)) = x$ , against $x < f(x)$ . If $f(x) < x$ , then $f(f(x)) \le f(x)$ , that is, $x \le f(x)$ , against $f(x) < x$ . So $f(x) = x$ for every $x$ , and $f = \operatorname{id}_{\mathbb{R}}$ by Sets and Functions, Proposition 2.1.1.

Without “increasing” the claim fails: $x \mapsto -x$ satisfies $-(-x) = x$ for every $x$ , and it is not the identity, since it sends $1$ to $-1$ .

What a complete answer needs. (10 points) - 2: fixes an arbitrary $x$ and compares $x$ with $f(x)$ (the three cases of trichotomy, or the two of totality). - 3: in one case ( $f(x) > x$ , or $x \le f(x)$ ), applies the monotonicity of $f$ to the inequality between $x$ and $f(x)$ and uses $f(f(x)) = x$ to reach a contradiction or, by antisymmetry, $f(x) = x$ . - 2: handles the other case, or says that it is symmetric with the reason. - 1: concludes $f = \operatorname{id}_{\mathbb{R}}$ from equality at every point. - 2: gives a function with $f \circ f = \operatorname{id}_{\mathbb{R}}$ , $f \ne \operatorname{id}_{\mathbb{R}}$ , and checks both properties.

Slips that lose the step. Applying $f$ to a strict inequality and keeping it strict, when $f$ is only increasing. Assuming $f$ is injective or continuous without proof.