---
title: "Limits and Continuity"
book: "University Mathematics — Year 1"
subject: math
language: en
chapter: 13
exercises: 12
source: https://one-course.com/books/math/3/en/chapter/13-limits-and-continuity
---

# Chapter 13 — Limits and Continuity

The intermediate value theorem and the extreme value theorem were used at High School level on visual faith. With sequences ([Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq)) and the topology of $\R$ ([Chapter 12](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#ch-b1-topology)) in hand, this chapter proves them — and completes the theory with the monotone bijection theorem (which legitimizes $\arcsin$, $\operatorname{arcosh}$ and their kin from [Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions)) and Heine’s theorem on [uniform continuity](#def-b1-continuity-uniform).

Throughout, $I$ is an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) and $f \colon I \to \R$; “$x_0 \in
\overline I$” allows limits at endpoints.

## 13.1 Limits of functions

**Definition 13.1 (Limit at a point).**

Let $x_0 \in \overline{I}$ and $\ell \in \R$. Then $f(x) \to \ell$ as $x \to x_0$ when

$$
\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x \in I,
\qquad \abs{x - x_0} \leq \delta \implies \abs{f(x) - \ell} \leq
\varepsilon .
$$

Limits at $\pm\infty$ and infinite limits are defined by the same pattern ($\abs{x - x_0} \leq \delta$ becomes $x \geq M$; $\abs{f(x) -
\ell} \leq \varepsilon$ becomes $f(x) \geq M'$). One-sided limits restrict $x$ to $x > x_0$ (written $x \to x_0^+$) or $x < x_0$. The limit is unique when it exists (same proof as [Proposition 11.4](https://one-course.com/books/math/3/en/chapter/11-sequences#prop-b1-seq-first)).

**Example 13.2 (A limit through the squeeze).**

Compute $\lim_{x \to 0} x\,\bigl\lfloor \frac1x \bigr\rfloor$. The [floor](https://one-course.com/books/math/3/en/chapter/10-real-numbers#thm-b1-reals-floor) bracketing $\frac1x - 1 < \lfloor \frac1x \rfloor \leq
\frac1x$ gives, after multiplying by $x$ (mind the sign!):

$$
1 - x < x\Bigl\lfloor \frac1x \Bigr\rfloor \leq 1 \quad (x > 0),
\qquad
1 \leq x\Bigl\lfloor \frac1x \Bigr\rfloor < 1 - x \quad (x < 0),
$$

and both one-sided squeezes close on $1$: the limit is $1$. Note what happened: $\lfloor \frac1x\rfloor$ alone has wild jumps near $0$, but the factor $x$ tames each jump ($x$ times a unit jump is small), and only the bracketing survives. The closing insight: limits of products of a small factor with a bounded-oscillation factor are squeeze problems, never operation-theorem problems — the operations theorem needs *both* factors to converge.

**Theorem 13.3 (Sequential characterization).**

$f(x) \to \ell$ as $x \to x_0$ if and only if: for *every* sequence $(u_n)$ of points of $I$ with $u_n \to x_0$, one has $f(u_n) \to \ell$.

**Proof.** ($\Rightarrow$) Let $u_n \to x_0$ and $\varepsilon > 0$. Take $\delta$ from the definition, then $N$ with $\abs{u_n - x_0} \leq
\delta$ for $n \geq N$: beyond $N$, $\abs{f(u_n) - \ell} \leq
\varepsilon$.

($\Leftarrow$) Contrapositive. If $f \not\to \ell$: some $\varepsilon_0 > 0$ defeats every $\delta$; choosing $\delta =
\frac{1}{n+1}$ produces $u_n \in I$ with $\abs{u_n - x_0} \leq
\frac{1}{n+1}$ and $\abs{f(u_n) - \ell} > \varepsilon_0$. Then $u_n
\to x_0$ but $f(u_n) \not\to \ell$. ∎

**Corollary 13.4 (Operations, composition, order).**

Sums, products, quotients (nonzero limit downstairs) of limits behave as for sequences; if $f \to \ell$ at $x_0$ and $g \to m$ at $\ell$, and $g$ is defined around $\ell$ with $g(\ell) = m$ or $f \neq \ell$ near $x_0$, then $g \circ f \to m$ at $x_0$; limits preserve wide inequalities, and the squeeze theorem holds.

**Proof.** Each [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) transfers through [Theorem 13.3](#thm-b1-continuity-seqchar) to its sequence analogue (Theorems [11.5](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-operations) and [11.7](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-order)). For composition, the direct chaining is worth writing once: given $\varepsilon >
0$, the limit of $g$ at $\ell$ provides $\eta > 0$ with

$$
\abs{y - \ell} \leq \eta \implies \abs{g(y) - m} \leq
\varepsilon \quad (y \text{ in the domain of } g),
$$

where the case $y = \ell$ is covered because $g(\ell) = m$; then the limit of $f$ at $x_0$ provides $\delta > 0$ with $\abs{x -
x_0} \leq \delta \implies \abs{f(x) - \ell} \leq \eta$; chaining, $\abs{x - x_0} \leq \delta$ gives $\abs{g(f(x)) - m} \leq
\varepsilon$. In the alternative hypothesis ($f \neq \ell$ near $x_0$), the value $y = \ell$ is never fed to $g$ and its value there is irrelevant. ∎

**Example 13.5 (Why the composition proviso exists).**

Let $g(y) = 0$ for $y \neq 0$ and $g(0) = 1$, and let $f$ be identically $0$. Then $f(x) \to 0$ as $x \to 0$, and $g(y) \to 0$ as $y \to 0$; yet $g(f(x)) = g(0) = 1$ for every $x$, so $g
\circ f \to 1 \neq 0$. The inner function sits *exactly on* the forbidden value $\ell = 0$ forever, and $g$’s limit at $0$ ignores what $g$ does at $0$. The proviso of [Corollary 13.4](#cor-b1-continuity-operations) — either $g(\ell) = m$ (i.e. $g$ [continuous](#def-b1-continuity-continuous) at $\ell$), or $f \neq \ell$ near $x_0$ — is precisely what rules this out. The closing insight: in practice one composes *[continuous](#def-b1-continuity-continuous)* functions and the proviso is free; it bites only when limits are taken along punctured [neighborhoods](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open), which is why the definition of $\lim_{x \to x_0}$ used in this book includes the point when it is in the domain.

## 13.2 Continuity

**Definition 13.6.**

$f$ is *continuous at $x_0 \in I$* when $f(x)
\to f(x_0)$ as $x \to x_0$; *continuous on $I$* when it is continuous at every point. By [Theorem 13.3](#thm-b1-continuity-seqchar): $f$ is continuous at $x_0$ iff $f(u_n) \to f(x_0)$ for every sequence $u_n \to x_0$ in $I$.

**Example 13.7 (A taxonomy of discontinuities).**

Three ways to fail at a point, in increasing severity. *Removable*: $f(x) = \frac{\sin x}{x}$ on $\R^*$ has limit $1$ at $0$; defining $f(0) = 1$ repairs it — the discontinuity was a hole, not a feature. *Jump*: $\lfloor x \rfloor$ at an integer has distinct one-sided limits ($n - 1$ and $n$); no choice of value can reconcile them, but both half-limits exist. *Essential*: $\sin\frac1x$ at $0$ has no one-sided limit at all ([Exercise 13.1](#exo-b1-continuity-1)) — oscillation without settlement. Monotone functions can only produce the middle kind (their one-sided limits always exist, being suprema and infima), which is why their discontinuity [sets](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) are at most countable — one rational per jump. Derivatives, by Darboux’s theorem ([Exercise 14.10](https://one-course.com/books/math/3/en/chapter/14-differentiation#exo-b1-derivative-10)), can only produce the last kind: a function with a jump discontinuity is never the derivative of anything.

**Proposition 13.8.**

Sums, products, quotients (where defined) and compositions of [continuous](#def-b1-continuity-continuous) functions are [continuous](#def-b1-continuity-continuous). [Polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def), [rational fractions](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#def-b1-fractions-field) (off their [poles](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#def-b1-fractions-field)), $\abs{\,\cdot\,}$, $\exp$, $\ln$, the trigonometric and [hyperbolic functions](https://one-course.com/books/math/3/en/chapter/4-standard-functions#def-b1-functions-hyperbolic) and their inverses ([Chapter 4](https://one-course.com/books/math/3/en/chapter/4-standard-functions#ch-b1-functions)) are [continuous](#def-b1-continuity-continuous) on their domains.

**Proof.** Operations: [Corollary 13.4](#cor-b1-continuity-operations). Constants and the identity are [continuous](#def-b1-continuity-continuous) directly from the definition ($\delta = \varepsilon$ serves the identity, any $\delta$ the constants); products of [continuous](#def-b1-continuity-continuous) functions being [continuous](#def-b1-continuity-continuous), each monomial $a_k x^k$ follows by induction on $k$, and sums finish the [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def); a [rational fraction](https://one-course.com/books/math/3/en/chapter/9-rational-fractions#def-b1-fractions-field) is a quotient of two [polynomials](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def), [continuous](#def-b1-continuity-continuous) wherever the denominator does not vanish. $\abs{\,\cdot\,}$: reverse triangle inequality, $\bigl|\abs{f(x)} - \abs{f(x_0)}\bigr| \leq \abs{f(x) -
f(x_0)}$, so the same $\delta$ works. For the classical functions we grant [continuity](#def-b1-continuity-continuous) here; differentiability (proved in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)) is stronger. ∎

**Example 13.9 (Max and min of continuous functions).**

If $f, g$ are [continuous](#def-b1-continuity-continuous), so are $\max(f, g)$ and $\min(f, g)$: no case analysis needed, thanks to the identities

$$
\max(f, g) = \frac{f + g + \abs{f - g}}{2},
\qquad
\min(f, g) = \frac{f + g - \abs{f - g}}{2},
$$

and the [continuity](#def-b1-continuity-continuous) of sums and of $\abs{\,\cdot\,}$ ([Proposition 13.8](#prop-b1-continuity-algebra)). In particular $f^+ =
\max(f, 0)$ and $f^- = \max(-f, 0)$ are [continuous](#def-b1-continuity-continuous) with $f =
f^+ - f^-$: the sign-splitting used for series ([Chapter 17](https://one-course.com/books/math/3/en/chapter/17-numerical-series#ch-b1-series)) and, at full scale, in the integration theory of the Year 3 volume, costs nothing in regularity.

**Theorem 13.10 (Intermediate value theorem).**

Let $f$ be [continuous](#def-b1-continuity-continuous) on $\intcc{a}{b}$ with $f(a) \leq 0 \leq
f(b)$. Then $f(c) = 0$ for some $c \in \intcc{a}{b}$. Consequently, a [continuous](#def-b1-continuity-continuous) function on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) takes every value between any two of its values: $f(I)$ is an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals).

**Proof.** Dichotomy. Set $a_0 = a$, $b_0 = b$. Given $\intcc{a_n}{b_n}$ with $f(a_n) \leq 0 \leq f(b_n)$, let $m$ be the midpoint: if $f(m) \leq
0$ keep $\intcc{m}{b_n}$, otherwise keep $\intcc{a_n}{m}$; the sign conditions persist. The sequences $(a_n), (b_n)$ are adjacent ($b_n -
a_n = \frac{b-a}{2^n}$), with common limit $c$ ([Theorem 11.11](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent)). By [continuity](#def-b1-continuity-continuous) and [Theorem 11.7](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-order): $f(c) = \lim f(a_n) \leq 0$ and $f(c) =
\lim f(b_n) \geq 0$, so $f(c) = 0$.

For the consequence: given values $f(u) < v < f(w)$, apply the above to $x \mapsto f(x) - v$ on the segment with endpoints $u$ and $w$; thus $f(I)$ is convex, i.e. an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) ([Proposition 10.19](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals)). ∎

![A continuous function with f(a) < 0 < f(b) must cross the axis: the dichotomy proof of traps a crossing point between adjacent sequences.](https://one-course.com/images/onecourse/chapters/math-3/b1-continuity/fig-65449770e506.svg)

*A [continuous](#def-b1-continuity-continuous) function with $f(a) < 0 < f(b)$ must cross the axis: the dichotomy proof of [Theorem 13.10](#thm-b1-continuity-ivt) traps a crossing point between [adjacent sequences](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-adjacent).*

**Example 13.11 (One equation, the full protocol).**

Solve $\eu^x = 3 - x$ over $\R$: existence, uniqueness, location. Set $g(x) = \eu^x + x - 3$, [continuous](#def-b1-continuity-continuous). Location and existence: $g(0) = -2 < 0$ and $g(1) = \eu - 2 > 0$, so the intermediate value theorem plants a solution in $\intoo{0}{1}$. Uniqueness: $g$ is a sum of the strictly increasing $\eu^x$ and $x - 3$, hence strictly increasing on $\R$; a strictly monotone function takes each value at most once, so the solution is unique on all of $\R$ (not just in the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) probed). The protocol — rearrange to $g = 0$, sign change for existence, monotonicity for uniqueness — settles most “how many solutions” questions in three lines, and [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)’s variation tables extend it to non-monotone $g$ by cutting $\R$ into monotone branches.

**Example 13.12 (Dichotomy as an algorithm).**

The proof of [Theorem 13.10](#thm-b1-continuity-ivt) computes. Take $f(x) =
x^3 + x - 1$: $f(0) = -1 < 0 < 1 = f(1)$, so a root lies in $\intoo{0}{1}$. Halving:

$$
f(0.5) = -0.375 < 0, \qquad
f(0.75) = 0.171875 > 0, \qquad
f(0.625) = -0.130859375 < 0 ,
$$

so the root is successively trapped in $\intoo{0.5}{1}$, then $\intoo{0.5}{0.75}$, then $\intoo{0.625}{0.75}$ (true value: $c \approx 0.6823$). After $n$ steps the error is at most $\frac{b - a}{2^n}$: ten steps give three decimals, twenty give six. The closing insight: the intermediate value theorem is not only an existence [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) — its dichotomy proof is a guaranteed, if slow, root-finding algorithm, against which the fast but local Newton method of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) should be measured.

**Theorem 13.13 (Extreme value theorem).**

A [continuous](#def-b1-continuity-continuous) function on a segment $\intcc{a}{b}$ is bounded and attains its bounds: there are $c, d \in \intcc{a}{b}$ with

$$
f(c) = \inf_{\intcc{a}{b}} f,
\qquad
f(d) = \sup_{\intcc{a}{b}} f .
$$

Combined with [Theorem 13.10](#thm-b1-continuity-ivt): *the [continuous](#def-b1-continuity-continuous) image of a segment is a segment* $\intcc{f(c)}{f(d)}$.

**Proof.** *Bounded above:* otherwise pick $u_n$ with $f(u_n) \geq n$. By compactness of the segment ([Theorem 12.19](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#thm-b1-topology-compact)), a [subsequence](https://one-course.com/books/math/3/en/chapter/11-sequences#def-b1-seq-subsequence) $u_{\varphi(n)} \to x \in \intcc{a}{b}$; [continuity](#def-b1-continuity-continuous) gives $f(u_{\varphi(n)}) \to f(x)$, but $f(u_{\varphi(n)}) \geq \varphi(n)
\to +\infty$: contradiction.

*Sup attained:* let $M = \sup f$ and choose $v_n$ with $f(v_n) >
M - \frac{1}{n+1}$ ([Proposition 10.4](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-epsilon)). Extract $v_{\varphi(n)} \to d \in \intcc{a}{b}$: then $f(d) = \lim
f(v_{\varphi(n)}) = M$ by the squeeze. The [infimum](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) is handled by $-f$. ∎

**Example 13.14 (Positive minimum on a segment).**

Let $f$ be [continuous](#def-b1-continuity-continuous) on $\intcc{0}{1}$ with $f(x) > 0$ for every $x$. Then $\inf f > 0$: by the extreme value theorem the [infimum](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) is a *value* $f(c)$, and $f(c) > 0$ by hypothesis. So a [continuous](#def-b1-continuity-continuous) positive function on a segment is bounded away from $0$ — a two-line argument used a dozen times in the coming chapters (denominators under control, step-function framings, error bounds). On a non-compact [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) this fails spectacularly: $f(x) = x$ on $\intoc{0}{1}$ is [continuous](#def-b1-continuity-continuous) and positive with $\inf f = 0$, not attained. The closing insight: “positive” upgrades to “uniformly positive” exactly when the domain is compact; every hypothesis of the extreme value theorem is load-bearing.

**Remark 13.15 (Common pitfalls around the three theorems).**

(i) *[Continuous](#def-b1-continuity-continuous) images*: only *segments* are robust. The [continuous](#def-b1-continuity-continuous) image of an [open](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) need not be [open](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) ($x \mapsto x^2$ [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\intoo{-1}{1}$ onto $\intco{0}{1}$), the image of a [closed set](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closed) need not be [closed](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closed) ($\arctan$ [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) the [closed](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closed) $\R$ onto the [open](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) $\intoo{-\frac\pi2}{\frac\pi2}$); but the image of a segment is a segment ([Theorem 13.13](#thm-b1-continuity-evt)). (ii) *The intermediate value theorem needs an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals)*: the function $x
\mapsto \frac1x$, [continuous](#def-b1-continuity-continuous) on $\intco{-1}{0} \cup
\intoc{0}{1}$, takes the values $-1$ and $1$ yet never vanishes — its domain is two disjoint pieces, and the value $0$ falls into the gap; always name the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) on which the theorem is applied. (iii) *[Uniform continuity](#def-b1-continuity-uniform) is a property of the pair (function, [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets))*: $x^2$ is [uniformly continuous](#def-b1-continuity-uniform) on every segment yet not on $\R$ ([Example 13.20](#ex-b1-continuity-notuniform)) — the words “[uniformly continuous](#def-b1-continuity-uniform)” without a domain are meaningless. (iv) *[Continuity](#def-b1-continuity-continuous) of the inverse is not formal*: it holds on [intervals](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) via monotonicity ([Theorem 13.16](#thm-b1-continuity-bijection)), but a [continuous](#def-b1-continuity-continuous) bijection between unions of [intervals](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) can have a discontinuous inverse — the remark after that theorem is there because students quote it without the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) hypothesis.

## 13.3 Monotone functions and inverse functions

**Theorem 13.16 (Monotone bijection theorem).**

Let $f$ be [continuous](#def-b1-continuity-continuous) and strictly monotonic on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $I$. Then

1. $f$ is a bijection from $I$ onto the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $J = f(I)$ ;
2. the inverse $f^{-1} \colon J \to I$ is strictly monotonic (in the same direction) and *[continuous](#def-b1-continuity-continuous)* .

**Proof.** Say $f$ is strictly increasing. (1) Injectivity is immediate from strict monotonicity; surjectivity onto $f(I)$ is trivial, and $f(I)$ is an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) by [Theorem 13.10](#thm-b1-continuity-ivt).

(2) $f^{-1}$ is strictly increasing: if $y < y'$ in $J$ but $f^{-1}(y)
\geq f^{-1}(y')$, applying the increasing $f$ gives $y \geq y'$, absurd. [Continuity](#def-b1-continuity-continuous) of $f^{-1}$ at $y_0 = f(x_0) \in J$: let $\varepsilon > 0$. Suppose first $x_0$ [interior](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) to $I$, and shrink $\varepsilon$ so that $x_0 \pm \varepsilon \in I$: their images satisfy $f(x_0 - \varepsilon) < y_0 < f(x_0 +
\varepsilon)$. Take $\delta = \min\bigl(y_0 - f(x_0 - \varepsilon),\, f(x_0 +
\varepsilon) - y_0\bigr) > 0$: for $\abs{y - y_0} \leq \delta$, monotonicity of $f^{-1}$ squeezes $f^{-1}(y)$ between $x_0 -
\varepsilon$ and $x_0 + \varepsilon$. If $x_0$ is, say, the left endpoint of $I$, only $x_0 + \varepsilon$ is available: then $y_0 = \min J$ (monotonicity), every $y \in J$ with $y - y_0
\leq f(x_0 + \varepsilon) - y_0$ satisfies $x_0 \leq f^{-1}(y)
\leq x_0 + \varepsilon$, and the one-sided estimate is exactly [continuity](#def-b1-continuity-continuous) at an endpoint; the right endpoint is symmetric. (Note: [continuity](#def-b1-continuity-continuous) of $f^{-1}$ was *not* deduced from that of $f$ by symmetry — it is monotonicity on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) that does the work.) ∎

**Remark 13.17.**

This theorem is what [Definition 4.9](https://one-course.com/books/math/3/en/chapter/4-standard-functions#def-b1-functions-arc) and [Proposition 4.21](https://one-course.com/books/math/3/en/chapter/4-standard-functions#prop-b1-functions-invhyp) silently used: $\arcsin$, $\arccos$, $\arctan$, $\operatorname{arsinh}$, … are [continuous](#def-b1-continuity-continuous). A complement ([Exercise 13.10](#exo-b1-continuity-10)): a [continuous](#def-b1-continuity-continuous) *[injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj)* function on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) is automatically strictly monotonic, so the monotonicity hypothesis costs nothing.

**Example 13.18 (Roots of all orders).**

For $n \in \N^*$, the function $f(x) = x^n$ is [continuous](#def-b1-continuity-continuous) and strictly increasing on $\intco{0}{+\infty}$, with $f(0) = 0$ and $f(x) \to +\infty$: its image is $\intco{0}{+\infty}$ ([Theorem 13.10](#thm-b1-continuity-ivt) for the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) structure). The monotone bijection theorem then delivers, in one stroke, a strictly increasing *[continuous](#def-b1-continuity-continuous)* inverse

$$
x \mapsto x^{1/n} \colon \intco{0}{+\infty} \to
\intco{0}{+\infty} :
$$

existence, uniqueness and [continuity](#def-b1-continuity-continuous) of $n$-th roots, with no computation. Compare with [Exercise 10.12](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-12), which built $\sqrt y$ by hand from the [supremum](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds): one chapter of theory has compressed that page of work into two lines, and the same two lines legitimized $\arcsin$, $\arctan$ and $\operatorname{arsinh}$ before them. The closing insight: a good theorem is stored labor.

## 13.4 Uniform continuity

**Definition 13.19.**

$f \colon I \to \R$ is *uniformly continuous* when

$$
\forall \varepsilon > 0,\ \exists \delta > 0,\ \forall x, y \in I,
\qquad \abs{x - y} \leq \delta \implies \abs{f(x) - f(y)} \leq
\varepsilon .
$$

The point: $\delta$ depends only on $\varepsilon$, not on the location in $I$. Uniform [continuity](#def-b1-continuity-continuous) implies [continuity](#def-b1-continuity-continuous); a *Lipschitz* function ($\abs{f(x) - f(y)} \leq k \abs{x - y}$) is uniformly [continuous](#def-b1-continuity-continuous) ($\delta = \varepsilon/k$).

**Example 13.20.**

$x \mapsto x^2$ is [continuous](#def-b1-continuity-continuous) on $\R$ but *not* [uniformly continuous](#def-b1-continuity-uniform): $\abs{(n + \frac 1n)^2 - n^2} = 2 + \frac{1}{n^2} \geq 2$ although the arguments are at distance $\frac 1n \to 0$. On any bounded [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) it is Lipschitz, hence [uniformly continuous](#def-b1-continuity-uniform) — consistent with Heine’s theorem below.

**Example 13.21 (Uniform moduli, explicitly).**

On a segment, Heine guarantees a uniform $\delta$; often one can also *compute* it. For $f(x) = x^2$ on $\intcc{0}{10}$:

$$
\abs{x^2 - y^2} = \abs{x + y}\,\abs{x - y} \leq 20\,\abs{x - y},
$$

so $\delta = \frac{\varepsilon}{20}$ works uniformly (a Lipschitz [modulus](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-field), linear in $\varepsilon$). For $\sqrt x$ on $\intcc{0}{1}$: $\abs{\sqrt x - \sqrt y} \leq \sqrt{\abs{x -
y}}$ ([Exercise 13.9](#exo-b1-continuity-9)), so $\delta = \varepsilon^2$ works — uniform but *not* linear: near $0$ the square root is steep, and the price appears in the exponent of $\varepsilon$, not in a failure of uniformity. The closing insight: [uniform continuity](#def-b1-continuity-uniform) is a spectrum, not a yes/no — the function $\delta(\varepsilon)$, called the [modulus](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-field), measures how expensive the uniformity is, and Lipschitz is simply its best grade.

**Theorem 13.22 (Heine).**

A [continuous](#def-b1-continuity-continuous) function on a *segment* $\intcc{a}{b}$ is [uniformly continuous](#def-b1-continuity-uniform).

**Proof.** By contradiction: suppose some $\varepsilon_0 > 0$ defeats every $\delta$. With $\delta = \frac{1}{n+1}$, pick $x_n, y_n \in
\intcc{a}{b}$ with $\abs{x_n - y_n} \leq \frac{1}{n+1}$ and $\abs{f(x_n) - f(y_n)} > \varepsilon_0$. Compactness ([Theorem 12.19](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#thm-b1-topology-compact)) extracts $x_{\varphi(n)} \to c \in
\intcc{a}{b}$; then $y_{\varphi(n)} \to c$ too (squeeze on $\abs{x -
y}$). [Continuity](#def-b1-continuity-continuous) at $c$ gives $f(x_{\varphi(n)}) \to f(c)$ and $f(y_{\varphi(n)}) \to f(c)$, so $\abs{f(x_{\varphi(n)}) -
f(y_{\varphi(n)})} \to 0 < \varepsilon_0$: contradiction. ∎

**Example 13.23 (Bounded, continuous, yet not uniformly so).**

The function $f(x) = \sin(x^2)$ is [continuous](#def-b1-continuity-continuous) and bounded on $\R$, but not [uniformly continuous](#def-b1-continuity-uniform). Take

$$
x_n = \sqrt{2\pi n}, \qquad y_n = \sqrt{2\pi n + \tfrac\pi2}:
\qquad
y_n - x_n = \frac{\pi/2}{\sqrt{2\pi n + \frac\pi2} +
\sqrt{2\pi n}} \longrightarrow 0 ,
$$

yet $f(y_n) - f(x_n) = \sin\bigl(2\pi n + \frac\pi2\bigr) -
\sin(2\pi n) = 1 - 0 = 1$ for every $n$: no single $\delta$ can serve $\varepsilon = \frac12$ everywhere. Geometrically, the oscillations of $\sin(x^2)$ *accelerate*: the graph completes a full wave over shorter and shorter windows, so the horizontal scale a given $\varepsilon$ requires shrinks to zero as $x$ grows. The closing insight: boundedness does not buy uniformity (this example), and unboundedness does not preclude it ($\sqrt x$, [Exercise 13.9](#exo-b1-continuity-9)); what decides is the *[modulus](https://one-course.com/books/math/3/en/chapter/3-complex-numbers#def-b1-complex-field) of oscillation*, and Heine’s theorem says compact domains discipline it automatically.

**Example 13.24 (The calculator experiment, explained).**

Type any number into a calculator and press $\cos$ repeatedly: the display locks onto $0.7390851\dots$ Why? After one press the value lies in $\intcc{-1}{1}$, after two in $\intcc{\cos 1}{1}
\subseteq \intcc{0.54}{1}$, a stable [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) for $\cos$. On it, $\abs{\cos a - \cos b} \leq \sin(1)\,\abs{a - b}$ with $\sin 1 =
0.841\dots < 1$ (the product-to-sum bound of [Problem 11.1](https://one-course.com/books/math/3/en/chapter/11-sequences#pb-b1-seq-1), or the mean value inequality of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)): the iteration contracts, so by the error-control step of [Method 11.23](https://one-course.com/books/math/3/en/chapter/11-sequences#met-b1-seq-recurrent),

$$
\abs{u_n - c} \leq (0.842)^{\,n-2}\,\abs{u_2 - c}
\longrightarrow 0 ,
$$

where $c$ is the unique fixed point $\cos c = c$ ([Exercise 13.6](#exo-b1-continuity-6)). About $40$ presses buy three decimals ($0.842^{40} \approx 10^{-3}$) — a geometric rate, milder than dichotomy’s $2^{-n}$ per step, but each press costs one keystroke while each dichotomy step costs a full sign evaluation. The closing insight: the fixed-point picture of [Chapter 11](https://one-course.com/books/math/3/en/chapter/11-sequences#ch-b1-seq) and the existence theorems of this chapter are two halves of one story — IVT finds $c$, contraction reaches it.

**Remark 13.25 (Where these theorems work next).**

The three pillars of this chapter each power a later one. The intermediate value theorem feeds every existence-of-solutions argument and the monotone bijection theorem; the extreme value theorem turns optimization problems into theorems (Rolle and the mean value theorem in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) start exactly there); Heine’s theorem is the reason [continuous](#def-b1-continuity-continuous) functions on segments can be integrated in [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) — the uniform $\delta$ is what makes Riemann sums converge. In the Year 2 volume the same trio reappears in normed vector spaces, with compactness doing the work that segments do here.

**Remark 13.26 (Perspectives inside this volume).**

[Continuity](#def-b1-continuity-continuous) is about to be outranked but never retired. [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) strengthens it to differentiability and returns the favor (differentiable implies [continuous](#def-b1-continuity-continuous)); [Chapter 15](https://one-course.com/books/math/3/en/chapter/15-integration-on-a-segment#ch-b1-integration) rests on it twice, through Heine for the construction and through the fundamental theorem, whose central object $x \mapsto \int_a^x f$ upgrades a merely [continuous](#def-b1-continuity-continuous) $f$ to a $C^1$ primitive. In [Chapter 25](https://one-course.com/books/math/3/en/chapter/25-functions-of-two-variables#ch-b1-multivar), [continuity](#def-b1-continuity-continuous) in two variables holds a trap worth previewing: the function $\frac{xy}{x^2 + y^2}$ (extended by $0$) is [continuous](#def-b1-continuity-continuous) in $x$ for each fixed $y$ and in $y$ for each fixed $x$, yet not [continuous](#def-b1-continuity-continuous) at the origin — along the diagonal $x = y$ it is constantly $\frac12$. Separate [continuity](#def-b1-continuity-continuous) is strictly weaker than [continuity](#def-b1-continuity-continuous): the sequential characterization survives the move to $\R^2$, but sequences must be allowed to approach from *every* direction, not only along the axes.

## 13.5 Exercises

**Exercise 13.1 ★.**

Using the sequential characterization, prove that $x \mapsto
\sin\frac 1x$ has no limit at $0^+$ *(exhibit two sequences)*. Does $x \mapsto x \sin\frac 1x$ have one?

**Solution of Exercise 13.1.**

Take $u_n = \frac{1}{2\pi n + \pi/2}$ and $v_n = \frac{1}{2\pi n}$: both tend to $0^+$, yet $\sin\frac{1}{u_n} = 1$ and $\sin\frac{1}{v_n} = 0$. Two sequences, two different limits of images: by [Theorem 13.3](#thm-b1-continuity-seqchar), no limit at $0^+$.

$x \sin\frac1x$: squeezed by $\abs{x\sin\frac1x} \leq \abs x \to 0$, so the limit at $0$ exists and equals $0$.

**Exercise 13.2 ★.**

Study the [continuity](#def-b1-continuity-continuous) on $\R$ of $f(x) = \lfloor x \rfloor$ and of $g(x) = x - \lfloor x \rfloor$, and of $h(x) = \lfloor x \rfloor +
(x - \lfloor x\rfloor)^2$.

**Solution of Exercise 13.2.**

$f = \lfloor\cdot\rfloor$ is [continuous](#def-b1-continuity-continuous) on $\R \setminus \Z$ (locally constant) and discontinuous at each $n \in \Z$: left limit $n - 1$, value $n$.

$g(x) = x - \lfloor x\rfloor$: same discontinuity points (the identity is [continuous](#def-b1-continuity-continuous), so $g$ inherits the jumps of $f$); at $n \in
\Z$, left limit $1 \neq 0 = g(n)$.

$h(x) = \lfloor x\rfloor + (x - \lfloor x\rfloor)^2$: on $\intco{n}{n
+ 1}$, $h(x) = n + (x - n)^2$, [continuous](#def-b1-continuity-continuous) there; at $x = n$, the left limit is $(n-1) + 1 = n = h(n)$: the jumps cancel. $h$ is [continuous](#def-b1-continuity-continuous) on $\R$ (and strictly increasing).

**Exercise 13.3 ★.**

Prove that the equation $x^5 - 3x + 1 = 0$ has at least three real solutions *(evaluate at well-chosen points and apply [Theorem 13.10](#thm-b1-continuity-ivt) on three disjoint segments)*.

**Solution of Exercise 13.3.**

$P(x) = x^5 - 3x + 1$: $P(-2) = -32 + 6 + 1 = -25 < 0$; $P(0) = 1 >
0$; $P(1) = -1 < 0$; $P(2) = 27 > 0$. Three sign changes on the disjoint segments $\intcc{-2}{0}$, $\intcc{0}{1}$, $\intcc{1}{2}$: by [Theorem 13.10](#thm-b1-continuity-ivt), at least three roots. (Being of degree $5$, $P$ has at most five; a variation study would show exactly three.)

**Exercise 13.4 ★.**

Prove that every [polynomial](https://one-course.com/books/math/3/en/chapter/8-polynomials#def-b1-poly-def) of odd degree has a real root.

**Solution of Exercise 13.4.**

Let $P = a_{2m+1}X^{2m+1} + \dots$ with $a_{2m+1} > 0$ (else replace $P$ by $-P$). Factoring the dominant term, $P(x) = a_{2m+1}
x^{2m+1}\bigl(1 + o(1)\bigr)$ as $x \to \pm\infty$: so $P(x) \to
+\infty$ at $+\infty$ and $-\infty$ at $-\infty$. Pick $a$ with $P(a) < 0$ and $b$ with $P(b) > 0$: the intermediate value theorem on $\intcc{a}{b}$ provides a root.

**Exercise 13.5 ★★.**

(Fixed point) Let $f \colon \intcc{0}{1} \to \intcc{0}{1}$ be [continuous](#def-b1-continuity-continuous). Prove that $f$ has a fixed point: $f(c) = c$ for some $c$. Illustrate that neither [continuity](#def-b1-continuity-continuous) nor the segment can be dropped.

**Solution of Exercise 13.5.**

Let $g(x) = f(x) - x$, [continuous](#def-b1-continuity-continuous) on $\intcc{0}{1}$. Since $f$ [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) into $\intcc{0}{1}$: $g(0) = f(0) \geq 0$ and $g(1) = f(1) - 1 \leq
0$. By [Theorem 13.10](#thm-b1-continuity-ivt), $g(c) = 0$ for some $c$: a fixed point.

Necessity of the hypotheses: on $\intcc{0}{1}$, the discontinuous [map](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $f(x) = 1$ for $x \leq \frac12$, $f(x) = 0$ for $x > \frac12$ has no fixed point; on the [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $\intoo{0}{1}$ (not a segment), $f(x) =
\frac x2$ is [continuous](#def-b1-continuity-continuous) into $\intoo{0}{1}$ with no fixed point (the candidate $0$ is missing); on $\R$, $f(x) = x + 1$.

**Exercise 13.6 ★★.**

Prove that the equation $\cos x = x$ has exactly one real solution, and that it lies in $\intoo{0}{1}$.

**Solution of Exercise 13.6.**

$g(x) = \cos x - x$ is [continuous](#def-b1-continuity-continuous), $g(0) = 1 > 0$, $g(1) = \cos 1 -
1 < 0$: a solution exists in $\intoo{0}{1}$ ([Theorem 13.10](#thm-b1-continuity-ivt)). Uniqueness: $g$ is strictly decreasing on $\R$ — for $x \leq 0$, $g(x) \geq 1 - x > 0$ has no zero anyway; and $g'(x) = -\sin x - 1 \leq 0$ with equality only at isolated points ($x \equiv -\frac\pi2 \bmod 2\pi$), so $g$ is strictly decreasing ([Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative); alternatively: on $\intcc{0}{1}$, $\cos$ is strictly decreasing and $-x$ too, so $g$ is). A strictly monotone function vanishes at most once.

**Exercise 13.7 ★★.**

Let $f \colon \R \to \R$ be [continuous](#def-b1-continuity-continuous) with $f(x) \to +\infty$ as $x
\to \pm\infty$. Prove that $f$ attains a global minimum on $\R$. *(Reduce to a segment containing a sublevel [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets).)*

**Solution of Exercise 13.7.**

Fix $M = f(0) + 1$. There is $A > 0$ with $f(x) \geq M$ for $\abs x
\geq A$ (definition of the two infinite limits; take the larger threshold). On the segment $\intcc{-A}{A}$, the extreme value theorem ([Theorem 13.13](#thm-b1-continuity-evt)) provides $c$ with $f(c) =
\inf_{\intcc{-A}{A}} f \leq f(0)$. For $\abs x \geq A$: $f(x) \geq M
> f(0) \geq f(c)$. So $f(c)$ is the global minimum.

**Exercise 13.8 ★★.**

Let $f \colon \R \to \R$ be [continuous](#def-b1-continuity-continuous) and periodic (of period $T >
0$). Prove that $f$ is bounded and attains its bounds, and that there exists $c$ with $f(c + \frac T2) = f(c)$. *(For the second point, study $g(x) = f(x + \frac T2) - f(x)$ over one period.)*

**Solution of Exercise 13.8.**

On the segment $\intcc{0}{T}$, $f$ is bounded and attains its bounds ([Theorem 13.13](#thm-b1-continuity-evt)); by periodicity, these are the bounds on all of $\R$, still attained.

Let $g(x) = f(x + \frac T2) - f(x)$, [continuous](#def-b1-continuity-continuous). Then

$$
g(0) + g\bigl(\tfrac T2\bigr)
= \bigl(f(\tfrac T2) - f(0)\bigr) + \bigl(f(T) - f(\tfrac T2)\bigr)
= f(T) - f(0) = 0 :
$$

$g(0)$ and $g(\frac T2)$ have opposite signs (or one vanishes), so the intermediate value theorem on $\intcc{0}{T/2}$ gives $c$ with $g(c) = 0$, i.e. $f(c + \frac T2) = f(c)$.

**Exercise 13.9 ★★.**

Prove that $x \mapsto \sqrt x$ is [uniformly continuous](#def-b1-continuity-uniform) on $\intco{0}{+\infty}$, although it is not Lipschitz near $0$. *(Prove and use $\abs{\sqrt x - \sqrt y} \leq \sqrt{\abs{x -
y}}$.)*

**Solution of Exercise 13.9.**

First the inequality: for $0 \leq y \leq x$,

$$
\bigl(\sqrt y + \sqrt{x - y}\bigr)^2 = x + 2\sqrt{y(x-y)} \geq x,
$$

so $\sqrt x \leq \sqrt y + \sqrt{x - y}$, i.e. $\sqrt x - \sqrt y
\leq \sqrt{x - y}$. Hence $\abs{\sqrt x - \sqrt y} \leq
\sqrt{\abs{x - y}}$ for all $x, y \geq 0$.

[Uniform continuity](#def-b1-continuity-uniform): given $\varepsilon > 0$, take $\delta =
\varepsilon^2$; then $\abs{x - y} \leq \delta$ implies $\abs{\sqrt x
- \sqrt y} \leq \sqrt\delta = \varepsilon$.

Not Lipschitz near $0$: $\frac{\sqrt x - \sqrt 0}{x - 0} =
\frac{1}{\sqrt x} \to +\infty$ as $x \to 0^+$, so no constant $k$ can dominate all difference quotients.

**Exercise 13.10 ★★★.**

Let $f$ be [continuous](#def-b1-continuity-continuous) and [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj) on an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $I$. Prove that $f$ is strictly monotonic. *Hint: if not, there are $a < b < c$ with, say, $f(b) > f(a)$ and $f(b) > f(c)$; apply the intermediate value theorem to a value between $\max(f(a), f(c))$ and $f(b)$ on both sides of $b$.*

**Solution of Exercise 13.10.**

Suppose $f$ is [injective](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-inj), [continuous](#def-b1-continuity-continuous), and not strictly monotonic. Then there are $a < b < c$ in $I$ with $f(b)$ not between $f(a)$ and $f(c)$ — indeed, if for *all* triples the middle value were between the outer ones, $f$ would be monotonic (compare any two pairs; a short case check). Say $f(b) > \max(f(a), f(c))$ (the other case is symmetric, replace $f$ by $-f$). Choose $v$ with $\max(f(a),
f(c)) < v < f(b)$. By the intermediate value theorem applied on $\intcc{a}{b}$ and on $\intcc{b}{c}$, there are $u_1 \in
\intoo{a}{b}$ and $u_2 \in \intoo{b}{c}$ with $f(u_1) = v = f(u_2)$: two distinct points with equal images, contradicting injectivity.

**Exercise 13.11 ★★★.**

(Cauchy’s functional equation, [continuous](#def-b1-continuity-continuous) case) Let $f \colon \R \to
\R$ be [continuous](#def-b1-continuity-continuous) with $f(x + y) = f(x) + f(y)$ for all $x, y$. Prove that $f(x) = f(1)\,x$ for all $x$: first on $\N$, $\Z$, $\Q$ (by additivity alone), then on $\R$ by [continuity](#def-b1-continuity-continuous) and [density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) ([Theorem 10.14](https://one-course.com/books/math/3/en/chapter/10-real-numbers#thm-b1-reals-density)).

**Solution of Exercise 13.11.**

$f(0) = f(0+0) = 2f(0)$ gives $f(0) = 0$; $f(-x) = -f(x)$ from $0 =
f(x - x)$. Set $\alpha = f(1)$. Induction: $f(n) = n\alpha$ for $n
\in \N$, then for $n \in \Z$ by oddness. For $q \in \N^*$: $q\,f(\frac pq) = f(p) = p\alpha$ (add $\frac pq$ to itself $q$ times), so $f(\frac pq) = \alpha\frac pq$: $f = \alpha\,
\mathrm{id}$ on $\Q$.

Let now $x \in \R$ and $(r_n)$ a sequence of rationals with $r_n \to
x$ ([density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense), [Theorem 10.14](https://one-course.com/books/math/3/en/chapter/10-real-numbers#thm-b1-reals-density), applied in nested [intervals](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals); or $r_n = \frac{\lfloor nx\rfloor}{n}$). [Continuity](#def-b1-continuity-continuous): $f(x) = \lim f(r_n) = \lim \alpha r_n = \alpha x$.

**Exercise 13.12 ★★★.**

Let $f \colon \intco{0}{+\infty} \to \R$ be [continuous](#def-b1-continuity-continuous) with $f(x) \to \ell \in \R$ as $x \to +\infty$. Prove that $f$ is *uniformly* [continuous](#def-b1-continuity-continuous) on $\intco{0}{+\infty}$. *(Cut at a large $A$: Heine on $\intcc{0}{A+1}$, the limit beyond $A$; make the two regimes overlap.)*

**Solution of Exercise 13.12.**

Let $\varepsilon > 0$. By the limit at $+\infty$, there is $A$ with $\abs{f(x) - \ell} \leq \frac\varepsilon2$ for $x \geq A$; hence for $x, y \geq A$: $\abs{f(x) - f(y)} \leq \varepsilon$ (no closeness needed).

On the segment $\intcc{0}{A + 1}$, Heine’s theorem ([Theorem 13.22](#thm-b1-continuity-heine)) gives $\delta_0 > 0$ for this $\varepsilon$; set $\delta = \min(\delta_0, 1)$.

Now take any $x, y \geq 0$ with $\abs{x - y} \leq \delta$, say $x
\leq y$. If $y \leq A + 1$: both lie in the segment, and the Heine $\delta_0$ applies. Otherwise $y > A + 1$, and then $x \geq y - 1 >
A$: both lie in $\intco{A}{+\infty}$, where the limit argument applies. In both cases $\abs{f(x) - f(y)} \leq \varepsilon$: [uniform continuity](#def-b1-continuity-uniform).

## 13.6 Problem: Cauchy’s functional equation and its sisters

**Problem 13.1.**

Weekend problem — $f(x + y) = f(x) + f(y)$: regularity forces linearity, and the portrait of the monsters

Which functions satisfy $f(x + y) = f(x) + f(y)$ for all real $x,
y$? Cauchy raised the question in 1821; the answer is a paradigm. [Exercise 13.11](#exo-b1-continuity-11) shows that such an *additive* function is linear on $\Q$ and that full [continuity](#def-b1-continuity-continuous) forces $f(x)
= cx$. This problem sharpens the hypothesis dramatically — [continuity](#def-b1-continuity-continuous) at a *single* point, or monotonicity, or mere boundedness on one small [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals), each suffices — then paints the portrait of a hypothetical nonlinear solution (its graph fills the plane), solves the sister equations characterizing $\eu^{cx}$, $c\ln x$, $x^c$ and $cx^2$, and closes with Jensen’s equation and the theorem *midpoint-convex $+$ [continuous](#def-b1-continuity-continuous) $\implies$ convex*. Throughout, *additive* means: $f(x + y) = f(x) + f(y)$ for all $x, y \in \R$.

**Part I — $\Q$-linearity, and one point of [continuity](#def-b1-continuity-continuous).**

1. Let $f$ be additive. From [Exercise 13.11](#exo-b1-continuity-11) , $f(r) = f(1)\,r$ for rational $r$ . Prove the finer [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) used below: for *every* $x \in \R$ and $r  \in \Q$ , $f(rx) = r\,f(x)$ ( $f$ is $\Q$ - *linear* ).
2. Suppose the additive $f$ is [continuous](#def-b1-continuity-continuous) at one single point $x_0$ . Show that $f$ is [continuous](#def-b1-continuity-continuous) everywhere *(compute $f(x + h) - f(x)$ in terms of $f(x_0 + h)  - f(x_0)$)* , hence $f(x) = f(1)\,x$ .
3. Show that an additive function is determined by its restriction to any [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) : if two additive functions agree on $\Z + \sqrt2\,\Z$ and both are [continuous](#def-b1-continuity-continuous) , they are equal — while without [continuity](#def-b1-continuity-continuous) , prescribing $f(1) = 0$ and $f(\sqrt 2) = 1$ is consistent with $\Q$ -linearity on the [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) . Compute $f(m +  n\sqrt2)$ for this prescription.
4. Let $f$ be additive and bounded above by $M$ on some [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $\intcc{a}{b}$ with $a < b$ . Show that $f$ is bounded above on $\intcc{0}{\ell}$ , $\ell = b - a$ *(translate by $a$)* .

**Part II — The regularity ladder.**

5. Continuing question 4: using $f(\ell - t) + f(t) =  f(\ell)$ , show that $f$ is also bounded *below* on $\intcc{0}{\ell}$ : $\abs f \leq C$ there.
6. Show that $\abs{f(t)} \leq \frac{C}{n}$ for $t \in  \intcc{0}{\ell/n}$ , and deduce that $f$ is [continuous](#def-b1-continuity-continuous) at $0$ *(oddness handles the left side)* , hence everywhere (question 2): an additive function bounded on one [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) is linear.
7. Deduce the monotone case: an additive function nondecreasing on some $\intcc{a}{b}$ ( $a < b$ ) is $f(x)  = cx$ with $c \geq 0$ .
8. Assemble the *regularity ladder* : for an additive $f$ , the following are equivalent — (a) $f(x) = cx$ ; (b) $f$ [continuous](#def-b1-continuity-continuous) ; (c) $f$ [continuous](#def-b1-continuity-continuous) at one point; (d) $f$ monotone on some nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) ; (e) $f$ bounded on some nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) . Arrange the implications so that each is either trivial or already proved.
9. Verify that questions 4–6 consumed only a bound *above* : an additive function bounded above on one nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) is already linear. Deduce the mirror [statement](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) for a bound below, and record the strongest form of the ladder’s rung (e) thus obtained.

**Part III — Portrait of a monster.** Suppose now $f$ is additive but *not* linear.

10. Show there are nonzero reals $u, v$ with $\dfrac{f(u)}{u} \neq \dfrac{f(v)}{v}$ , and that the vectors $(u, f(u))$ and $(v, f(v))$ span the plane (their determinant $u f(v) - v f(u)$ is nonzero).
11. Show that the graph of $f$ contains all points $$r\,(u, f(u)) + s\,(v, f(v)), \qquad r, s \in \Q ,$$ and deduce that the graph is *[dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\R^2$*: for every point $(x_0, y_0)$ of the plane and every $\varepsilon > 0$, some $(x, f(x))$ is within $\varepsilon$ of it *(solve the $2\times2$ real system, then approximate the real coefficients by rationals)*.
12. Deduce from question 11 the full portrait: a nonlinear additive function is unbounded on every nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) , discontinuous at every point, monotone on no [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) , and its image of any [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) is [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\R$ . Reconcile with question 8.
13. Monsters exist — on a [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) , constructively: on $G = \Z + \sqrt2\,\Z$ define $f(m + n\sqrt2) = n$ . Show $f$ is well defined and additive on $G$ , and that $f$ is unbounded on $G \cap \intoo{0}{\varepsilon}$ for every $\varepsilon > 0$ *(for fixed $N$, only finitely many $g = m + n\sqrt2 \in \intoo{0}{1}$ have $\abs n \leq N$; yet $G \cap \intoo{0}{\varepsilon}$ is infinite)* . Explain in one paragraph why extending such an $f$ to all of $\R$ requires a basis of $\R$ as a $\Q$ -vector space (a *Hamel basis* ), whose existence is an axiom-of-choice matter beyond this volume.

**Part IV — The sister equations.** All functions here are [continuous](#def-b1-continuity-continuous).

14. Let $f \colon \R \to \R$ be [continuous](#def-b1-continuity-continuous) , not identically $0$ , with $f(x + y) = f(x)f(y)$ . Show $f(x) =  f\bigl(\frac x2\bigr)^2 \geq 0$ , then $f > 0$ everywhere, then $f(x) = \eu^{cx}$ for some $c$ : the exponentials are exactly the [continuous](#def-b1-continuity-continuous) [morphisms](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-morphism) from $(\R, +)$ to $(\R^*, \times)$ .
15. Let $f \colon \intoo{0}{+\infty} \to \R$ be [continuous](#def-b1-continuity-continuous) with $f(xy) = f(x) + f(y)$ . Show $f(x) = c\ln x$ *(transport by $\exp$)* .
16. Let $f \colon \intoo{0}{+\infty} \to \intoo{0}{+\infty}$ be [continuous](#def-b1-continuity-continuous) with $f(xy) = f(x)f(y)$ . Show $f(x) = x^c$ .
17. Find all [continuous](#def-b1-continuity-continuous) $f \colon \R \to \R$ with $$f(x + y) = f(x) + f(y) + f(x)f(y) .$$ *(Study $h = 1 + f$; treat the degenerate case separately.)*
18. (Parallelogram) Find all [continuous](#def-b1-continuity-continuous) $f \colon \R \to \R$ with $f(x + y) + f(x - y) = 2f(x) + 2f(y)$ : show $f$ is even, $f(0) = 0$ , $f(nx) = n^2 f(x)$ by induction, then $f(x) = f(1)\,x^2$ . (This equation is the fingerprint of quadratic forms — the parallelogram law that detects, in the Year 2 volume, which norms come from an inner product.)

**Part V — Jensen and midpoint convexity.**

19. (Jensen’s equation) Let $f \colon \R \to \R$ be [continuous](#def-b1-continuity-continuous) with $f\bigl(\frac{x+y}{2}\bigr) = \frac{f(x)  + f(y)}{2}$ . Show that $g = f - f(0)$ satisfies $g\bigl(\frac x2\bigr) = \frac{g(x)}{2}$ , deduce that $g$ is additive, and conclude $f(x) = cx + d$ .
20. Suppose now only the *inequality*: $f$ [continuous](#def-b1-continuity-continuous) with $$f\Bigl(\frac{x + y}{2}\Bigr) \leq \frac{f(x) +  f(y)}{2} \qquad (x, y \in \R).$$ Prove by induction on $n$ that for all dyadic weights $\lambda = \frac{k}{2^n} \in \intcc{0}{1}$: $$f\bigl(\lambda x + (1 - \lambda)y\bigr) \leq \lambda  f(x) + (1 - \lambda) f(y) .$$
21. Extend by [continuity](#def-b1-continuity-continuous) and [density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) of the dyadics ( [Exercise 10.8](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-8) ) to every $\lambda \in  \intcc{0}{1}$ : a [continuous](#def-b1-continuity-continuous) midpoint-convex function satisfies the full convexity inequality (the notion studied systematically in [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative) ).
22. Show [continuity](#def-b1-continuity-continuous) cannot be dropped: a nonlinear additive $f$ satisfies the *midpoint equality* of question 19 yet no convexity inequality on any [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) (question 12). Moral: midpoint convexity is a countable-stage property (dyadics), convexity a continuum one; [continuity](#def-b1-continuity-continuous) is the bridge — exactly as in Parts I–II.

**Part VI — Last variations and synthesis.**

23. Find all [continuous](#def-b1-continuity-continuous) $f \colon \R \to \R$ with $f(x + y)  = f(x) + f(y) + xy$ *(subtract the particular solution $\frac{x^2}{2}$)* .
24. Prove: if $f \colon \R \to \R$ is [continuous](#def-b1-continuity-continuous) and additive merely on a *[dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup)* $G$ (i.e. $f(g + g') = f(g) + f(g')$ for $g, g' \in G$ ), then $f$ is additive on $\R$ . More generally, two [continuous](#def-b1-continuity-continuous) functions that agree on a [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) subset of $\R$ are equal.
25. Synthesis, one sentence each: (i) state the regularity ladder of question 8 from memory; (ii) explain why “graph [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in the plane” is the right mental image for the failure of regularity; (iii) list the five classical functions characterized in Part IV–V and the single method that caught them all; (iv) name the two places where [density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) of $\Q$ (or of the dyadics) in $\R$ carried the argument, and the place where it could not (question 13).

**Solution of Problem 13.1.**

**1.** For $n \in \N$: $f(nx) = n f(x)$ by induction ($f((n+1)x) = f(nx) + f(x)$). Also $f(0) = 2f(0)$ gives $f(0) =
0$, and $0 = f(x - x) = f(x) + f(-x)$ gives oddness, so $f(nx) =
nf(x)$ for $n \in \Z$. For $r = \frac pq$: $q\,f\bigl(\tfrac
pq x\bigr) = f(px) = p\,f(x)$, so $f(rx) = r f(x)$: $f$ is $\Q$-linear.

**2.** Additivity gives, for all $x$ and $h$:

$$
f(x + h) - f(x) = f(h) = f(x_0 + h) - f(x_0) .
$$

As $h \to 0$, the right side tends to $0$ by [continuity](#def-b1-continuity-continuous) at $x_0$; hence $f(x + h) \to f(x)$: [continuity](#def-b1-continuity-continuous) at every $x$. Then [Exercise 13.11](#exo-b1-continuity-11) yields $f(x) = f(1)\,x$.

**3.** Two [continuous](#def-b1-continuity-continuous) additive functions are of the form $cx$ and $c'x$ (question 2); if they agree on the [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) $\Z + \sqrt2\,\Z$ ([Exercise 10.9](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-9)), then $c\,g = c'g$ for some $g \neq 0$ in it: $c = c'$, the functions are equal. Without [continuity](#def-b1-continuity-continuous): $\Q$-linearity only ties values at $\Q$-combinations, and $1, \sqrt2$ are $\Q$-independent ($\sqrt2
\notin \Q$), so $f(1) = 0$, $f(\sqrt2) = 1$ is consistent and forces, on the [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup),

$$
f(m + n\sqrt2) = m\,f(1) + n\,f(\sqrt2) = n .
$$

**4.** For $t \in \intcc{0}{\ell}$: $a + t \in
\intcc{a}{b}$, so $f(t) = f(a + t) - f(a) \leq M - f(a) =: M'$.

**5.** For $t \in \intcc{0}{\ell}$, also $\ell - t \in
\intcc{0}{\ell}$, and additivity gives $f(t) = f(\ell) - f(\ell
- t) \geq f(\ell) - M'$. Hence $\abs{f} \leq C$ on $\intcc{0}{\ell}$ with $C = \max\bigl(\abs{M'}, \abs{f(\ell) -
M'}\bigr)$.

**6.** For $t \in \intcc{0}{\ell/n}$: $nt \in
\intcc{0}{\ell}$ and $f(t) = \frac{f(nt)}{n}$ (question 1), so $\abs{f(t)} \leq \frac Cn$. Given $\varepsilon > 0$, choose $n >
\frac C\varepsilon$: for $0 \leq h \leq \frac\ell n$, $\abs{f(h)} \leq \varepsilon$, and for negative $h$ use $f(h) =
-f(-h)$. So $f(h) \to 0 = f(0)$ as $h \to 0$: [continuity](#def-b1-continuity-continuous) at $0$, hence everywhere (question 2), hence $f(x) = f(1)x$.

**7.** If $f$ is nondecreasing on $\intcc{a}{b}$, then $f(a) \leq f(x) \leq f(b)$ there: bounded, so linear by question 6, $f(x) = cx$; and $c(b - a) = f(b) - f(a) \geq 0$ forces $c
\geq 0$.

**8.** (a)$\Rightarrow$(b)$\Rightarrow$(c): trivial. (c)$\Rightarrow$(a): question 2. (a)$\Rightarrow$(d): a linear function is monotone everywhere. (d)$\Rightarrow$(e): a monotone function on $\intcc{a}{b}$ is bounded there by its endpoint values. (e)$\Rightarrow$(a): questions 4–6. The five [statements](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-statement) are equivalent — the regularity ladder.

**9.** Question 4 used only the [upper bound](https://one-course.com/books/math/3/en/chapter/10-real-numbers#def-b1-reals-bounds) $M$; question 5 *derived* the lower bound from the upper one via the reflection $f(t) = f(\ell) - f(\ell - t)$; question 6 then ran on $\abs f \leq C$. So: additive and bounded *above* on one nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) already implies linear. For a bound below, apply this to $-f$ (additive, bounded above). Strongest rung (e): a one-sided bound on one [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) suffices.

**10.** If $\frac{f(x)}{x}$ were one constant $c$ for all $x \neq 0$, $f$ would be linear; so there are nonzero $u, v$ with $\frac{f(u)}u \neq \frac{f(v)}v$, i.e. $u f(v) - v f(u)
\neq 0$: the determinant of the vectors $(u, f(u))$, $(v,
f(v))$ is nonzero, and they span $\R^2$.

**11.** For $r, s \in \Q$: $f(ru + sv) = r f(u) + s f(v)$ (question 1 twice plus additivity), so the graph contains

$$
\bigl(ru + sv,\; r f(u) + s f(v)\bigr)
= r\,(u, f(u)) + s\,(v, f(v)) .
$$

Given $(x_0, y_0)$ and $\varepsilon > 0$: the $2 \times 2$ system $a(u, f(u)) + b(v, f(v)) = (x_0, y_0)$ has a (unique) real solution $(a, b)$ since the determinant is nonzero. Pick rationals $r_n \to a$, $s_n \to b$: then $r_n(u, f(u)) +
s_n(v, f(v)) \to (x_0, y_0)$ coordinatewise, and each of these points lies on the graph: the graph is [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\R^2$.

**12.** Let $I$ be a nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals), $x_0$ its midpoint, $M$ arbitrary: [density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) provides a graph point within $\min\bigl(\frac{\abs I}{2}, 1\bigr)$ of $(x_0, M + 1)$, i.e. $x \in I$ with $f(x) > M$: unbounded on $I$, hence (question 8) discontinuous at every point and monotone on no [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals); and for any target $y_0$, graph points near $(x_0, y_0)$ give $f(x)$ arbitrarily close to $y_0$ with $x \in I$: $f(I)$ is [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\R$. This is question 8 read backwards: since all rungs are equivalent, a nonlinear additive function must fail *all* of them, everywhere.

**13.** Well defined: $m + n\sqrt2 = m' + n'\sqrt2$ forces $(n - n')\sqrt2 = m' - m \in \Z$, so $n = n'$ (else $\sqrt2 \in
\Q$) and $m = m'$. Additivity on $G$ is then clear coordinate by coordinate. Unboundedness near $0^+$: fix $\varepsilon \in
\intoo{0}{1}$ and $N \in \N$. For each fixed $n$ with $\abs n
\leq N$, the condition $m + n\sqrt2 \in \intoo{0}{1}$ pins $m$ inside an [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) of length $1$: at most one integer $m$ per $n$, so at most $2N + 1$ elements of $G \cap \intoo{0}{1}$ have $\abs{f} \leq N$. But $G \cap \intoo{0}{\varepsilon}$ is infinite ($G$ is [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense), [Exercise 10.9](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-9)); hence it contains some $g$ with $\abs{f(g)} > N$: $f$ is unbounded on every right [neighborhood](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-open) of $0$. Extending $f$ to an additive function on $\R$ means choosing values coherently on a family of reals that is $\Q$-linearly independent and spans $\R$ over $\Q$ — a Hamel basis; producing one requires the axiom of choice, and no explicit formula can do it: constructively we own the monster only on $G$.

**14.** $f(x) = f(\frac x2 + \frac x2) = f(\frac x2)^2 \geq
0$. If $f(x_0) = 0$, then $f(x) = f(x - x_0)f(x_0) = 0$ for all $x$: excluded. So $f > 0$ and $g = \ln \circ f$ is [continuous](#def-b1-continuity-continuous) ([Proposition 13.8](#prop-b1-continuity-algebra)) with $g(x + y) = g(x) +
g(y)$: by [Exercise 13.11](#exo-b1-continuity-11), $g(x) = cx$, so $f(x) =
\eu^{cx}$. Conversely each $\eu^{cx}$ works: the [continuous](#def-b1-continuity-continuous) [morphisms](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-morphism) $(\R, +) \to (\R^*, \times)$ are exactly the exponentials.

**15.** $g(u) = f(\eu^u)$ is [continuous](#def-b1-continuity-continuous) and $g(u + v) =
f(\eu^u \eu^v) = g(u) + g(v)$: $g(u) = cu$, and every $x > 0$ writes $x = \eu^u$ with $u = \ln x$: $f(x) = c\ln x$.

**16.** $h = \ln \circ f$ is [continuous](#def-b1-continuity-continuous) on $\intoo{0}{+\infty}$ with $h(xy) = h(x) + h(y)$: by question 15, $h(x) = c\ln x$, so $f(x) = \eu^{c\ln x} = x^c$.

**17.** $x = y = 0$: $f(0) = 2f(0) + f(0)^2$, so $f(0)(1 +
f(0)) = 0$. If $f(0) = -1$: setting $y = 0$, $f(x) = f(x) +
f(0) + f(x)f(0) = -1$ for all $x$: the constant $f \equiv -1$ (which indeed satisfies the equation). Otherwise $f(0) = 0$; $h
= 1 + f$ is [continuous](#def-b1-continuity-continuous), $h(0) = 1$, and

$$
h(x + y) = 1 + f(x) + f(y) + f(x)f(y) = h(x)\,h(y) :
$$

by question 14, $h(x) = \eu^{cx}$, i.e. $f(x) = \eu^{cx} - 1$ (the case $c = 0$ giving $f \equiv 0$). Complete list: $f
\equiv -1$ and $f(x) = \eu^{cx} - 1$, $c \in \R$.

**18.** $x = y = 0$: $2f(0) = 4f(0)$, so $f(0) = 0$. $x =
0$: $f(y) + f(-y) = 2f(y)$, so $f$ is even. $y = x$: $f(2x) =
4f(x)$. Induction using $(x, y) \to (nx, x)$:

$$
f((n{+}1)x) = 2f(nx) + 2f(x) - f((n{-}1)x)
= (2n^2 + 2 - (n-1)^2) f(x) = (n+1)^2 f(x) .
$$

Then $f(x) = f\bigl(q\cdot\frac xq\bigr) = q^2 f\bigl(\frac
xq\bigr)$ gives $f\bigl(\frac pq x\bigr) = \frac{p^2}{q^2}f(x)$: $f(r) = r^2 f(1)$ on $\Q$ (evenness handles signs). The two [continuous](#def-b1-continuity-continuous) functions $f$ and $x \mapsto f(1)x^2$ agree on the [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) $\Q$, hence everywhere (question 24): $f(x) =
f(1)\,x^2$; every $cx^2$ satisfies the equation.

**19.** $g = f - f(0)$ is [continuous](#def-b1-continuity-continuous), $g(0) = 0$, and satisfies Jensen’s equation (the constants cancel). Taking $y =
0$: $g\bigl(\frac x2\bigr) = \frac{g(x)}{2}$. Then for all $x,
y$:

$$
\frac{g(x + y)}{2} = g\Bigl(\frac{x+y}{2}\Bigr)
= \frac{g(x) + g(y)}{2} ,
$$

so $g$ is additive and [continuous](#def-b1-continuity-continuous): $g(x) = cx$ (question 2), and $f(x) = cx + d$ with $d = f(0)$. All affine functions satisfy Jensen: the list is complete.

**20.** Induction on $n$. For $n = 0$: $\lambda \in \{0,
1\}$, trivial. Assume the inequality for all weights $\frac{k}{2^n}$. A weight $\lambda = \frac{k}{2^{n+1}}$ with $k$ even reduces to level $n$; for $k = 2j + 1$, $\lambda$ is the midpoint of $\lambda_1 = \frac{j}{2^n}$ and $\lambda_2 =
\frac{j+1}{2^n}$. With $z_i = \lambda_i x + (1 - \lambda_i)y$: $\lambda x + (1-\lambda)y = \frac{z_1 + z_2}{2}$, so

$$
f\bigl(\lambda x + (1{-}\lambda)y\bigr)
\leq \frac{f(z_1) + f(z_2)}{2}
\leq \frac{(\lambda_1 + \lambda_2)f(x) + (2 - \lambda_1 -
\lambda_2)f(y)}{2}
= \lambda f(x) + (1 - \lambda)f(y) .
$$

**21.** Fix $x, y$. The [maps](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-map) $\lambda \mapsto f(\lambda x +
(1 - \lambda)y)$ and $\lambda \mapsto \lambda f(x) + (1 -
\lambda)f(y)$ are [continuous](#def-b1-continuity-continuous) on $\intcc{0}{1}$ (composition and algebra, [Proposition 13.8](#prop-b1-continuity-algebra)). The inequality holds on the dyadic weights, which are [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) in $\intcc{0}{1}$ ([Exercise 10.8](https://one-course.com/books/math/3/en/chapter/10-real-numbers#exo-b1-reals-8)); for arbitrary $\lambda$ take dyadics $\lambda_n \to \lambda$ and pass to the limit ([Theorem 13.3](#thm-b1-continuity-seqchar) and [Theorem 11.7](https://one-course.com/books/math/3/en/chapter/11-sequences#thm-b1-seq-order)): the convexity inequality holds for every $\lambda \in
\intcc{0}{1}$ — midpoint convexity plus [continuity](#def-b1-continuity-continuous) equals convexity (the notion of [Chapter 14](https://one-course.com/books/math/3/en/chapter/14-differentiation#ch-b1-derivative)).

**22.** A nonlinear additive $f$ satisfies $f\bigl(\frac{x+y}{2}\bigr) = \frac{f(x) + f(y)}{2}$ exactly (question 1 with $r = \frac12$, then additivity): it is midpoint-convex, even midpoint-affine. If it satisfied the full convexity inequality on some [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $\intcc{x}{y}$, then for $\lambda \in \intcc{0}{1}$: $f(\lambda x + (1-\lambda)y) \leq
\max(f(x), f(y))$ — bounded above on a nondegenerate [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals), hence linear by question 9: contradiction. So [continuity](#def-b1-continuity-continuous) in question 21 is not a luxury: without it, midpoint convexity controls only the countable dyadic skeleton, and the continuum in between runs wild.

**23.** $p(x) = \frac{x^2}{2}$ satisfies $p(x+y) = p(x) +
p(y) + xy$. If $f$ is any [continuous](#def-b1-continuity-continuous) solution, $g = f - p$ is [continuous](#def-b1-continuity-continuous) and additive, so $g(x) = cx$:

$$
f(x) = \frac{x^2}{2} + cx , \qquad c \in \R ,
$$

and each of these is a solution: the list is complete.

**24.** General principle: if $u, v$ are [continuous](#def-b1-continuity-continuous) and agree on a [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) $D \subseteq \R$, then for $x \in \R$ choose $d_n \in D$ with $d_n \to x$ ([Proposition 12.11](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#prop-b1-topology-closureprops)); $u(x) = \lim u(d_n) =
\lim v(d_n) = v(x)$. Now let $f$ be [continuous](#def-b1-continuity-continuous) and additive on the [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [subgroup](https://one-course.com/books/math/3/en/chapter/7-algebraic-structures#def-b1-structures-subgroup) $G$. Fix $x, y \in \R$ and take $g_n \to x$, $g'_n \to y$ with $g_n, g'_n \in G$; then $g_n + g'_n \to x + y$ and, by sequential [continuity](#def-b1-continuity-continuous) at $x + y$, $x$ and $y$:

$$
f(x + y) = \lim f(g_n + g'_n) = \lim\bigl(f(g_n) +
f(g'_n)\bigr) = f(x) + f(y) :
$$

$f$ is additive on all of $\R$ (hence linear, by question 2).

**25.** (i) For additive $f$: linear $\iff$ [continuous](#def-b1-continuity-continuous) $\iff$ [continuous](#def-b1-continuity-continuous) at one point $\iff$ monotone on some [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals) $\iff$ bounded (even one-sidedly) on some [interval](https://one-course.com/books/math/3/en/chapter/10-real-numbers#prop-b1-reals-intervals). (ii) [Density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) of the graph in the plane shows the failure is not a local defect but a global explosion: above every subinterval the values smear over all of $\R$, so every regularity property fails everywhere at once. (iii) The catch: $cx$, $\eu^{cx}$, $c\ln x$, $x^c$, $cx^2$, and $cx + d$ — six characterizations, one method: transport the equation to Cauchy’s, prove the $\Q$-skeleton by induction, upgrade to $\R$ by [density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) plus [continuity](#def-b1-continuity-continuous). (iv) [Density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) of $\Q$ (or the dyadics) carried the upgrades in [Exercise 13.11](#exo-b1-continuity-11) and in question 21; it carried nothing in question 13, because without [continuity](#def-b1-continuity-continuous) values do not propagate from a [dense](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) [set](https://one-course.com/books/math/3/en/chapter/1-logic-sets-and-maps#def-b1-logic-sets) to its [closure](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-closure) — [density](https://one-course.com/books/math/3/en/chapter/12-topology-of-the-real-line#def-b1-topology-dense) transfers information only along [continuity](#def-b1-continuity-continuous).