---
title: "Complete Spaces: Baire, Ascoli, Stone–Weierstrass"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 7
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass
---

# Chapter 7 — Complete Spaces: Baire, Ascoli, Stone–Weierstrass

[Completeness](#def-b3-complete-complete) — every Cauchy sequence converges — is the property that lets analysis *produce* objects: fixed points of contractions, sums of series, solutions of equations obtained as limits. This chapter assembles the three great existence machines of the metric theory. *Baire’s theorem* shows that a [complete](#def-b3-complete-complete) space cannot be a countable union of negligible pieces, and conjures objects ([continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) [nowhere differentiable functions](#thm-b3-complete-nowherediff)!) out of pure cardinality-style reasoning. *Arzelà–Ascoli* identifies the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) subsets of $\mathcal C(K)$ and is analysis’ compactness workhorse — the weekend problem uses it to prove Peano’s existence theorem for differential equations. *Stone–Weierstrass* shows polynomials, and much else, are dense in $\mathcal C(K)$: approximation becomes an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) verification. Along the way we construct [completions](#thm-b3-complete-completion) and prove the extension theorem for uniformly [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), the daily bread of Chapters [12](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#ch-b3-lp), [13](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#ch-b3-hilbert) and [14](https://one-course.com/books/math/5/en/chapter/14-the-fourier-transform#ch-b3-fouriertransform).

## 7.1 Complete spaces, completions, extensions

**Definition 7.1.**

A metric space is *complete* if every Cauchy sequence converges (Year 2: $\R^n$ is complete; $\mathcal C(\intcc01)$ with $\norm\cdot_\infty$ is complete). A closed subset of a complete space is complete; a complete subset of any metric space is closed.

**Proof.** For the two statements: a Cauchy sequence of the closed $F$ converges in $X$, and its limit, adherent to $F$, lies in $F$; a convergent-in-$X$ sequence of a [complete](#def-b3-complete-complete) $A$ is Cauchy, so converges in $A$, and limits are unique. ∎

**Theorem 7.2 (Extension of uniformly continuous maps).**

Let $D \subseteq X$ be dense, $Y$ [complete](#def-b3-complete-complete), and $f \colon D \to
Y$ uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). Then $f$ extends *uniquely* to a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $\bar f \colon X \to Y$, and $\bar f$ is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity).

**Proof.** Uniqueness: two [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) extensions agree on the dense $D$, hence everywhere (the agreement set $\{g = h\}$ is closed: preimage of the closed diagonal under $x\mapsto(g(x), h(x))$). Existence: for $x \in X$ pick $d_n \to x$, $d_n \in D$. The sequence $(f(d_n))$ is Cauchy: given $\varepsilon$, uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) provides $\delta$ with $d(u,v) < \delta \Rightarrow
d(f(u), f(v)) < \varepsilon$, and $(d_n)$ is Cauchy. Define $\bar f(x) = \lim f(d_n)$; the limit does not depend on the chosen sequence (interlace two of them). $\bar f$ extends $f$ (constant sequences) and inherits the modulus of [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): if $d(x, x') < \delta$, approximating both by points of $D$ at distance $< \frac{\delta - d(x,x')}2$ gives $d(\bar f(x), \bar
f(x')) \leq \varepsilon$ in the limit — $\bar f$ is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). ∎

**Theorem 7.3 (Completion).**

Every metric space $X$ embeds isometrically as a dense subset of a [complete metric space](#def-b3-complete-complete) $\hat X$, unique up to isometry fixing $X$ pointwise: its *completion*.

**Proof.** *Existence.* Let $\mathcal X$ be the set of Cauchy sequences of $X$, with the pseudo-distance

$$
D\bigl((x_n), (y_n)\bigr) = \lim_n d(x_n, y_n),
$$

the limit existing because $\abs{d(x_n, y_n) - d(x_m, y_m)}
\leq d(x_n, x_m) + d(y_n, y_m)$ makes the real sequence Cauchy. Let $\hat X = \mathcal X/{\sim}$, identifying sequences at $D$-distance $0$; $D$ descends to a distance. Embed $X$ by constant sequences: an isometry, with dense image (a Cauchy sequence is $D$-approximated by the constants built on its own terms: $D\bigl((x_n), (x_k)_{\rm
const}\bigr) = \lim_n d(x_n, x_k) \to 0$ as $k \to \infty$ by Cauchyness). [Completeness](#def-b3-complete-complete) of $\hat X$: let $(\xi^k)$ be Cauchy in $\hat X$; by density pick $x_k \in X$ with $D(\xi^k, x_k)
\leq 2^{-k}$; then $(x_k)$ is Cauchy in $X$ (triangle inequality through the $\xi$’s), defines a point $\xi \in \hat
X$, and $D(\xi^k, \xi) \leq 2^{-k} + D(x_k, \xi) \to 0$ (the distance from the constant $x_k$ to the class of $(x_j)_j$ is $\lim_j d(x_k, x_j)$, small for large $k$).

*Uniqueness*: two [completions](#thm-b3-complete-completion) $\hat X_1, \hat X_2$ contain $X$ densely; the identity of $X$, an isometry, is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), so extends to $\hat X_1 \to \hat X_2$ ([Theorem 7.2](#thm-b3-complete-extension)), still an isometry on a [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) hence everywhere; symmetrically in the other direction, and the composites fix the dense $X$: they are the identities. ∎

**Theorem 7.4 (Banach fixed point).**

Let $X$ be [complete](#def-b3-complete-complete), nonempty, and $f \colon X \to X$ a *contraction*: $d(f(x), f(y)) \leq k\,d(x,y)$ with $k < 1$. Then $f$ has a unique fixed point $x^*$, and every [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) converges to it, with the explicit rate $d(x_n, x^*) \leq
\frac{k^n}{1-k}\,d(x_1, x_0)$.

**Proof.** (Year 2 proved this; we re-record the two-line argument for self-containedness.) The [orbit](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-action) $x_{n+1} = f(x_n)$ has $d(x_{n+1},
x_n) \leq k^nd(x_1, x_0)$, hence is Cauchy (geometric series); its limit $x^*$ is fixed ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$), unique since two fixed points satisfy $d \leq k\,d$. The rate: sum the geometric tail. ∎

**Example 7.5 (Perturbing the identity).**

Let $g \colon \R^d \to \R^d$ be $k$-Lipschitz with $k < 1$. Then $\varphi = \mathrm{id} + g$ is a *[homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $\R^d$ onto $\R^d$*. Injectivity, with a quantitative modulus:

$$
\norm{\varphi(x) - \varphi(y)} \geq \norm{x - y} -
\norm{g(x) - g(y)} \geq (1 - k)\norm{x - y} .
$$

Surjectivity is the fixed point theorem: solving $\varphi(x) = y$ means $x = y - g(x)$, and $x \mapsto y -
g(x)$ is a $k$-contraction of the [complete](#def-b3-complete-complete) $\R^d$ — a unique solution $x = \psi(y)$ exists for every $y$. The displayed inequality makes the inverse $\psi$ Lipschitz with constant $\frac1{1-k}$: a [homeomorphism](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), with explicit bounds on both moduli. This innocuous-looking statement is the engine inside the inverse function theorem ([Chapter 20](https://one-course.com/books/math/5/en/chapter/20-submanifolds-of-rn#ch-b3-submanifolds)): near a point where $Df$ is invertible, $f$ *is* an invertible linear map plus a small Lipschitz perturbation, and today’s example does the rest. It also quantifies numerical robustness: a system perturbed by less than the inverse’s margin remains [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived), with the solution moving by at most $\frac{1}{1-k}$ times the perturbation.

## 7.2 Baire’s theorem

**Theorem 7.6 (Baire).**

In a [complete metric space](#def-b3-complete-complete), a countable intersection of dense [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) is dense. Equivalently: if $X = \bigcup_{n}F_n$ with each $F_n$ closed, then some $F_n$ has nonempty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior).

**Proof.** Let $(U_n)$ be dense opens and $B_0 = B(x_0, r_0)$ any open ball; we find a point of $\bigcap U_n$ in $B_0$. Inductively: $U_{n}$ being dense and open, it meets the open ball $B_{n-1}$ in an [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology), which contains a closed ball $\bar B(x_n, r_n)$ with $0 < r_n \leq r_{n-1}/2$ and $\bar B(x_n, r_n) \subseteq
B_{n-1}\cap U_n$. The [centers](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) form a Cauchy sequence ($x_m \in
B_n$ for $m \geq n$, radii $\to 0$); the limit $x$ lies in every $\bar B(x_n, r_n)$ (closedness), hence in every $U_n$ and in $B_0$. For the second form: if no $F_n$ has [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior), the $U_n = X \setminus F_n$ are open and dense, and a point of $\bigcap U_n$ escapes $\bigcup F_n = X$: absurd. ∎

**Remark 7.7.**

Vocabulary: a set is *nowhere dense* if its [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) has empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior), *meagre* (first category) if it is a countable union of nowhere [dense sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior). Baire: *a [complete metric space](#def-b3-complete-complete) is not [meagre](#rem-b3-complete-meagre) in itself*, and the complement of a [meagre set](#rem-b3-complete-meagre) is dense. “[Meagre](#rem-b3-complete-meagre)” is a notion of smallness orthogonal to measure ([Chapter 9](https://one-course.com/books/math/5/en/chapter/9-measure-theory#ch-b3-measure) will produce [meagre sets](#rem-b3-complete-meagre) of full measure), and Baire arguments prove *existence by abundance*: to exhibit one object without property P, show the P-objects form a [meagre set](#rem-b3-complete-meagre).

**Corollary 7.8.**

(a) $\R$ is uncountable. (b) $\Q$ is not a countable intersection of open subsets of $\R$, and a nonempty [complete metric space](#def-b3-complete-complete) without isolated points is uncountable.

**Proof.** (a) $\R = \bigcup_{x}\{x\}$ over a countable set would make some singleton have [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior). (b) If $\Q = \bigcap_n V_n$ with $V_n$ open (necessarily dense, as $\supseteq \Q$), then the sets $V_n$ and the complements $\R\setminus\{q\}$, $q \in \Q$, form a countable family of dense opens with empty intersection — contradicting Baire. If $X$ is [complete](#def-b3-complete-complete) without isolated points and countable, $X = \bigcup_{x \in X}\{x\}$ exhibits it as a countable union of closed sets with empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) (no isolated points): Baire again. ∎

**Theorem 7.9 (Weierstrass’s monsters exist).**

There exist [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions on $\intcc01$ differentiable at no point. Indeed, the set of $f \in \mathcal C(\intcc01)$ having a (finite) derivative at even one point is [meagre](#rem-b3-complete-meagre) in $(\mathcal C(\intcc01), \norm\cdot_\infty)$.

**Proof.** For $n \geq 1$ let

$$
F_n = \Bigl\{f : \exists x \in \intcc01,\
\forall h \neq 0 \text{ with } x + h \in \intcc01,\
\abs{f(x+h) - f(x)} \leq n\abs h \Bigr\}.
$$

If $f$ is differentiable at $x$, then $f \in F_n$ for some $n$: the quotient $\abs{f(x+h)-f(x)}/\abs h$ is bounded for $\abs h
\leq \delta$ (differentiability: it tends to $\abs{f'(x)}$) and bounded by $2\norm f_\infty/\delta$ for $\abs h \geq \delta$. So $\bigcup F_n$ contains all somewhere-differentiable functions, and it suffices to show each $F_n$ is closed with empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior).

*Closed*: let $f_k \to f$ uniformly, $f_k \in F_n$ with witnesses $x_k \to x$ (compactness, after extraction). For $h$ with $x + h \in \intcc01$: choose $h_k \to h$ with $x_k + h_k
\in \intcc01$ (e.g. $h_k = h + x - x_k$ clipped); then $\abs{f(x + h) - f(x)} = \lim \abs{f_k(x_k + h_k) - f_k(x_k)}
\leq \lim n\abs{h_k} = n\abs h$, using uniform convergence and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$ at the relevant points: $f \in F_n$.

*Empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior)*: given $f \in F_n$ and $\varepsilon > 0$, we find $g$ with $\norm{g - f}_\infty \leq \varepsilon$ and $g
\notin F_n$. First approximate $f$ within $\varepsilon/2$ by a piecewise affine $\varphi$ (uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): interpolate on a fine grid), of slopes bounded by some $M$. Add a small sawtooth: $g = \varphi + \frac\varepsilon2\,s_N$, where $s_N(x)$ is the $\frac1N$-periodic zigzag of amplitude $1$ and slope $\pm 2N$. At every $x$, on one side there is $h$ arbitrarily small with the sawtooth contributing slope $\pm 2N\cdot\frac\varepsilon2 =
\pm\varepsilon N$ over $[x, x+h]$: the difference quotient of $g$ exceeds $\varepsilon N - M > n$ for $N$ large. So $g \notin
F_n$, at any uniform distance $\leq \varepsilon$ from $f$. Conclusion: $\bigcup F_n$ is [meagre](#rem-b3-complete-meagre); by Baire its complement — made of [nowhere differentiable functions](#thm-b3-complete-nowherediff) — is dense in $\mathcal C(\intcc01)$: such functions exist *in abundance*. ∎

## 7.3 Arzelà–Ascoli

Throughout, $K$ is a *[compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)* metric space and $\mathcal
C(K) = \mathcal C(K, \R^d)$, with $\norm f_\infty = \sup_K
\norm{f(x)}$: a [complete](#def-b3-complete-complete) space (uniform limits of [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) — Year 2).

**Definition 7.10.**

A family $\mathcal F \subseteq \mathcal C(K)$ is *equicontinuous* if for every $\varepsilon > 0$ there is $\delta > 0$ such that

$$
d(x, y) < \delta \implies \norm{f(x) - f(y)} < \varepsilon
\quad\text{for \emph{all} } f \in \mathcal F
$$

(one $\delta$ for the whole family — e.g. any family with a common Lipschitz constant, or a common Hölder modulus), and *pointwise bounded* if $\sup_{f}\norm{f(x)} < \infty$ for each $x$.

**Theorem 7.11 (Arzelà–Ascoli).**

A subset $\mathcal F \subseteq \mathcal C(K)$ is relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) (has [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior)) iff it is [equicontinuous](#def-b3-complete-equicontinuous) and pointwise bounded. In particular, every [equicontinuous](#def-b3-complete-equicontinuous), pointwise bounded *sequence* has a uniformly convergent subsequence.

**Proof.** ($\Leftarrow$) Let $(f_k)$ be a sequence of $\mathcal F$. $K$ [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) metric is *separable*: for each $n$, finitely many balls of radius $\frac1n$ cover $K$ ([Theorem 6.16](https://one-course.com/books/math/5/en/chapter/6-general-topology#thm-b3-topology-metriccompact)); their [centers](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) form a countable [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) $D = \{x_1, x_2, \dots\}$. By pointwise boundedness and Bolzano–Weierstrass, extract successively subsequences converging at $x_1$, then also at $x_2$, etc., and take the *diagonal* subsequence $(g_j)$: it converges at every point of $D$. [Equicontinuity](#def-b3-complete-equicontinuous) upgrades this to uniform Cauchy: given $\varepsilon$, take $\delta$ as in the definition, cover $K$ by finitely many balls $B(x_i, \delta)$ with $x_i \in D$ ($i \leq m$), and pick $J$ so large that $\norm{g_j(x_i) - g_{j'}(x_i)} < \varepsilon$ for $j, j' \geq
J$, $i \leq m$. For arbitrary $x \in B(x_i, \delta)$:

$$
\norm{g_j(x) - g_{j'}(x)}
\leq \norm{g_j(x) - g_j(x_i)} + \norm{g_j(x_i) - g_{j'}(x_i)}
+ \norm{g_{j'}(x_i) - g_{j'}(x)} < 3\varepsilon .
$$

So $(g_j)$ is uniformly Cauchy, and converges in the [complete](#def-b3-complete-complete) $\mathcal C(K)$. Thus every sequence of $\mathcal F$ has a convergent subsequence: $\bar{\mathcal F}$ is (sequentially, hence by [Theorem 6.16](https://one-course.com/books/math/5/en/chapter/6-general-topology#thm-b3-topology-metriccompact)) [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact).

($\Rightarrow$) If $\bar{\mathcal F}$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): pointwise boundedness is clear (evaluation is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). For [equicontinuity](#def-b3-complete-equicontinuous), cover $\mathcal F$ by finitely many balls $B(f_i, \varepsilon)$ of $\mathcal C(K)$; each $f_i$ is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (Heine, [Corollary 6.17](https://one-course.com/books/math/5/en/chapter/6-general-topology#cor-b3-topology-heineborel)), giving a common $\delta$ for $i \leq m$; then for $f \in B(f_i, \varepsilon)$ and $d(x,y)
< \delta$: $\norm{f(x)-f(y)} \leq 2\varepsilon +
\norm{f_i(x)-f_i(y)} < 3\varepsilon$. ∎

**Example 7.12.**

The closed unit ball of $\mathcal C(\intcc01)$ is *not* [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ($f_n(x) = x^n$ has no uniformly convergent subsequence: the pointwise limit is discontinuous), and indeed $(x^n)$ is not [equicontinuous](#def-b3-complete-equicontinuous) at $1$. By contrast $\{f : \norm
f_\infty \leq 1,\ \operatorname{Lip}(f) \leq 1\}$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): bounded and $1$-Lipschitz-equicontinuous, and closed. Ascoli explains *why* compactness fails in infinite dimension (Riesz, Year 2) and what to add to restore it: a uniform modulus of [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity).

## 7.4 Stone–Weierstrass

**Lemma 7.13 (Dini).**

Let $K$ be [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and $(f_n)$ a *monotone* sequence of [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) real functions converging *pointwise* to a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$. Then the convergence is uniform.

**Proof.** Say $f_n \uparrow f$; let $g_n = f - f_n \downarrow 0$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). Given $\varepsilon$, the [open sets](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $U_n = \{g_n <
\varepsilon\}$ increase and cover $K$ (pointwise convergence); extract a finite subcover: $K = U_{n_0}$ for some $n_0$ (increasing family), i.e. $0 \leq g_n < \varepsilon$ everywhere for $n \geq n_0$. ∎

**Lemma 7.14.**

There is a sequence of *polynomials* $u_n$ with $u_n(t) \to
\sqrt t$ uniformly on $\intcc01$.

**Proof.** Define $u_0 = 0$, $u_{n+1}(t) = u_n(t) + \frac12\bigl(t -
u_n(t)^2\bigr)$: polynomials. By induction $0 \leq u_n(t) \leq
\sqrt t$ on $\intcc01$: granting it for $n$,

$$
\sqrt t - u_{n+1}(t) = \bigl(\sqrt t - u_n(t)\bigr)
\Bigl(1 - \tfrac12\bigl(\sqrt t + u_n(t)\bigr)\Bigr) \geq 0,
$$

since $\sqrt t + u_n \leq 2$; and $u_{n+1} \geq u_n \geq 0$. So $(u_n(t))$ is nondecreasing, bounded by $\sqrt t$: it converges pointwise, and the limit $\ell(t)$ satisfies $\ell =
\ell + \frac12(t - \ell^2)$: $\ell(t) = \sqrt t$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). Dini ([Lemma 7.13](#lem-b3-complete-dini)) upgrades to uniform. ∎

**Theorem 7.15 (Stone–Weierstrass, real version).**

Let $K$ be a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ([Hausdorff](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-hausdorff)) space and $\mathcal A \subseteq
\mathcal C(K, \R)$ a *subalgebra* (stable under sums, products, scalar multiples) that *contains the constants* and *separates points* (for $x \neq y$, some $f \in
\mathcal A$ has $f(x) \neq f(y)$). Then $\mathcal A$ is dense in $(\mathcal C(K,\R), \norm\cdot_\infty)$.

**Proof.** Let $\bar{\mathcal A}$ be the [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior), again an algebra (products of uniform limits on bounded sets converge).

*Step 1: $\bar{\mathcal A}$ is a lattice*, i.e. stable under $\max$ and $\min$. Since $\max(f,g) = \frac{f + g +
\abs{f-g}}2$ and $\min$ likewise, it suffices that $f \in
\bar{\mathcal A} \Rightarrow \abs f \in \bar{\mathcal A}$: with $M = \norm f_\infty > 0$, $\abs f = M\sqrt{(f/M)^2}$, and [Lemma 7.14](#lem-b3-complete-sqrt) gives polynomials $u_n$ with $u_n\bigl((f/M)^2\bigr) \to \abs f/M$ uniformly; polynomials in members of the algebra (with constant term: constants are there) stay in $\bar{\mathcal A}$.

*Step 2: two-point interpolation.* For $x \neq y$ and $a,
b \in \R$, some $g \in \mathcal A$ has $g(x) = a$, $g(y) = b$: take $h$ separating $x, y$ and set $g = a + (b -
a)\frac{h - h(x)}{h(y) - h(x)}$.

*Step 3.* Let $f \in \mathcal C(K)$, $\varepsilon > 0$. For each pair $x, y$ pick $g_{x,y} \in \mathcal A$ with $g_{x,y}(x) = f(x)$, $g_{x,y}(y) = f(y)$ (Step 2; for $x = y$ take the constant function $g_{x,x} = f(x)$). Fix $x$: for each $y$, the [open set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) $V_y = \{g_{x,y} < f +
\varepsilon\}$ contains $y$; compactness extracts $y_1, \dots,
y_m$ with $K = \bigcup V_{y_j}$, and $h_x = \min_j g_{x, y_j}
\in \bar{\mathcal A}$ (Step 1) satisfies $h_x < f +
\varepsilon$ everywhere, $h_x(x) = f(x)$. Now vary $x$: $W_x =
\{h_x > f - \varepsilon\}$ is open, contains $x$; extract $x_1,
\dots, x_l$ covering $K$, and $h = \max_i h_{x_i} \in
\bar{\mathcal A}$ satisfies $f - \varepsilon < h < f +
\varepsilon$: $\norm{f - h}_\infty \leq \varepsilon$. Hence $f \in \bar{\mathcal A}$. ∎

**Corollary 7.16.**

(a) (*Weierstrass*) Polynomials are dense in $\mathcal
C(\intcc ab, \R)$; polynomials in $n$ variables are dense in $\mathcal C(K, \R)$ for [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K \subseteq \R^n$. (b) (*Complex version*) If $\mathcal A \subseteq \mathcal
C(K, \C)$ is a subalgebra containing constants, separating points, and *stable under conjugation*, it is dense. (c) (*Trigonometric version*) Trigonometric polynomials $\sum_{\abs n \leq N}c_n\eu^{\iu n t}$ are dense in the space of [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $2\pi$-periodic functions with $\norm\cdot_\infty$.

**Proof.** (a) Polynomials form an algebra with constants; the coordinate functions separate points of $\R^n$. (b) The real and imaginary parts $\frac{f + \bar f}2$, $\frac{f - \bar f}{2\iu}$ of members of $\mathcal A$ form a real algebra $\mathcal A_\R \subseteq \mathcal C(K, \R)$ with constants; it separates points ($f(x) \ne f(y)$ forces $\operatorname{Re}f$ or $\operatorname{Im}f$ to separate). Apply the real theorem and recombine. (c) View $2\pi$-periodic functions as $\mathcal C(S^1, \C)$ ($S^1 = \R/2\pi\Z$, [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): [Exercise 6.5](https://one-course.com/books/math/5/en/chapter/6-general-topology#exo-b3-topology-5)); the algebra generated by $\eu^{\iu t}, \eu^{-\iu t}$ and constants is stable under conjugation and separates points of the circle ($\eu^{\iu t}$ is injective on it). Apply (b). ∎

**Remark 7.17.**

The trigonometric version repairs, and vastly generalizes, the gap left in Year 2’s Fourier chapter: density of trigonometric polynomials in $(\mathcal C(S^1), \norm\cdot_2)$ follows a fortiori ($\norm\cdot_2 \leq \norm\cdot_\infty$ up to the normalizing constant), which will make the Fourier system an orthonormal *basis* in [Chapter 13](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#ch-b3-hilbert), proving Parseval in full generality at last.

## 7.5 Exercises

**Exercise 7.1 ★.**

(a) Show that $\mathcal C(\intcc01, \R)$ with $\norm f_1 = \int_0^1\abs f$ is *not* [complete](#def-b3-complete-complete): the functions $f_n$ equal to $0$ on $\intcc0{\frac12}$, $1$ on $\intcc{\frac12 + \frac1n}1$, affine between, are $\norm\cdot_1$-Cauchy with no [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) limit. (b) Show that a normed space in which every absolutely convergent series converges is [complete](#def-b3-complete-complete). *(Extract from a Cauchy sequence a subsequence with $\norm{x_{n_{k+1}} -
x_{n_k}} \leq 2^{-k}$.)*

**Solution of Exercise 7.1.**

(a) For $m \geq n$, $f_n - f_m$ vanishes outside an interval of length $\frac1n$ and is bounded by $1$: $\norm{f_n - f_m}_1
\leq \frac1n \to 0$: Cauchy. If $f_n \to f$ in $\norm\cdot_1$ with $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): for fixed $\alpha <
\frac12$, $\int_0^{\alpha}\abs f = \lim \int_0^\alpha\abs{f -
f_n + f_n} \leq \lim\bigl(\norm{f - f_n}_1 + 0\bigr) = 0$ ($f_n \equiv 0$ there for large $n$), so $f \equiv 0$ on $[0, \frac12)$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)); likewise $f \equiv 1$ on $(\frac12, 1]$: no [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) function does both. So the space is incomplete — the [completion](#thm-b3-complete-completion) is $L^1$, built in [Chapter 12](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#ch-b3-lp).

(b) Let $(x_n)$ be Cauchy; choose $n_1 < n_2 < \cdots$ with $\norm{x_{n_{k+1}} - x_{n_k}} \leq 2^{-k}$. The series $\sum_k (x_{n_{k+1}} - x_{n_k})$ converges absolutely, hence converges; its partial sums are $x_{n_{k+1}} - x_{n_1}$, so $(x_{n_k})$ converges, and a Cauchy sequence with a convergent subsequence converges.

**Exercise 7.2 ★.**

Using Baire: (a) show that a [complete](#def-b3-complete-complete) normed space has no countable ([algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic)) basis — deduce that the space of polynomials is [complete](#def-b3-complete-complete) for *no* norm; (b) show that if a sequence of [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f_n \colon \R \to \R$ converges pointwise to $f$, the set of [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) points of $f$ is dense. *(For (b): admit or prove that $\Omega_\delta = \{x :
\operatorname{osc}_x f < \delta\}$ is open, and show it is dense using $F_{N} = \{x: \abs{f_n(x) - f_m(x)} \leq \delta/3\
\forall n,m \geq N\}$, closed sets covering $\R$; work in an arbitrary closed ball to apply Baire.)*

**Solution of Exercise 7.2.**

(a) Suppose $E$ [complete](#def-b3-complete-complete) with [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) basis $(e_n)_{n\in\N}$ and let $F_n = \operatorname{Vect}(e_1, \dots, e_n)$: closed (finite-dimensional subspaces are [complete](#def-b3-complete-complete), hence closed — Year 2), with empty [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior): if $B(x, r) \subseteq F_n$, take $v \notin F_n$; then $x + \frac{r}{2\norm v}v \in B(x,r)
\setminus F_n$, absurd. But $E = \bigcup F_n$ (every vector is a finite combination): contradicts Baire ([Theorem 7.6](#thm-b3-complete-baire)). The space $\R[X]$ has the countable basis $(X^n)$, so no norm makes it [complete](#def-b3-complete-complete).

(b) Fix $\delta > 0$ and a nonempty open ball $B$; we find in $B$ a point of $\Omega_\delta = \{x : \operatorname{osc}_xf <
\delta\}$, where $\operatorname{osc}_xf = \inf_{V \ni
x}\operatorname{diam} f(V)$. ($\Omega_\delta$ is open: if $\operatorname{diam}f(V) < \delta$ for an open $V \ni x$, every $y \in V$ has oscillation $< \delta$.) The sets

$$
F_N = \bigl\{x : \abs{f_n(x) - f_m(x)} \leq \tfrac\delta3\
\ \forall\, n, m \geq N \bigr\}
$$

are closed (intersections of preimages of closed sets) and cover $\R$ (pointwise convergence makes $(f_n(x))$ Cauchy). Applying Baire inside the [complete](#def-b3-complete-complete) $\bar B$: some $F_N \cap
\bar B$ contains a ball $B' = B(x_0, \rho)$. Letting $m \to
\infty$: $\abs{f_N - f} \leq \frac\delta3$ on $B'$. By [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f_N$ at $x_0$, shrink to $B'' \ni x_0$ where $\abs{f_N - f_N(x_0)} \leq \frac\delta3$; then for $x \in B''$,

$$
\abs{f(x) - f(x_0)} \leq \abs{f - f_N}(x) + \abs{f_N(x) -
f_N(x_0)} + \abs{f_N - f}(x_0) \leq \delta:
$$

$\operatorname{diam} f(B'') \leq 2\delta$, so $x_0 \in
\Omega_{3\delta} \cap B$. Thus each $\Omega_\delta$ is open and dense; $\bigcap_k\Omega_{1/k}$ — the set of [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) points — is dense by Baire.

**Exercise 7.3 ★★.**

(a) Show that $f(x) = \frac12\bigl(x + \frac ax\bigr)$ ($a >
1$) is a contraction of $[\sqrt a, +\infty)$ and identify its fixed point — Heron’s method. Estimate the number of iterations for $10^{-12}$-accuracy starting from $x_0 = a$, for $a = 2$. (b) (Kepler’s equation) For $0 \leq e < 1$ and $m \in \R$, show that $x = m + e\sin x$ has a unique solution, depending [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $m$.

**Solution of Exercise 7.3.**

(a) $f(x) = \frac12(x + \frac ax)$ maps $[\sqrt a, \infty)$ to itself (AM–GM: $f(x) \geq \sqrt{x \cdot \frac ax} = \sqrt a$), and $f'(x) = \frac12(1 - \frac a{x^2}) \in [0, \frac12)$ there: a $\frac12$-Lipschitz contraction of a closed ([complete](#def-b3-complete-complete)) set. Fixed point: $x = f(x) \iff x^2 = a$: $x^* = \sqrt a$. Rate ([Theorem 7.4](#thm-b3-complete-banach)): $d(x_n, \sqrt a) \leq
2^{-n+1}\,d(x_1, x_0)$. For $a = 2$, $x_0 = 2$: $d(x_1, x_0) =
\frac12$, so $n = 40$ guarantees $2^{-40} < 10^{-12}$. (In reality Newton’s method converges quadratically: a handful of iterations suffice; the contraction estimate is pessimistic but free.)

(b) $f_m(x) = m + e\sin x$ is $e$-Lipschitz with $e < 1$ on the [complete](#def-b3-complete-complete) $\R$: unique fixed point $x(m)$. For two parameters:

$$
\abs{x(m) - x(m')} = \abs{f_m(x(m)) - f_{m'}(x(m'))}
\leq \abs{m - m'} + e\abs{x(m) - x(m')},
$$

so $\abs{x(m) - x(m')} \leq \frac{\abs{m - m'}}{1 - e}$: even Lipschitz in $m$.

**Exercise 7.4 ★★.**

(a) Let $X$ be [complete](#def-b3-complete-complete) and $f \colon X \to X$ such that some iterate $f^p$ is a contraction. Show that $f$ has a unique fixed point. Application: the integral operator $T$ on $\mathcal C(\intcc0a)$, $Tf(x) = \int_0^x f$, satisfies $\norm{T^p} \leq a^p/p!$ — deduce that $u = g + \lambda Tu$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) for every $\lambda$. (b) (Edelstein) Let $K$ be *[compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)* and $f\colon K \to K$ with $d(f(x), f(y)) < d(x, y)$ for $x \neq y$. Show that $f$ has a unique fixed point, but that the contraction rate can be lost: on $X = [1, +\infty)$ ([complete](#def-b3-complete-complete), not [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)), $f(x) = x
+ \frac1x$ has no fixed point despite strictly decreasing distances.

**Solution of Exercise 7.4.**

(a) Let $x^*$ be the unique fixed point of $f^p$. Then $f^p(f(x^*)) = f(f^p(x^*)) = f(x^*)$: $f(x^*)$ is a fixed point of $f^p$, so $f(x^*) = x^*$. A fixed point of $f$ is one of $f^p$: uniqueness transfers. For $T$: by induction $\abs{T^pf(x)}
\leq \norm f_\infty x^p/p!$ (each integration adds a factor $\frac xk$), so $\norm{T^p} \leq a^p/p!$. The map $S(u) = g +
\lambda Tu$ satisfies $S^p(u) - S^p(v) = \lambda^pT^p(u - v)$, of norm $\leq \abs\lambda^pa^p/p!\,\norm{u - v} \to 0$: some $S^p$ is a contraction, and $S$ has a unique fixed point: the Volterra equation $u = g + \lambda Tu$ is uniquely [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) for *every* $\lambda$.

(b) $\varphi(x) = d(x, f(x))$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $K$: it attains its minimum at some $x_0$. If $f(x_0) \neq x_0$, then $\varphi(f(x_0)) = d(f(x_0), f^2(x_0)) < d(x_0, f(x_0)) =
\min\varphi$: absurd. Uniqueness: two fixed points $x \ne y$ would give $d(x,y) = d(f(x), f(y)) < d(x,y)$. Without compactness: $f(x) = x + \frac1x$ on $[1, \infty)$ satisfies, for $x < y$, $f(y) - f(x) = (y - x)\bigl(1 - \frac1{xy}\bigr) <
y - x$, yet $f(x) > x$ always: no fixed point — strict distance decrease is weaker than a uniform contraction factor.

**Exercise 7.5 ★★.**

(a) Two [continuous maps](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) into a [Hausdorff space](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-hausdorff) agreeing on a dense subset agree everywhere; where was this used in the chapter? (b) Let $D \subseteq X$ dense and $f \colon D \to Y$ an isometric bijection onto a dense subset of a [complete](#def-b3-complete-complete) $Y$, with $X$ [complete](#def-b3-complete-complete). Show that $f$ extends to an isometric bijection $X \to Y$. Deduce again the uniqueness of [completions](#thm-b3-complete-completion).

**Solution of Exercise 7.5.**

(a) The set $A = \{g = h\}$ is the preimage of the diagonal $\Delta_Y$ under $x \mapsto (g(x), h(x))$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity); $\Delta_Y$ is closed because $Y$ is [Hausdorff](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-hausdorff) (for $y_1 \neq y_2$, disjoint open [neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) give an open box around $(y_1,
y_2)$ disjoint from the diagonal, so the complement of $\Delta_Y$ is open): so $A$ is closed, contains a [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior), equals $X$. Used: uniqueness in [Theorem 7.2](#thm-b3-complete-extension), hence in the uniqueness of [completions](#thm-b3-complete-completion).

(b) $f$, an isometry, is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): it extends to $F\colon X \to Y$ ([Theorem 7.2](#thm-b3-complete-extension)), still isometric (the relation $d(F(x), F(x')) = d(x, x')$ holds on a [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) of pairs and both sides are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)). Likewise $f^{-1}\colon f(D) \to X$ extends to $G \colon Y \to X$. The composite $G \circ F$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and fixes the dense $D$: it is $\mathrm{id}_X$ (part (a)); symmetrically $F \circ G =
\mathrm{id}_Y$. So $F$ is an isometric bijection. Uniqueness of [completions](#thm-b3-complete-completion): apply this to $D = X$ sitting densely in two [completions](#thm-b3-complete-completion).

**Exercise 7.6 ★★.**

Which of the following families are [equicontinuous](#def-b3-complete-equicontinuous), pointwise bounded, relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) in $\mathcal C(\intcc01)$?

$$
\{x \mapsto \sin(nx)\}_{n},\qquad
\{x \mapsto x^n\}_n,
$$

$$
\Bigl\{f \in \mathcal C^1 : \norm f_\infty \leq 1,\
\norm{f'}_\infty \leq 1\Bigr\},\qquad
\{f : \operatorname{Lip}(f)\leq 1,\ f(0) = 0\}.
$$

Justify each answer with Ascoli or a counterexample sequence.

**Solution of Exercise 7.6.**

$\{\sin(nx)\}$: pointwise bounded by $1$; *not* [equicontinuous](#def-b3-complete-equicontinuous): at $x = 0$, $\sin(n\cdot\frac{\pi}{2n}) = 1$ with $\frac\pi{2n} \to 0$, violating any common $\delta$ for $\varepsilon = \frac12$. Not relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) (Ascoli’s necessity, [Theorem 7.11](#thm-b3-complete-ascoli)).

$\{x^n\}$: bounded; not [equicontinuous](#def-b3-complete-equicontinuous) at $1$: $1 - (1 -
\delta)^n \to 1$ as $n \to \infty$ for fixed $\delta$. Not relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — consistently, its pointwise limit is discontinuous, so no subsequence converges uniformly.

$\{\norm f_\infty \leq 1, \norm{f'}_\infty \leq 1\}$: the mean value inequality makes the family $1$-Lipschitz, hence [equicontinuous](#def-b3-complete-equicontinuous); bounded: relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) by Ascoli. (Not [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): it is not closed — uniform limits need not be $\mathcal C^1$; its [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) is the $1$-Lipschitz functions of norm $\leq 1$.)

$\{\operatorname{Lip}(f) \leq 1, f(0) = 0\}$: [equicontinuous](#def-b3-complete-equicontinuous); pointwise bounded ($\abs{f(x)} \leq x \leq 1$); and closed under uniform limits (the Lipschitz inequality and the value at $0$ pass to limits): [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact).

**Exercise 7.7 ★★★.**

([Compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) integral operators) Let $k \in \mathcal C(\intcc01^2)$ and, for $f \in \mathcal C(\intcc01)$, $Tf(x) =
\int_0^1k(x,y)f(y)\,\dd y$. (a) Show that $T$ maps the unit ball of $\mathcal C(\intcc01)$ to an [equicontinuous](#def-b3-complete-equicontinuous), uniformly bounded set; conclude that $T$ is a *[compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) operator*: images of bounded sets are relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact). (b) Deduce that if $(f_n)$ is bounded, $(Tf_n)$ has a uniformly convergent subsequence, and that $T$ cannot be a bijection with [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) inverse. *(The image of the unit ball would be a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $0$ in $\mathcal C(\intcc01)$: forbidden by Riesz’s theorem from Year 2.)*

**Solution of Exercise 7.7.**

(a) For $\norm f_\infty \leq 1$: $\abs{Tf(x)} \leq
\norm k_\infty$, and

$$
\abs{Tf(x) - Tf(x')} \leq \int_0^1\abs{k(x,y) -
k(x',y)}\,\dd y \leq \omega_k\bigl(\abs{x - x'}\bigr),
$$

where $\omega_k$ is a modulus of uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $k$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) square (Heine): the image of the unit ball is uniformly bounded and [equicontinuous](#def-b3-complete-equicontinuous), hence relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) (Ascoli). By linearity every bounded set has relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) image: $T$ is a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) operator.

(b) The subsequence statement is the definition of relative compactness applied to $\{Tf_n\}$. If $T$ were bijective with [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) inverse, then $T(B(0,1)) \supseteq B(0,
\varepsilon)$ for some $\varepsilon > 0$ ($T^{-1}$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $0$); the closed ball $\bar B(0,\varepsilon)$, a closed subset of the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $\overline{T(B(0,1))}$, would be [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — impossible in the infinite-dimensional $\mathcal
C(\intcc01)$ by Riesz’s theorem (Year 2).

**Exercise 7.8 ★★.**

Prove or disprove, for $f_n \colon \intcc01 \to \R$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): (a) $f_n \downarrow 0$ pointwise $\Rightarrow$ uniformly (Dini — reprove it); (b) same without monotonicity; (c) same with monotonicity but $f$ discontinuous; (d) same with monotonicity, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) limit, but on $\intoo01$.

**Solution of Exercise 7.8.**

(a) Dini: see [Lemma 7.13](#lem-b3-complete-dini) — the covering argument. (b) False: the moving bump $f_n(x) = \max(0, 1 -
\abs{nx - 1})$ tends to $0$ pointwise (for $x > 0$, $f_n(x) =
0$ once $n > 2/x$; $f_n(0) = 0$) but $\norm{f_n}_\infty = 1$. (c) False: $f_n(x) = x^n$ decreases to the discontinuous $\mathbf 1_{\{1\}}$; $\sup_{[0,1]}\abs{f_n - f} = \sup_{[0,1)}
x^n = 1$. (d) False: $x^n \downarrow 0$ pointwise on the noncompact $\intoo01$, with $\sup_{(0,1)}x^n = 1$. Each hypothesis of Dini is needed.

**Exercise 7.9 ★★.**

(a) (Moments determine) Let $f \in \mathcal C(\intcc01)$ with $\int_0^1 f(x)\,x^n\,\dd x = 0$ for all $n \in \N$. Show $f =
0$. *(Approximate $f$ uniformly by polynomials and compute $\int f^2$.)* (b) Show that the even polynomials are dense in $\mathcal
C(\intcc01)$ but *not* in $\mathcal C(\intcc{-1}1)$; where does Stone–Weierstrass’s hypothesis fail? (c) Is the algebra generated by $x \mapsto \eu^{\iu x}$ alone (without $\eu^{-\iu x}$) dense in $\mathcal C(S^1, \C)$? *(Consider $\int_0^{2\pi}f(t)\,\eu^{\iu t}\dd t$.)*

**Solution of Exercise 7.9.**

(a) By linearity $\int_0^1 fP = 0$ for every polynomial $P$. Choose $P_n \to f$ uniformly ([Corollary 7.16](#cor-b3-complete-weierstrass)): $\int_0^1 f^2 =
\lim\int_0^1 fP_n = 0$, and the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f^2 \geq 0$ with zero integral vanishes identically.

(b) On $\intcc01$: the polynomials in $x^2$ form an algebra with constants, separating points ($x \mapsto x^2$ is injective on $\intcc01$): dense by Stone–Weierstrass. On $\intcc{-1}1$: $x^2$ takes equal values at $\pm x$, and so does every polynomial in $x^2$: a uniform limit of such is an even function. If even functions $g_n$ converged uniformly to the identity, then $x = \lim g_n(x) = \lim g_n(-x) = -x$ for all $x$: absurd — not dense. The separation hypothesis fails at the pairs $\{x, -x\}$.

(c) No. For $f$ in the algebra $\mathcal A$ generated by constants and $\eu^{\iu t}$ — linear combinations of $\eu^{\iu nt}$, $n \geq 0$ — one has $\Lambda(f) =
\int_0^{2\pi}f(t)\,\eu^{\iu t}\,\dd t = 0$ (each $\int_0^{2\pi}\eu^{\iu(n+1)t}\dd t = 0$). $\Lambda$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) for $\norm\cdot_\infty$ ($\abs{\Lambda(f)}\leq
2\pi\norm f_\infty$), so $\Lambda$ vanishes on $\bar{\mathcal
A}$; but $\Lambda(\eu^{-\iu t}) = 2\pi \neq 0$: $\eu^{-\iu t}
\notin \bar{\mathcal A}$. (Stone–Weierstrass does not apply: $\mathcal A$ is not stable under conjugation — and the obstruction is precisely the one that holomorphic function theory will systematize in [Chapter 16](https://one-course.com/books/math/5/en/chapter/16-holomorphic-functions#ch-b3-holomorphic).)

**Exercise 7.10 ★★★.**

(Uniform boundedness, metric version) Let $X$ be a [complete metric space](#def-b3-complete-complete) and $\mathcal F \subseteq \mathcal C(X, \R)$ a family that is *pointwise* bounded: $\sup_{f \in \mathcal
F}\abs{f(x)} < \infty$ for each $x$. Show that there is a nonempty open $U \subseteq X$ on which $\mathcal F$ is *uniformly* bounded: $\sup_{f}\sup_U \abs f < \infty$. *(Consider $F_n = \{x : \abs{f(x)} \leq n\ \forall f\}$.)* This is the engine behind Banach–Steinhaus in [Chapter 8](https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems#ch-b3-banach).

**Solution of Exercise 7.10.**

$F_n = \bigcap_{f \in \mathcal F}\{x : \abs{f(x)} \leq n\}$ is an intersection of closed sets: closed. Pointwise boundedness gives $X = \bigcup_n F_n$. Baire ([Theorem 7.6](#thm-b3-complete-baire)) provides $n_0$ with $U = \mathring F_{n_0} \neq \varnothing$: on $U$, $\abs f \leq n_0$ for every $f \in \mathcal F$ simultaneously.

**Exercise 7.11 ★★.**

($\mathcal C^1$ needs its own norm) On $E = \mathcal
C^1(\intcc01, \R)$ consider $\norm f_{\mathcal C^1} = \norm
f_\infty + \norm{f'}_\infty$. (a) Show that $(E, \norm\cdot_{\mathcal C^1})$ is [complete](#def-b3-complete-complete) *(a $\mathcal C^1$-Cauchy sequence has $f_n \to f$ and $f_n' \to g$ uniformly; identify $g = f'$ by passing to the limit in $f_n(x) = f_n(0) + \int_0^xf_n'$)*. (b) Show that $(E, \norm\cdot_\infty)$ is *not* [complete](#def-b3-complete-complete): exhibit a uniform limit of $\mathcal C^1$ functions that is not differentiable (e.g. smooth approximations of $\abs{x - \tfrac12}$). (c) Deduce from (a), (b) and the open mapping circle of ideas — or directly — that no constant $C$ satisfies $\norm{f'}_\infty \leq C\,\norm f_\infty$ on $E$: exhibit a sequence witnessing it. Differentiation is unbounded; this is the cliff behind [Theorem 7.9](#thm-b3-complete-nowherediff).

**Solution of Exercise 7.11.**

(a) A Cauchy sequence for $\norm\cdot_{\mathcal C^1}$ is uniformly Cauchy together with its derivatives: $f_n \to f$ and $f_n' \to g$ uniformly, with $f, g$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). Passing to the limit (uniform convergence allows it under the integral) in $f_n(x) = f_n(0) + \int_0^xf_n'(t)\,\dd t$ gives $f(x) = f(0) + \int_0^xg$: $f$ is $\mathcal C^1$ with $f' = g$, and $\norm{f_n - f}_{\mathcal C^1} \to 0$. [Complete](#def-b3-complete-complete).

(b) $h(x) = \abs{x - \frac12}$ is a uniform limit of $\mathcal C^1$ functions, e.g. $h_n(x) = \sqrt{(x -
\frac12)^2 + \frac1n}$ ($\abs{h_n - h} \leq \frac1{\sqrt n}$ by the conjugate-quantity bound), yet $h \notin E$: the $\sup$-norm on $E$ is not [complete](#def-b3-complete-complete) — its [completion](#thm-b3-complete-completion) is $\mathcal C(\intcc01)$.

(c) $f_n(x) = \sin(2\pi nx)$ has $\norm{f_n}_\infty = 1$ and $\norm{f_n'}_\infty = 2\pi n \to \infty$: no $C$ exists. (Conceptually: if differentiation were bounded for the sup norm, the two norms of (a)–(b) would be equivalent, making $(E, \norm\cdot_\infty)$ [complete](#def-b3-complete-complete) — contradicting (b). This unboundedness is exactly why generic [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functions can fail to be differentiable anywhere, [Theorem 7.9](#thm-b3-complete-nowherediff).)

**Exercise 7.12 ★★★.**

(Croft’s lemma) Let $f\colon\intoo0\infty\to\R$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) and suppose that for every $x > 0$, $f(nx) \to 0$ as the integer $n \to \infty$. Show that $f(t) \to 0$ as $t \to
+\infty$. *(Fix $\varepsilon > 0$; the sets $$F_N = \{x > 0 : \abs{f(nx)} \leq \varepsilon\ \ \forall n
\geq N\}$$ are closed and cover $\intoo0\infty$; Baire in some interval $\intcc ab$ gives $N$ and a subinterval $\intcc{a'}{b'}
\subseteq F_N$; then the dilates $\bigl[na', nb'\bigr]$, $n
\geq N$, cover a whole [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $+\infty$ once $n(b' - a') \geq a'$.)* Where is the hypothesis “for every $x$” (not just rational $x$) used?

**Solution of Exercise 7.12.**

Fix $\varepsilon > 0$. Each $F_N$ is an intersection over $n
\geq N$ of preimages of the closed $\intcc{-\varepsilon}
\varepsilon$ under the [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $x \mapsto f(nx)$: closed. The hypothesis says every $x > 0$ lies in some $F_N$. By Baire applied inside the [complete](#def-b3-complete-complete) $\intcc ab$ (any $0 < a <
b$), some $F_N$ is dense in a subinterval; being closed, it contains an interval $\intcc{a'}{b'}$ with $0 < a' < b'$. Then for every $n \geq \max(N, \frac{a'}{b' - a'})$, the intervals $\intcc{na'}{nb'}$ and $\intcc{(n+1)a'}{(n+1)b'}$ overlap ($nb' \geq (n+1)a'$), so

$$
\bigcup_{n \geq n_0}\intcc{na'}{nb'} \supseteq
\intco{n_0a'}{+\infty},
$$

and every $t \geq n_0a'$ is $t = nx$ with $n \geq N$, $x \in
\intcc{a'}{b'} \subseteq F_N$: $\abs{f(t)} \leq \varepsilon$. Hence $\limsup_{t\to\infty}\abs f \leq \varepsilon$ for all $\varepsilon$: $f \to 0$. The full hypothesis is needed because $F_N$ must *cover* an interval-worth of $x$’s — with only rational $x$ the union of the $F_N$ is countable and Baire gives nothing; indeed there are [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) counterexamples vanishing along all rational rays but not at infinity.

## 7.6 Problem: Peano’s existence theorem

**Problem 7.1.**

Weekend problem — existence of solutions of $x' = f(t, x)$ without Lipschitz

Cauchy–Lipschitz (Year 2; re-proved in [Chapter 19](https://one-course.com/books/math/5/en/chapter/19-ordinary-differential-equations#ch-b3-ode)) demands $f$ Lipschitz in $x$. Peano (1890): *[continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$ already yields existence* — though not uniqueness. We prove it with Euler polygons and Ascoli. Setting: $f \colon R
\to \R^d$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the rectangle $R = \intcc{t_0 - a}{t_0
+ a}\times \bar B(x_0, b) \subseteq \R\times\R^d$, $M =
\sup_R\norm f$, and

$$
T = \min\bigl(a,\ b/M\bigr)
\qquad (\text{if } M = 0 \text{ the problem is trivial}).
$$

**Part I — Euler polygons.** For $n \geq 1$, subdivide $[t_0, t_0 + T]$ by $t_k = t_0 + kT/n$ ($0 \leq k
\leq n$) and define $\varphi_n$ piecewise affinely: $\varphi_n(t_0) = x_0$ and, on $[t_k, t_{k+1}]$,

$$
\varphi_n(t) = \varphi_n(t_k) + (t -
t_k)\,f\bigl(t_k, \varphi_n(t_k)\bigr).
$$

1. Show by induction that $\varphi_n$ is well defined, with $\norm{\varphi_n(t) - x_0} \leq M(t - t_0) \leq b$ on $[t_0, t_0 + T]$ — so the evaluation points stay in $R$ . *(This is where $T \leq b/M$ enters.)*
2. Show that each $\varphi_n$ is $M$ -Lipschitz.
3. Deduce from Arzelà–Ascoli ( [Theorem 7.11](#thm-b3-complete-ascoli) ) that a subsequence $(\varphi_{n_j})$ converges uniformly on $[t_0, t_0 +  T]$ to some $\varphi$ , itself $M$ -Lipschitz with $\varphi(t_0) = x_0$ .

**Part II — The limit solves the equation.** Define the *defect* $\Delta_n(t) = \varphi_n'(t) -
f\bigl(t, \varphi_n(t)\bigr)$ at non-grid points.

4. Show that $f$ is uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $R$ , and deduce: for every $\varepsilon > 0$ there is $n_0$ such that for $n \geq n_0$ and every non-grid $t$ , $\norm{\Delta_n(t)} \leq \varepsilon$ . *(On $(t_k,  t_{k+1})$, $\varphi_n'(t) = f(t_k, \varphi_n(t_k))$, and $(t, \varphi_n(t))$ is within distance $(1 + M)\,T/n$ of $(t_k, \varphi_n(t_k))$.)*
5. Establish the integral form: for all $t$, $$\varphi_n(t) = x_0 + \int_{t_0}^{t}  f\bigl(s, \varphi_n(s)\bigr)\dd s +  \int_{t_0}^t \Delta_n(s)\,\dd s,$$ the middle integrand being piecewise [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity).
6. Pass to the limit along $(n_j)$: show $f(s, \varphi_{n_j}(s)) \to f(s, \varphi(s))$ *uniformly* (uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$ again), and conclude $$\varphi(t) = x_0 + \int_{t_0}^t  f\bigl(s, \varphi(s)\bigr)\dd s .$$
7. Deduce that $\varphi$ is $\mathcal C^1$ on $[t_0, t_0 +  T]$ and solves $x' = f(t, x)$ , $x(t_0) = x_0$ ; extend the construction to $[t_0 - T, t_0]$ (time reversal). *This is Peano’s theorem.*

**Part III — Uniqueness genuinely fails.** Consider $x' = 2\sqrt{\abs x}$, $x(0) = 0$, on $\R$.

8. Check that $f(x) = 2\sqrt{\abs x}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) but not Lipschitz on any [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $0$ .
9. Verify that $x \equiv 0$ and, for every $c \geq 0$, $$x_c(t) = \begin{cases} 0 & t \leq c,\\ (t - c)^2 & t  \geq c,\end{cases}$$ are all solutions through $(0,0)$: a continuum of distinct solutions.
10. Where does the Picard iteration argument (Banach fixed point) break down for this $f$ ?

**Part IV — Limits of the method.**

11. Show that Peano’s theorem fails in infinite dimension: we admit (or you may take on faith) the classical example of Dieudonné in the space $c_0$ of null sequences; instead, prove the finite-dimensional ingredient that fails there: the closed unit ball of $c_0$ (sup norm) is not [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — exhibit a bounded sequence with no convergent subsequence, and explain which step of Part I breaks.
12. Summarize: which hypotheses give existence? existence and uniqueness? State precisely the two theorems (Peano; Cauchy–Lipschitz) side by side.

**Part V — Osgood: uniqueness beyond Lipschitz.** Let $\omega\colon\intoo0\infty\to\intoo0\infty$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), nondecreasing, with

$$
\int_0^1\frac{\dd r}{\omega(r)} = +\infty
\qquad(\text{divergence at } 0),
$$

and suppose $f$ satisfies $\norm{f(t, x) - f(t, y)} \leq
\omega\bigl(\norm{x - y}\bigr)$ on $R$.

13. Check that $\omega(r) = Lr$ qualifies (Lipschitz), that $\omega(r) = r\log\frac1r$ (extended by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) , for $r$ small) qualifies although it is *not* $O(r)$ , and that $\omega(r) = 2\sqrt r$ does not. Compute the integral in each case.
14. Let $x_1, x_2$ solve the equation on $[t_0, t_0 + T]$ with $x_1(t_0) = x_2(t_0)$, and $\delta(t) =  \norm{x_1(t) - x_2(t)}$. Show, from the integral forms alone, that for $t_0 \leq s \leq t$: $$\delta(t) \leq \delta(s) +  \int_s^t\omega\bigl(\delta(v)\bigr)\dd v .$$
15. (Osgood’s theorem) Suppose $\delta(t_1) > 0$ for some $t_1$, and let $\tau = \sup\{t \leq t_1 : \delta(t) =  0\}$. For $s \in \intoo\tau{t_1}$, set $u(t) = \delta(s)  + \int_s^t\omega(\delta(v))\,\dd v$. Show $\delta \leq  u$, $u' = \omega(\delta) \leq \omega(u)$, and deduce $$\int_{\delta(s)}^{u(t_1)}\frac{\dd r}{\omega(r)}  \leq t_1 - s \leq t_1 - \tau .$$ Let $s \downarrow \tau$ and derive a contradiction with the divergence of the integral. Conclude: *solutions through a common initial condition coincide* — uniqueness under Osgood’s condition.
16. Draw the consequences: Cauchy–Lipschitz uniqueness is the case $\omega(r) = Lr$ ; the equation $x' =  x\log\frac1{\abs x}$ (extended by $0$ at $0$ ) has unique solutions although its right side is not Lipschitz at $0$ ; and for $x' = 2\sqrt{\abs x}$ the convergence of $\int_0\frac{\dd r}{2\sqrt r}$ is exactly what lets a solution leave $0$ in finite time — match the value of the integral $\int_0^{h^2}\frac{\dd r}{2\sqrt r} = h$ with the escape behaviour of $x_c$ .

**Part VI — Rates, schemes, funnels.**

17. (Integral Grönwall lemma) Let $e, \eta \geq 0$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on $[t_0, t_0+T]$ with $e(t) \leq \int_{t_0}  ^t\bigl(L\,e(s) + \eta(s)\bigr)\dd s$ for all $t$. Show $$e(t) \leq \frac{\sup\eta}{L}\bigl(\eu^{L(t - t_0)} -  1\bigr)$$ *(set $G(t) = \int_{t_0}^t(Le + \eta)$, note $G'  \leq LG + \sup\eta$, and differentiate $\eu^{-Lt}G(t)$)*.
18. (Euler converges with a rate) Suppose now $f$ is $L$-Lipschitz in $x$ and $L'$-Lipschitz in $t$ on $R$. Combining the defect bound of question 4 (made quantitative: $\norm{\Delta_n} \leq (L' + LM)T/n$) with question 17, prove $$\sup_{[t_0, t_0+T]}\norm{\varphi_n - \varphi}  \;\leq\; \frac{(L' + LM)\,T}{n}\cdot  \frac{\eu^{LT} - 1}{L} = O\Bigl(\frac1n\Bigr),$$ where $\varphi$ is *the* solution: with Lipschitz data the whole sequence converges, with an explicit rate — no subsequences needed. Why does uniqueness upgrade subsequential convergence to full convergence even without this computation?
19. (The scheme chooses) For $x' = 2\sqrt{\abs x}$ , $x(0) =  0$ : show that every Euler polygon is identically zero, so the scheme converges to the solution $x \equiv 0$ ; but started at $x(0) = \varepsilon > 0$ it converges (as $n \to \infty$ , then $\varepsilon \to 0$ ) to $t  \mapsto t^2$ , a *different* solution through the origin. Non-uniqueness resurfaces as sensitivity of the numerical scheme to perturbations.
20. Show that the set $\mathcal S$ of *all* solutions of $x' = f(t,x)$ , $x(t_0) = x_0$ on $[t_0, t_0+T]$ (values in $\bar B(x_0, b)$ ) is nonempty (Part II), uniformly $M$ -Lipschitz, and closed in $\bigl(\mathcal C([t_0, t_0+T], \R^d),  \norm\cdot_\infty\bigr)$ ; conclude by Ascoli that $\mathcal S$ is *[compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact)* . (Kneser’s theorem adds that $\mathcal S$ is [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) ; we shall not prove it.)
21. Verify Kneser’s phenomenon on the example: for $x' =  2\sqrt{\abs x}$ , $x(0) = 0$ , on $[0, 1]$ , show $\mathcal S = \{x_c : c \in \intcc0\infty\}$ with $x_\infty \equiv 0$ *(for any solution, let $c =  \sup\{t : x(t) = 0\}$ and integrate $(\sqrt x)' = 1$ on $\{x > 0\}$)* , that $c \mapsto x_c$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) from $\intcc0\infty$ ( [one-point compactification](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-locallycompact) , i.e. $c = \infty$ glued as limit) to $\mathcal C([0,1])$ , and conclude that $\mathcal S$ is indeed [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) — a segment-shaped funnel.
22. (Reachable sets) Deduce from question 20 that for each fixed $t$ , the *reachable set* $\mathcal S(t) =  \{x(t) : x \in \mathcal S\}$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) ; compute it for the example of question 21 and check it is also [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) : $\mathcal S(t) = \intcc0{t^2}$ — every intermediate state is attained by some solution.

**Part VII — Complements: dependence, optimality, and a scheme computed by hand.**

23. ([Continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) dependence) Suppose $f$ is $L$-Lipschitz in $x$ on $R$, and let $x, y$ be two solutions with initial values $x_0, y_0$ at $t_0$. Adapting the proof of question 17 to the inequality $e(t) \leq \norm{x_0  - y_0} + \int_{t_0}^tL\,e(s)\dd s$, show $$\norm{x(t) - y(t)} \leq \norm{x_0 - y_0}\,  \eu^{L(t - t_0)},$$ and check on $x' = Lx$ that the bound is attained: Grönwall is sharp. Deduce again uniqueness ($x_0 =  y_0$), and that the flow map $x_0 \mapsto x(t; x_0)$ is Lipschitz, with constant $\eu^{LT}$, wherever it is defined.
24. (Osgood is optimal) Conversely, let $\omega$ be [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), nondecreasing, positive on $\intoo0\infty$, with $\omega(r) \to 0$ as $r \to 0^+$ but $$\int_0^1\frac{\dd r}{\omega(r)} < +\infty,$$ and extend $f(x) = \omega(\abs x)$ by $f(0) = 0$. Show that $\Omega(x) = \int_0^x\frac{\dd r}{\omega(r)}$ is an increasing bijection from $\intoc0{1}$ onto $\intoc0{\Omega(1)}$, that its inverse $g$ solves $g'  = \omega(g)$ with $g(0^+) = 0$, and that $g$, extended by $0$ for $t \leq 0$, is a $\mathcal C^1$ solution of $x' = f(x)$ through $(0, 0)$ distinct from $x \equiv  0$ *(for $g'(0) = 0$, bound $\frac{g(t)}t$ by $\omega(g(t))$)*. Conclude: the divergence hypothesis of question 15 is not a convenience but the exact frontier of uniqueness; recover Part III from $\omega(r) = 2\sqrt r$, $\Omega(x) = \sqrt x$.
25. (Euler computed by hand) For $x' = x$, $x(0) = 1$ on $[0, 1]$: show that the Euler polygon satisfies $\varphi_n(1) = \bigl(1 + \frac1n\bigr)^n$. Prove the expansion $$\Bigl(1 + \frac1n\Bigr)^{n}  = \eu\Bigl(1 - \frac1{2n} + O\Bigl(\frac1{n^2}\Bigr)  \Bigr),$$ so the error at $t = 1$ is $\frac{\eu}{2n} +  O(n^{-2})$: question 18’s $O(\frac1n)$ rate, with the exact constant. Check numerically for $n = 10$: $1.1^{10} = 2.59374$ against $\eu \approx 2.71828$, an error $0.12454$ to compare with $\frac{\eu}{20}  \approx 0.13591$.

**Solution of Problem 7.1.**

**1.** Induction on $k$: if $\norm{\varphi_n(t_k) - x_0}
\leq M(t_k - t_0) \leq MT \leq b$, the point $(t_k,
\varphi_n(t_k))$ lies in $R$, so the slope $f(t_k,
\varphi_n(t_k))$ is defined, of norm $\leq M$; then for $t \in
[t_k, t_{k+1}]$, $\norm{\varphi_n(t) - x_0} \leq
\norm{\varphi_n(t_k) - x_0} + M(t - t_k) \leq M(t - t_0) \leq
MT \leq b$.

**2.** Each affine piece has slope of norm $\leq M$; a piecewise affine function with slopes bounded by $M$ is $M$-Lipschitz (chain through the grid points).

**3.** The family $(\varphi_n)$ is pointwise bounded (values in $\bar B(x_0, b)$) and [equicontinuous](#def-b3-complete-equicontinuous) (common Lipschitz constant $M$): Ascoli ([Theorem 7.11](#thm-b3-complete-ascoli)) extracts $\varphi_{n_j} \to
\varphi$ uniformly on $[t_0, t_0 + T]$. The bounds pass to the limit: $\varphi$ is $M$-Lipschitz, $\varphi(t_0) = x_0$.

**4.** $R$ is [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): uniformly [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (Heine, [Corollary 6.17](https://one-course.com/books/math/5/en/chapter/6-general-topology#cor-b3-topology-heineborel)); let $\delta(\varepsilon)$ be a modulus. For non-grid $t \in (t_k,
t_{k+1})$: $\varphi_n'(t) = f(t_k, \varphi_n(t_k))$, and the two evaluation points of $f$ differ by $\abs{t - t_k} \leq T/n$ in time and $\norm{\varphi_n(t) - \varphi_n(t_k)} \leq MT/n$ in space. For $n \geq n_0$ with $(1 + M)T/n_0 < \delta(\varepsilon)$: $\norm{\Delta_n(t)} \leq \varepsilon$.

**5.** On each $[t_k, t_{k+1}]$, $\varphi_n$ is affine, so $\varphi_n(t_{k+1}) - \varphi_n(t_k) = \int_{t_k}^{t_{k+1}}
\varphi_n'(s)\dd s$, the derivative being the constant slope; summing over pieces (and cutting the last one at $t$): $\varphi_n(t) = x_0 + \int_{t_0}^t\varphi_n'(s)\,\dd s$. Writing $\varphi_n' = f(s, \varphi_n(s)) + \Delta_n(s)$ (piecewise [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) integrands, finitely many jumps) gives the display.

**6.** Given $\varepsilon$: for $j$ large, $\norm{\varphi_{n_j} - \varphi}_\infty < \delta(\varepsilon)$, so $\norm{f(s, \varphi_{n_j}(s)) - f(s, \varphi(s))} \leq
\varepsilon$ for all $s$: uniform convergence of the integrands, and $\int_{t_0}^t f(s,\varphi_{n_j}(s))\dd s \to
\int_{t_0}^tf(s, \varphi(s))\dd s$ uniformly in $t$. Also $\norm{\int_{t_0}^t\Delta_{n_j}} \leq T\sup\norm{\Delta_{n_j}}
\to 0$ (question 4). Passing to the limit in question 5’s identity: $\varphi(t) = x_0 + \int_{t_0}^tf(s,
\varphi(s))\,\dd s$.

**7.** The integrand $s \mapsto f(s, \varphi(s))$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), so the right side is $\mathcal C^1$ in $t$ with derivative $f(t, \varphi(t))$: $\varphi$ solves the Cauchy problem on $[t_0, t_0 + T]$. For the left half, set $g(t, x) =
-f(2t_0 - t, x)$, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on the reflected rectangle with the same bound $M$; a solution $\psi$ of $y' = g(t,y)$, $y(t_0) = x_0$ on $[t_0, t_0 + T]$ yields $\varphi(t) =
\psi(2t_0 - t)$ solving the original equation on $[t_0 - T,
t_0]$; the two halves glue to a $\mathcal C^1$ solution (both one-sided derivatives at $t_0$ equal $f(t_0, x_0)$). — *Peano’s theorem*: a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ admits a local solution through every initial condition.

**8.** [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) is clear. Lipschitz near $0$ fails: $\abs{f(h) - f(0)} = 2\sqrt h$, and $2\sqrt h \leq L h$ is false for $h < 4/L^2$.

**9.** For $t \leq c$: $x_c \equiv 0$ solves. For $t \geq
c$: $x_c'(t) = 2(t - c) = 2\sqrt{(t-c)^2} = 2\sqrt{\abs{x_c}}$. At $t = c$ the one-sided derivatives are both $0$: $x_c$ is $\mathcal C^1$ and solves globally, with $x_c(0) = 0$ for every $c \geq 0$ — together with $x \equiv 0$, a continuum of solutions through the origin.

**10.** Picard’s iteration sets up $\Phi(x)(t) = x_0 +
\int_0^t f(x(s))\dd s$ and needs $\norm{\Phi(x) - \Phi(y)}
\leq k\norm{x - y}$ with $k < 1$ on a suitable ball — which follows from a Lipschitz bound on $f$, transferred under the integral. Here $\sqrt\cdot$ admits no Lipschitz bound near $0$, and no choice of interval or ball repairs it. And indeed no proof of uniqueness could succeed: uniqueness is *false* (question 9).

**11.** In $c_0$, the unit vectors $e_n = (0, \dots, 0, 1,
0, \dots)$ satisfy $\norm{e_n - e_m}_\infty = 1$ for $n \neq
m$: no subsequence is Cauchy, so the closed unit ball is not [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact). The step that breaks is the extraction (question 3): Ascoli for $\mathcal C([t_0, t_0+T], E)$ requires the values to live in a space where bounded sets are relatively [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — true in $\R^d$ (Bolzano–Weierstrass), false in $c_0$; pointwise extraction is no longer available (indeed Dieudonné’s example has no local solution at all).

**12.** *Peano*: $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on a [neighborhood](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) of $(t_0, x_0)$ in $\R\times\R^d$ $\Rightarrow$ there exists a $\mathcal C^1$ solution of $x' = f(t,x)$, $x(t_0) = x_0$, on some $[t_0 - T, t_0 + T]$. *Cauchy–Lipschitz* ([Chapter 19](https://one-course.com/books/math/5/en/chapter/19-ordinary-differential-equations#ch-b3-ode)): if moreover $f$ is locally Lipschitz in the $x$-variable, the solution is *unique* (any two agree on their common interval) — existence and uniqueness. The pair $(x' = 2\sqrt{\abs x},\ x(0) = 0)$ separates the two theorems.

**13.** $\omega(r) = Lr$: $\int_0^1\frac{\dd r}{Lr} =
+\infty$: qualifies. $\omega(r) = r\log\frac1r$ (near $0$): $\int\frac{\dd r}{r\log(1/r)} = \bigl[-\log\log\frac1r\bigr]
\to +\infty$ as $r \to 0$: qualifies — yet $\omega(r)/r =
\log\frac1r \to \infty$: not Lipschitz. $\omega(r) = 2\sqrt
r$: $\int_0^1\frac{\dd r}{2\sqrt r} = \bigl[\sqrt r\bigr]_0^1
= 1 < \infty$: fails.

**14.** Subtract the two integral forms:

$$
x_1(t) - x_2(t) = x_1(s) - x_2(s) + \int_s^t\bigl(f(v,
x_1(v)) - f(v, x_2(v))\bigr)\dd v,
$$

take norms and use the Osgood modulus: $\delta(t) \leq
\delta(s) + \int_s^t\omega(\delta(v))\,\dd v$.

**15.** $\tau$ is well defined ($\delta(t_0) = 0$) with $\delta(\tau) = 0$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)) and $\delta > 0$ on $\intoc\tau{t_1}$. Fix $s \in \intoo\tau{t_1}$. Then $u$ is $\mathcal C^1$, $u(s) = \delta(s) > 0$, $u \geq \delta$ on $[s, t_1]$ (question 14), and $u' = \omega(\delta) \leq
\omega(u)$ ($\omega$ nondecreasing, $u > 0$). Divide and integrate:

$$
\int_{\delta(s)}^{u(t_1)}\frac{\dd r}{\omega(r)}
= \int_s^{t_1}\frac{u'(v)}{\omega(u(v))}\,\dd v \leq t_1 - s
\leq t_1 - \tau .
$$

Although $u$ depends on $s$, the bound $u(t_1) \geq
\delta(t_1)$ holds for every $s$, so the left side is at least $\int_{\delta(s)}^{\delta(t_1)}\frac{\dd r}{\omega(r)}$, which tends to $+\infty$ as $s \downarrow \tau$ (then $\delta(s) \to \delta(\tau) = 0$, and the integral diverges at $0$): the bounded right side is contradicted. Hence $\delta \equiv 0$: uniqueness.

**16.** Lipschitz is $\omega = Lr$: uniqueness recovered. For $x' = x\log\frac1{\abs x}$: the right side satisfies the Osgood modulus $\omega(r) = r\log\frac1r$ near $0$ (mean value inequality on $x \mapsto x\log\frac1x$, whose derivative $\log\frac1x - 1$ is unbounded — Lipschitz fails, Osgood holds): unique solutions; note $x \equiv 0$ is one, so no other solution can touch $0$. For $\omega(r) = 2\sqrt r$: $\int_0^{h^2}\frac{\dd r}{2\sqrt r} = h$ — the “Osgood budget” for climbing from $0$ to height $h^2$ is exactly the time $h$, and indeed $x_c(c + h) = h^2$: the solution spends time $h$ doing precisely what the convergent integral permits. Divergence of the integral is the impossibility of leaving $0$ in finite time; convergence is the escape route.

**17.** $G(t) = \int_{t_0}^t(Le(s) + \eta(s))\dd s$ is $\mathcal C^1$ with $G' = Le + \eta \leq LG + \sup\eta$ (hypothesis $e \leq G$). Then $\bigl(\eu^{-Lt}(G +
\tfrac{\sup\eta}L)\bigr)' = \eu^{-Lt}(G' - LG - \sup\eta)
\leq 0$: the bracket decreases, so $G(t) + \frac{\sup\eta}L
\leq \eu^{L(t - t_0)}\bigl(G(t_0) + \frac{\sup\eta}L\bigr) =
\eu^{L(t-t_0)}\frac{\sup\eta}L$, i.e. $e(t) \leq G(t) \leq
\frac{\sup\eta}L(\eu^{L(t-t_0)} - 1)$.

**18.** Quantitative defect: on $(t_k, t_{k+1})$, $\Delta_n(t) = f(t_k, \varphi_n(t_k)) - f(t, \varphi_n(t))$ with $\abs{t - t_k} \leq T/n$ and $\norm{\varphi_n(t) -
\varphi_n(t_k)} \leq MT/n$, so $\norm{\Delta_n} \leq L'T/n +
LMT/n$. Subtracting the integral identities for $\varphi_n$ (question 5) and $\varphi$ and using the Lipschitz bound:

$$
\norm{\varphi_n(t) - \varphi(t)} \leq
\int_{t_0}^t L\,\norm{\varphi_n - \varphi}(s)\,\dd s +
\int_{t_0}^t\norm{\Delta_n(s)}\,\dd s,
$$

and question 17 with $\eta = \norm{\Delta_n}$ gives the stated $O(1/n)$ bound, $\varphi$ being unique by Cauchy–Lipschitz (or Osgood). Even without rates: every subsequence of the equibounded, equi-Lipschitz $(\varphi_n)$ has a sub-subsequence converging (Ascoli + Part II) to *a* solution, which uniqueness forces to be $\varphi$: a sequence all of whose subsequences have subsubsequences with the same limit converges.

**19.** From $\varphi_n(t_k) = 0$: the slope $2\sqrt0 =
0$ gives $\varphi_n(t_{k+1}) = 0$; induction: $\varphi_n
\equiv 0$, converging to the zero solution. From $\varepsilon > 0$: on $[\varepsilon', \infty)$ with $\varepsilon' < \varepsilon$ the function $2\sqrt x$ is Lipschitz, so question 18 applies and Euler converges to the unique solution through $(0, \varepsilon)$, namely $x(t) =
(t + \sqrt\varepsilon)^2$ (check: $x' = 2(t + \sqrt
\varepsilon) = 2\sqrt x$). As $\varepsilon \to 0$, $(t +
\sqrt\varepsilon)^2 \to t^2$ uniformly on $[0,1]$: the double limit lands on $x_0(t) = t^2$, not on $0$. An arbitrarily small perturbation of the initial datum redirects the scheme from one solution to another: non-uniqueness read as numerical instability.

**20.** Nonempty: Part II. Every solution satisfies $\norm{x'} = \norm{f(t, x)} \leq M$: $\mathcal S$ is uniformly $M$-Lipschitz and uniformly bounded (values in $\bar B(x_0, b)$). Closed: if $x_n \in \mathcal S \to x$ uniformly, pass to the limit in $x_n(t) = x_0 +
\int_{t_0}^tf(s, x_n(s))\dd s$ (the integrands converge uniformly by uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $f$ on the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) $R$): $x \in \mathcal S$. Ascoli: $\mathcal S$ is a closed, bounded, [equicontinuous](#def-b3-complete-equicontinuous) subset of $\mathcal C([t_0, t_0+T],
\R^d)$: [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact).

**21.** Any solution is nondecreasing ($x' = 2\sqrt{\abs
x} \geq 0$) with $x(0) = 0$, hence $x \geq 0$. Let $c =
\sup\{t \in [0,1] : x(t) = 0\}$ (possibly $c = \infty$ if $x \equiv 0$, in which case $x = x_\infty$). For $t > c$: $x
> 0$ (monotonicity plus definition of $c$), and there $(\sqrt
x)' = \frac{x'}{2\sqrt x} = 1$, so $\sqrt{x(t)} = t - c$ ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) at $c$): $x = x_c$. [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $c \mapsto
x_c$: for $c, c' \in [0, 1]$, $\sup_{[0,1]}\abs{(t - c)_+^2 -
(t - c')_+^2} \leq 2\abs{c - c'}$ (the map $c \mapsto
(t-c)_+^2$ is $2$-Lipschitz uniformly in $t \in [0,1]$), and since $x_c \equiv 0$ on $[0,1]$ for every $c \geq 1$, the family reduces to $\mathcal S = \{x_c : c \in [0,1]\}$ (with $x_1 = 0 = x_\infty$). So $\mathcal S$ is the image of the [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) $[0,1]$ under the [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $c
\mapsto x_c$: [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) and [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected). The funnel of solutions is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) segment running from $t^2$ (immediate escape) down to $0$ (eternal rest).

**22.** The evaluation $\operatorname{ev}_t\colon
\mathcal C([0,1]) \to \R$, $x \mapsto x(t)$, is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) ($\abs{x(t) - y(t)} \leq \norm{x - y}_\infty$), so $\mathcal
S(t) = \operatorname{ev}_t(\mathcal S)$ is a [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) image of a [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact): [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — and of a [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected) set: [connected](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-connected). For the example: $x_c(t) = (t - c)_+^2$ sweeps, as $c$ runs through $[0, 1]$, all values from $t^2$ (at $c = 0$) down to $0$ (at $c \geq t$), [continuously](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $\mathcal S(t) =
\intcc0{t^2}$. At each instant, the funnel’s cross-section is a full segment: between resting and maximal escape, every compromise is realized by an actual solution.

**23.** Subtracting the integral forms $x(t) = x_0 +
\int_{t_0}^tf(s, x(s))\dd s$ and its analogue for $y$, and setting $e(t) = \norm{x(t) - y(t)}$, $d = \norm{x_0 - y_0}$:

$$
e(t) \leq d + \int_{t_0}^tL\,e(s)\dd s = G(t).
$$

Then $G(t_0) = d$ and $G' = Le \leq LG$, so $\bigl(\eu^{-L(t
- t_0)}G\bigr)' \leq 0$ and $G(t) \leq d\,\eu^{L(t - t_0)}$; hence $e(t) \leq d\,\eu^{L(t-t_0)}$. Sharpness: for $x' =
Lx$, the solutions through $x_0$ and $y_0$ are $x_0\eu^{L(t -
t_0)}$ and $y_0\eu^{L(t-t_0)}$, whose distance is exactly $d\,\eu^{L(t-t_0)}$. With $d = 0$, $e \equiv 0$: Cauchy–Lipschitz uniqueness, re-derived in two lines. And for fixed $t \in [t_0, t_0 + T]$, $\norm{x(t; x_0) - x(t;
y_0)} \leq \eu^{LT}\norm{x_0 - y_0}$: the flow is Lipschitz in the initial condition — deterministic dependence, at a controlled exponential price.

**24.** On $\intoc01$, $\Omega$ is well defined (the integral converges at $0$ by hypothesis), $\mathcal C^1$ with $\Omega' = \frac1\omega > 0$: an increasing bijection onto $\intoc0{\Omega(1)}$, with $\Omega(x) \to 0$ as $x \to 0^+$. Its inverse $g \colon \intoc0{\Omega(1)} \to \intoc01$ is $\mathcal C^1$ with

$$
g'(t) = \frac1{\Omega'(g(t))} = \omega\bigl(g(t)\bigr) > 0,
\qquad g(t) \xrightarrow[t \to 0^+]{} 0 .
$$

Extend $g$ by $0$ on $t \leq 0$: [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) is clear, and at $t = 0$, for $t > 0$,

$$
\frac{g(t)}t = \frac1t\int_0^tg'(s)\dd s
= \frac1t\int_0^t\omega\bigl(g(s)\bigr)\dd s
\leq \omega\bigl(g(t)\bigr) \xrightarrow[t\to0^+]{} 0
$$

($\omega$ nondecreasing, $g$ increasing, $\omega(0^+) = 0$): $g'(0) = 0 = f(g(0))$, and $g' = f(g)$ holds on both sides of $0$. So $x \equiv 0$ and $g$ are two distinct $\mathcal
C^1$ solutions through $(0, 0)$: when $\int_0\frac{\dd r}
{\omega(r)}$ converges, uniqueness fails — question 15’s divergence is exactly the frontier. For $\omega(r) = 2\sqrt
r$: $\Omega(x) = \int_0^x\frac{\dd r}{2\sqrt r} = \sqrt x$, $g(t) = t^2$, and time translations give the whole family $x_c$ of Part III.

**25.** With step $h = \frac1n$: $\varphi_n(t_{k+1}) =
\varphi_n(t_k) + h\,\varphi_n(t_k) = (1 + h)\varphi_n(t_k)$, so $\varphi_n(1) = (1 + \frac1n)^n$ after $n$ steps. Expansion:

$$
n\log\Bigl(1 + \frac1n\Bigr)
= n\Bigl(\frac1n - \frac1{2n^2} + O\Bigl(\frac1{n^3}\Bigr)
\Bigr) = 1 - \frac1{2n} + O\Bigl(\frac1{n^2}\Bigr),
$$

and exponentiating, $(1 + \frac1n)^n = \eu\,\eu^{-1/(2n) +
O(n^{-2})} = \eu\bigl(1 - \frac1{2n} + O(n^{-2})\bigr)$. The error at $t = 1$ is therefore $\eu - \varphi_n(1) =
\frac{\eu}{2n} + O(n^{-2})$: the $O(\frac1n)$ of question 18, here with its exact constant $\frac\eu2$. Numerically, $n =
10$: $1.1^{10} = 2.5937424601$ ($1.1^2 = 1.21$, $1.1^4 =
1.4641$, $1.1^8 = 2.14358881$, times $1.21$), and $\eu -
2.59374 \approx 0.12454$, against the asymptotic prediction $\frac{\eu}{20} \approx 0.13591$: agreement to within the $O(n^{-2})$ correction, whose leading term here lowers the prediction toward the observed value.