8.3 Bounded Functions and Their Suprema

Number Systems proved that $\mathbb{R}$ is a complete ordered field (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2). By the definition of completeness (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Definition 5.5.1), every non-empty set of real numbers that is bounded above has a supremum. A supremum is a least upper bound, as in Sets and Functions (Sets and Functions, Definition 5.3.5). A real function $f \colon D \to \mathbb{R}$ hands us such sets, one for each $A \subseteq D$ : the set $f(A)$ of its values on $A$ . This section asks when $f(A)$ is bounded, names its supremum and infimum, and fixes one method for computing them: bound every value, then find a witness. It then gives rules for comparing suprema without computing them, measures the spread of values, and shows that non-constant polynomials are unbounded on $\mathbb{R}$ .

8.3.1. A least upper bound that is never reached

Let $f(x) = x/(1 + x)$ on $[0, \infty)$ . A few values: $f(0) = 0 , \qquad f(1) = \frac12 , \qquad f(9) = \frac{9}{10} , \qquad f(99) = \frac{99}{100} .$ Every value is below $1$ , since $x < 1 + x$ and $1 + x > 0$ . The values creep toward $1$ and never reach it.

“Its maximum is $1$ ” is false: a maximum is a value, and $1$ is not a value of $f$ . “Its least upper bound is $1$ ” is true, and proving it takes two halves. First, $1$ bounds every value. Second, no number below $1$ bounds every value. The first half is one line. The second half is the real work, and the epsilon test below turns it into a search for a single point.

8.3.2. The infimum and the epsilon test

Suprema come with a mirror image. Sets and Functions defined the infimum of a set as its greatest lower bound (Sets and Functions, Definition 5.3.5), but completeness speaks only of suprema. Reflection in $0$ turns one into the other. For $A \subseteq \mathbb{R}$ , write $-A \mathrel{:=} \{ -a : a \in A \}$ , the image of $A$ under $x \mapsto -x$ (Sets and Functions, Example (Reflections, scalings and sums as images)).

A small case first. Let $A = \{ 1, 2, 3 \}$ , so $-A = \{ -1, -2, -3 \}$ . The lower bound $1$ of $A$ becomes the upper bound $-1$ of $-A$ . The supremum of $-A$ is $-1$ , and $\inf A = 1 = -(-1)$ .

Number Systems set part (b) of the next proposition as an exercise (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Exercise (B1)). We prove it here so that later sections can cite it.

Proposition 8.3.1 (Infimum by Reflection)

Let $A \subseteq \mathbb{R}$ .

(a) A real number $l$ is a lower bound of $A$ if and only if $-l$ is an upper bound of $-A$ .
(b) If $A$ is non-empty and bounded below, then $A$ has an infimum, and $\inf A = -\sup(-A)$ .

Proof.

(a) For $a \in A$ , $l \le a$ holds if and only if $-a \le -l$ , by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (a) together with the case of equality. The elements of $-A$ are exactly the numbers $-a$ with $a \in A$ , so $l \le a$ for every $a \in A$ if and only if $b \le -l$ for every $b \in -A$ .
(b) Let $l$ be a lower bound of $A$ . By (a), $-l$ is an upper bound of $-A$ , and $-A$ is non-empty because $A$ is. Since $\mathbb{R}$ is complete (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 5.5.2), $s \mathrel{:=} \sup(-A)$ exists. By (a), applied to $-s$ , the number $-s$ is a lower bound of $A$ , since $-(-s) = s$ is an upper bound of $-A$ . Let $l'$ be any lower bound of $A$ . Then $-l'$ is an upper bound of $-A$ by (a). A supremum is at most every upper bound, (S2), so $s \le -l'$ , that is, $l' \le -s$ . So $-s$ is the greatest lower bound of $A$ .

Uses: Sets and Functions, Definition 5.3.5 (S2: least upper bound).

Every statement about suprema thus has a twin about infima, obtained by reflection.

The second half of “least upper bound” says that nothing smaller is an upper bound. In $\mathbb{R}$ “smaller” can be measured by a positive gap $\varepsilon$ , and “not an upper bound” means that some element sticks out above. That gives a test with an explicit witness.

Here is the test on a small case. Take $A = [0, 1)$ and the upper bound $s = 1$ . For the gap $\varepsilon = 1/10$ , the element $0.95$ of $A$ lies above $1 - 1/10 = 0.9$ . For $\varepsilon = 1/100$ , the element $0.995$ lies above $0.99$ . Every gap, however small, has such an element.

Proposition 8.3.2 (The Epsilon Test for a Supremum)

Let $A \subseteq \mathbb{R}$ be non-empty and let $s$ be an upper bound of $A$ . Then $s = \sup A$ if and only if, for every $\varepsilon > 0$ , there exists $a \in A$ with $a > s - \varepsilon$ . In symbols: $\forall \varepsilon > 0 \ \ \exists a \in A \ \ a > s - \varepsilon .$ Likewise, if $i$ is a lower bound of $A$ , then $i = \inf A$ if and only if for every $\varepsilon > 0$ there exists $a \in A$ with $a < i + \varepsilon$ .

Proof. $(\Rightarrow)$ Let $s = \sup A$ and $\varepsilon > 0$ . Then $s - \varepsilon < s$ . If $s - \varepsilon$ were an upper bound of $A$ , the least-upper-bound property (S2) would give $s \le s - \varepsilon$ , which is false. So some $a \in A$ fails $a \le s - \varepsilon$ ; by totality of the order (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (i)), $a > s - \varepsilon$ .

$(\Leftarrow)$ Assume the condition, and let $u$ be an upper bound of $A$ . Suppose $u < s$ , and put $\varepsilon \mathrel{:=} s - u > 0$ . The condition gives $a \in A$ with $a > s - \varepsilon = u$ , against $u$ being an upper bound. So $s \le u$ by totality. Hence $s$ is an upper bound, (S1), and is at most every upper bound, (S2): it is the supremum.

For the infimum, apply the supremum statement to $-A$ and the upper bound $-i$ (Proposition 8.3.1 (a)). By the proof of Proposition 8.3.1 (b), $i = \inf A$ if and only if $-i = \sup(-A)$ ; and $-a > -i - \varepsilon$ if and only if $a < i + \varepsilon$ .

Uses: Sets and Functions, Definition 5.3.5 (S1, S2).

The element $a$ is a witness for $\varepsilon$ . It may depend on $\varepsilon$ , and it usually does: a smaller $\varepsilon$ asks for an element closer to the top.

Quick check. Is $0.99$ the supremum of $[0, 1)$ ? Use the test.

Solution. No, and the test does not even apply: it is stated for an upper bound $s$ , and $0.99$ is not one, since $0.995 \in [0, 1)$ and $0.995 > 0.99$ . The supremum is $1$ : it bounds $[0, 1)$ , and for $\varepsilon > 0$ the element $\max(0, 1 - \varepsilon/2)$ of $[0, 1)$ exceeds $1 - \varepsilon$ .

8.3.3. Bounded functions and their suprema

A real function (Definition 8.1.1) on $D$ has a set of values on each $A \subseteq D$ , and the words of the previous subsection apply to that set.

A small case first. Let $f(x) = x^2$ and $A = [-1, 2]$ . For $x \in A$ , $\left| x \right| \le 2$ , so $f(x) \le 4$ ; and $f(x) \ge 0$ . One number, $4$ , lies above every value on $A$ , and $0$ lies below every value.

A function is bounded on $A$ when its values on $A$ stay inside a fixed window. That is, some fixed interval contains every value on $A$ .

Definition 8.3.1 (Bounded Function)

Let $f \colon D \to \mathbb{R}$ and $A \subseteq D$ . The function $f$ is bounded above on $A$ if the set $f(A) = \{ f(x) : x \in A \}$ (Sets and Functions, Definition 2.5.1) is bounded above: there exists $M \in \mathbb{R}$ with $f(x) \le M$ for every $x \in A$ . It is bounded below on $A$ if $f(A)$ is bounded below, and bounded on $A$ if there exists $M \in \mathbb{R}$ with $\left| f(x) \right| \le M$ for every $x \in A$ . When $A = D$ , “on $D$ ” may be dropped.

In words: one number $M$ , chosen before $x$ , works for all $x \in A$ . By Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (d), $\left| f(x) \right| \le M$ says $-M \le f(x) \le M$ , so a bounded function is bounded above and below. Conversely, if $L \le f(x) \le U$ on $A$ , then $M \mathrel{:=} \max(\left| L \right|, \left| U \right|)$ gives $f(x) \le \left| U \right| \le M$ and $-f(x) \le -L \le \left| L \right| \le M$ , so $\left| f(x) \right| \le M$ . The two meanings of “bounded” agree.

Examples. The function $x/(1 + x)$ from the start of the section is bounded on $[0, \infty)$ , with $0 \le f(x) < 1$ . A constant function $c$ is bounded on every $A$ , by $\left| c \right|$ . The function $1/x$ on $(0, 1]$ is bounded below by $0$ but not above: given $M$ , the point $x \mathrel{:=} 1/(\left| M \right| + 1)$ lies in $(0, 1]$ and $1/x = \left| M \right| + 1 > M$ . Change only the set to $[1, 2]$ and the same formula is bounded, with $1/2 \le 1/x \le 1$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (c)). Boundedness belongs to the pair $(f, A)$ , not to the formula.

Before the next definition, one more small case. On $A = \{ 0, 1, 2 \}$ the function $x^2$ takes the values $0, 1, 4$ . The least upper bound of these values is $4$ , and it is a value, at $x = 2$ . The greatest lower bound is $0$ , at $x = 0$ .

Definition 8.3.2 (Supremum and Infimum of a Function)

Let $f \colon D \to \mathbb{R}$ and let $A \subseteq D$ be non-empty. If $f$ is bounded above on $A$ , the supremum of $f$ on $A$ is $\sup_A f \mathrel{:=} \sup f(A) ,$ also written $\sup_{x \in A} f(x)$ . If $f$ is bounded below on $A$ , its infimum on $A$ is $\inf_A f \mathrel{:=} \inf f(A)$ . The supremum is attained if $\sup_A f \in f(A)$ , that is, $\sup_A f = f(x_0)$ for some $x_0 \in A$ ; it is then the maximum of $f$ on $A$ , written $\max_A f$ . Likewise an attained infimum is the minimum $\min_A f$ .

Both exist under the stated hypotheses: $f(A)$ is non-empty because $A$ is, so completeness gives $\sup f(A)$ , and Proposition 8.3.1 (b) gives $\inf f(A)$ . The set $A$ must be non-empty, since every real number bounds the empty set and no least one exists. An attained supremum $f(x_0)$ is an upper bound of $f(A)$ that lies in $f(A)$ , so it is the largest value. Conversely, a largest value $f(x_0)$ is an upper bound, and every upper bound is at least $f(x_0)$ . So it is the supremum.

Warning. The supremum need not be a value. A proof that starts “let $x_0$ be the point where $f$ reaches $\sup_A f$ ” has assumed that the supremum is attained. For $x/(1 + x)$ on $[0, \infty)$ no such point exists. The supremum also depends on $A$ : $\sup_{[0, 1]} x^2 = 1$ and $\sup_{[0, 2]} x^2 = 4$ .

The two halves we used for $x/(1 + x)$ at the start of the section now become a general method.

Example (Method: bound every value, then find a witness)

How it goes. To prove $\sup_A f = M$ for a non-empty $A$ and a candidate $M \in \mathbb{R}$ :

$(\le)$ Bound every value. Let $x \in A$ be arbitrary and prove $f(x) \le M$ .
$(\ge)$ Find a witness. Let $\varepsilon > 0$ be arbitrary and produce $x \in A$ , depending on $\varepsilon$ , with $f(x) > M - \varepsilon$ .

The first half shows that $f$ is bounded above on $A$ and that $M$ is an upper bound of $f(A)$ ; then $M = \sup_A f$ by Proposition 8.3.2. For an infimum, reverse the inequalities: $f(x) \ge m$ for every $x \in A$ , and for each $\varepsilon > 0$ some $x$ with $f(x) < m + \varepsilon$ .

We know	We want
$f$ on $A \ne \emptyset$ ; a candidate $M$	$(\le)$ $\forall x \in A \ \ f(x) \le M$ ; $(\ge)$ $\forall \varepsilon > 0 \ \exists x \in A \ \ f(x) > M - \varepsilon$

When to use it. Whenever a supremum or infimum is to be computed. An inequality $\sup_A f \le M$ needs only the first half, since the supremum is the least upper bound; an equality needs both.

Common mistakes. Proving only $(\le)$ shows that $M$ is an upper bound, not that it is the least one. A witness for one $\varepsilon$ , such as $1/10$ , is not enough: the second half needs a witness for every $\varepsilon$ . A witness for $f$ must not be paired with a different witness for $g$ , as if $f(x) + g(y)$ were a value of $f + g$ . Exercise B4 shows where that mistake leads.

8.3.4. Two worked examples

The first example returns to $x/(1 + x)$ . Its witness has to move with $\varepsilon$ .

Example (A supremum that is not attained)

Let $f(x) = x/(1 + x)$ on $A = [0, \infty)$ . Prove that $\sup_A f = 1$ and that it is not attained. Find $\inf_A f$ .

Idea (Scratch work). The bounding half: $x/(1 + x) \le 1$ means $x \le 1 + x$ , after multiplying by $1 + x > 0$ , and that holds, even strictly.

The witness half is the gap. Given $\varepsilon > 0$ , we want $x \ge 0$ with $\frac{x}{1 + x} > 1 - \varepsilon .$ For $\varepsilon \ge 1$ the right side is at most $0$ and any $x > 0$ works. For $0 < \varepsilon < 1$ , work backwards: multiply by $1 + x$ , collect $x$ : $x > (1 - \varepsilon)(1 + x) \iff \varepsilon x > 1 - \varepsilon \iff x > \frac{1}{\varepsilon} - 1 .$ So $x \mathrel{:=} 1/\varepsilon$ is a clean choice, a little larger than needed. Does it also cover $\varepsilon \ge 1$ ? At $x = 1/\varepsilon$ the value is $1/(1 + \varepsilon)$ , and $1/(1 + \varepsilon) > 1 - \varepsilon$ means $1 > (1 - \varepsilon)(1 + \varepsilon) = 1 - \varepsilon^2$ , true for every $\varepsilon > 0$ . One choice, no cases; the restriction $\varepsilon < 1$ was needed only while solving backwards.

We know	We want
$x \in [0, \infty)$	$x/(1 + x) \le 1$
$\varepsilon > 0$	some $x \ge 0$ with $x/(1 + x) > 1 - \varepsilon$
$x \mathrel{:=} 1/\varepsilon$ , value $1/(1 + \varepsilon)$	$1/(1 + \varepsilon) > 1 - \varepsilon$ , i.e. $1 > 1 - \varepsilon^2$ : the gap is closed

The proof runs forwards: bound, then choose $x$ from $\varepsilon$ , then check.

Solution. Bound every value. Let $x \ge 0$ . Then $1 + x > 0$ and $x < 1 + x$ ; multiplying by $1/(1 + x) > 0$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 3.2.2 (a), Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c)) gives $f(x) < 1$ .

Find a witness. Let $\varepsilon > 0$ and put $x \mathrel{:=} 1/\varepsilon$ , which lies in $A$ . Then $f(x) = \frac{1/\varepsilon}{1 + 1/\varepsilon} = \frac{1}{1 + \varepsilon} .$ Since $(1 - \varepsilon)(1 + \varepsilon) = 1 - \varepsilon^2 < 1$ and $1 + \varepsilon > 0$ , dividing by $1 + \varepsilon$ gives $1 - \varepsilon < 1/(1 + \varepsilon) = f(x)$ .

By the method of Example (Method: bound every value, then find a witness), $\sup_A f = 1$ . It is not attained, because $f(x) < 1$ for every $x \in A$ .

For the infimum, $f(x) \ge 0$ for $x \ge 0$ , as a quotient of $x \ge 0$ by $1 + x > 0$ , and $f(0) = 0$ . So $\inf_A f = 0$ , attained: $\min_A f = 0$ .

Quick check. For $\varepsilon = 2$ , which $x$ does the proof choose, and is its value above $1 - \varepsilon$ ?

Solution. $x = 1/2$ , with $f(1/2) = (1/2)/(3/2) = 1/3$ , and $1/3 > -1 = 1 - \varepsilon$ . The proof needs no separate case for large $\varepsilon$ .

In the second example the candidate is a value, and the witness half costs nothing.

Example (A supremum that is attained)

Find $\sup$ and $\inf$ of $g(x) = x(1 - x)$ on $[0, 1]$ .

Solution. Completing the square, $x(1 - x) = \frac14 - \bigl( x - \frac12 \bigr)^2$ .

Bound every value. Since $\bigl( x - \frac12 \bigr)^2 \ge 0$ , $g(x) \le \frac14$ for every $x$ .

Find a witness. $g(1/2) = 1/4$ , and $\frac14 > \frac14 - \varepsilon$ for every $\varepsilon > 0$ . So $x = 1/2$ is a witness for every $\varepsilon$ at once.

Hence $\sup_{[0, 1]} g = 1/4$ , attained at $1/2$ : it is $\max_{[0, 1]} g$ . For the infimum, on $[0, 1]$ both $x \ge 0$ and $1 - x \ge 0$ , so $g(x) \ge 0$ , and $g(0) = 0$ : $\min_{[0, 1]} g = 0$ .

An attained value is its own witness, which is why the witness in the second solution does not depend on $\varepsilon$ .

Quick check. Find $\sup$ and $\inf$ of $x^2$ on $(-1, 2)$ , and of $\mathbf{1}_{\mathbb{Q}} \colon \mathbb{R} \to \{0, 1\}$ (Sets and Functions, Definition 2.3.2) on $[0, 1]$ . Which are attained?

Solution. For $x^2$ : the values on $(-1, 2)$ form $[0, 4)$ , so $\sup = 4$ , not attained ( $x^2 = 4$ needs $x = \pm 2$ ), and $\inf = 0$ , attained at $0$ .

For $\mathbf{1}_{\mathbb{Q}}$ : $\sup = 1$ , attained at any rational point such as $0$ , and $\inf = 0$ , attained at any irrational point of $[0, 1]$ . One is $\sqrt2/2$ , with $\sqrt2$ from Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 6.1.4: it lies in $[0, 1]$ since $0 \le \sqrt 2 \le 2$ . First, $\sqrt 2$ is irrational. If $\sqrt 2 = q \in \mathbb{Q}$ , then $q^2 = 2$ in $\mathbb{R}$ . The copy of $\mathbb{Q}$ in $\mathbb{R}$ respects products and is injective (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 4.2.2), so $q^2 = 2$ would hold in $\mathbb{Q}$ . That is impossible, since no rational number has square $2$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Theorem 3.3.1). Then $\sqrt2/2$ is irrational too: $\sqrt2/2 = q \in \mathbb{Q}$ would make $\sqrt 2 = 2q$ rational.

8.3.5. Rules for suprema of functions

Suprema are compared more often than they are computed. Each comparison below, except (e), which is reflection, is one or both halves of the method.

A small case first. On $A = \{ 1, 2 \}$ let $f$ take the values $f(1) = 1$ , $f(2) = 3$ , and let $g$ take $g(1) = 2$ , $g(2) = 2$ . Then $\sup f = 3$ and $\sup g = 2$ . The sum $f + g$ takes the values $3$ and $5$ , so $\sup (f + g) = 5 = 3 + 2$ . The function $2f$ takes $2$ and $6$ , so $\sup (2f) = 6 = 2 \cdot 3$ .

Theorem 8.3.3 (Rules for Suprema of Functions)

Let $f, g \colon D \to \mathbb{R}$ , let $A \subseteq D$ be non-empty, let $f$ and $g$ be bounded above on $A$ , and let $c \in \mathbb{R}$ . All suprema are on $A$ , and $f + g$ , $cf$ , $\left| f \right|$ are as in Definition 8.1.2.

(a) If $f(x) \le g(x)$ for every $x \in A$ , then $\sup f \le \sup g$ .
(b) $\sup (f + g) \le \sup f + \sup g$ .
(c) $\sup (f + c) = \sup f + c$ .
(d) If $c \ge 0$ , then $\sup (cf) = c \sup f$ .
(e) If $f$ is bounded below on $A$ , then $-f$ is bounded above on $A$ and $\sup (-f) = -\inf f$ .
(f) If $B \subseteq A$ is non-empty, then $\sup_B f \le \sup_A f$ .
(g) If $f$ and $g$ are bounded (above and below) on $A$ , then $\bigl| \sup f - \sup g \bigr| \le \sup \left| f - g \right|$ .

Idea. Parts (a), (b) and (f) are inequalities between suprema, so only the bounding half is needed: the right side bounds every value of the left side, and the left supremum is the least such bound. Parts (c) and (d) are equalities and take both halves; (d) treats $c = 0$ apart, since a witness for $cf$ comes from $\varepsilon/c$ . Part (e) is Proposition 8.3.1 applied to the set of values. Part (g) uses that $f$ lies between $g$ shifted down and $g$ shifted up by the constant $\sup \left| f - g \right|$ .

Proof. Write $S_f \mathrel{:=} \sup_A f$ and $S_g \mathrel{:=} \sup_A g$ .

(a) For $x \in A$ , $f(x) \le g(x) \le S_g$ . So $S_g$ is an upper bound of $f(A)$ . A supremum is at most every upper bound, (S2), so $S_f \le S_g$ .
(b) Bound every value. For $x \in A$ , $f(x) \le S_f$ and $g(x) \le S_g$ . Adding the two inequalities, $f(x) + g(x) \le S_f + S_g$ . So $f + g$ is bounded above on $A$ , and $S_f + S_g$ is an upper bound of $(f + g)(A)$ . Least upper bound. The supremum is at most every upper bound, so $\sup (f + g) \le S_f + S_g$ .
(c) For $x \in A$ , $f(x) + c \le S_f + c$ . For $\varepsilon > 0$ , Proposition 8.3.2 gives $x \in A$ with $f(x) > S_f - \varepsilon$ , so $f(x) + c > (S_f + c) - \varepsilon$ . By the method, $\sup (f + c) = S_f + c$ .
(d) If $c = 0$ , then $cf$ is the constant $0$ on the non-empty set $A$ , so $\sup (cf) = 0 = 0 \cdot S_f$ . Let $c > 0$ . Bound every value. For $x \in A$ , $f(x) \le S_f$ , so $cf(x) \le c S_f$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (c), with equality allowed). Find a witness. Let $\varepsilon > 0$ . Then $\varepsilon/c > 0$ , and Proposition 8.3.2 gives $x \in A$ with $f(x) > S_f - \varepsilon/c$ ; multiplying by $c > 0$ , $cf(x) > c S_f - \varepsilon$ . By the method, $\sup (cf) = c S_f$ .
(e) Let $l$ be a lower bound of $f(A)$ . Then $-f(x) \le -l$ for $x \in A$ , so $-f$ is bounded above. The set of values of $-f$ on $A$ is $-f(A)$ , so $\sup (-f) = \sup(-f(A)) = -\inf f(A) = -\inf f$ by Proposition 8.3.1 (b).
(f) $f(B) \subseteq f(A)$ , so $S_f$ is an upper bound of $f(B)$ , and $f(B)$ is non-empty. Hence $f$ is bounded above on $B$ , and $\sup_B f \le S_f$ , since a supremum is at most every upper bound.
(g) If $\left| f(x) \right| \le M$ and $\left| g(x) \right| \le N$ on $A$ , then $\left| f(x) - g(x) \right| \le M + N$ by the triangle inequality (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (e)), so $d \mathrel{:=} \sup \left| f - g \right|$ exists. For $x \in A$ , $f(x) = g(x) + \bigl( f(x) - g(x) \bigr) \le g(x) + \left| f(x) - g(x) \right| \le g(x) + d ,$ since a number is at most its absolute value. By (a) and then (c), $S_f \le \sup (g + d) = S_g + d$ . Swapping $f$ and $g$ , and using $\left| g - f \right| = \left| f - g \right|$ , gives $S_g \le S_f + d$ . So $-d \le S_f - S_g \le d$ , which is $\left| S_f - S_g \right| \le d$ .

Uses: Sets and Functions, Definition 5.3.5 (S2: least upper bound); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (b) (adding inequalities); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (b) ( $\left| g - f \right| = \left| f - g \right|$ ), (c) ( $t \le \left| t \right|$ ), (d) ( $\left| t \right| \le d$ as two inequalities).

Rule (b) bounds the supremum of a sum, and rule (g) says that two functions close at every point have close suprema.

Warning. Rule (b) is an inequality, and it can be strict. On $[0, 1]$ let $f(x) = x$ and $g(x) = 1 - x$ . Then $\sup f = \sup g = 1$ , while $f + g$ is the constant $1$ , so $\sup (f + g) = 1 < 2$ . The two suprema are reached at different points, $1$ and $0$ .

Quick check. Which half of the method does rule (f) use, and why is the other half not needed?

Solution. Only $(\le)$ , bounding every value: every value of $f$ on $B$ is a value on $A$ , so $\sup_A f$ bounds $f(B)$ , and the least upper bound of $f(B)$ is at most it. No witness is needed because the claim is an inequality, not an equality.

8.3.6. How far apart can two values be?

For a bounded function, the largest gap between two of its values measures how much the function varies on $A$ . The obvious guess is “the top minus the bottom”, and the next result confirms it, with the top and bottom read as supremum and infimum.

A small case first. On $A = \{ 1, 2, 3 \}$ let $f$ take the values $1, 4, 2$ . The gaps $\left| f(x) - f(y) \right|$ are $0, 1, 2, 3$ , and the largest is $3 = 4 - 1$ , the largest value minus the smallest.

Proposition 8.3.4 (The Spread of Values)

Let $A$ be non-empty and let $f \colon D \to \mathbb{R}$ be bounded (above and below) on $A \subseteq D$ . Then $\sup \bigl\{ \left| f(x) - f(y) \right| : x, y \in A \bigr\} = \sup_A f - \inf_A f .$

Try it first (ten minutes): which two points make $\left| f(x) - f(y) \right|$ almost $\sup_A f - \inf_A f$ ?

Idea. The natural attempt takes $x$ where $f$ is largest and $y$ where it is smallest. That attempt fails at once: neither point need exist, as $x/(1 + x)$ on $[0, \infty)$ showed. What does exist, for each $\varepsilon > 0$ , is a point $x$ with $f(x)$ within $\varepsilon/2$ of the supremum and a point $y$ with $f(y)$ within $\varepsilon/2$ of the infimum (Proposition 8.3.2). Their gap falls short of $\sup_A f - \inf_A f$ by less than $\varepsilon/2 + \varepsilon/2 = \varepsilon$ . The bounding half needs two inequalities, one for each sign of $f(x) - f(y)$ .

Before reading on: why $\varepsilon/2$ , and not $\varepsilon$ ?

Proof. Let $S \mathrel{:=} \sup_A f$ and $I \mathrel{:=} \inf_A f$ , which exist by Definition 8.3.2, and let $T \mathrel{:=} \{ \left| f(x) - f(y) \right| : x, y \in A \}$ , a non-empty set.

Bound every value. Let $x, y \in A$ . Then $I \le f(x) \le S$ and $I \le f(y) \le S$ , so $f(x) - f(y) \le S - I$ and $f(y) - f(x) \le S - I$ . Thus $-(S - I) \le f(x) - f(y) \le S - I$ , and $\left| f(x) - f(y) \right| \le S - I$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (d). So $S - I$ is an upper bound of $T$ .

Find a witness. Let $\varepsilon > 0$ . Then $\varepsilon/2 > 0$ . By Proposition 8.3.2 there are $x \in A$ with $f(x) > S - \varepsilon/2$ and $y \in A$ with $f(y) < I + \varepsilon/2$ . A number is at most its absolute value, so $\left| f(x) - f(y) \right| \ge f(x) - f(y) > \Bigl( S - \frac{\varepsilon}{2} \Bigr) - \Bigl( I + \frac{\varepsilon}{2} \Bigr) = (S - I) - \varepsilon .$ By Proposition 8.3.2, $\sup T = S - I$ .

Uses: Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (b) ( $\varepsilon/2 > 0$ ); Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (c) ( $t \le \left| t \right|$ ).

With $\varepsilon$ in place of each $\varepsilon/2$ , the last display would end in $(S - I) - 2\varepsilon$ , and the test asks for $-\varepsilon$ ; any two positive tolerances adding up to $\varepsilon$ would do. The proof never used a point where $f$ is largest. The result holds for $x/(1 + x)$ on $[0, \infty)$ , where the spread is $1 - 0 = 1$ and is not attained.

8.3.7. Polynomials are unbounded

On a set like $[0, 1]$ many functions are bounded. On all of $\mathbb{R}$ , no polynomial function of degree at least $1$ is. Chapter 7 proved how fast a complex polynomial grows (Lemma 7.6.1), and a real polynomial is a complex one whose coefficients happen to be real. Nothing new has to be estimated.

A small case first. Let $p(x) = x^2 - 3x$ . Then $p(10) = 70$ and $p(100) = 9700$ . Once $\left| x \right|$ is large, the top term $x^2$ outweighs the rest, and the values grow without limit.

Example (Polynomials of positive degree are unbounded)

Let $p = \sum_{j=0}^{d} a_j x^j \in \mathbb{R}[x]$ with $d = \deg p \ge 1$ , and put $K \mathrel{:=} \left| a_0 \right| + \dots + \left| a_{d-1} \right| , \qquad R \mathrel{:=} \max\Bigl\{ 1, \frac{2K}{\left| a_d \right|} \Bigr\} .$ Show:

(a) $\left| p(x) \right| \ge \tfrac12 \left| a_d \right| \left| x \right|$ for every real $x$ with $\left| x \right| \ge R$ ;
(b) every real root of $p$ lies in $(-R, R)$ ;
(c) the polynomial function of $p$ on $\mathbb{R}$ (Definition 8.1.3) is not bounded.

Solution. Regard $p$ as an element of $\mathbb{C}[x]$ , with coefficients $a_j \in \mathbb{R} \subseteq \mathbb{C}$ (Proposition 3.1.1). The inclusion respects sums and products. So, for real $x$ , the value of $p$ at $x$ computed in $\mathbb{C}$ is the real number $p(x)$ . The modulus of a real number is its absolute value (Proposition 3.2.1). So the numbers $K$ and $R$ above are the $A$ and $R$ of Lemma 7.6.1, with $n = d$ .

(a) For real $x$ with $\left| x \right| \ge R$ , the lemma gives $\left| p(x) \right| \ge \tfrac12 \left| a_d \right| \left| x \right|^d \ge \tfrac12 \left| a_d \right| \left| x \right|$ .
(b) If $\left| x \right| \ge R$ , then $\left| x \right| \ge 1$ and $a_d \ne 0$ , so (a) gives $\left| p(x) \right| \ge \tfrac12 \left| a_d \right| > 0$ , and $x$ is not a root. So every real root has $\left| x \right| < R$ , that is, $-R < x < R$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (d)).
(c) Let $M \in \mathbb{R}$ and put $x \mathrel{:=} R + 2\left| M \right|/\left| a_d \right|$ . Then $x \ge R > 0$ , so $\left| x \right| = x \ge R$ , and (a) gives $\left| p(x) \right| \ge \tfrac12 \left| a_d \right| \, x = \tfrac12 \left| a_d \right| R + \left| M \right| > \left| M \right| \ge M ,$ since $\tfrac12 \left| a_d \right| R > 0$ . So no $M$ satisfies $\left| p(x) \right| \le M$ for every $x \in \mathbb{R}$ .

The growth bound says where roots can be, not that one exists: $x^2 + 1$ has no real root. That a real polynomial of odd degree has a real root came from factoring, in Chapter 7 (Corollary 7.6.11), and (b) now adds that such a root lies in $(-R, R)$ . For example $x^3 - 4x + 1$ has $K = 5$ and $R = 10$ , so its roots lie in $(-10, 10)$ .

Quick check. Is $x^3 - 100x$ bounded on $[-20, 20]$ ? On $\mathbb{R}$ ?

Solution. On $[-20, 20]$ , yes: $\left| x^3 - 100x \right| \le \left| x \right|^3 + 100\left| x \right| \le 8000 + 2000 = 10000$ , by the triangle inequality and $\left| x \right| \le 20$ . On $\mathbb{R}$ , no, by Example (Polynomials of positive degree are unbounded) (c). Boundedness depends on the set, as it did for $1/x$ on $(0, 1]$ and on $[1, 2]$ .

8.3.8. Suprema over a product

A function of two variables can be bounded in two stages: first over one variable with the other frozen, then over the other. The order of the stages does not matter. Here $A$ and $B$ are any sets, not necessarily of numbers, and $\sup_{A \times B} f$ means $\sup f(A \times B)$ , the supremum of the set of values, as in Definition 8.3.2; likewise for the inner suprema.

A small case first. Let $A = B = \{ 1, 2 \}$ , and write the values of $f$ as a table with rows $a = 1, 2$ and columns $b = 1, 2$ : the first row is $1, 5$ and the second is $3, 2$ . The row tops are $5$ and $3$ , and the larger is $5$ . The column tops are $3$ and $5$ , and the larger is again $5$ , the largest entry.

Proposition 8.3.5 (Iterated Suprema)

Let $A$ and $B$ be non-empty sets and let $f \colon A \times B \to \mathbb{R}$ be bounded above. Then $\sup_{A \times B} f = \sup_{a \in A} \, \sup_{b \in B} f(a, b) = \sup_{b \in B} \, \sup_{a \in A} f(a, b) .$

In words: to find the top of a table of numbers, take the top of each row and then the top of those, or the same with columns.

Second readingSuprema over a product

Proof. Let $S \mathrel{:=} \sup_{A \times B} f$ , which exists since $A \times B$ is non-empty. For fixed $a \in A$ , every $f(a, b)$ is a value of $f$ , so $f(a, b) \le S$ for every $b \in B$ . Hence $b \mapsto f(a, b)$ is bounded above on the non-empty $B$ . A supremum is at most every upper bound, (S2), so $\varphi(a) \mathrel{:=} \sup_{b \in B} f(a, b) \le S$ . So $\varphi$ is bounded above on $A$ by $S$ , and $\sup_A \varphi \le S$ for the same reason.

Conversely, for $(a, b) \in A \times B$ , $f(a, b) \le \varphi(a) \le \sup_A \varphi$ . So $\sup_A \varphi$ is an upper bound of the values of $f$ , and $S \le \sup_A \varphi$ . Hence $S = \sup_A \varphi$ , the first equality. Exchanging the roles of $A$ and $B$ gives the second.

If $B = \{ b_0 \}$ is a singleton, the inner supremum is $f(a, b_0)$ , and the statement says only that $\sup_{A \times \{b_0\}} f = \sup_{a \in A} f(a, b_0)$ : the same set of values, written twice. The hypothesis that $A$ and $B$ are non-empty cannot be dropped, since an empty factor makes $A \times B$ empty and leaves nothing to take the supremum of.

Remark (Looking ahead). Integration, Series, and Function Spaces, Chapter 2, calls the number $\sup_A f - \inf_A f$ of Proposition 8.3.4 the oscillation of $f$ on $A$ , and measures integrability with it.

8.3.9. Exercises

A. Check your understanding

Exercise (A1)

Parts (a)–(c) are about the proof of Proposition 8.3.4. Parts (d)–(f) are true or false: give a reason or a counterexample (here $A$ is non-empty and $f$ bounded on $A$ ).

(a) Run the statement on $f(x) = x^2$ on $A = [-1, 2)$ : find $\sup_A f$ , $\inf_A f$ and the supremum of $\left| f(x) - f(y) \right|$ . Is that supremum attained?
(b) Which line of the proof uses that $f$ is bounded below on $A$ ?
(c) State the idea of the proof in two sentences.
(d) $\sup_A f \in f(A)$ .
(e) $\inf_A f \le \sup_A f$ .
(f) $\sup_A \left| f \right| = \max\bigl( \left| \sup_A f \right|, \left| \inf_A f \right| \bigr)$ .

Solution.

(a) The values of $x^2$ on $[-1, 2)$ form $[0, 4)$ , so $\sup_A f = 4$ and $\inf_A f = 0$ , and the supremum of $\left| f(x) - f(y) \right|$ is $4 - 0 = 4$ . It is not attained: both values lie in $[0, 4)$ , so $\left| f(x) - f(y) \right| < 4$ , and equality would need $f(x) = 4$ , that is, $x = \pm 2$ , neither of which lies in $A$ .
(b) The opening sentence, first: $I = \inf_A f$ exists only because $f$ is bounded below on $A$ (Proposition 8.3.1 (b)). It is then used in the choice of $y$ with $f(y) < I + \varepsilon/2$ . The infimum version of Proposition 8.3.2 then gives $y$ . (The bounding half also uses $I \le f(y)$ .)
(c) A largest and a smallest value need not exist, so take points whose values are within $\varepsilon/2$ of the supremum and of the infimum. Their gap is within $\varepsilon$ of $\sup_A f - \inf_A f$ , which also bounds every gap.
(d) False: Example (A supremum that is not attained).
(e) True: pick any $x \in A$ , possible since $A \ne \emptyset$ ; then $\inf_A f \le f(x) \le \sup_A f$ .
(f) True. Write $S = \sup_A f$ , $I = \inf_A f$ , $m = \max(\left| S \right|, \left| I \right|)$ . Bound every value: $I \le f(x) \le S$ gives $f(x) \le \left| S \right| \le m$ and $-f(x) \le -I \le \left| I \right| \le m$ , so $\left| f(x) \right| \le m$ . Find a witness, for $\varepsilon > 0$ . If $S \ge 0$ and $m = \left| S \right| = S$ , take $x$ with $f(x) > S - \varepsilon$ ; then $\left| f(x) \right| \ge f(x) > m - \varepsilon$ . If $I \le 0$ and $m = \left| I \right| = -I$ , take $x$ with $f(x) < I + \varepsilon$ ; then $\left| f(x) \right| \ge -f(x) > m - \varepsilon$ . No other case occurs: $S < 0$ forces $I \le S < 0$ , so $m = \left| I \right| = -I$ ; and $I > 0$ forces $S \ge I > 0$ , so $m = \left| S \right| = S$ .

B. Practice

Exercise (B1)

Let $A$ be non-empty and $f, g$ bounded below on $A$ . Here is a proof that $\inf_A (f + g) \ge \inf_A f + \inf_A g$ , with its last step left out.

For each $x \in A$ , $f(x) \ge \inf_A f$ and $g(x) \ge \inf_A g$ , so $f(x) + g(x) \ge \inf_A f + \inf_A g$ .

Say which method this is, then complete it.

Solution. The method is the bounding half of Example (Method: bound every value, then find a witness), mirrored for an infimum, followed by “the infimum is the greatest lower bound”. The displayed line shows that $\inf_A f + \inf_A g$ is a lower bound of $(f + g)(A)$ ; in particular $f + g$ is bounded below on $A$ , so $\inf_A (f + g)$ exists (Definition 8.3.2). The infimum is greater than or equal to every lower bound (Sets and Functions, Definition 5.3.5), so $\inf_A (f + g) \ge \inf_A f + \inf_A g$ .

Exercise (B2: Prove or disprove)

Let $A$ be non-empty and $f, g$ bounded on $A$ . Prove or disprove each statement; all suprema and infima are on $A$ .

(a) $\sup (f + g) = \sup f + \sup g$ .
(b) $\sup \max(f, g) = \max(\sup f, \sup g)$ .
(c) If $f(x) > 0$ and $g(x) > 0$ for every $x \in A$ , then $\sup (fg) = \sup f \cdot \sup g$ .
(d) $\sup f = -\inf (-f)$ .

Solution.

(a) False. On $[0, 1]$ , $f(x) = x$ and $g(x) = 1 - x$ give $\sup (f + g) = 1$ and $\sup f + \sup g = 2$ , as in the warning after Theorem 8.3.3.
(b) True. Let $m \mathrel{:=} \max(\sup f, \sup g)$ . Bound every value: $\max(f(x), g(x))$ is $f(x) \le \sup f \le m$ or $g(x) \le \sup g \le m$ . So $\sup \max(f, g) \le m$ . Other half: $f \le \max(f, g)$ and $g \le \max(f, g)$ on $A$ , so rule (a) of Theorem 8.3.3 gives $\sup f \le \sup \max(f, g)$ and $\sup g \le \sup \max(f, g)$ , hence $m \le \sup \max(f, g)$ .
(c) False. On $(0, 1)$ let $f(x) = x$ and $g(x) = 1 - x$ , both positive. Then $\sup f = \sup g = 1$ , but $fg = x(1 - x) \le 1/4$ by Example (A supremum that is attained), with $\sup (fg) = 1/4$ attained at $1/2 \in (0, 1)$ . So $1/4 \ne 1$ .
(d) True. $-f$ is bounded (above and below) on $A$ , and rule (e) of Theorem 8.3.3 applied to $-f$ gives $\sup (-(-f)) = -\inf (-f)$ , that is, $\sup f = -\inf(-f)$ .

Exercise (B3)

Find $\sup$ and $\inf$ of $h(x) = x/(x^2 + 1)$ on $\mathbb{R}$ , and say whether each is attained. Hence $\left| h(x) \right| \le \tfrac12$ for every real $x$ .

Hint: compare $h(x)$ with $\tfrac12$ over a common denominator.

Solution. For every real $x$ , $x^2 + 1 > 0$ , and $\frac12 - \frac{x}{x^2 + 1} = \frac{x^2 - 2x + 1}{2(x^2 + 1)} = \frac{(x - 1)^2}{2(x^2 + 1)} \ge 0 , \qquad \frac{x}{x^2 + 1} + \frac12 = \frac{(x + 1)^2}{2(x^2 + 1)} \ge 0 .$ So $-\tfrac12 \le h(x) \le \tfrac12$ for every $x$ : every value is bounded. Witnesses: $h(1) = \tfrac12$ and $h(-1) = -\tfrac12$ are values, each its own witness for every $\varepsilon$ . Hence $\sup_{\mathbb{R}} h = \tfrac12 = \max_{\mathbb{R}} h$ and $\inf_{\mathbb{R}} h = -\tfrac12 = \min_{\mathbb{R}} h$ , both attained. The two-sided bound is $\left| h(x) \right| \le \tfrac12$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (d).

Exercise (B4: Find the error)

Here is an argument for “ $\sup_A (f + g) = \sup_A f + \sup_A g$ for $f, g$ bounded above on a non-empty $A$ .”

(i) The inequality $\le$ is rule (b) of Theorem 8.3.3.
(ii) For $\ge$ , let $\varepsilon > 0$ . Choose $x \in A$ with $f(x) > \sup f - \varepsilon/2$ and $y \in A$ with $g(y) > \sup g - \varepsilon/2$ .
(iii) Then $f(x) + g(y) > \sup f + \sup g - \varepsilon$ .
(iv) So $\sup f + \sup g$ passes the epsilon test for the values of $f + g$ , and equality holds.

Is the conclusion true? Which line fails, and why?

Solution. The conclusion is false: Exercise (B2: Prove or disprove) (a) gives $f(x) = x$ , $g(x) = 1 - x$ on $[0, 1]$ , with $1 \ne 2$ .

Lines (i)–(iii) are correct. Line (iv) fails: the epsilon test needs an element of the set of values $(f + g)(A) = \{ f(z) + g(z) : z \in A \}$ , and $f(x) + g(y)$ with $x \ne y$ is not of that form. In the counterexample, $x = 1$ and $y = 0$ (each a witness for every $\varepsilon$ ) give $f(1) + g(0) = 2$ , while every value of $f + g$ is $1$ . This is the mistake of pairing a witness for $f$ with a different witness for $g$ , listed in Example (Method: bound every value, then find a witness).

C. Going deeper

Exercise (C1: Chapter check: the supremum of a square)

Let $A$ be non-empty and let $f \colon D \to \mathbb{R}$ be bounded on $A \subseteq D$ , with $f(x) \ge 0$ for every $x \in A$ . Prove that $\sup_A f^2 = \bigl( \sup_A f \bigr)^2$ .

Solution (Hint 1). Use both halves of Example (Method: bound every value, then find a witness), with $s \mathrel{:=} \sup_A f$ .

Solution (Hint 2). For the witness, aim at $f(x) > s - \delta$ and expand $(s - \delta)^2$ ; then choose $\delta$ from $\varepsilon$ and $s$ .

Solution. Let $s \mathrel{:=} \sup_A f$ . Since $A \ne \emptyset$ and $f \ge 0$ on $A$ , $s \ge f(x) \ge 0$ for any $x \in A$ .

Bound every value. For $x \in A$ , $0 \le f(x) \le s$ , so $f(x)^2 \le s^2$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.2 (d), applied with both factors equal. So $f^2$ is bounded above on $A$ by $s^2$ .

Find a witness. Let $\varepsilon > 0$ . If $s = 0$ , then $f(x) = 0$ on $A$ , so $f^2$ is the constant $0$ and any $x \in A$ has $f(x)^2 = 0 > s^2 - \varepsilon$ . Let $s > 0$ and put $\delta \mathrel{:=} \varepsilon/(2s) > 0$ , chosen from $\varepsilon$ and $s$ . By Proposition 8.3.2 there is $x \in A$ with $f(x) > s - \delta$ .

Case $s - \delta \ge 0$ . Then $0 \le s - \delta < f(x)$ , so $(s - \delta)^2 < f(x)^2$ by Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g), and $(s - \delta)^2 = s^2 - 2s\delta + \delta^2 = s^2 - \varepsilon + \delta^2 > s^2 - \varepsilon .$ So $f(x)^2 > s^2 - \varepsilon$ .

Case $s - \delta < 0$ . Then $s < \delta = \varepsilon/(2s)$ , so $s^2 < \varepsilon/2$ , and $s^2 - \varepsilon < -\varepsilon/2 < 0 \le f(x)^2$ .

In every case $f(x)^2 > s^2 - \varepsilon$ . By the method, $\sup_A f^2 = s^2$ .

The hypothesis $f \ge 0$ is needed: $f(x) = x$ on $[-2, 1]$ has $\sup f = 1$ , but $\sup f^2 = 4$ .

What a complete answer needs. (10 points)

2: names $s = \sup_A f$ and states both halves as the goal (every $f(x)^2 \le s^2$ ; for every $\varepsilon > 0$ some $f(x)^2 > s^2 - \varepsilon$ ).
3: proves the bounding half, naming the nonnegativity of $f(x)$ (and of $s$ ) as what allows squaring the inequality.
3: chooses the witness tolerance from $\varepsilon$ (and $s$ ) before taking the witness, and verifies $f(x)^2 > s^2 - \varepsilon$ .
2: handles the cases the choice leaves open ( $s = 0$ , and $s - \delta < 0$ or its equivalent in another route).

A route that writes $f(x)^2 - s^2 = (f(x) - s)(f(x) + s)$ and bounds $\left| f(x) + s \right| \le 2s$ earns the same points for the same steps. A route that bounds $s$ by $\sqrt{\sup_A f^2}$ , using the least-upper-bound property and squaring non-negative numbers (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 2.4.3 (g)), earns the 3 + 2 points of the witness bullets for those steps.

Slips that lose the step (0 points for it): squaring an inequality between numbers that may be negative; a witness that does not depend on $\varepsilon$ ; dividing by $s$ without treating $s = 0$ .

Exercise (C2)

Let $a < b$ , $L \ge 0$ , and let $f \colon [a, b] \to \mathbb{R}$ satisfy $\left| f(x) - f(y) \right| \le L \left| x - y \right|$ for all $x, y \in [a, b]$ . Here is the first half of a proof that $f$ is bounded and $\sup f - \inf f \le L(b - a)$ , all suprema on $[a, b]$ .

For $x \in [a, b]$ , $\left| f(x) \right| \le \left| f(a) \right| + \left| f(x) - f(a) \right| \le \left| f(a) \right| + L(x - a) \le \left| f(a) \right| + L(b - a)$ , so $f$ is bounded. By Proposition 8.3.4, $\sup f - \inf f = \sup \{ \left| f(x) - f(y) \right| : x, y \in [a, b] \}$ , so it suffices to bound this supremum.

Say which method this is, then complete it.

Solution. The method is the bounding half of Example (Method: bound every value, then find a witness), followed by “the supremum is the least upper bound”.

Bound every value. Let $x, y \in [a, b]$ . Both lie between $a$ and $b$ , so $-(b - a) \le x - y \le b - a$ , that is, $\left| x - y \right| \le b - a$ (Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (d)). Since $L \ge 0$ , $\left| f(x) - f(y) \right| \le L \left| x - y \right| \le L(b - a) .$ So $L(b - a)$ is an upper bound of $\{ \left| f(x) - f(y) \right| : x, y \in [a, b] \}$ , and its supremum is at most $L(b - a)$ , since a supremum is at most every upper bound, (S2). Hence $\sup f - \inf f \le L(b - a)$ .

(The first half used the triangle inequality, Number Systems: From $\mathbb{N}$ to $\mathbb{R}$ , Proposition 4.1.3 (e), and $0 \le x - a \le b - a$ .) The bound can be an equality: $f(x) = Lx$ has $\sup f - \inf f = Lb - La$ .