---
title: "Normed Vector Spaces"
book: "University Mathematics — Year 2"
subject: math
language: en
chapter: 5
exercises: 12
source: https://one-course.com/books/math/4/en/chapter/5-normed-vector-spaces
---

# Chapter 5 — Normed Vector Spaces

When the [metric space](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-def) is a vector space and the distance comes from a [norm](#def-b2-nvs-norm), topology and linear algebra begin to interact: linear maps are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) exactly when they are bounded on the unit ball, finite dimension forces all [norms](#def-b2-nvs-norm) to agree, and [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) turns absolutely convergent series into convergent ones. The finite/infinite dimension divide — crystallized in Riesz’s theorem — is the deepest lesson of the chapter.

Throughout, $E, F$ are vector spaces over $K = \R$ or $\C$.

## 5.1 Norms

**Definition 5.1.**

A *norm* on $E$ is a map $\norm{\,\cdot\,} \colon E
\to \R_+$ with, for all $x, y \in E$, $\lambda \in K$:

$$
\norm x = 0 \iff x = 0,
\qquad
\norm{\lambda x} = \abs\lambda\,\norm x,
\qquad
\norm{x + y} \leq \norm x + \norm y .
$$

Then $d(x, y) = \norm{x - y}$ is a distance, and all of [Chapter 4](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#ch-b2-metric) applies. The reverse triangle inequality $\bigl|\norm x - \norm y\bigr| \leq \norm{x - y}$ makes the norm itself $1$-Lipschitz; addition and scalar multiplication are [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) (estimates $\norm{(x + y) - (x' + y')} \leq \norm{x -
x'} + \norm{y - y'}$, etc.).

**Example 5.2.**

On $K^n$:

$$
\norm{x}_1 = \sum_i \abs{x_i},
\qquad
\norm{x}_2 = \Bigl(\sum_i \abs{x_i}^2\Bigr)^{1/2},
\qquad
\norm{x}_\infty = \max_i \abs{x_i}
$$

($\norm\cdot_2$ is a [norm](#def-b2-nvs-norm) by Cauchy–Schwarz, Year 1 volume). On $C(\intcc{a}{b})$:

$$
\norm f_\infty = \sup \abs f,
\qquad
\norm f_1 = \int_a^b \abs f,
\qquad
\norm f_2 = \Bigl(\int_a^b \abs f^2\Bigr)^{1/2},
$$

the last two being [norms](#def-b2-nvs-norm) thanks to strict positivity of the integral and integral Cauchy–Schwarz (Year 1 volume). On matrices: any [norm](#def-b2-nvs-norm) on $\mathcal M_n(K) \simeq K^{n^2}$; the [operator norms](#thm-b2-nvs-continuouslinear) below are the structurally important ones.

**Definition 5.3 (Equivalent norms).**

Two [norms](#def-b2-nvs-norm) $N_1, N_2$ on $E$ are *equivalent* when there are constants $c, C > 0$ with

$$
c\,N_1 \leq N_2 \leq C\, N_1 .
$$

Equivalent [norms](#def-b2-nvs-norm) have the same [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) sets, the same convergent and Cauchy sequences, the same [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) and [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) subsets: the same analysis.

**Example 5.4 (Non-equivalence in infinite dimension).**

On $C(\intcc{0}{1})$: $\norm f_1 \leq \norm f_\infty$ always, but no reverse bound holds: $f_n(x) = x^n$ has $\norm{f_n}_\infty = 1$ and $\norm{f_n}_1 = \frac{1}{n+1} \to 0$. So $f_n \to 0$ for $\norm\cdot_1$ but not for $\norm\cdot_\infty$: the two [norms](#def-b2-nvs-norm) disagree about convergence itself.

**Example 5.5 (Explicit constants in dimension nnn).**

On $K^n$ the three classical [norms](#def-b2-nvs-norm) are [equivalent](#def-b2-nvs-equivalent) with sharp constants:

$$
\norm x_\infty \leq \norm x_2 \leq \norm x_1
\leq \sqrt n\,\norm x_2 \leq n\,\norm x_\infty ,
$$

the middle bound $\norm x_1 \leq \sqrt n\norm x_2$ coming from Cauchy–Schwarz against the all-ones vector. Extremal vectors: $e_1$ makes the first two inequalities equalities, $(1, 1,
\dots, 1)$ the last two. The dimension $n$ sits visibly in the constants — the quantitative seed of the failure in infinite dimension: as $n \to \infty$ no uniform constant survives, which is exactly what [Example 5.4](#ex-b2-nvs-nonequivalent) exhibits on function spaces.

![The unit balls of the three classical norms of ℝ2, nested as the inequalities of dictate: smaller ball, larger norm. Roundness matters: the flat sides of the diamond and the square are exactly the failures of strict convexity exploited in ’s weekend problem and in this chapter’s (question 4).](https://one-course.com/images/onecourse/chapters/math-4/b2-nvs/fig-89dd87920119.svg)

*The unit balls of the three classical [norms](#def-b2-nvs-norm) of $\R^2$, nested as the inequalities of [Example 5.5](#ex-b2-nvs-constants) dictate: smaller ball, larger [norm](#def-b2-nvs-norm). Roundness matters: the flat sides of the diamond and the square are exactly the failures of strict convexity exploited in [Chapter 8](https://one-course.com/books/math/4/en/chapter/8-functions-of-a-real-variable#ch-b2-realfun)’s weekend problem and in this chapter’s (question 4).*

## 5.2 Continuous linear maps

**Theorem 5.6 (Characterization).**

For a linear map $u \colon E \to F$ between normed spaces, the following are [equivalent](#def-b2-nvs-equivalent):

1. $u$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ;
2. $u$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$ ;
3. $u$ is bounded on the closed unit ball: $\sup_{\norm x \leq  1} \norm{u(x)} < \infty$ ;
4. there is $C \geq 0$ with $\norm{u(x)} \leq C \norm x$ for all $x$ ;
5. $u$ is [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) .

The least such $C$ is the *operator norm* $\vertiii{u} = \sup_{\norm x \leq 1}\norm{u(x)} = \sup_{x \neq 0}
\frac{\norm{u(x)}}{\norm x}$; it makes the space $\mathcal{L}_c(E,
F)$ of [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) linear maps a normed space, with

$$
\vertiii{v \circ u} \leq \vertiii v\, \vertiii u .
$$

**Proof.** (1 $\Rightarrow$ 2) trivial. (2 $\Rightarrow$ 3): [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$ with $\varepsilon = 1$ gives $\delta$ with $\norm x \leq \delta
\Rightarrow \norm{u(x)} \leq 1$; homogeneity then scales any $x$ with $\norm x \leq 1$ down into that ball and back:

$$
\norm{u(x)} = \frac1\delta\,\norm{u(\delta x)} \leq
\frac1\delta ,
$$

since $\norm{\delta x} \leq \delta$. (3 $\Rightarrow$ 4): for $x \neq 0$, apply the bound to $\frac{x}{\norm x}$. (4 $\Rightarrow$ 5): $\norm{u(x) - u(y)} = \norm{u(x - y)} \leq
C\norm{x - y}$. (5 $\Rightarrow$ 1) known.

[Norm](#def-b2-nvs-norm) axioms for $\vertiii\cdot$: homogeneity and separation are clear ($\vertiii u = 0$ forces $u = 0$ on the ball, hence everywhere); triangle inequality from $\norm{(u + v)(x)} \leq
\norm{u(x)} + \norm{v(x)}$. Submultiplicativity: $\norm{v(u(x))}
\leq \vertiii v\,\norm{u(x)} \leq \vertiii v \vertiii u \norm x$. ∎

**Example 5.7.**

On $\bigl(C(\intcc{0}{1}), \norm\cdot_\infty\bigr)$: evaluation $f \mapsto f(0)$ has [operator norm](#thm-b2-nvs-continuouslinear) $1$; integration $f \mapsto
\int_0^1 f$ has [norm](#def-b2-nvs-norm) $1$; the map $f \mapsto \int_0^1 t f(t)\dd t$ has [norm](#def-b2-nvs-norm) $\int_0^1 t\,\dd t = \frac12$ (upper bound by the triangle inequality for integrals; attained at $f \equiv 1$). But differentiation, from $(C^1, \norm\cdot_\infty)$ to $(C^0,
\norm\cdot_\infty)$, is *not* [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity): $\norm{\sin(nx)}_\infty
= 1$ while the derivative has sup [norm](#def-b2-nvs-norm) $n$. Linear does not imply [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) in infinite dimension.

**Example 5.8 (Two norms, two verdicts on one sequence).**

On $C(\intcc01)$, let $g_n(x) = \sqrt{n}\,x^n$. Then

$$
\norm{g_n}_1 = \frac{\sqrt n}{n + 1} \longrightarrow 0,
\qquad
\norm{g_n}_2^2 = \frac{n}{2n + 1} \longrightarrow \frac12,
\qquad
\norm{g_n}_\infty = \sqrt n \longrightarrow \infty :
$$

one sequence, three [norms](#def-b2-nvs-norm), three behaviours — convergence to zero, no convergence ([norms](#def-b2-nvs-norm) stabilize at $\frac1{\sqrt2}$ but the pointwise limit is $0$), and explosion. Mass concentrating near $x = 1$ is invisible to $\norm\cdot_1$, half-visible to $\norm\cdot_2$, dominant for $\norm\cdot_\infty$. In infinite dimension, “does it converge?” is not a question about a sequence: it is a question about a sequence *and* a [norm](#def-b2-nvs-norm).

**Method 5.9 (Computing an operator norm).**

Always in two moves. *Upper bound:* estimate $\norm{u(x)}$ by $C\norm x$ using triangle inequalities, Cauchy–Schwarz, or integral bounds — this proves $\vertiii u
\leq C$. *Witness:* exhibit either a specific $x_0 \neq 0$ with $\norm{u(x_0)} = C\norm{x_0}$ (the bound is attained), or a sequence of unit vectors $x_n$ with $\norm{u(x_n)} \to C$ (the bound is approached). Both moves are mandatory: an upper bound alone gives only $\vertiii u \leq C$, a witness alone only $\vertiii u \geq C$. In infinite dimension the witness may have to be a sequence — the supremum need not be attained ([Exercise 5.8](#exo-b2-nvs-8)).

**Example 5.10 (Diagonal operators see every norm alike).**

For $D = \operatorname{diag}(d_1, \dots, d_n)$ on $K^n$ with *any* of the [norms](#def-b2-nvs-norm) $\norm\cdot_1, \norm\cdot_2,
\norm\cdot_\infty$: from $\abs{d_ix_i} \leq
\bigl(\max_j\abs{d_j}\bigr)\abs{x_i}$ coordinatewise, $\norm{Dx} \leq \max_j\abs{d_j}\,\norm x$; and $x = e_{j_0}$ (a maximizing index) attains it. So $\vertiii D =
\max_j\abs{d_j}$ in all three cases: for diagonal maps, all reasonable [norms](#def-b2-nvs-norm) tell the same story, the largest stretch factor. Everything difficult about [operator norms](#thm-b2-nvs-continuouslinear) is about *non*-diagonal behaviour — which is why the adapted [norms](#def-b2-nvs-norm) of this chapter’s weekend problem (question 22) work by forcing a matrix to become diagonal first.

**Example 5.11 (Column sums: the 111-norm twin of Exercise 5.4).**

On $(\R^n, \norm\cdot_1)$, the [operator norm](#thm-b2-nvs-continuouslinear) of a matrix $A$ is the largest absolute *column* sum. Run the method: for $\norm x_1 \leq 1$,

$$
\norm{Ax}_1 = \sum_i\Bigl|\sum_j a_{ij}x_j\Bigr|
\leq \sum_j \abs{x_j}\sum_i\abs{a_{ij}}
\leq \Bigl(\max_j\sum_i\abs{a_{ij}}\Bigr)\norm x_1 ,
$$

and the bound is attained at $x = e_{j_0}$ for a maximizing column $j_0$ — the tidiest witness imaginable. So for $A =
\left(\begin{smallmatrix}1 & -2\\ 3 & 1\end{smallmatrix}\right)$: $\vertiii A_1 = \max(1 + 3,\ 2 + 1) = 4$, while $\vertiii A_\infty = 4$ too (rows) — a coincidence here, not a law: [transpose](https://one-course.com/books/math/4/en/chapter/2-linear-algebra#def-b2-linalg-transpose) the matrix entries asymmetrically and the two [norms](#def-b2-nvs-norm) part company. Rows for $\norm\cdot_\infty$, columns for $\norm\cdot_1$: the mnemonic is that each [norm](#def-b2-nvs-norm)’s unit vectors (sign patterns, resp. basis vectors) pick out the matching sums.

**Proposition 5.12 (Bilinear maps).**

A bilinear map $b \colon E \times F \to G$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) iff $\norm{b(x,y)} \leq C\norm x\,\norm y$ for some $C$; then it is [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on bounded sets. (Same proof pattern; the product $(u, v) \mapsto v \circ u$ and matrix multiplication are the key examples.)

**Proof.** If the bound holds:

$$
b(x,y) - b(x_0,y_0) = b(x - x_0,\, y) + b(x_0,\, y - y_0),
$$

so $\norm{b(x,y) - b(x_0,y_0)} \leq C\norm{x - x_0}\norm y +
C\norm{x_0}\norm{y - y_0}$: [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $(x_0, y_0)$, and a [Lipschitz](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) bound where $\norm x, \norm y \leq R$. Conversely, [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $(0,0)$ gives $\delta$ with $\norm{b(x,y)} \leq 1$ on $\norm x, \norm y \leq \delta$; scale both variables. ∎

## 5.3 Finite dimension

**Theorem 5.13 (Equivalence of norms in finite dimension).**

On a finite-dimensional space, *all [norms](#def-b2-nvs-norm) are [equivalent](#def-b2-nvs-equivalent)*. Consequently, in finite dimension: convergence, openness, [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact), [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) are norm-independent notions; [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) $=$ closed and bounded; the space is [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete); and every linear (or multilinear) map *from* a finite-dimensional space is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity).

**Proof.** Fix a basis and identify $E \simeq K^n$; it suffices to compare any [norm](#def-b2-nvs-norm) $N$ with $\norm\cdot_\infty$.

*One direction is algebra:* $N(x) = N(\sum x_i e_i) \leq
\sum \abs{x_i} N(e_i) \leq C \norm x_\infty$ with $C = \sum N(e_i)$. This also shows $N$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on $(K^n, \norm\cdot_\infty)$ (it is $C$-Lipschitz: $\abs{N(x) - N(y)} \leq N(x - y)$).

*The other is topology:* the unit sphere $S = \{x :
\norm{x}_\infty = 1\}$ is closed and bounded in $(K^n,
\norm\cdot_\infty)$, hence [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) ([Theorem 4.16](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-compactprops) (2), valid for $\C^n \simeq
\R^{2n}$). The [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) function $N$ attains its minimum $c$ on $S$; $c > 0$ since $N$ vanishes only at $0 \notin S$. Homogeneity spreads the bound: $N(x) \geq c \norm{x}_\infty$ for all $x$.

Consequences: all statements reduce to $(K^n,
\norm\cdot_\infty)$, where they are known ([Theorem 4.9](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-rncomplete), [Theorem 4.16](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-compactprops)); a linear $u$ from finite-dimensional $E$ satisfies $\norm{u(x)} \leq \sum\abs{x_i}
\norm{u(e_i)} \leq C'\norm{x}_\infty$: bound (4) of [Theorem 5.6](#thm-b2-nvs-continuouslinear). ∎

**Corollary 5.14.**

A finite-dimensional subspace of any normed space is closed.

**Proof.** It is [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) for the induced [norm](#def-b2-nvs-norm) ([Theorem 5.13](#thm-b2-nvs-finitedim)), and [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) subsets are closed ([Definition 4.7](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete)). ∎

**Example 5.15 (A best approximation computed by symmetry).**

In $\bigl(C(\intcc{-1}{1}), \norm\cdot_\infty\bigr)$, how far is $f(x) = \abs x$ from the (closed, two-dimensional) subspace of affine functions $a + bx$? By symmetry, replacing $a + bx$ by $a - bx$ leaves $\norm{f - (a \pm bx)}_\infty$ unchanged, and the midpoint $a$ does at least as well (triangle inequality on the average): it suffices to consider constants. For a constant $a$:

$$
\norm{\abs x - a}_\infty = \max\,(1 - a,\ a)
\geq \frac12 ,
$$

minimized at $a = \frac12$: the distance is $\frac12$, attained by the constant $\frac12$. Note the error curve $\abs x -
\frac12$: it reaches $\pm\frac12$ alternately at $x = -1, 0,
1$ — three extrema of alternating sign for a best approximation from a two-parameter family. That *equioscillation* pattern is no accident; it is the signature of optimality that this chapter’s weekend problem turns into Chebyshev’s theorem.

**Example 5.16 (Closed versus dense subspaces).**

In $E = \bigl(C(\intcc{0}{1}), \norm\cdot_\infty\bigr)$: each $\R_n[X]$ (polynomials of degree $\leq n$, restricted to $\intcc01$) is a finite-dimensional, hence *closed*, subspace — a uniform limit of polynomials of degree $\leq n$ is one. But the union $\R[X]$ of all of them is *dense* in $E$ (the Weierstrass approximation theorem, proved in [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq)), and dense proper subspaces are as non-closed as can be. The moral: closedness of subspaces is a finite-dimensional privilege; stacking closed floors can build a dense skyscraper.

**Theorem 5.17 (Riesz).**

The closed unit ball of a normed space $E$ is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) *if and only if* $\dim E < \infty$.

**Proof.** Finite dimension: closed and bounded suffices ([Theorem 5.13](#thm-b2-nvs-finitedim)).

Conversely, suppose $\dim E = \infty$. *Riesz’s lemma:* for every proper closed subspace $F \subsetneq E$ and $\varepsilon \in
\intoo{0}{1}$, there is a unit vector $x$ with $d(x, F) \geq 1 -
\varepsilon$. Proof: pick $y \notin F$, let $\delta = d(y, F) > 0$ ($F$ closed), choose $f \in F$ with $\norm{y - f} \leq
\frac{\delta}{1 - \varepsilon}$, and set $x = \frac{y - f}{\norm{y -
f}}$: for any $g \in F$,

$$
\norm{x - g} = \frac{\norm{y - (f + \norm{y-f}\,g)}}{\norm{y - f}}
\geq \frac{\delta}{\norm{y-f}} \geq 1 - \varepsilon ,
$$

the numerator being a distance from $y$ to a point of $F$.

Now build unit vectors $x_1, x_2, \dots$ inductively: $F_k =
\operatorname{Vect}(x_1, \dots, x_k)$ is finite-dimensional, hence closed ([Corollary 5.14](#cor-b2-nvs-closedsubspace)) and proper; Riesz’s lemma with $\varepsilon = \frac12$ provides a unit $x_{k+1}$ with $d(x_{k+1},
F_k) \geq \frac12$. The sequence satisfies $\norm{x_p - x_q} \geq
\frac12$ for $p \neq q$: no convergent subsequence — the unit ball is not [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact). ∎

**Example 5.18 (Riesz as a dimension detector).**

Is $C(\intcc01)$ finite-dimensional? Riesz answers without exhibiting any explicit infinite free family: the sequence $f_n(x) = x^n$ lies in the closed unit ball and satisfies, for $m > n$, $\norm{f_n - f_m}_\infty \geq f_n(x_0) - f_m(x_0) > 0$ at suitable points — quantified cleanly in this chapter’s weekend problem (question 16), where a subsequence stays at mutual distance $\geq \frac14$. No convergent subsequence, so the ball is not [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact), so $\dim C(\intcc01) = \infty$ by [Theorem 5.17](#thm-b2-nvs-riesz). [Compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) of the unit ball is a perfect dichotomy: it holds in finite dimension, fails in infinite dimension, with no middle ground — geometry alone reads off the dimension type.

## 5.4 Banach spaces

**Definition 5.19.**

A *Banach space* is a [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) normed space. Examples: every finite-dimensional normed space ([Theorem 5.13](#thm-b2-nvs-finitedim)); $\bigl(C(\intcc{a}{b}),
\norm\cdot_\infty\bigr)$ ([Theorem 4.9](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-rncomplete)); $\mathcal{L}_c(E, F)$ for $F$ Banach (same proof pattern as for [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) functions). Non-example: $\bigl(C(\intcc{0}{1}),
\norm\cdot_1\bigr)$ ([Exercise 5.7](#exo-b2-nvs-7)).

**Example 5.20 (The operator norm of integration).**

On $\bigl(C(\intcc01), \norm\cdot_\infty\bigr)$, let $T(f)(x) =
\int_0^x f(t)\,\dd t$ (an endomorphism: $T(f)$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)). Run [Method 5.9](#met-b2-nvs-opnorm). Upper bound:

$$
\abs{T(f)(x)} \leq \int_0^x\abs f \leq x\,\norm f_\infty \leq
\norm f_\infty ,
$$

so $\vertiii T \leq 1$. Witness: $f \equiv 1$ gives $T(f)(x) =
x$ and $\norm{T(f)}_\infty = 1 = \norm f_\infty$: attained, $\vertiii T = 1$. But note $\vertiii{T^2} = \frac12 \neq
\vertiii T^2$: indeed $T^2(f)(x) = \int_0^x(x - t)f(t)\dd t$ has $\abs{T^2(f)(x)} \leq \frac{x^2}2\norm f_\infty$, attained again at $f \equiv 1$; and generally $\vertiii{T^n} = \frac1{n!}$ — the submultiplicative bound $\vertiii T^n = 1$ is off by a factorial. This is exactly the phenomenon the iterate trick of [Chapter 4](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#ch-b2-metric)’s weekend problem converts into global solvability of linear differential equations.

**Theorem 5.21 (Absolute convergence in Banach spaces).**

In a [Banach space](#def-b2-nvs-banach), if $\sum \norm{u_n} < \infty$ then $\sum u_n$ converges, and $\norm{\sum u_n} \leq \sum\norm{u_n}$. (The full theory of series in normed spaces is [Chapter 7](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#ch-b2-series).)

**Proof.** Partial sums $S_N$: for $q > p$, $\norm{S_q - S_p} \leq
\sum_{n=p+1}^{q}\norm{u_n}$, which tends to $0$ (Cauchy criterion for the real series of [norms](#def-b2-nvs-norm)): $(S_N)$ is Cauchy, hence convergent. The inequality passes to the limit from the finite triangle inequality. ∎

**Example 5.22 (Matrix exponential, first contact).**

The *matrix exponential*: $\mathcal{M}_n(K)$ with any submultiplicative [norm](#def-b2-nvs-norm) ($\vertiii{AB}
\leq \vertiii A \vertiii B$) is Banach (finite dimension). Then, for every $A$,

$$
\eu^A = \sum_{k=0}^{\infty} \frac{A^k}{k!}
$$

converges absolutely ($\vertiii{A^k/k!} \leq \vertiii A^k /k!$, summable): well defined. [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq) exploits it systematically.

**Example 5.23 (A Neumann series that terminates).**

For $A = \left(\begin{smallmatrix}0 & \frac12\\ 0 &
0\end{smallmatrix}\right)$: $\vertiii A < 1$ in any [operator norm](#thm-b2-nvs-continuouslinear) built on [Example 5.2](#ex-b2-nvs-examples)’s [norms](#def-b2-nvs-norm), and $A^2 = 0$, so the geometric series collapses:

$$
(I - A)^{-1} = \sum_{k \geq 0} A^k = I + A =
\begin{pmatrix}1 & \tfrac12\\ 0 & 1\end{pmatrix},
$$

verified by $(I - A)(I + A) = I - A^2 = I$. Nilpotency truncates the series exactly as it truncated the exponential in [Chapter 3](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#ch-b2-reduction); and the example calibrates expectations: the Neumann inverse is an infinite series in general, a polynomial precisely when the perturbation is nilpotent, and the error after $N$ terms is always bounded by the geometric tail $\vertiii A^{N+1}/(1 - \vertiii A)$.

**Example 5.24 (The exponential of a rotation generator).**

Take $A = \left(\begin{smallmatrix}0 & -\theta\\ \theta &
0\end{smallmatrix}\right)$. Then $A^2 = -\theta^2 I$, so the powers [cycle](https://one-course.com/books/math/4/en/chapter/1-sets-and-structures#def-b2-structures-sn) with period four, and the series splits into even and odd parts:

$$
\eu^{A} = \sum_{k}\frac{A^k}{k!}
= \Bigl(\sum_{j}\frac{(-1)^j\theta^{2j}}{(2j)!}\Bigr) I
+ \Bigl(\sum_{j}\frac{(-1)^j\theta^{2j+1}}{(2j+1)!}\Bigr)
\frac{A}{\theta}
= \begin{pmatrix}
\cos\theta & -\sin\theta\\
\sin\theta & \cos\theta
\end{pmatrix},
$$

all rearrangements being licensed by absolute convergence. The exponential of an antisymmetric generator is a rotation — computed here purely from the series, three chapters before the differential equation $x' = Ax$ ([Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq)) explains *why*: $\eu^{tA}$ is uniform circular motion. The closing insight: identities among matrix series are proved exactly like scalar ones, once a submultiplicative [norm](#def-b2-nvs-norm) certifies absolute convergence.

**Example 5.25 (Sup norm means uniform: the dictionary).**

The statement $\norm{f_n - f}_\infty \to 0$ *is* uniform convergence: one number, $\sup_x\abs{f_n(x) - f(x)}$, bounds the error at every point simultaneously. The dictionary in action on $f_n(x) = x^n$ over $\intcc{0}{1}$: pointwise, $f_n
\to 0$ on $\intco{0}{1}$ and $f_n(1) = 1$; in [norm](#def-b2-nvs-norm), $\norm{f_n - 0}_\infty = 1 \not\to 0$, and indeed the pointwise limit is discontinuous, hence out of reach of a $\norm\cdot_\infty$-limit in $C(\intcc01)$ (which is closed under uniform limits, [Theorem 4.9](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-rncomplete)). On $\intcc{0}{a}$, $a < 1$: $\norm{f_n}_\infty = a^n \to 0$ — uniform convergence restored by shrinking the domain. Every convergence statement of [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq) is a statement about this one [norm](#def-b2-nvs-norm); keeping the dictionary in mind halves that chapter.

**Remark 5.26 (Common pitfalls).**

(i) An [operator norm](#thm-b2-nvs-continuouslinear) depends on *both* chosen [norms](#def-b2-nvs-norm): the same matrix has $\vertiii\cdot_\infty$ given by row sums ([Exercise 5.4](#exo-b2-nvs-4)) and a different $\vertiii\cdot_1$ (column sums); quoting “the” [norm](#def-b2-nvs-norm) of a matrix without naming the underlying [norms](#def-b2-nvs-norm) is meaningless. (ii) $\vertiii{AB} \leq
\vertiii A\,\vertiii B$ is an inequality, usually strict — powers can shrink much faster than the bound $\vertiii A^k$ suggests, which is the whole point of adapted [norms](#def-b2-nvs-norm) (this chapter’s weekend problem, question 22). (iii) “Linear implies [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)” is a finite-dimensional privilege: differentiation on polynomials is linear and unbounded ([Example 5.7](#ex-b2-nvs-operatornorms)). (iv) Absolute convergence of $\sum u_n$ only helps when the space is [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) ([Exercise 7.9](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#exo-b2-series-9) builds the counterexample). (v) In infinite dimension a supremum over the unit ball is a genuine supremum: do not assume it is attained ([Exercise 5.8](#exo-b2-nvs-8)).

**Remark 5.27 (Perspectives within this volume).**

Three appointments are now fixed. With [Chapter 7](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#ch-b2-series): in a [Banach space](#def-b2-nvs-banach), absolutely convergent series converge, so the geometric and exponential series of operators become everyday tools — inverting $I - A$, defining $\eu^{A}$ ([Example 7.2](https://one-course.com/books/math/4/en/chapter/7-sequences-and-series#ex-b2-series-neumann)). With [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq) and [Chapter 11](https://one-course.com/books/math/4/en/chapter/11-power-series#ch-b2-powerseries): convergence of function sequences and power series is convergence in $\bigl(C, \norm\cdot_\infty\bigr)$ ([Example 5.25](#ex-b2-nvs-uniformdictionary)), and the radius of convergence is a statement about which geometric series dominate. With [Chapter 14](https://one-course.com/books/math/4/en/chapter/14-fourier-series#ch-b2-fourier): the [norms](#def-b2-nvs-norm) $\norm\cdot_2$ and $\norm\cdot_\infty$ genuinely disagree on $C(\intcc{0}{1})$ ([Example 5.8](#ex-b2-nvs-sqrtnxn)), which is exactly why mean-square convergence of Fourier series and uniform convergence are two different theorems with two different prices.

**Remark 5.28 (Where this chapter is used).**

[Operator norms](#thm-b2-nvs-continuouslinear) and the geometric series drive the perturbation arguments of [Chapter 15](https://one-course.com/books/math/4/en/chapter/15-differential-calculus#ch-b2-diffcalc) (inverse function theorem) and the [matrix exponential](#ex-b2-nvs-matrixexp) of [Chapter 16](https://one-course.com/books/math/4/en/chapter/16-differential-equations#ch-b2-diffeq); [equivalence of norms](#thm-b2-nvs-finitedim) silently authorizes every “choose your favourite [norm](#def-b2-nvs-norm)” argument in [Chapter 10](https://one-course.com/books/math/4/en/chapter/10-sequences-and-series-of-functions#ch-b2-funcseq) and beyond; and the finite/infinite divide of Riesz’s theorem — made quantitative in this chapter’s weekend problem — is the reason the Year 3 volume needs new tools (weak convergence, Arzelà–Ascoli, Hilbert space projections) where this volume could still extract convergent subsequences.

## 5.5 Exercises

**Exercise 5.1 ★.**

On $\R^2$, draw the unit balls of $\norm\cdot_1$, $\norm\cdot_2$, $\norm\cdot_\infty$, and prove the inequalities $\norm x_\infty
\leq \norm x_2 \leq \norm x_1 \leq 2\norm x_\infty$ with the best constants in dimension $2$.

**Solution of Exercise 5.1.**

Unit balls: a diamond ($\norm\cdot_1$), a disk ($\norm\cdot_2$), a square ($\norm\cdot_\infty$), nested in that order. Inequalities: $\norm x_\infty \leq \norm x_2$ (one square is at most the sum); $\norm x_2 \leq \norm x_1$ (squaring: $x_1^2 + x_2^2 \leq (\abs{x_1}
+ \abs{x_2})^2$); $\norm x_1 \leq 2\norm x_\infty$ (two terms, each $\leq \max$). Sharpness: $(1, 0)$ makes the first two equalities; $(1, 1)$ makes $\norm x_1 = 2\norm x_\infty$ and also shows $\norm x_2 = \sqrt2 \norm x_\infty$ and $\norm x_1 = \sqrt2 \norm
x_2$ are the extreme ratios in the other direction.

**Exercise 5.2 ★.**

Is $N(f) = \abs{f(0)} + \norm{f'}_\infty$ a [norm](#def-b2-nvs-norm) on $C^1(\intcc{0}{1})$? Compare it with $\norm{f}_\infty$: one inequality holds, the other fails (exhibit).

**Solution of Exercise 5.2.**

[Norm](#def-b2-nvs-norm) axioms: homogeneity and triangle inequality are inherited termwise; separation: $N(f) = 0$ forces $f' = 0$ (so $f$ constant) and $f(0) = 0$: $f = 0$. It is a [norm](#def-b2-nvs-norm).

Comparison: $\norm f_\infty \leq N(f)$, since $\abs{f(x)} \leq
\abs{f(0)} + \abs{\int_0^x f'} \leq \abs{f(0)} +
\norm{f'}_\infty$. The converse fails: take $f_n(x) = \frac1n
\sin(nx)$: then $\norm{f_n}_\infty \leq \frac1n \to 0$ while $N(f_n) = 0 + \norm{\cos(nx)}_\infty = 1$. No constant $C$ gives $N \leq C\norm\cdot_\infty$.

**Exercise 5.3 ★.**

Compute the [operator norm](#thm-b2-nvs-continuouslinear) of $u(f) = \int_0^1 f(t)\,\eu^t\,\dd t$ on $\bigl(C(\intcc{0}{1}), \norm\cdot_\infty\bigr) \to \R$, and of the shift $S(x_1, x_2, \dots, x_n) = (x_2, \dots, x_n, 0)$ on $(K^n, \norm\cdot_\infty)$.

**Solution of Exercise 5.3.**

$\abs{u(f)} \leq \norm f_\infty \int_0^1 \eu^t\,\dd t = (\eu -
1)\norm f_\infty$, with equality for $f \equiv 1$: $\vertiii u =
\eu - 1$.

Shift: $\norm{S(x)}_\infty = \max(\abs{x_2}, \dots, \abs{x_n}) \leq
\norm x_\infty$, equality at $x = e_2$: $\vertiii S = 1$ (for $n
\geq 2$).

**Exercise 5.4 ★★.**

On $(\R^n, \norm\cdot_\infty)$, prove that the [operator norm](#thm-b2-nvs-continuouslinear) of a matrix $A$ is $\vertiii A_\infty = \max_i \sum_j \abs{a_{ij}}$ (the largest absolute row sum). Compute it for $\begin{pmatrix} 1 & -2\\
3 & 1\end{pmatrix}$.

**Solution of Exercise 5.4.**

Upper bound: for $\norm x_\infty \leq 1$,

$$
\abs{(Ax)_i} = \Bigl|\sum_j a_{ij}x_j\Bigr| \leq \sum_j
\abs{a_{ij}},
$$

so $\norm{Ax}_\infty \leq \max_i \sum_j \abs{a_{ij}}$. Attained: let $i_0$ realize the max and take $x_j = \operatorname{sign}(a_{i_0
j})$ (modulus $1$ entries): then $(Ax)_{i_0} = \sum_j \abs{a_{i_0
j}}$. Hence the formula. For the given matrix: row sums $3$ and $4$: $\vertiii A_\infty = 4$.

**Exercise 5.5 ★★.**

Prove that $GL_n(K)$ is [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) in $\mathcal{M}_n(K)$ and that $A
\mapsto A^{-1}$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) on it. *Hint: for openness, if $\vertiii H < \frac{1}{\vertiii{A^{-1}}}$ then $A + H = A(I +
A^{-1}H)$ with $\vertiii{A^{-1}H} < 1$, and $I + B$ is invertible for $\vertiii B < 1$ by the geometric series ([Theorem 5.21](#thm-b2-nvs-absoluteconvergence)); for [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), bound $(A+H)^{-1} - A^{-1}$ using the same series.*

**Solution of Exercise 5.5.**

*Geometric series:* for $\vertiii B < 1$, the series $\sum
B^k$ converges absolutely in the Banach $\mathcal{M}_n(K)$ ([Theorem 5.21](#thm-b2-nvs-absoluteconvergence), $\vertiii{B^k} \leq
\vertiii B^k$), and

$$
(I - B)\sum_{k=0}^{K} B^k = I - B^{K+1} \longrightarrow I :
$$

by [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of the product ([Proposition 5.12](#prop-b2-nvs-bilinear)), $(I - B)\sum_{k\geq0} B^k = I$, so $I - B$ is invertible with inverse the sum (and $\vertiii{(I-B)^{-1}} \leq \frac{1}{1 - \vertiii B}$).

*Openness:* for $A$ invertible and $\vertiii H <
\frac{1}{\vertiii{A^{-1}}}$: $A + H = A(I + A^{-1}H)$ with $\vertiii{A^{-1}H} \leq \vertiii{A^{-1}}\vertiii H < 1$: invertible. So a ball around $A$ stays in $GL_n$.

*[Continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of inversion:* with $B = -A^{-1}H$,

$$
(A + H)^{-1} - A^{-1} = \bigl((I - B)^{-1} - I\bigr)A^{-1}
= \Bigl(\sum_{k \geq 1} B^k\Bigr) A^{-1},
$$

of [norm](#def-b2-nvs-norm) $\leq \frac{\vertiii B}{1 - \vertiii B}\vertiii{A^{-1}}
\to 0$ as $H \to 0$.

**Exercise 5.6 ★★.**

Let $\varphi$ be a linear form on a normed space $E$. Prove that $\varphi$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) if and only if $\ker\varphi$ is closed. *(If $\ker\varphi$ is closed and $\varphi \neq 0$, pick $a$ with $\varphi(a) = 1$ and $r > 0$ with $B(a, r) \cap \ker\varphi =
\emptyset$; deduce $\abs{\varphi(h)} \leq \frac{1}{r}\norm h$ by a scaling argument on $a - \frac{h}{\varphi(h)}$.)*

**Solution of Exercise 5.6.**

[Continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $\Rightarrow$ closed kernel: preimage of the closed $\{0\}$ ([Theorem 4.6](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-globalcontinuity)).

Conversely, suppose $\ker\varphi$ closed and $\varphi \neq 0$. Pick $a$ with $\varphi(a) = 1$; since $a \notin \ker\varphi$ and the kernel is closed, some ball $B(a, r)$ misses it. Now let $h \in E$ with $\varphi(h) \neq 0$: the vector $a - \frac{h}{\varphi(h)}$ lies in $\ker\varphi$, hence outside $B(a, r)$:

$$
\Bigl\Vert \frac{h}{\varphi(h)} \Bigr\Vert \geq r
\quad\Longrightarrow\quad
\abs{\varphi(h)} \leq \frac{\norm h}{r},
$$

an inequality trivially true when $\varphi(h) = 0$ as well: bound (4) of [Theorem 5.6](#thm-b2-nvs-continuouslinear): [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity).

**Exercise 5.7 ★★.**

Prove that $\bigl(C(\intcc{0}{1}), \norm\cdot_1\bigr)$ is not [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete): show that the functions $f_n$, affine ramps from $0$ to $1$ over $\bigl[\frac12 - \frac1n, \frac12\bigr]$ (value $0$ before, $1$ after), form a Cauchy sequence with no [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $\norm\cdot_1$-limit.

**Solution of Exercise 5.7.**

Let $f_n$ be $0$ on $\intcc{0}{\frac12 - \frac1n}$, affine up to value $1$ at $\frac12$, then $1$ on $\intcc{\frac12}{1}$. For $m
\geq n$, $f_m - f_n$ is supported on an interval of length $\frac1n$ with values in $\intcc{-1}{1}$: $\norm{f_m - f_n}_1 \leq
\frac1n$: Cauchy.

Suppose $f_n \to f$ in $\norm\cdot_1$ with $f$ [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity). On $\intcc{0}{\frac12 - \delta}$ (fixed $\delta$): $\int \abs{f} =
\int\abs{f - f_n} \leq \norm{f - f_n}_1 \to 0$ for $n >
\frac1\delta$, so $\int_0^{1/2 - \delta}\abs f = 0$, and by strict positivity $f = 0$ there — for every $\delta$: $f = 0$ on $\intoo{0}{\frac12}$. Similarly $f = 1$ on $\intcc{\frac12}{1}$ (the $f_n$ all equal $1$ there). By [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $\frac12$: $0 =
1$, absurd. No limit exists: the space is not [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete).

**Exercise 5.8 ★★★.**

On $E = C(\intcc{0}{1})$ with $\norm\cdot_\infty$, consider

$$
\varphi(f) = \sum_{n \geq 1} (-1)^n\, 2^{-n} f\bigl(\tfrac1n\bigr).
$$

Prove that $\varphi$ is a well-defined [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) linear form with $\vertiii\varphi = 1$, but that the supremum defining $\vertiii\varphi$ is *not attained* on the closed unit ball. *(Upper bound: triangle inequality. [Norm](#def-b2-nvs-norm) $= 1$: build [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f_K$ with $\norm{f_K}_\infty \leq 1$ and $f_K(\frac1n)
= (-1)^n$ for $n \leq K$ — the points $\frac1n$ are isolated from each other. Non-attainment: equality would force $f(\frac1n) =
(-1)^n$ for every $n$, incompatible with [continuity](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) of $f$ at $0$ since $\frac1n \to 0$.)*

**Solution of Exercise 5.8.**

Well defined and [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity): $\abs{\varphi(f)} \leq \sum 2^{-n}
\norm f_\infty = \norm f_\infty$, so $\varphi$ is a linear form with $\vertiii\varphi \leq 1$ (the series converges absolutely for each $f$).

[Norm](#def-b2-nvs-norm) $1$: fix $K$; the points $1, \frac12, \dots, \frac1K$ are pairwise distinct, so there is a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f_K$, $\norm{f_K}_\infty
\leq 1$, with $f_K(\frac1n) = (-1)^n$ for $n \leq K$ (piecewise affine interpolation, constant near $0$). Then

$$
\varphi(f_K) \geq \sum_{n=1}^{K} 2^{-n} - \sum_{n > K} 2^{-n}
= 1 - 2^{-K+1} \xrightarrow[K \to \infty]{} 1 .
$$

Not attained: if $\norm f_\infty \leq 1$ and $\varphi(f) = 1$, each term must contribute its maximum: $(-1)^n f(\frac1n) = 1$ for every $n$ (otherwise the strict deficit of one term cannot be compensated, all terms being $\leq 2^{-n}$). So $f(\frac1n) = (-1)^n$; but $\frac1n \to 0$ and $f$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) at $0$, forcing the contradictory convergence of $(-1)^n$. Hence the supremum is not a maximum — impossible in finite dimension, where the closed unit ball is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact).

**Exercise 5.9 ★★★.**

Let $E$ be a normed space in which the closed unit ball is [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact). Re-derive, without quoting [Theorem 5.17](#thm-b2-nvs-riesz), that every bounded sequence has a convergent subsequence, and prove that every linear form on $E$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) if and only if $\dim E < \infty$. *(For infinite dimension, build a discontinuous form by defining it freely on a linearly independent normalized sequence and extending — admitting the existence of an algebraic complement.)*

**Solution of Exercise 5.9.**

*Bounded sequences:* a bounded sequence lies in some closed ball $\overline B(0, R) = R\,\overline B(0,1)$, [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) (image of the [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) unit ball under the homeomorphism $x \mapsto Rx$): extract there.

*Linear forms:* if $\dim E < \infty$, every linear map from $E$ is [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Theorem 5.13](#thm-b2-nvs-finitedim)). Conversely, suppose $\dim E = \infty$ (which, by [Theorem 5.17](#thm-b2-nvs-riesz), is actually excluded by the [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) hypothesis — the point of this question is the implication between the two properties in general normed spaces): choose a normalized linearly independent sequence $(e_n)$, complete it into an algebraic basis (admitted), and define $\varphi(e_n) = n$, $\varphi = 0$ on the other basis vectors, extended linearly. Then $\abs{\varphi(e_n)} = n$ with $\norm{e_n} = 1$: unbounded on the unit ball, discontinuous. So “all forms [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity)” forces finite dimension.

**Exercise 5.10 ★★.**

On $C(\intcc{0}{1})$, prove $\norm f_1 \leq \norm f_2 \leq \norm
f_\infty$ *(Cauchy–Schwarz for the first)*, and show with the family $f_n(x) = x^n$ that neither inequality can be reversed up to a constant: the three [norms](#def-b2-nvs-norm) are pairwise [non-equivalent](#def-b2-nvs-equivalent).

**Solution of Exercise 5.10.**

Cauchy–Schwarz with the constant function $1$: $\norm f_1 =
\int_0^1 \abs f\cdot 1 \leq \bigl(\int_0^1
f^2\bigr)^{1/2}\bigl(\int_0^1 1\bigr)^{1/2} = \norm f_2$. And $\norm f_2^2 = \int f^2 \leq \norm f_\infty^2$. For $f_n(x) =
x^n$:

$$
\norm{f_n}_1 = \frac1{n+1}, \qquad
\norm{f_n}_2 = \frac1{\sqrt{2n+1}}, \qquad
\norm{f_n}_\infty = 1 .
$$

Then $\norm{f_n}_2/\norm{f_n}_1 = \frac{n+1}{\sqrt{2n+1}} \to
\infty$ and $\norm{f_n}_\infty/\norm{f_n}_2 = \sqrt{2n+1} \to
\infty$: no reverse inequalities, no pair is [equivalent](#def-b2-nvs-equivalent).

**Exercise 5.11 ★★.**

(Distance to a hyperplane) Let $\varphi$ be a nonzero [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) linear form on a normed space $E$. Prove that

$$
d\bigl(x, \ker\varphi\bigr) =
\frac{\abs{\varphi(x)}}{\vertiii\varphi}
\qquad (x \in E),
$$

and check on [Exercise 5.8](#exo-b2-nvs-8) that the infimum need not be attained by any point of the hyperplane.

**Solution of Exercise 5.11.**

*Lower bound for the distance:* for $h \in \ker\varphi$, $\abs{\varphi(x)} = \abs{\varphi(x - h)} \leq
\vertiii\varphi\,\norm{x - h}$; take the infimum over $h$: $d(x, \ker\varphi) \geq \abs{\varphi(x)}/\vertiii\varphi$.

*Upper bound:* we may assume $\varphi(x) \neq 0$. Given $\varepsilon > 0$, pick a unit $u$ with $\abs{\varphi(u)} \geq
\vertiii\varphi - \varepsilon > 0$ and set $h = x -
\frac{\varphi(x)}{\varphi(u)}\,u$: then $\varphi(h) = 0$ and

$$
\norm{x - h} = \frac{\abs{\varphi(x)}}{\abs{\varphi(u)}}
\leq \frac{\abs{\varphi(x)}}{\vertiii\varphi - \varepsilon}.
$$

Let $\varepsilon \to 0$: $d(x, \ker\varphi) \leq
\abs{\varphi(x)}/\vertiii\varphi$; equality.

*Non-attainment:* take $\varphi$ from [Exercise 5.8](#exo-b2-nvs-8) ($\vertiii\varphi = 1$, not attained) and any $x$ with $\varphi(x) \neq 0$. If some $h \in \ker\varphi$ realized $\norm{x - h} = \abs{\varphi(x)}$, the unit vector $v = (x -
h)/\norm{x - h}$ would satisfy $\abs{\varphi(v)} =
\abs{\varphi(x)}/\norm{x - h} = 1 = \vertiii\varphi$: the [operator norm](#thm-b2-nvs-continuouslinear) would be attained — contradiction.

**Exercise 5.12 ★★★.**

On $E = \R[X]$ (all polynomials), let $N_1(P) =
\sup_{\intcc{0}{1}}\abs P$ and $N_2(P) =
\sup_{\intcc{0}{2}}\abs P$. Show that $N_1 \leq N_2$ but that $N_1$ and $N_2$ are *not* [equivalent](#def-b2-nvs-equivalent); deduce that the identity $(E, N_2) \to (E, N_1)$ is a [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) linear bijection whose inverse is discontinuous. Show finally that $(E, N_1)$ is not [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) *(Taylor partial sums of $\eu^x$)*. All three phenomena are impossible in finite dimension — say why.

**Solution of Exercise 5.12.**

$N_1 \leq N_2$ is monotonicity of the sup in the domain, so the identity $(E, N_2) \to (E, N_1)$ is $1$-Lipschitz. For $P_n(x) =
(x/2)^n$: $N_2(P_n) = 1$ (attained at $x = 2$) while $N_1(P_n) =
2^{-n}$: a bound $N_2 \leq CN_1$ would give $1 \leq C2^{-n}$ for all $n$: impossible. So the [norms](#def-b2-nvs-norm) are not [equivalent](#def-b2-nvs-equivalent) and the inverse identity is a discontinuous linear bijection.

*Incompleteness:* let $S_n = \sum_{k=0}^{n}\frac{X^k}{k!}$. For $m > n$, $N_1(S_m - S_n) \leq \sum_{k>n}\frac1{k!} \to 0$: Cauchy for $N_1$. If $S_n \to P$ in $(E, N_1)$, then pointwise $P(x) = \lim S_n(x) = \eu^x$ on $\intcc{0}{1}$; but a polynomial of degree $d$ cannot equal $\eu^x$ on an interval (differentiate $d + 1$ times: the left side dies, $\eu^x$ does not). No limit in $E$: not [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete).

In finite dimension all three phenomena are impossible: all [norms](#def-b2-nvs-norm) are [equivalent](#def-b2-nvs-equivalent), every normed space is [complete](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete), and the inverse of a linear bijection is linear from a finite-dimensional space, hence [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) ([Theorem 5.13](#thm-b2-nvs-finitedim)).

## 5.6 Problem: Best Approximation and Chebyshev’s Theorem

How well can a function be approximated by polynomials of a given degree, and which polynomial does it best? On the existence side, the answer belongs to this chapter: [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) in finite dimension makes best approximations exist. On the explicit side, one nontrivial case can be solved completely with bare hands — among all *monic* polynomials of degree $n$, the one of smallest sup [norm](#def-b2-nvs-norm) on $\intcc{-1}{1}$ is the (normalized) Chebyshev polynomial, of [norm](#def-b2-nvs-norm) $2^{1-n}$: *Chebyshev’s extremal theorem*. The problem proves both sides, then measures how badly [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) fails in infinite dimension: the unit ball of $C(\intcc{0}{1})$ contains infinite constellations of points at mutual distance $1$.

**Problem 5.1.**

Weekend problem — Chebyshev’s extremal theorem and the geometry of the unit ball

[Norms](#def-b2-nvs-norm) without subscript are sup [norms](#def-b2-nvs-norm) on the indicated segment.

**Part I — Best approximation in normed spaces.**

1. Let $F$ be a finite-dimensional subspace of a normed space $E$ and $x \in E$ . Prove that the distance $d(x,  F) = \inf_{f \in F}\norm{x - f}$ is *attained* *(reduce to a closed bounded subset of $F$ and use [Theorem 5.13](#thm-b2-nvs-finitedim))* .
2. A [norm](#def-b2-nvs-norm) is *strictly convex* when $\norm u = \norm v  = 1$ and $u \neq v$ imply $\bigl\Vert\frac{u +  v}2\bigr\Vert < 1$ . Show that $\norm\cdot_2$ on $\R^n$ is strictly convex *(parallelogram identity)* , and that $\norm\cdot_1$ and $\norm\cdot_\infty$ are not for $n \geq 2$ .
3. Prove that for a strictly convex [norm](#def-b2-nvs-norm) , the best approximation of question 1 is *unique* .
4. In $(\R^2, \norm\cdot_\infty)$ , compute all best approximations of $x = (0, 1)$ by the line $F =  \operatorname{Vect}\bigl((1,0)\bigr)$ : an interval of minimizers.
5. In $\bigl(C(\intcc{a}{b}), \norm\cdot_\infty\bigr)$ , show that the best approximation of $f$ by *constants* is unique, equal to $c^* = \frac{\max f  + \min f}{2}$ , with distance $\frac{\max f - \min  f}{2}$ ; compute both for $f(x) = x^2$ on $\intcc{0}{1}$ .

**Part II — Chebyshev polynomials.**

6. Show there is exactly one polynomial $T_n$ with $T_n(\cos\theta) = \cos n\theta$ for all $\theta$ *(recurrence $T_{n+1} = 2XT_n - T_{n-1}$ from the cosine addition formula)* , that $\deg T_n = n$ , and that its leading coefficient is $2^{n-1}$ for $n \geq 1$ .
7. Show $\abs{T_n} \leq 1$ on $\intcc{-1}{1}$ , with $T_n(\eta_k) = (-1)^k$ at the $n + 1$ points $\eta_k =  \cos\frac{k\pi}{n}$ ( $k = 0, \dots, n$ ), and that the roots of $T_n$ are the $n$ points $\cos\frac{(2k-1)\pi}{2n}$ , interlacing the $\eta_k$ .
8. Compute $T_2, T_3, T_4$ , and verify the alternation of $T_3$ at $\eta_0, \dots, \eta_3 = 1, \frac12, -\frac12,  -1$ by direct evaluation.
9. For $\abs x \geq 1$, prove $$T_n(x) = \frac{\bigl(x + \sqrt{x^2 - 1}\bigr)^n +  \bigl(x - \sqrt{x^2 - 1}\bigr)^n}{2},$$ and deduce $T_n(x) \sim \frac12\bigl(x + \sqrt{x^2 -  1}\bigr)^n \to \infty$ geometrically for fixed $x > 1$.
10. Prove the composition law $T_m \circ T_n = T_{mn}$ *(check on $\intcc{-1}{1}$ and invoke the rigidity of polynomials)* .

**Part III — Chebyshev’s extremal theorem.** Write $Q_n = 2^{1-n}T_n$ (monic, by question 6).

11. Let $P$ be monic of degree $n \geq 1$ with $\sup_{\intcc{-1}{1}}\abs P < 2^{1-n}$. By evaluating $D  = Q_n - P$ at the points $\eta_k$ and counting sign changes, derive a contradiction. Conclude: $$\sup_{\intcc{-1}{1}}\abs P \;\geq\; 2^{1-n}  \qquad\text{for every monic } P \text{ of degree } n.$$
12. (Equality case) Suppose $\sup_{\intcc{-1}{1}}\abs P =  2^{1-n}$ with $P$ monic of degree $n$ , and let $D = Q_n  - P \neq 0$ . Show $(-1)^kD(\eta_k) \geq 0$ for all $k$ ; show that each of the $n$ intervals $\intcc{\eta_{k}}{\eta_{k-1}}$ contains a zero of $D$ , and that a zero shared by two consecutive intervals is an *interior* point $\eta_k$ where $D' = 0$ as well. Conclude that $D$ has $n$ zeros counted with multiplicity, hence $D = 0$ : the minimizer is exactly $Q_n$ — *Chebyshev’s extremal theorem* .
13. Restate the theorem as a distance: on $\intcc{-1}{1}$, $$d_\infty\bigl(X^n,\ \R_{n-1}[X]\bigr) = 2^{1-n},$$ with unique best approximation $X^n - Q_n$; and show by the affine substitution $x = \frac{1+t}2$ that on $\intcc{0}{1}$ the distance becomes $2^{1-2n}$.
14. (Optimal interpolation nodes) For $n$ nodes $x_1, \dots,  x_n \in \intcc{-1}{1}$ , the node polynomial $\omega(x) = \prod_i(x - x_i)$ is monic of degree $n$ . Deduce from question 12 which choice of nodes minimizes $\sup_{\intcc{-1}{1}}\abs\omega$ , the node-dependent factor of the classical interpolation error bound, and give the minimal value.
15. Verify the case $n = 2$ of the theorem by hand (find $\inf_c \sup_{\intcc{-1}{1}}\abs{x^2 - c}$ directly), and compute numerically the distance of question 13 on $\intcc{0}{1}$ for $n = 10$ . What does its size say about the graph of $x^{10}$ ?

**Part IV — The unit ball of $C(\intcc{0}{1})$.**

16. Let $g_k(x) = x^{2^k}$ . Show $\norm{g_k}_\infty = 1$ and $\norm{g_k - g_j}_\infty \geq \frac14$ for $j > k$ *(evaluate at the point where $x^{2^k} =  \frac12$)* : an explicit bounded sequence with no convergent subsequence — the closed unit ball is not [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) , by bare hands.
17. (Riesz’s lemma, sharpened) Let $F$ be a *finite-dimensional* proper subspace of a normed space $E$ . Using question 1, produce a unit vector $x$ with $d(x, F) = 1$ exactly — not just $\geq 1 -  \varepsilon$ as in [Theorem 5.17](#thm-b2-nvs-riesz) ’s lemma.
18. Deduce: in every infinite-dimensional normed space there is a sequence of unit vectors with pairwise distances $\geq 1$ , and re-derive Riesz’s theorem from it.
19. In $C(\intcc{0}{1})$ , exhibit such a constellation explicitly: the tent functions $h_n$ supported on $\bigl[\frac1{n+1}, \frac1n\bigr]$ with peak value $1$ . Verify $\norm{h_n} = 1$ , $\norm{h_n - h_m} = 1$ for $n  \neq m$ , and note that $h_n \to 0$ pointwise but not uniformly.
20. (Total boundedness fails) Show that the closed unit ball of $C(\intcc{0}{1})$ cannot be covered by finitely many balls of radius $\frac13$ *(each such ball contains at most one $h_n$)* — contrast with the total-boundedness step in the proof of [Theorem 4.20](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-borellebesgue) .

**Part V — [Norms](#def-b2-nvs-norm) at work on matrices, and synthesis.**

21. Prove that every [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\lambda$ of $A \in  \mathcal{M}_n(\C)$ satisfies $\abs\lambda \leq  \vertiii A$ for every [operator norm](#thm-b2-nvs-continuouslinear) ; apply [Exercise 5.4](#exo-b2-nvs-4) to bound the [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) of $\left(\begin{smallmatrix}1 & -2\\ 3 & 1\end{smallmatrix}\right)$ and compare with their true modulus.
22. (Adapted [norms](#def-b2-nvs-norm) ) Let $A$ be [diagonalizable](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-diag) , $A =  P\,\mathrm{diag}(\lambda_1, \dots, \lambda_n)\,P^{-1}$ . Show $N_P(x) = \norm{P^{-1}x}_\infty$ is a [norm](#def-b2-nvs-norm) whose [operator norm](#thm-b2-nvs-continuouslinear) satisfies $\vertiii A_{N_P} =  \max_i\abs{\lambda_i}$ .
23. Deduce: for [diagonalizable](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-diag) $A$ , $A^k \to 0$ if and only if all [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) satisfy $\abs{\lambda_i} < 1$ — [equivalence of norms](#thm-b2-nvs-finitedim) makes the conclusion norm-independent. Check on $A =  \frac14\left(\begin{smallmatrix}1 & 2\\ 2 &  1\end{smallmatrix}\right)$ .
24. (Equivalence constants blow up) On $\R_n[X]$ , compare $N_c(P) = \max_k \abs{a_k}$ (coefficients) and $\norm{P}_{\intcc{0}{1}}$ : both are [norms](#def-b2-nvs-norm) , so they are [equivalent](#def-b2-nvs-equivalent) for each fixed $n$ ; but show, using the monic minimizer of question 13 on $\intcc{0}{1}$ , that the best constant $C_n$ in $N_c \leq  C_n\norm\cdot_{\intcc{0}{1}}$ satisfies $C_n \geq  2^{2n-1}$ . Conclude in one sentence why “all [norms](#def-b2-nvs-norm) are [equivalent](#def-b2-nvs-equivalent) ” dies in infinite dimension.
25. (Synthesis) One sentence each: where [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) of finite-dimensional balls worked (questions 1, 12); what strict convexity governs; what the constellation of questions 18–19 destroys; and how question 24 quantifies the failure. Name the summit (Chebyshev’s extremal theorem) and state where best approximation finds its modern home (the projection theorem on Hilbert spaces, Year 3 volume, where [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) replaces [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) ).

**Solution of Problem 5.1.**

**1.** The candidates worth considering form $K = \{f \in F
: \norm{x - f} \leq \norm x\}$: nonempty ($0 \in K$), closed (preimage of a closed interval under the [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) $f \mapsto
\norm{x - f}$, intersected with the closed $F$, [Corollary 5.14](#cor-b2-nvs-closedsubspace)), bounded ($\norm f \leq
\norm{f - x} + \norm x \leq 2\norm x$). In the finite-dimensional $F$, closed and bounded means [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) ([Theorem 5.13](#thm-b2-nvs-finitedim)); the [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity) function $f
\mapsto \norm{x - f}$ attains on $K$ its infimum, which equals the infimum over all of $F$ (any $f \notin K$ gives $\norm{x -
f} > \norm x \geq \inf$).

**2.** Parallelogram identity in $(\R^n,
\norm\cdot_2)$: $\norm{u + v}^2 + \norm{u - v}^2 = 2\norm u^2 +
2\norm v^2$ (expand the squares of sums of coordinates). For unit $u \neq v$:

$$
\Bigl\Vert\frac{u+v}2\Bigr\Vert^2 = 1 - \frac{\norm{u -
v}^2}{4} < 1 .
$$

Not strictly convex: for $\norm\cdot_\infty$, take $u = (1, 1, 0,
\dots)$, $v = (1, -1, 0, \dots)$: unit vectors with midpoint $(1, 0, \dots)$ of [norm](#def-b2-nvs-norm) $1$; for $\norm\cdot_1$, take $u = (1,
0, \dots)$, $v = (0, 1, 0, \dots)$: midpoint $(\frac12,
\frac12, 0, \dots)$ of [norm](#def-b2-nvs-norm) $1$.

**3.** Let $d = d(x, F)$. If $d = 0$: $x \in \overline F =
F$ and the only minimizer is $x$. If $d > 0$ and $f_1 \neq f_2$ both minimize: $u = \frac{x - f_1}{d}$ and $v = \frac{x -
f_2}{d}$ are distinct unit vectors, so

$$
\Bigl\Vert x - \frac{f_1 + f_2}2\Bigr\Vert
= d\,\Bigl\Vert\frac{u + v}2\Bigr\Vert < d ,
$$

with $\frac{f_1 + f_2}2 \in F$: contradicts the definition of $d$. Unique minimizer.

**4.** $\norm{(0,1) - t(1,0)}_\infty = \max(\abs t, 1)
\geq 1$, with equality iff $\abs t \leq 1$: the minimizers form the segment $\{t(1, 0) : t \in \intcc{-1}{1}\}$, all at distance $1$ — uniqueness fails exactly because the square ball has flat sides (question 2).

**5.** Let $M = \max f$, $m = \min f$ (attained: [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact)). For any constant $c$: $\sup\abs{f - c} \geq
\max(M - c,\, c - m) \geq \frac{M - m}2$, the last inequality because the two quantities average to $\frac{M-m}2$; equality in both forces $M - c = c - m$, i.e. $c = c^* = \frac{M + m}2$. Conversely $\sup\abs{f - c^*} = \max(M - c^*, c^* - m) =
\frac{M - m}2$. Unique best constant. For $f(x) = x^2$ on $\intcc01$: $c^* = \frac12$, distance $\frac12$.

**6.** From $\cos(n{+}1)\theta + \cos(n{-}1)\theta =
2\cos\theta\cos n\theta$: the polynomials defined by $T_0 = 1$, $T_1 = X$, $T_{n+1} = 2XT_n - T_{n-1}$ satisfy $T_n(\cos\theta)
= \cos n\theta$ by induction. Uniqueness: two polynomials agreeing on $\intcc{-1}{1}$ (infinitely many points) are equal. Induction again: $\deg T_n = n$ with leading coefficient $2^{n-1}$ for $n \geq 1$ ($T_1$: coefficient $1 = 2^0$; the recurrence doubles it).

**7.** Every $x \in \intcc{-1}{1}$ is $\cos\theta$, and $\abs{\cos n\theta} \leq 1$. At $\eta_k = \cos\frac{k\pi}n$: $T_n(\eta_k) = \cos k\pi = (-1)^k$, and $1 = \eta_0 > \eta_1 >
\dots > \eta_n = -1$. Roots: $\cos n\theta = 0$ iff $\theta =
\frac{(2k-1)\pi}{2n}$: the $n$ distinct points $\cos\frac{(2k-1)\pi}{2n}$, and since $\frac{(k-1)\pi}n <
\frac{(2k-1)\pi}{2n} < \frac{k\pi}n$, each root lies strictly between two consecutive extrema.

**8.** $T_2 = 2X^2 - 1$, $T_3 = 4X^3 - 3X$, $T_4 = 8X^4 -
8X^2 + 1$. For $T_3$: $T_3(1) = 1$, $T_3(\tfrac12) = \tfrac12 -
\tfrac32 = -1$, $T_3(-\tfrac12) = 1$, $T_3(-1) = -1$: perfect alternation.

**9.** Let $u_\pm = x \pm \sqrt{x^2 - 1}$ for $x \geq 1$: the roots of $z^2 - 2xz + 1$, with $u_+u_- = 1$. The sequence $s_n = \frac{u_+^n + u_-^n}2$ satisfies $s_{n+1} = 2x\,s_n -
s_{n-1}$ (Newton-type recurrence from the quadratic), $s_0 = 1$, $s_1 = x$: the same recurrence and initial values as $n \mapsto
T_n(x)$, so $s_n = T_n(x)$ for all $n$. Since $0 < u_- \leq 1
\leq u_+$ with $u_+ > 1$ for $x > 1$: $T_n(x) \geq
\frac{u_+^n}2 \to \infty$ and $T_n(x) \sim \frac12\bigl(x +
\sqrt{x^2-1}\bigr)^n$. (For $x \leq -1$ use the parity $T_n(-x)
= (-1)^nT_n(x)$, clear from the recurrence.)

**10.** For every $\theta$: $T_m\bigl(T_n(\cos\theta)\bigr)
= T_m(\cos n\theta) = \cos mn\theta =
T_{mn}(\cos\theta)$. The polynomials $T_m \circ T_n$ and $T_{mn}$ agree on $\intcc{-1}{1}$, hence are equal.

**11.** $D = Q_n - P$ has degree $\leq n - 1$ (the monic leading terms cancel). At the extrema: $(-1)^kD(\eta_k) =
2^{1-n} - (-1)^kP(\eta_k) \geq 2^{1-n} - \abs{P(\eta_k)} > 0$ by hypothesis. So $D$ takes nonzero values of alternating signs at the $n + 1$ decreasing points $\eta_0 > \dots > \eta_n$: by the intermediate value theorem it has at least $n$ distinct roots, one in each [open](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-topology) interval $\intoo{\eta_{k}}{\eta_{k-1}}$. A nonzero polynomial of degree $\leq n - 1$ cannot have $n$ roots; and $D = 0$ contradicts the strict signs. Contradiction: $\sup\abs P \geq 2^{1-n}$ for every monic $P$ of degree $n$.

**12.** Now $(-1)^kD(\eta_k) = 2^{1-n} - (-1)^kP(\eta_k)
\geq 0$ since $\abs{P} \leq 2^{1-n}$. On each $\intcc{\eta_k}{\eta_{k-1}}$ ($k = 1, \dots, n$) the endpoint values of $D$ have opposite weak signs: the intermediate value theorem yields a zero $z_k$ in the closed interval. If the $z_k$ can be chosen pairwise distinct, $D \neq 0$ of degree $\leq n-1$ has $n$ roots: contradiction. Two consecutive intervals can only share the zero $z_k = z_{k+1} = \eta_k$ with $0 < k < n$ (interior). There, $D(\eta_k) = 0$ means $P(\eta_k) =
(-1)^k2^{1-n}$, an extremal value of $P$ on $\intcc{-1}{1}$ attained at an *interior* point: $P'(\eta_k) = 0$; and $\eta_k$ is an interior extremum of $T_n$ too: $Q_n'(\eta_k) =
0$. Hence $D'(\eta_k) = 0$: $\eta_k$ is a root of multiplicity $\geq 2$, compensating the shared interval. In all cases $D$ has at least $n$ roots counted with multiplicity, degree $\leq n -
1$, so $D = 0$: $P = Q_n$. Chebyshev’s extremal theorem is proved: the unique monic minimizer is $2^{1-n}T_n$, of sup [norm](#def-b2-nvs-norm) $2^{1-n}$.

**13.** Monic polynomials of degree $n$ are exactly the $X^n - R$ with $R \in \R_{n-1}[X]$, so

$$
d_\infty\bigl(X^n, \R_{n-1}[X]\bigr) = \min_{P \text{
monic}}\ \sup_{\intcc{-1}{1}}\abs P = 2^{1-n},
$$

uniquely at $R^* = X^n - Q_n$. Substitution $x = \frac{1+t}2$: if $P$ is monic of degree $n$ on $\intcc01$, then $t \mapsto
2^nP\bigl(\frac{1+t}2\bigr)$ is monic on $\intcc{-1}{1}$ with sup equal to $2^n\sup_{\intcc01}\abs P$: hence $\sup_{\intcc01}\abs P \geq 2^{-n}\cdot2^{1-n} = 2^{1-2n}$, with equality exactly for $P^*(x) = 2^{-n}Q_n(2x - 1)$: on $\intcc{0}{1}$ the distance is $2^{1-2n}$.

**14.** $\omega$ is monic of degree $n$, so $\sup_{\intcc{-1}{1}}\abs\omega \geq 2^{1-n}$ with equality iff $\omega = Q_n = 2^{1-n}T_n$, i.e. iff the nodes are the $n$ roots of $T_n$: $x_k = \cos\frac{(2k-1)\pi}{2n}$ — the *Chebyshev nodes*. Minimal value: $2^{1-n}$. Equidistant nodes are strictly worse; the interpolation error factor is minimized by clustering nodes near the endpoints.

**15.** $n = 2$ by hand: $x^2$ ranges over $\intcc01$, so $\sup_{\intcc{-1}{1}}\abs{x^2 - c} = \max(\abs c, \abs{1 - c})
\geq \frac12$, minimized at $c = \frac12$: minimal monic quadratic $x^2 - \frac12 = \frac12(2x^2 - 1) = Q_2$, value $\frac12 = 2^{1-2}$. For $n = 10$ on $\intcc{0}{1}$: $2^{1-20} =
2^{-19} \approx 1.9\cdot10^{-6}$. Some polynomial of degree $9$ stays within two millionths of $x^{10}$ on the whole of $\intcc01$: at this scale the two graphs are indistinguishable — the flatness of $x^{10}$ near $0$ lets lower degrees do all the work.

**16.** $\norm{g_k}_\infty = g_k(1) = 1$. For $j > k$ set $a = 2^k$, $b = 2^j \geq 2a$, and evaluate at $x_0 = 2^{-1/a}$ (so $x_0^a = \frac12$):

$$
g_k(x_0) - g_j(x_0) = \frac12 - \Bigl(\frac12\Bigr)^{b/a}
\geq \frac12 - \frac14 = \frac14 .
$$

So $\norm{g_k - g_j}_\infty \geq \frac14$ for all $j \neq k$: no subsequence is Cauchy, none converges. The closed unit ball of $\bigl(C(\intcc01), \norm\cdot_\infty\bigr)$ is not [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact).

**17.** $F$ is closed ([Corollary 5.14](#cor-b2-nvs-closedsubspace)) and proper: pick $y \notin F$, so $\delta = d(y, F) > 0$. By question 1 the distance is attained at some $f^* \in F$. Set $x
= \frac{y - f^*}{\delta}$, a unit vector ($\norm{y - f^*} =
\delta$). For every $g \in F$:

$$
\norm{x - g} = \frac{\norm{y - (f^* + \delta g)}}{\delta}
\geq \frac{\delta}{\delta} = 1 ,
$$

since $f^* + \delta g \in F$. Hence $d(x, F) \geq 1$; and $d(x,
F) \leq \norm{x - 0} = 1$: exactly $1$.

**18.** In an infinite-dimensional $E$, build unit vectors inductively: $x_1$ arbitrary; given $x_1, \dots, x_k$, the subspace $F_k = \operatorname{Vect}(x_1, \dots, x_k)$ is finite-dimensional, hence proper, and question 17 gives a unit $x_{k+1}$ with $d(x_{k+1}, F_k) = 1$: in particular $\norm{x_{k+1} - x_i} \geq 1$ for $i \leq k$. The sequence has pairwise distances $\geq 1$: the unit ball contains a sequence with no convergent subsequence, hence is not [compact](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) — Riesz’s theorem, with the sharp constant $1$.

**19.** Let $h_n$ be affine on each half of $\bigl[\frac1{n+1}, \frac1n\bigr]$, from $0$ up to $1$ at the midpoint and back to $0$, and zero elsewhere: [continuous](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-continuity), $\norm{h_n}_\infty = 1$. For $n \neq m$ the supports meet at most at a common endpoint, where both vanish; at the peak of $h_n$, $h_m = 0$: $\norm{h_n - h_m}_\infty = 1$ exactly. For fixed $x > 0$: $h_n(x) = 0$ as soon as $\frac1n < x$, and $h_n(0) = 0$ always: $h_n \to 0$ pointwise; but $\norm{h_n - 0}_\infty = 1$: not uniformly. An explicit constellation at mutual distance $1$ inside the unit ball.

**20.** A ball of radius $\frac13$ has diameter $\leq
\frac23 < 1$, so it contains at most one of the $h_n$ (two are at distance $1$). Finitely many such balls contain finitely many of the infinitely many $h_n$: they cannot cover the unit ball. Total boundedness — which [compact metric spaces](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) enjoy, by the proof of [Theorem 4.20](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#thm-b2-metric-borellebesgue) — fails as badly as possible.

**21.** If $Ax = \lambda x$ with $x \neq 0$: $\abs\lambda\,\norm x = \norm{Ax} \leq \vertiii A\,\norm x$, so $\abs\lambda \leq \vertiii A$. For $\left(\begin{smallmatrix}1 & -2\\ 3 & 1\end{smallmatrix}\right)$: $\vertiii A_\infty = \max(1 + 2,\ 3 + 1) = 4$ ([Exercise 5.4](#exo-b2-nvs-4)), so every [eigenvalue](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) has modulus $\leq
4$; in fact $\chi_A = X^2 - 2X + 7$ gives $\lambda = 1 \pm
\iu\sqrt6$ of modulus $\sqrt7 \approx 2.65$: the bound is valid, not sharp.

**22.** $N_P$ is a [norm](#def-b2-nvs-norm): $N_P(x) = 0$ forces $P^{-1}x = 0$, so $x = 0$; homogeneity and triangle inequality are inherited from $\norm\cdot_\infty$ through the linear $P^{-1}$. [Operator norm](#thm-b2-nvs-continuouslinear): with $y = P^{-1}x$ and $D = \mathrm{diag}(\lambda_i)$,

$$
N_P(Ax) = \norm{P^{-1}AP\,y}_\infty = \norm{Dy}_\infty,
$$

so $\vertiii A_{N_P}$ is the $\norm\cdot_\infty$-operator [norm](#def-b2-nvs-norm) of $D$, which is its largest absolute row sum ([Exercise 5.4](#exo-b2-nvs-4)): $\max_i\abs{\lambda_i}$.

**23.** If all $\abs{\lambda_i} < 1$: $N_P(A^kx) \leq
\rho^kN_P(x)$ with $\rho = \max\abs{\lambda_i} < 1$, so $A^kx
\to 0$ for every $x$, and $A^k \to 0$ in any [norm](#def-b2-nvs-norm) on $\mathcal{M}_n$ (all [equivalent](#def-b2-nvs-equivalent) in finite dimension, [Theorem 5.13](#thm-b2-nvs-finitedim); convergence of $A^ke_j$ for each $j$ is entrywise convergence). If some $\abs{\lambda} \geq 1$ with [eigenvector](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $x$: $\norm{A^kx} = \abs\lambda^k\norm x
\not\to 0$. For $A = \frac14\left(\begin{smallmatrix}1 & 2\\ 2 &
1\end{smallmatrix}\right)$: [eigenvalues](https://one-course.com/books/math/4/en/chapter/3-reduction-of-endomorphisms#def-b2-reduction-eigen) $\frac14(1 \pm 2) =
\frac34, -\frac14$, both of modulus $< 1$: $A^k \to 0$.

**24.** Both are [norms](#def-b2-nvs-norm) on the finite-dimensional $\R_n[X]$, hence [equivalent](#def-b2-nvs-equivalent) for each $n$. Take the minimal monic $P^*_n$ of question 13 on $\intcc01$: its coefficient of $X^n$ is $1$, so $N_c(P^*_n) \geq 1$, while $\norm{P^*_n}_{\intcc01} = 2^{1-2n}$. Hence

$$
C_n \geq \frac{N_c(P^*_n)}{\norm{P^*_n}_{\intcc01}}
\geq 2^{2n-1} .
$$

The equivalence constants explode with the dimension: on the union $\R[X]$ no single constant serves, which is exactly the non-equivalence seen in [Exercise 5.12](#exo-b2-nvs-12) — “all [norms](#def-b2-nvs-norm) are [equivalent](#def-b2-nvs-equivalent)” is a theorem about one dimension at a time, and infinite dimension is where it dies.

**25.** [Compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) of closed bounded sets in finite dimension produced existence of best approximations (question 1) and powered the zero-counting at the $n + 1$ extremal points (questions 11–12, via attained sups). Strict convexity governs uniqueness of best approximation — round balls give one minimizer, flat-sided balls give segments of them (questions 2–4). The constellation of unit vectors at mutual distance $1$ (questions 17–19) destroys [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) of the unit ball and total boundedness with it (question 20). Question 24 quantifies the collapse: the constants relating two [norms](#def-b2-nvs-norm) on $\R_n[X]$ grow like $4^n$, so no uniform comparison survives the passage to $\R[X]$. The summit is Chebyshev’s extremal theorem (questions 11–12): the unique monic minimizer $2^{1-n}T_n$. Best approximation finds its modern home in Hilbert spaces, where the projection theorem replaces [compactness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-compact) by [completeness](https://one-course.com/books/math/4/en/chapter/4-topology-of-metric-spaces#def-b2-metric-complete) plus the parallelogram identity — proved honestly in the Year 3 volume.