---
title: "Banach Spaces and the Fundamental Theorems"
book: "University Mathematics — Year 3"
subject: math
language: en
chapter: 8
exercises: 12
source: https://one-course.com/books/math/5/en/chapter/8-banach-spaces-and-the-fundamental-theorems
---

# Chapter 8 — Banach Spaces and the Fundamental Theorems

Functional analysis studies infinite-dimensional normed spaces through the operators and functionals living on them. Its founding discovery is that *[completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete)*, via Baire’s theorem, forces strong uniformity: pointwise bounded families of operators are norm-bounded (Banach–Steinhaus), [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) bijections have [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) inverses (open mapping), and graphs detect [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) (closed graph). The other pillar, *Hahn–Banach*, needs no [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) at all — only Zorn’s lemma — and guarantees that [dual spaces](#def-b3-banach-operator) are rich enough to see every vector. This chapter proves all four theorems and tests them on the classical sequence spaces $\ell^p$, on concrete [dual](#def-b3-banach-operator) computations, and on a genuinely surprising application: there exist [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $2\pi$-periodic functions whose Fourier series *diverges* at a point — resolving in the negative a question Year 2 left open.

Throughout, $E, F$ are normed spaces over $K = \R$ or $\C$; *Banach space* means [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) normed space.

## 8.1 Bounded operators; sequence spaces

**Definition 8.1.**

$\mathcal L(E, F)$ denotes the space of *bounded* (= [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), Year 2) linear maps with the *operator norm* $\vertiii T = \sup_{\norm x \leq
1}\norm{Tx}$; it is submultiplicative: $\vertiii{ST} \leq
\vertiii S\,\vertiii T$. The *dual* is $E' = \mathcal L(E, K)$.

**Proposition 8.2.**

If $F$ is a Banach space, so is $\mathcal L(E, F)$; in particular $E'$ is always a Banach space.

**Proof.** Let $(T_n)$ be Cauchy for $\vertiii\cdot$. For each $x$, $\norm{T_nx - T_mx} \leq \vertiii{T_n - T_m}\norm x$: $(T_nx)$ is Cauchy in $F$, convergent; call the limit $Tx$. $T$ is linear (limits of linear identities); passing to the limit in $\norm{T_nx - T_mx} \leq \varepsilon\norm x$ ($n, m \geq N$) gives $\norm{T_nx - Tx} \leq \varepsilon\norm x$: $T_n \to T$ in [operator norm](#def-b3-banach-operator), and $\vertiii T \leq \vertiii{T_N} +
\varepsilon < \infty$. ∎

**Definition 8.3.**

The classical sequence spaces (over $K$, indexed by $\N$):

$$
\ell^p = \Bigl\{x = (x_n) : \norm x_p = \Bigl(\sum_n
\abs{x_n}^p\Bigr)^{1/p} < \infty\Bigr\}\quad (1 \leq p <
\infty),
$$

$$
\ell^\infty = \{x : \norm x_\infty = \sup\abs{x_n} < \infty\},
$$

and $c_0 = \{x : x_n \to 0\}$ with $\norm\cdot_\infty$. That $\norm\cdot_p$ is a norm follows from the Minkowski inequality, proved in the discrete case exactly as in [Chapter 12](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#ch-b3-lp) (or by summing the finite-dimensional inequality of Year 2). All are Banach spaces, and $c_0$ is a closed subspace of $\ell^\infty$ ([Exercise 8.3](#exo-b3-banach-3)).

![Unit balls of the p-norms in the plane, nested as p grows from 1 (diamond) through 2 (disc) and 4 (superellipse) to ∈fty (square). Convexity of every ball is the Minkowski inequality; the corners at p = 1 and p = ∈fty are where strict convexity, uniqueness of best approximations, and the equality cases of all degenerate at once.](https://one-course.com/images/onecourse/chapters/math-5/b3-banach/fig-ab47c70fb946.svg)

*Unit balls of the $p$-norms in the plane, nested as $p$ grows from $1$ (diamond) through $2$ (disc) and $4$ (superellipse) to $\infty$ (square). Convexity of every ball *is* the Minkowski inequality; the corners at $p = 1$ and $p = \infty$ are where strict convexity, uniqueness of best approximations, and the equality cases of [Exercise 12.12](https://one-course.com/books/math/5/en/chapter/12-the-lp-spaces#exo-b3-lp-12) all degenerate at once.*

**Proposition 8.4 (Neumann series).**

Let $E$ be Banach and $T \in \mathcal L(E) = \mathcal L(E,E)$ with $\vertiii T < 1$. Then $I - T$ is invertible in $\mathcal
L(E)$, with $(I - T)^{-1} = \sum_{n\geq0}T^n$ (convergent in [operator norm](#def-b3-banach-operator)). Consequently the set of invertible operators is open, and inversion is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) on it.

**Proof.** The series converges absolutely ($\vertiii{T^n} \leq \vertiii
T^n$, geometric) in the Banach $\mathcal L(E)$ ([Proposition 8.2](#prop-b3-banach-llcomplete); [Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1)(b)). Telescoping, $(I - T)\sum_{n \leq
N}T^n = I - T^{N+1} \to I$, and similarly on the other side. For openness: if $S$ is invertible and $\vertiii{H} <
1/\vertiii{S^{-1}}$, then $S + H = S(I + S^{-1}H)$ with $\vertiii{S^{-1}H} < 1$: invertible. [Continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of inversion: the series expression gives $\vertiii{(S+H)^{-1} - S^{-1}} =
O(\vertiii H)$ locally. ∎

**Example 8.5 (A Volterra equation, solved by Neumann).**

On $E = \mathcal C(\intcc01)$, consider the integral equation

$$
u(x) = 1 + \lambda\int_0^xu(t)\,\dd t,
\qquad\text{i.e.}\qquad u = \mathbf 1 + \lambda Tu,
\quad (Tu)(x) = \int_0^xu .
$$

Here $\vertiii T \leq 1$, so for $\abs\lambda < 1$ the [Neumann series](#prop-b3-banach-neumann) applies directly: $u = (I - \lambda T)^{-1}\mathbf 1 =
\sum_n\lambda^nT^n\mathbf 1$. Computing, $T^n\mathbf 1 =
\frac{x^n}{n!}$, so

$$
u(x) = \sum_{n\geq0}\frac{(\lambda x)^n}{n!} =
\eu^{\lambda x} ,
$$

as differentiation confirms. Better: $\vertiii{T^n} \leq
\frac1{n!}$ (the iterated kernel shrinks factorially), so $\sum\lambda^nT^n$ converges for *every* $\lambda$ — the operator $I - \lambda T$ is invertible for all $\lambda
\in \C$, even though $\vertiii{\lambda T} \geq 1$ eventually: what matters is the spectral decay of the powers, not the first norm. Volterra operators have this factorial decay built in ([Exercise 7.4](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-4)(a) exploited exactly this), which is why initial value problems never suffer the resonance phenomena of boundary value problems ([Chapter 15](https://one-course.com/books/math/5/en/chapter/15-compact-operators-and-the-spectral-theorem#ch-b3-spectral)).

## 8.2 Hahn–Banach

**Theorem 8.6 (Hahn–Banach, analytic form).**

Let $E$ be a *real* vector space, $p \colon E \to \R$ *sublinear* ($p(x + y) \leq p(x) + p(y)$ and $p(tx) =
tp(x)$ for $t \geq 0$), $F \subseteq E$ a subspace and $f
\colon F \to \R$ linear with $f \leq p$ on $F$. Then $f$ extends to a linear $\tilde f \colon E \to \R$ with $\tilde f \leq p$ on $E$.

**Proof.** *One-step extension.* Let $x_0 \notin F$; we extend $f$ to $F \oplus \R x_0$ by choosing $\alpha = \tilde f(x_0)$ correctly: we need, for all $y \in F$, $t > 0$,

$$
f(y) + t\alpha \leq p(y + tx_0)
\quad\text{and}\quad
f(y) - t\alpha \leq p(y - tx_0),
$$

which after dividing by $t$ (sublinearity) reduce to

$$
\sup_{v \in F}\ \bigl[f(v) - p(v - x_0)\bigr]
\;\leq\; \alpha \;\leq\;
\inf_{u \in F}\ \bigl[p(u + x_0) - f(u)\bigr].
$$

Such $\alpha$ exists iff every left member is $\leq$ every right member: indeed $f(v) + f(u) = f(u + v) \leq p(u + v) \leq
p(u + x_0) + p(v - x_0)$, i.e. $f(v) - p(v - x_0) \leq p(u +
x_0) - f(u)$.

*Zorn.* Order the extensions of $f$ dominated by $p$ (pairs: subspace, functional) by extension; a chain has the union as upper bound; a maximal element must be defined on all of $E$, else the one-step extension contradicts maximality. ∎

**Corollary 8.7.**

Let $E$ be a normed space ($K = \R$ or $\C$).

1. Every $f \in F'$ ( $F$ a subspace) extends to $\tilde f  \in E'$ with $\norm{\tilde f}_{E'} = \norm f_{F'}$ .
2. For every $x \neq 0$ there is $f \in E'$ with $\norm f  = 1$ and $f(x) = \norm x$ . In particular $E'$ separates the points of $E$ , and $\norm x = \sup_{\norm f \leq  1}\abs{f(x)}$ .
3. For a closed subspace $F$ and $x \notin F$ , there is $f  \in E'$ vanishing on $F$ with $f(x) = d(x, F)$ and $\norm f \leq 1$ .

**Proof.** (1) Real case: apply [Theorem 8.6](#thm-b3-banach-hahnbanach) with $p(x)
= \norm f_{F'}\,\norm x$ (sublinear); the extension satisfies $\pm\tilde f(x) = \tilde f(\pm x) \leq p(x)$, so $\norm{\tilde f} \leq \norm f$, and $\geq$ is restriction. Complex case: let $u = \operatorname{Re}f$, a real functional with $\abs u \leq \norm f\norm\cdot$; note $f(x) = u(x) -
\iu\,u(\iu x)$ (check on real and imaginary parts: $\operatorname{Im}f(x) = -\operatorname{Re}f(\iu x)$). Extend $u$ real-linearly with the same bound, and set $\tilde f(x) =
\tilde u(x) - \iu\tilde u(\iu x)$: $\C$-linear (direct check on multiplication by $\iu$), extends $f$; norm: for given $x$ write $\tilde f(x) = r\eu^{\iu\theta}$, then $\abs{\tilde f(x)}
= \tilde f(\eu^{-\iu\theta}x) = \tilde u(\eu^{-\iu\theta}x)
\leq \norm f\,\norm x$.

(2) On $F = Kx$ define $f(tx) = t\norm x$: norm $1$ on $F$; extend by (1). The duality formula: $\leq$ is clear, $\geq$ by this $f$.

(3) On $F \oplus Kx$ define $f(y + tx) = t\,d(x, F)$; then for $t \neq 0$, $\norm{y + tx} = \abs t\,\norm{x + y/t} \geq \abs
t\,d(x, F) = \abs{f(y + tx)}$: $\norm f \leq 1$ on the subspace; extend by (1). ∎

**Remark 8.8.**

By (2), the canonical map $J \colon E \to E''$, $J(x)(f) =
f(x)$, is an isometry ([Exercise 8.10](#exo-b3-banach-10)): every normed space sits inside its bidual. Spaces with $J$ surjective are called *reflexive*; the weekend problem shows $\ell^p$ ($1 < p < \infty$) is [reflexive](#rem-b3-banach-bidual) while $\ell^1$ is not.

## 8.3 The Baire trilogy

**Theorem 8.9 (Banach–Steinhaus, uniform boundedness).**

Let $E$ be a *Banach* space, $F$ normed, and $(T_i)_{i\in
I} \subseteq \mathcal L(E, F)$ a family with $\sup_i
\norm{T_ix} < \infty$ for every $x \in E$. Then $\sup_i
\vertiii{T_i} < \infty$.

**Proof.** The sets $F_n = \{x : \sup_i\norm{T_ix} \leq n\}$ are closed (intersections of preimages of closed balls) and cover $E$. Baire ([Theorem 7.6](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-baire)) gives $n_0$ and a ball $B(x_0, r) \subseteq F_{n_0}$. For $\norm z < r$: $\norm{T_iz}
\leq \norm{T_i(x_0 + z)} + \norm{T_ix_0} \leq 2n_0$, so $\vertiii{T_i} \leq 2n_0/r$ for every $i$. ∎

**Corollary 8.10.**

If $E$ is Banach and $T_n \in \mathcal L(E,F)$ converge *pointwise* ($T_nx \to Tx$ for each $x$), then $\sup_n
\vertiii{T_n} < \infty$, $T \in \mathcal L(E, F)$, and $\vertiii T \leq \liminf \vertiii{T_n}$.

**Proof.** Convergent sequences are bounded: pointwise boundedness; Banach–Steinhaus bounds the norms by some $M$; then $\norm{Tx}
= \lim\norm{T_nx} \leq M\norm x$ ($T$ is linear as a pointwise limit), and the sharper bound by passing to $\liminf$ in $\norm{T_nx} \leq \vertiii{T_n}\norm x$. ∎

**Theorem 8.11 (Divergent Fourier series).**

There exist [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $2\pi$-periodic functions $f$ whose Fourier series diverges at $0$: $\sup_N\abs{S_N(f)(0)} =
\infty$. Indeed such $f$ form a dense subset of $\mathcal
C(S^1)$.

**Proof.** Work in $E = (\mathcal C(S^1), \norm\cdot_\infty)$, a Banach space, with the functionals $\Lambda_N(f) = S_N(f)(0) =
\frac1{2\pi}\int_{-\pi}^{\pi}f(t)\,D_N(t)\,\dd t$ (Dirichlet kernel, Year 2). Each $\Lambda_N$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) with

$$
\norm{\Lambda_N} = \frac1{2\pi}\int_{-\pi}^\pi\abs{D_N(t)}\,\dd
t \;=\; L_N .
$$

($\leq$ is clear; $\geq$: take $f$ [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), $\norm f_\infty
\leq 1$, approximating $\operatorname{sign}D_N$ — the sign has finitely many jumps; smoothing each jump on an interval of length $\varepsilon$ changes the integral by $O(N\varepsilon)$.) The *Lebesgue constants* $L_N$ tend to infinity:

$$
L_N = \frac1{2\pi}\int_{-\pi}^{\pi}
\frac{\abs{\sin\bigl((N{+}\tfrac12)t\bigr)}}{\abs{\sin(t/2)}}
\,\dd t
\geq \frac{2}{\pi}\int_0^{\pi}
\frac{\abs{\sin\bigl((N{+}\tfrac12)t\bigr)}}{t}\,\dd t
= \frac{2}{\pi}\int_0^{(N+\frac12)\pi}\frac{\abs{\sin
u}}{u}\,\dd u
$$

(using $\abs{\sin(t/2)} \leq t/2$ on $[0, \pi]$, then substituting $u = (N + \tfrac12)t$). Cutting into arches:

$$
\int_{(k-1)\pi}^{k\pi}\frac{\abs{\sin u}}u\,\dd u
\geq \frac1{k\pi}\int_{(k-1)\pi}^{k\pi}\abs{\sin u}\,\dd u =
\frac{2}{k\pi},
\qquad\text{so}\qquad
L_N \geq \frac{4}{\pi^2}\sum_{k=1}^N\frac1k
\xrightarrow[N\to\infty]{} \infty .
$$

If every [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ had $\sup_N
\abs{\Lambda_N(f)} < \infty$, Banach–Steinhaus would force $\sup_N\norm{\Lambda_N} < \infty$: contradiction. So some $f$ — in fact a nonmeagre, [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) of $f$ (the complement of $\bigcup_M\{f: \sup_N\abs{\Lambda_Nf}\leq M\}$, a countable union of closed sets which, having no [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) by the above applied in any ball, is [meagre](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#rem-b3-complete-meagre)) — has $\sup_N\abs{S_Nf(0)} =
\infty$. ∎

**Theorem 8.12 (Open mapping).**

Let $E, F$ be Banach spaces and $T \in \mathcal L(E, F)$ *surjective*. Then $T$ is open: $T(B_E(0,1)) \supseteq
B_F(0, c)$ for some $c > 0$. Consequently a bijective bounded operator between Banach spaces has a bounded inverse.

**Proof.** Write $B = B_E(0,1)$. Surjectivity gives $F = \bigcup_n
\overline{T(nB)} = \bigcup_n n\,\overline{T(B)}$; Baire ([Theorem 7.6](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-baire)) gives [interior](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) to $\overline{T(B)}$: some $B_F(y_0, 4c) \subseteq
\overline{T(B)}$. [Re-center](https://one-course.com/books/math/5/en/chapter/1-group-theory#ex-b3-groups-actions) at $0$: for $\norm y < 4c$, both $y_0 + y$ and $y_0$ are limits of images $Tu_k$, $Tv_k$ with $u_k, v_k \in B$, so $y = \lim T(u_k - v_k)$ with $u_k - v_k
\in 2B$: $B_F(0, 4c) \subseteq \overline{T(2B)}$, i.e. $B_F(0, 2c) \subseteq \overline{T(B)}$.

*Removing the [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior)* (here [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) of $E$ enters): let $\norm y < c$. Pick $x_1 \in \frac12 B$ with $\norm{y -
Tx_1} < c/2$ ($B_F(0,2c) \subseteq \overline{T(B)}$ scaled by $\frac12$); inductively $x_k \in 2^{-k}B$ with $\norm{y -
T(x_1 + \dots + x_k)} < c\,2^{-k}$. The series $\sum x_k$ converges absolutely in the Banach $E$, to $x \in B$ (norm $<
\sum 2^{-k} = 1$), and $Tx = y$ by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $B_F(0, c)
\subseteq T(B)$. Openness of $T$ on arbitrary opens follows by translation and scaling; for the corollary, openness of $T$ means $T^{-1}$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). ∎

**Corollary 8.13 (Equivalent norms).**

If a vector space is [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) for two comparable norms ($\norm\cdot_a \leq C\norm\cdot_b$), the norms are equivalent.

**Proof.** The identity $(E, \norm\cdot_b) \to (E, \norm\cdot_a)$ is bounded and bijective between Banach spaces: its inverse is bounded. ∎

**Theorem 8.14 (Closed graph).**

Let $E, F$ be Banach and $T \colon E \to F$ linear. If the graph $\Gamma = \{(x, Tx)\}$ is closed in $E \times F$ (i.e. $x_n \to x$ and $Tx_n \to y$ imply $y = Tx$), then $T$ is bounded.

**Proof.** $E \times F$ with $\norm{(x,y)} = \norm x + \norm y$ is Banach; $\Gamma$, a closed subspace, is Banach. The projection $\pi_E\colon \Gamma \to E$ is bounded and bijective, so its inverse $x \mapsto (x, Tx)$ is bounded ([Theorem 8.12](#thm-b3-banach-openmapping)): $\norm{Tx} \leq \norm{(x,
Tx)} \leq C\norm x$. ∎

**Method 8.15.**

When to reach for which theorem. *Hahn–Banach*: to produce a functional with prescribed behavior (norming a vector, vanishing on a subspace, extending from a subspace) — no [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) needed. *Banach–Steinhaus*: to convert pointwise information into uniform bounds — typically to show a limit operation is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), or (contrapositive) to prove *divergence* for some element, as for Fourier series. *Open mapping / closed graph*: to get [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) for free from [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) bijectivity or from a [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) property of the graph — typical use: comparing two [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) norms, or proving automatic [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity). All three Baire theorems require [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) *of the source*; counterexamples otherwise ([Exercise 8.7](#exo-b3-banach-7)).

## 8.4 Dual spaces, concretely

**Theorem 8.16.**

Isometrically: $(c_0)' \cong \ell^1$ and $(\ell^1)' \cong
\ell^\infty$, via the pairing $\langle x, y\rangle = \sum_n
x_ny_n$.

**Proof.** We prove $(c_0)' \cong \ell^1$; the second identification is [Exercise 8.5](#exo-b3-banach-5). To $y \in \ell^1$ associate $\Lambda_y(x)
= \sum x_ny_n$ ($x \in c_0$): absolutely convergent, with $\abs{\Lambda_y(x)} \leq \norm x_\infty\norm y_1$, so $\norm{\Lambda_y} \leq \norm y_1$. Conversely let $\Lambda \in
(c_0)'$; set $y_n = \Lambda(e_n)$ ($e_n$ the unit sequences). For any $N$, test $x^{(N)} = \sum_{n \leq N}
\operatorname{sign}(\overline{y_n})\,e_n \in c_0$ (norm $\leq
1$; in the complex case use unimodular factors $\bar y_n/\abs
{y_n}$): $\Lambda(x^{(N)}) = \sum_{n\leq N}\abs{y_n} \leq
\norm\Lambda$. So $y \in \ell^1$ with $\norm y_1 \leq
\norm\Lambda$. Finally $\Lambda = \Lambda_y$: both agree on the $e_n$, hence on finite sequences, dense in $c_0$ (truncation: $\norm{x - \sum_{n \leq N}x_ne_n}_\infty = \sup_{n > N}\abs{x_n}
\to 0$ precisely because $x_n \to 0$); [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) functionals agreeing on a [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) are equal. The correspondence is linear, bijective, and isometric ($\norm{\Lambda_y} = \norm y_1$ from the two inequalities). ∎

## 8.5 Exercises

**Exercise 8.1 ★.**

Compute the [operator norms](#def-b3-banach-operator): (a) the shifts $S(x_1, x_2, \dots) = (0, x_1, x_2, \dots)$ and $S^*(x_1, x_2, \dots) = (x_2, x_3, \dots)$ on $\ell^2$; (b) the multiplication operator $M_a x = (a_nx_n)$ on $\ell^2$, for $a \in \ell^\infty$; (c) the functional $\Lambda(f) = \int_0^{1/2}f -
\int_{1/2}^1f$ on $\mathcal C(\intcc01)$ — show $\norm\Lambda = 1$ and that the norm is *not attained*.

**Solution of Exercise 8.1.**

(a) $\norm{Sx}_2 = \norm x_2$: $S$ is an isometry, $\vertiii S =
1$. For the backward shift: $\norm{S^*x}_2^2 = \sum_{n \geq
2}\abs{x_n}^2 \leq \norm x_2^2$, with equality for $x = e_2$: $\vertiii{S^*} = 1$.

(b) $\norm{M_ax}_2^2 = \sum\abs{a_n}^2\abs{x_n}^2 \leq \norm
a_\infty^2\norm x_2^2$; testing $x = e_n$ gives $\vertiii{M_a}
\geq \abs{a_n}$ for every $n$: $\vertiii{M_a} = \norm a_\infty$.

(c) $\abs{\Lambda(f)} \leq \int_0^1\abs f \leq \norm f_\infty$: $\norm\Lambda \leq 1$. For $\varepsilon > 0$ let $f_\varepsilon$ be $1$ on $[0, \frac12 - \varepsilon]$, $-1$ on $[\frac12 +
\varepsilon, 1]$, affine between: $\norm{f_\varepsilon}_\infty =
1$ and $\Lambda(f_\varepsilon) \geq 1 - 2\varepsilon$: $\norm\Lambda = 1$. Not attained: $\Lambda(f) = 1$ with $\norm
f_\infty \leq 1$ forces $\int_0^{1/2}f = \frac12$ and $\int_{1/2}^1 f = -\frac12$, i.e. ([continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), $\abs f \leq 1$) $f \equiv 1$ on $[0, \frac12]$ and $f \equiv -1$ on $[\frac12,
1]$: contradiction at $\frac12$.

**Exercise 8.2 ★.**

Let $E$ be Banach, $T \in \mathcal L(E)$ invertible, and $S$ with $\vertiii{S - T} < 1/\vertiii{T^{-1}}$. Show that $S$ is invertible and estimate $\vertiii{S^{-1} - T^{-1}}$. Application: if a linear system $Tx = b$ is [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived) with $T$ invertible, a sufficiently small perturbation of $T$ keeps it uniquely [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived), with a quantitative bound on the change of solution.

**Solution of Exercise 8.2.**

Write $S = T\bigl(I - T^{-1}(T - S)\bigr)$ with $\vertiii{T^{-1}(T-S)} \leq \vertiii{T^{-1}}\,\vertiii{T - S} =
\theta < 1$: by [Proposition 8.4](#prop-b3-banach-neumann), $S$ is invertible with $S^{-1} = \sum_{n\geq0}\bigl(T^{-1}(T -
S)\bigr)^nT^{-1}$, whence

$$
\vertiii{S^{-1} - T^{-1}} \leq
\sum_{n\geq1}\theta^n\,\vertiii{T^{-1}}
= \frac{\vertiii{T^{-1}}^2\,\vertiii{T - S}}{1 - \theta}.
$$

For the linear system: $x_T = T^{-1}b$ and $x_S = S^{-1}b$ differ by at most that bound times $\norm b$ — small perturbations of an invertible system remain uniquely [solvable](https://one-course.com/books/math/5/en/chapter/1-group-theory#def-b3-groups-derived), with Lipschitz dependence of the solution on the operator.

**Exercise 8.3 ★★.**

(a) Prove that $\ell^1$, $\ell^\infty$ and $c_0$ are Banach spaces, and that $c_0$ is the [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) in $\ell^\infty$ of the space of finite sequences. (b) Show $\ell^p \subseteq \ell^q$ with $\norm\cdot_q \leq
\norm\cdot_p$ for $1 \leq p \leq q \leq \infty$, and that the inclusion is strict.

**Solution of Exercise 8.3.**

(a) $\ell^1$: let $(x^{(k)})$ be Cauchy. Each coordinate is Cauchy ($\abs{x^{(k)}_n - x^{(l)}_n} \leq \norm{x^{(k)} -
x^{(l)}}_1$): let $x_n = \lim_kx^{(k)}_n$. Given $\varepsilon$, for $k, l \geq K$: $\sum_{n \leq N}\abs{x^{(k)}_n - x^{(l)}_n}
\leq \varepsilon$ for every $N$; let $l \to \infty$, then $N \to
\infty$: $\norm{x^{(k)} - x}_1 \leq \varepsilon$, and $x =
x^{(k)} - (x^{(k)} - x) \in \ell^1$. $\ell^\infty$: Cauchy for $\norm\cdot_\infty$ is uniformly Cauchy: converges uniformly to a bounded sequence. $c_0$ is closed in $\ell^\infty$: if $x^{(k)} \to x$ uniformly with $x^{(k)}_n \to_n 0$, then $\abs{x_n} \leq \norm{x - x^{(k)}}_\infty + \abs{x^{(k)}_n}$ gives $\limsup_n\abs{x_n} \leq \varepsilon$: $x \in c_0$; a closed subspace of a Banach space is Banach. Finite sequences: their [closure](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) contains every $x \in c_0$ (truncations converge: $\sup_{n>N}\abs{x_n} \to 0$) and is contained in the closed $c_0$.

(b) By homogeneity assume $\norm x_p = 1$: then $\abs{x_n} \leq
1$ for all $n$, so $\abs{x_n}^q \leq \abs{x_n}^p$ and $\norm
x_q \leq 1 = \norm x_p$; for $q = \infty$, $\abs{x_n} \leq
\norm x_p$ directly. Strictness: $x_n = n^{-\alpha}$ with $\frac1q < \alpha \leq \frac1p$ lies in $\ell^q \setminus
\ell^p$ (Riemann series).

**Exercise 8.4 ★★.**

Let $F \subseteq E$ be a closed subspace and $x \notin F$. Using [Corollary 8.7](#cor-b3-banach-hbnormed), prove the duality formula

$$
d(x, F) = \max\bigl\{\abs{f(x)} : f \in E',\ \norm f \leq 1,\
f\restriction_F = 0\bigr\}
$$

(note: a *maximum*). Deduce that $F = \bigcap\{\ker f : f
\in E',\ f\restriction_F = 0\}$: closed subspaces are exactly the intersections of kernels of functionals.

**Solution of Exercise 8.4.**

($\leq$) If $\norm f \leq 1$ and $f\restriction_F = 0$: for every $y \in F$, $\abs{f(x)} = \abs{f(x - y)} \leq \norm{x -
y}$; take the infimum. ($\geq$, attained) [Corollary 8.7](#cor-b3-banach-hbnormed)(3) produces $f$ with $f\restriction_F = 0$, $\norm f \leq 1$, $f(x) = d(x, F)$: the supremum is a maximum. Consequence: $F \subseteq \bigcap\{\ker
f : f\restriction_F = 0\}$ trivially, and a point $x \notin F$ is excluded from the intersection by the functional above ($f(x) = d(x,F) > 0$, $F$ being closed).

**Exercise 8.5 ★★.**

Prove $(\ell^1)' \cong \ell^\infty$ isometrically, following the scheme of [Theorem 8.16](#thm-b3-banach-duals) (finite sequences are dense in $\ell^1$). Where does the argument break for $(\ell^\infty)'$?

**Solution of Exercise 8.5.**

For $y \in \ell^\infty$: $\abs{\Lambda_y(x)} =
\abs{\sum x_ny_n} \leq \norm y_\infty\norm x_1$, so $\norm{\Lambda_y} \leq \norm y_\infty$; testing on $e_n$: $\abs{y_n} = \abs{\Lambda_y(e_n)} \leq \norm{\Lambda_y}$: equality. Conversely, given $\Lambda \in (\ell^1)'$, set $y_n =
\Lambda(e_n)$: $\abs{y_n} \leq \norm\Lambda$, so $y \in
\ell^\infty$; $\Lambda$ and $\Lambda_y$ agree on finite sequences, which are dense in $\ell^1$ ($\norm{x - \sum_{n\leq
N}x_ne_n}_1 = \sum_{n>N}\abs{x_n} \to 0$): $\Lambda =
\Lambda_y$. The map $y \mapsto \Lambda_y$ is linear, isometric, onto. For $(\ell^\infty)'$ the same start produces a sequence $y_n = \Lambda(e_n)$, but finite sequences are *not* dense in $\ell^\infty$ (the constant sequence $\mathbf 1$ is at distance $1$ from all of them), so $\Lambda$ is not determined by the $y_n$ — and indeed $(\ell^\infty)' \neq \ell^1$ ([Problem 8.1](#pb-b3-banach-1)).

**Exercise 8.6 ★★.**

Let $E, F, G$ be normed with $E$ Banach, and $B \colon E \times
F \to G$ bilinear, [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) in each variable separately. Show that $B$ is (jointly) [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity): $\norm{B(x,y)} \leq
C\norm x\norm y$. *(Apply Banach–Steinhaus to the family $(B(\cdot, y))_{\norm y \leq 1}$.)*

**Solution of Exercise 8.6.**

For each fixed $x$, $y \mapsto B(x, y)$ is [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) linear: $\sup_{\norm y \leq 1}\norm{B(x,y)} < \infty$. So the family $\{B(\cdot, y) : \norm y \leq 1\} \subseteq \mathcal L(E, G)$ (each member [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity), by [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) in $x$) is pointwise bounded on the Banach space $E$: Banach–Steinhaus ([Theorem 8.9](#thm-b3-banach-banachsteinhaus)) yields $C$ with $\norm{B(x,y)} \leq C\norm x$ for all $\norm y \leq 1$; homogeneity in $y$ finishes: $\norm{B(x,y)} \leq
C\norm x\norm y$.

**Exercise 8.7 ★★.**

(a) On $E = \mathcal C(\intcc01)$, compare $\norm\cdot_\infty$ and $\norm\cdot_1$: the identity $(E, \norm\cdot_\infty) \to
(E, \norm\cdot_1)$ is bounded and bijective but its inverse is unbounded. Which hypothesis of [Corollary 8.13](#cor-b3-banach-equivnorms) fails? (b) Exhibit a discontinuous linear map from a dense subspace of $\ell^2$ to $K$ (e.g. on finite sequences), and explain why this does not contradict the closed graph theorem.

**Solution of Exercise 8.7.**

(a) $\norm f_1 \leq \norm f_\infty$: the identity is bounded, and bijective. Its inverse is unbounded: $f_n(x) = x^n$ has $\norm{f_n}_1 = \frac1{n+1} \to 0$ but $\norm{f_n}_\infty = 1$. No contradiction with [Corollary 8.13](#cor-b3-banach-equivnorms): $(\mathcal
C(\intcc01), \norm\cdot_1)$ is *not* [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) ([Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1)); the corollary requires [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) on both sides.

(b) On the space $E_0$ of finite sequences (dense in $\ell^2$), $\varphi(x) = \sum_n n\,x_n$ is linear and unbounded ($\varphi(e_n) = n$ with $\norm{e_n}_2 = 1$). The closed graph theorem does not apply: $E_0$ is not [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) — and $\varphi$ has no [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) extension to $\ell^2$, illustrating that density without uniform [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) is powerless ([Theorem 7.2](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#thm-b3-complete-extension)).

**Exercise 8.8 ★★★.**

(Hellinger–Toeplitz) Let $T \colon \ell^2 \to \ell^2$ be linear (everywhere defined) and *symmetric*: $\langle Tx,
y\rangle = \langle x, Ty\rangle$ for all $x, y$, where $\langle
x, y \rangle = \sum x_n\bar y_n$. Show that $T$ is bounded. *(Closed graph: if $x_k \to x$ and $Tx_k \to z$, test against arbitrary $y$.)* Moral: unbounded symmetric operators — the Hamiltonians of quantum mechanics — can never be defined on the whole space.

**Solution of Exercise 8.8.**

We verify the closed-graph hypothesis. Let $x_k \to x$ and $Tx_k \to z$ in $\ell^2$. For every $y$:

$$
\langle z, y\rangle = \lim_k\langle Tx_k, y\rangle
= \lim_k \langle x_k, Ty\rangle = \langle x, Ty\rangle
= \langle Tx, y\rangle,
$$

using [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of the inner product in each slot (Cauchy–Schwarz) and symmetry twice. So $z - Tx$ is orthogonal to every $y$, in particular to itself: $z = Tx$. The graph is closed and $\ell^2$ is Banach: $T$ is bounded ([Theorem 8.14](#thm-b3-banach-closedgraph)). Hence a symmetric operator defined on *all* of $\ell^2$ is automatically bounded; genuinely unbounded symmetric operators (position, momentum, Hamiltonians) must live on proper dense subspaces.

**Exercise 8.9 ★★★.**

(Polya’s theorem on quadrature) For each $n$, let $\Lambda_n(f)
= \sum_{i=0}^{n} w_{i,n}f(x_{i,n})$ be a quadrature rule on $\mathcal C(\intcc01)$ ($x_{i,n} \in \intcc01$, $w_{i,n} \in
\R$). Show that $\Lambda_n(f) \to \int_0^1f$ for *every* [continuous](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) $f$ if and only if: (i) $\Lambda_n(P) \to \int_0^1P$ for every polynomial $P$, and (ii) $\sup_n\sum_i\abs{w_{i,n}} <
\infty$. *(Compute $\norm{\Lambda_n}$; use Banach–Steinhaus and Weierstrass.)* Check that rules with *positive* weights exact on constants satisfy (ii) automatically.

**Solution of Exercise 8.9.**

First, $\norm{\Lambda_n} = \sum_i\abs{w_{i,n}}$: $\leq$ is the triangle inequality; $\geq$ by testing a piecewise linear $f$ with $\norm f_\infty \leq 1$ and $f(x_{i,n}) =
\operatorname{sign}(w_{i,n})$ (interpolate linearly between the finitely many nodes; where nodes coincide the signs agree).

($\Rightarrow$) Pointwise convergence at every $f$ implies (i), and pointwise boundedness, so Banach–Steinhaus ([Theorem 8.9](#thm-b3-banach-banachsteinhaus)) on the Banach $\mathcal
C(\intcc01)$ gives (ii).

($\Leftarrow$) Let $M = \sup_n\norm{\Lambda_n} + 1$. Given $f$ and $\varepsilon$, choose a polynomial $P$ with $\norm{f -
P}_\infty < \varepsilon/(2M)$ ([Corollary 7.16](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#cor-b3-complete-weierstrass)); then

$$
\Bigl|\Lambda_n f - \int_0^1 f\Bigr|
\leq \abs{\Lambda_n(f - P)} + \Bigl|\Lambda_nP -
\int_0^1P\Bigr| + \Bigl|\int_0^1(P - f)\Bigr|
\leq \varepsilon + \Bigl|\Lambda_nP - \int_0^1P\Bigr| \to
\varepsilon .
$$

Positive weights, exactness on constants: $\sum_i\abs{w_{i,n}}
= \sum_iw_{i,n} = \Lambda_n(\mathbf 1) = \int_0^1 1 = 1$ for rules exact on constants — (ii) holds with constant $1$.

**Exercise 8.10 ★★.**

Show that $J \colon E \to E''$, $J(x)(f) = f(x)$, is a linear isometry (use [Corollary 8.7](#cor-b3-banach-hbnormed)(2)), and that it is surjective when $\dim E < \infty$. Show also that if $E'$ is separable then so is $E$. *(Pick $x_n$ nearly norming a dense sequence of $E'$ and show their closed span is $E$, via [Corollary 8.7](#cor-b3-banach-hbnormed)(3).)*

**Solution of Exercise 8.10.**

Linearity of $J$ is formal; $\norm{J(x)} = \sup_{\norm f \leq
1}\abs{f(x)} = \norm x$ by [Corollary 8.7](#cor-b3-banach-hbnormed)(2). If $\dim E = n$: $\dim E' =
n$ (a basis gives coordinate functionals), so $\dim E'' = n$, and the injective (isometric) $J$ is onto. Separability: let $(f_n)$ be dense in $E'$ and choose $\norm{x_n} = 1$ with $\abs{f_n(x_n)} \geq \frac12\norm{f_n}$. Let $F =
\overline{\operatorname{Vect}}(x_n)$; if $F \neq E$, take $g
\in E'$, $g \neq 0$, vanishing on $F$ ([Corollary 8.7](#cor-b3-banach-hbnormed)(3)); choose $f_{n_k} \to g$:

$$
\norm{f_{n_k} - g} \geq \abs{(f_{n_k} - g)(x_{n_k})}
= \abs{f_{n_k}(x_{n_k})} \geq \tfrac12\norm{f_{n_k}}
\geq \tfrac12\bigl(\norm g - \norm{f_{n_k} - g}\bigr),
$$

so $\norm{f_{n_k} - g} \geq \frac13\norm g > 0$: contradiction. Hence $F = E$, and rational (or $\Q + \iu\Q$) combinations of the $x_n$ form a countable [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior).

**Exercise 8.11 ★★.**

(Quotient spaces) Let $E$ be a Banach space and $F \subseteq
E$ a *closed* subspace. On $E/F$ define

$$
\norm{\bar x} \;=\; d(x, F) = \inf_{y\in F}\norm{x - y} .
$$

(a) Show this is a well-defined *norm* on $E/F$ (where does closedness of $F$ enter?), and that the projection $\pi
\colon E \to E/F$ has $\vertiii\pi \leq 1$ and maps the open unit ball onto the open unit ball. (b) Show that $E/F$ is [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete). *(Use the series criterion of [Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1)(b): given classes $\bar x_k$ with $\sum\norm{\bar x_k} < \infty$, lift each to $x_k \in E$ with $\norm{x_k} \leq \norm{\bar x_k} + 2^{-k}$ and sum in $E$.)* (c) Compute: for $E = c$ (convergent sequences) and $F =
c_0$, show $c/c_0 \cong K$ isometrically via $\bar x \mapsto
\lim_nx_n$.

**Solution of Exercise 8.11.**

(a) Well defined: $d(x, F)$ depends only on $\bar x$ (translating $x$ by $F$ does not change the distance). Homogeneity and triangle inequality pass from $\norm\cdot$ through the infimum. Separation needs closedness: $\norm{\bar
x} = 0$ means $d(x, F) = 0$, i.e. $x \in \bar F = F$, i.e. $\bar x = 0$. $\vertiii\pi \leq 1$: $\norm{\bar x} \leq
\norm x$. Open ball onto open ball: if $\norm{\bar x} < 1$, some representative has $\norm{x - y} < 1$; conversely $\pi(B_E(0,1)) \subseteq B_{E/F}(0,1)$ by the norm inequality — so $\pi$ is open, the model case of the open mapping theorem.

(b) Let $\sum_k\norm{\bar x_k} < \infty$ and lift with $\norm{x_k} \leq \norm{\bar x_k} + 2^{-k}$: then $\sum\norm{x_k} < \infty$, so $s = \sum_kx_k$ converges in the Banach $E$ ([Exercise 7.1](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#exo-b3-complete-1)(b)), and [continuity](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity) of $\pi$ gives $\sum_k\bar x_k = \bar s$: every absolutely convergent series of $E/F$ converges, which is equivalent to [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) (same exercise).

(c) The map $\lambda(x) = \lim_nx_n$ is linear $c \to K$, vanishes exactly on $c_0$, so it induces a linear bijection $c/c_0 \to K$. Isometry: $d(x, c_0) = \abs{\lambda(x)}$ — $\leq$: subtract from $x$ the sequence $x - \lambda(x)\mathbf
1 \in c_0$, leaving $\lambda(x)\mathbf 1$ of norm $\abs{\lambda(x)}$; $\geq$: for $y \in c_0$, $\norm{x -
y}_\infty \geq \limsup_n\abs{x_n - y_n} = \abs{\lambda(x)}$.

**Exercise 8.12 ★★.**

(Bounded projections and complemented subspaces) Let $E$ be a Banach space and $P \colon E \to E$ linear with $P^2 = P$ (an [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) projection), $V = \operatorname{im}P$, $W =
\ker P$. (a) Suppose $P$ is bounded. Show that $V$ and $W$ are closed and $E = V \oplus W$ with the decomposition $x = Px + (x -
Px)$. (b) Conversely, suppose $E = V \oplus W$ with *both* $V,
W$ closed, and let $P$ be the projection onto $V$ along $W$. Show that $P$ is bounded. *(Closed graph: if $x_n \to x$ and $Px_n \to z$, then $z \in V$, $x_n - Px_n \to x - z \in
W$, and uniqueness of the decomposition identifies $z =
Px$.)* (c) Deduce the *equivalence*: a subspace $V$ admits a bounded projection iff it is closed and has a closed [algebraic](https://one-course.com/books/math/5/en/chapter/4-field-extensions-and-galois-theory#def-b3-galois-algebraic) complement — and note (without proof) that closed subspaces without this property exist ($c_0$ inside $\ell^\infty$ is the classical example): Hilbert spaces, where $V^\perp$ always works ([Chapter 13](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#ch-b3-hilbert)), are the exception, not the rule.

**Solution of Exercise 8.12.**

(a) $W = \ker P$ is closed (preimage of $0$ under a [continuous map](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-continuity)); $V = \operatorname{im}P = \ker(I - P)$ (indeed $Px = x$ iff $x \in \operatorname{im}P$, using $P^2 =
P$), closed likewise. Every $x$ splits as $Px + (x - Px)$ with $Px \in V$, $x - Px \in W$, and $V \cap W = 0$ ($x = Px
= 0$): $E = V \oplus W$.

(b) The graph argument: let $x_n \to x$ and $Px_n \to z$. Then $z \in V$ ($V$ closed, $Px_n \in V$) and $x_n - Px_n \to
x - z \in W$ ($W$ closed). So $x = z + (x - z)$ with $z \in
V$, $x - z \in W$; by uniqueness of the decomposition, $z =
Px$. The graph of $P$ is closed, $E$ is Banach: $P$ is bounded ([Theorem 8.14](#thm-b3-banach-closedgraph)).

(c) (a) and (b) together are the equivalence. In a Hilbert space every closed $V$ has the closed complement $V^\perp$ ([Chapter 13](https://one-course.com/books/math/5/en/chapter/13-hilbert-spaces#ch-b3-hilbert)): every closed subspace is complemented. In general Banach spaces this fails — $c_0$ has no closed complement in $\ell^\infty$ (Phillips’ theorem, beyond our tools) — so bounded projections are a privilege, and the closed graph theorem is exactly the bookkeeping that converts geometric splittings into bounded operators.

## 8.6 Problem: the duality of the $\ell^p$ spaces

**Problem 8.1.**

Weekend problem — $(\ell^p)' = \ell^q$, reflexivity, and the strangeness of $\ell^\infty$

Fix $1 < p < \infty$ and let $q$ be the conjugate exponent, $\frac1p + \frac1q = 1$. The pairing throughout is $\langle x,
y\rangle = \sum_n x_ny_n$.

**Part I — Hölder and Minkowski for sequences.**

1. (Young’s inequality) For $a, b \geq 0$ show $ab \leq  \frac{a^p}p + \frac{b^q}q$ , using concavity of $\log$ or by studying $t \mapsto \frac{t^p}p + \frac1q - t$ .
2. (Hölder) Deduce: $\abs{\langle x, y\rangle} \leq  \norm x_p\norm y_q$ for $x \in \ell^p$ , $y \in \ell^q$ ; identify the equality case.
3. (Minkowski) Deduce the triangle inequality for $\norm\cdot_p$ . *(Write $\abs{x_n + y_n}^p \leq  \abs{x_n}\,\abs{x_n{+}y_n}^{p-1} +  \abs{y_n}\,\abs{x_n{+}y_n}^{p-1}$ and apply Hölder to each term.)*
4. Prove that $\ell^p$ is [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) and that the finite sequences are dense in it.

**Part II — The duality $(\ell^p)' =
\ell^q$.**

5. For $y \in \ell^q$ , show that $\Lambda_y(x) = \langle  x, y\rangle$ defines $\Lambda_y \in (\ell^p)'$ with $\norm{\Lambda_y} \leq \norm y_q$ , and, testing on $x_n = \abs{y_n}^{q-1}\operatorname{sign}(y_n)$ (suitably truncated and normalized), that $\norm{\Lambda_y} = \norm y_q$ .
6. Conversely, given $\Lambda \in (\ell^p)'$ , set $y_n =  \Lambda(e_n)$ ; show $y \in \ell^q$ with $\norm y_q \leq  \norm\Lambda$ *(test on truncations as in question 5 and let the truncation length grow)* , and conclude $\Lambda = \Lambda_y$ : the map $y \mapsto \Lambda_y$ is an isometric isomorphism $\ell^q \to (\ell^p)'$ .
7. Deduce that $\ell^p$ is *[reflexive](#rem-b3-banach-bidual)* for $1 < p <  \infty$ : composing the two dualities, every element of $(\ell^p)''$ comes from $\ell^p$ ; verify carefully that the composite is the canonical $J$ .

**Part III — $\ell^1$ and $\ell^\infty$ are different animals.**

8. Show that $\ell^p$ ( $1 \leq p < \infty$ ) and $c_0$ are separable, but $\ell^\infty$ is not. *(The uncountably many indicator sequences of subsets of $\N$ are pairwise at distance $1$.)*
9. Deduce from [Exercise 8.10](#exo-b3-banach-10) that $(\ell^1)'  \cong \ell^\infty$ but $(\ell^\infty)' \not\cong  \ell^1$ : $\ell^1$ is *not* [reflexive](#rem-b3-banach-bidual) . *(If $(\ell^\infty)'$ were $\ell^1$, it would be separable, forcing $\ell^\infty$ separable.)*
10. (A Banach limit, explicitly) On $\ell^\infty_\R$ , let $p(x) = \limsup_n \frac{x_1 + \dots + x_n}{n}$ . Show that $p$ is sublinear, and that on the subspace $c$ of convergent sequences, $\mathrm{LIM}(x) = \lim x$ satisfies $\mathrm{LIM} \leq p$ . Extend by Hahn–Banach to $\mathrm{LIM} \colon \ell^\infty_\R \to \R$ and show: $\mathrm{LIM}$ is positive ( $x \geq 0 \Rightarrow  \mathrm{LIM}(x) \geq 0$ ), shift-invariant ( $\mathrm{LIM}(x_2, x_3, \dots) = \mathrm{LIM}(x)$ ), extends the limit, and satisfies $\liminf x \leq  \mathrm{LIM}(x)\leq \limsup x$ .
11. Show that such a $\mathrm{LIM}$ , viewed in $(\ell^\infty)'$ , is *not* of the form $\Lambda_y$ for any $y \in \ell^1$ ; conclude again $(\ell^\infty)'  \neq \ell^1$ . *(Evaluate on the unit sequences $e_n$, then on the constant sequence $1$.)*
12. Evaluate $\mathrm{LIM}$ on $(0,1,0,1,\dots)$ , and show that no shift-invariant *multiplicative* extension of the limit can exist *(consider $x =  (0,1,0,1,\dots)$ and $x\cdot Sx$ where $S$ is the shift)* .

**Part IV — Epilogue: why reflexivity matters.**

13. Using [Corollary 8.10](#cor-b3-banach-bslimits) and question 6, show that every bounded sequence of $\ell^p$ ( $1 < p <  \infty$ ) has a subsequence $(x^{(k)})$ that converges *weakly* : $\Lambda(x^{(k)})$ converges for every $\Lambda \in (\ell^p)'$ . *(Diagonal extraction on the countably many coordinates; identify the weak limit in $\ell^p$ using uniform boundedness of norms and Hölder.)* Show by example ( $e_n$ in $\ell^1$ , against well-chosen elements of $\ell^\infty$ ) that this fails in $\ell^1$ : weak compactness is a privilege of [reflexive spaces](#rem-b3-banach-bidual) .

**Part V — The weak [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) at work, and Schur’s surprise.** Write $x^{(k)} \rightharpoonup x$ in a normed space $E$ (*weak convergence*) when $\Lambda(x^{(k)}) \to \Lambda(x)$ for every $\Lambda \in E'$.

14. [Complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) the census: show $(c_0)' \cong \ell^1$ isometrically, by the scheme of questions 5–6 (what replaces the test sequences?). Assemble the chain $c_0 \to \ell^1 \to \ell^\infty \to \dots$ of successive [duals](#def-b3-banach-operator) and mark where reflexivity fails.
15. Show that every weakly convergent sequence of a Banach space is bounded: view the $x^{(k)}$ through the canonical embedding $J$ as functionals on $E'$ and apply Banach–Steinhaus ( [Theorem 8.9](#thm-b3-banach-banachsteinhaus) ) — on which Banach space, and why is [completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete) available there?
16. Show that in $\ell^p$ , $1 < p < \infty$ : $x^{(k)}  \rightharpoonup x$ iff $\sup_k\norm{x^{(k)}}_p <  \infty$ and $x^{(k)}_n \to x_n$ for every coordinate $n$ *(one direction uses Banach–Steinhaus through the canonical embedding; for the other, approximate $y  \in \ell^q$ by finite sequences)* . Deduce $e_k  \rightharpoonup 0$ in $\ell^2$ while $\norm{e_k}_2 =  1$ : weak limits can lose mass.
17. Show that the norm is weakly lower semicontinuous: $x^{(k)} \rightharpoonup x$ implies $\norm x \leq  \liminf_k\,\norm{x^{(k)}}$ *(pick a norming functional for $x$, [Corollary 8.7](#cor-b3-banach-hbnormed))* .
18. (Radon–Riesz in $\ell^2$ ) Show that in $\ell^2$ , weak convergence together with convergence of norms implies norm convergence *(expand $\norm{x^{(k)} -  x}_2^2$)* . Give a counterexample to the same statement without the norm hypothesis.
19. (Schur, step 1) Let $x^{(k)} \rightharpoonup 0$ in $\ell^1$ and suppose, for contradiction, $\norm{x^{(k)}}_1 \geq \delta > 0$ along a subsequence. Show first that $x^{(k)}_n \to 0$ for each $n$ (which functionals?), then construct recursively indices $k_1 < k_2 < \cdots$ and integers $0 = N_0 < N_1 < N_2 < \cdots$ such that the mass of $x^{(k_j)}$ concentrates on the block $B_j =  \intoc{N_{j-1}}{N_j}$: $$\sum_{n \in B_j}\bigl|x^{(k_j)}_n\bigr| \geq  \norm{x^{(k_j)}}_1 - \frac\delta{10} .$$
20. (Schur, step 2) Define $y \in \ell^\infty$ by $y_n =  \operatorname{sign}\bigl(x^{(k_j)}_n\bigr)$ for $n \in  B_j$ . Show $\langle x^{(k_j)}, y\rangle \geq  \norm{x^{(k_j)}}_1 - \frac{2\delta}{10} \geq  \frac{8\delta}{10}$ and derive a contradiction with $x^{(k)} \rightharpoonup 0$ . Conclude *Schur’s theorem* : in $\ell^1$ , weakly convergent sequences converge in norm.
21. Deduce that $(e_k)$ has no weakly convergent subsequence in $\ell^1$ (its only candidate limit is $0$ , coordinatewise — then apply Schur), recovering question 13’s failure of weak compactness; and resolve the apparent paradox: in $\ell^1$ weak and norm convergence of *sequences* coincide, yet the weak and norm *topologies* differ and bounded sets still fail to be weakly sequentially [compact](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-compact) — no contradiction, only the failure of reflexivity.
22. (Synthesis table) For $E \in \{c_0,\ \ell^1,\ \ell^p\  (1{<}p{<}\infty),\ \ell^\infty\}$ , tabulate: the [dual](#def-b3-banach-operator) ; separability; reflexivity; whether bounded sequences admit weakly convergent subsequences; and one signature property of each space, justified in one line from this problem.

**Part VI — Complements: nearest points, averaged convergence, the value of a Banach limit.**

23. (Nearest points: a dividend of reflexivity) Let $F$ be a closed subspace of $\ell^p$ ( $1 < p < \infty$ ) and $x \in \ell^p$ . Show that $d = \operatorname{dist}(x,  F)$ is *attained* : extract from a minimizing sequence a weakly convergent subsequence (question 13), keep the weak limit inside $F$ by building, via Hahn–Banach, a functional vanishing on $F$ but not at a point outside it, and conclude with question 17. Then show the privilege is not universal: in $c_0$ , for $\Lambda(x) = \sum_n2^{-n}x_n$ , prove $\norm\Lambda = 1$ is not attained on the unit ball, establish the distance formula $\operatorname{dist}(x, \ker\Lambda) =  \abs{\Lambda(x)}$ , and deduce that no $x \notin  \ker\Lambda$ has a nearest point in the closed hyperplane $\ker\Lambda$ .
24. (Banach–Saks in $\ell^2$) Let $x^{(k)}  \rightharpoonup 0$ in $\ell^2_{\R}$ with $\norm{x^{(k)}}_2 \leq C$. Construct a subsequence $(y_j)$ with $\abs{\langle y_i, y_j\rangle} \leq  \frac1j$ for all $i < j$, and deduce $$\Bigl\lVert\frac{y_1 + \dots + y_m}m\Bigr\rVert_2^2  \leq \frac{C^2 + 2}m \longrightarrow 0 :$$ after extraction, the Cesàro means converge in *norm*. Check on $(e_k)$, whose means have norm $\frac1{\sqrt m}$: weak convergence, useless for the sequence itself (question 16), becomes norm convergence for averages.
25. (The value of a Banach limit) Let $L$ be any Banach limit (question 10) and $A_mx = \frac1m(x + Sx + \dots  + S^{m-1}x)$ . Show $L(A_mx) = L(x)$ and $\liminf A_mx  \leq L(x) \leq \limsup A_mx$ ; deduce that *all* Banach limits agree on periodic sequences, with value the mean over a period — $\frac13$ on $(1, 0, 0, 1,  0, 0, \dots)$ , consistent with question 12’s $\frac12$ . Then show agreement fails in general: for the block sequence $x$ equal to $1$ on $\intoc{3^{j-1}}{3^j}$ for even $j$ and $0$ elsewhere, show that the Cesàro means oscillate between $\leq  \frac13$ and $\geq \frac23$ , and build two Banach limits $L_\pm$ with $L_-(x) \leq \frac13 < \frac23  \leq L_+(x)$ *(extend from $c \oplus \R x$ with the extreme admissible values $\pm$: check that $\Lambda(y + tx) = \lim y + t\,p(x)$ is dominated by the sublinear $p$ of question 10)* .

**Solution of Problem 8.1.**

**1.** For $a, b > 0$: by concavity of $\log$, $\log\bigl(\tfrac{a^p}p + \tfrac{b^q}q\bigr) \geq \tfrac1p\log
a^p + \tfrac1q\log b^q = \log(ab)$; exponentiate. (If $ab = 0$ the inequality is trivial.) Equality iff $a^p = b^q$.

**2.** We may assume $\norm x_p = \norm y_q = 1$ (homogeneity; zero cases trivial). Then

$$
\abs{\langle x, y\rangle} \leq \sum_n\abs{x_n}\abs{y_n}
\leq \sum_n\Bigl(\frac{\abs{x_n}^p}p +
\frac{\abs{y_n}^q}q\Bigr) = \frac1p + \frac1q = 1 =
\norm x_p\norm y_q .
$$

Equality requires $\abs{x_n}^p = \abs{y_n}^q$ for all $n$ (Young’s equality case) and alignment of the phases of $x_ny_n$.

**3.** $\abs{x_n + y_n}^p \leq \bigl(\abs{x_n} +
\abs{y_n}\bigr)\abs{x_n + y_n}^{p-1}$; summing and applying Hölder ($p$ against $q$, noting $(p - 1)q = p$):

$$
\norm{x + y}_p^p \leq \bigl(\norm x_p + \norm
y_p\bigr)\,\Bigl(\sum_n\abs{x_n + y_n}^{p}\Bigr)^{1/q}
= \bigl(\norm x_p + \norm y_p\bigr)\,\norm{x+y}_p^{p/q};
$$

if $\norm{x + y}_p \neq 0$ (else trivial), divide by $\norm{x+y}_p^{p/q}$ and use $p - \frac pq = 1$. (Finiteness of $\norm{x+y}_p$ first: $\abs{x_n+y_n}^p \leq
2^p(\abs{x_n}^p + \abs{y_n}^p)$.)

**4.** [Completeness](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete): as for $\ell^1$ ([Exercise 8.3](#exo-b3-banach-3)), coordinatewise limits plus the uniform tail bound $\sum_{n\leq N}\abs{x^{(k)}_n - x^{(l)}_n}^p
\leq \varepsilon^p$, letting $l$ then $N$ tend to infinity. Density of finite sequences: $\norm{x - \sum_{n\leq
N}x_ne_n}_p^p = \sum_{n > N}\abs{x_n}^p \to 0$.

**5.** Hölder gives $\abs{\Lambda_y(x)} \leq \norm
x_p\norm y_q$: $\norm{\Lambda_y} \leq \norm y_q$. Testing: let $x^{(N)}_n = \abs{y_n}^{q-1}\overline{\operatorname{sign}}(y_n)$ for $n \leq N$, $0$ beyond (with $\operatorname{sign}$ the unimodular phase, so that $x_ny_n = \abs{y_n}^q$). Then $\Lambda_y(x^{(N)}) = \sum_{n\leq N}\abs{y_n}^q$ and $\norm{x^{(N)}}_p = \bigl(\sum_{n\leq
N}\abs{y_n}^{q}\bigr)^{1/p}$ (as $(q-1)p = q$), so

$$
\norm{\Lambda_y} \geq \Bigl(\sum_{n\leq
N}\abs{y_n}^q\Bigr)^{1 - 1/p} \xrightarrow[N\to\infty]{}
\norm y_q .
$$

**6.** Set $y_n = \Lambda(e_n)$. With the same test vectors, $\sum_{n \leq N}\abs{y_n}^q = \Lambda(x^{(N)}) \leq
\norm\Lambda\,\bigl(\sum_{n\leq N}\abs{y_n}^q\bigr)^{1/p}$, whence $\bigl(\sum_{n\leq N}\abs{y_n}^q\bigr)^{1/q} \leq
\norm\Lambda$ for every $N$: $y \in \ell^q$, $\norm y_q \leq
\norm\Lambda$. The functionals $\Lambda$ and $\Lambda_y$ agree on the dense finite sequences (question 4): $\Lambda =
\Lambda_y$. With question 5, $y \mapsto \Lambda_y$ is an isometric isomorphism $\ell^q \cong (\ell^p)'$.

**7.** Let $\xi \in (\ell^p)''$. Composing with the isometry $\ell^q \cong (\ell^p)'$ of question 6, $\xi$ defines an element of $(\ell^q)'$, which (question 6 with $p, q$ swapped) is $\Lambda_z$ for a unique $z \in \ell^p$: for every $y \in \ell^q$, $\xi(\Lambda_y) = \sum_nz_ny_n$. On the other hand $J(z)(\Lambda_y) = \Lambda_y(z) = \sum_ny_nz_n$: the same value. Since every element of $(\ell^p)'$ is some $\Lambda_y$, $\xi = J(z)$: $J$ is onto — $\ell^p$ is [reflexive](#rem-b3-banach-bidual).

**8.** Finite sequences with rational (real and imaginary) entries are countable and dense in $\ell^p$ ($p < \infty$) and in $c_0$. In $\ell^\infty$: the family $\{\mathbf 1_A : A
\subseteq \N\}$ is uncountable with $\norm{\mathbf 1_A -
\mathbf 1_B}_\infty = 1$ for $A \neq B$; the balls $B(\mathbf
1_A, \frac12)$ are pairwise disjoint, and a [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) must meet each: no countable [dense set](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-interior) exists.

**9.** If $\ell^1$ were [reflexive](#rem-b3-banach-bidual), then $(\ell^\infty)'
\cong ((\ell^1)')' = J(\ell^1)$ would be separable (isometric image of the separable $\ell^1$); by [Exercise 8.10](#exo-b3-banach-10), the separability of the [dual](#def-b3-banach-operator) $(\ell^\infty)'$ would force $\ell^\infty$ separable — contradicting question 8. So $\ell^1$ is not [reflexive](#rem-b3-banach-bidual) (and $(\ell^\infty)'$ is strictly larger than $\ell^1$, as question 11 makes concrete).

**10.** Homogeneity of $p$ is clear; subadditivity: averages are linear, and $\limsup(u_n + v_n) \leq \limsup u_n +
\limsup v_n$. On $c$: the Cesàro means of a convergent sequence converge to its limit, so $p(x) = \lim x =
\mathrm{LIM}(x)$ there; in particular $\mathrm{LIM} \leq p$ on $c$. Hahn–Banach ([Theorem 8.6](#thm-b3-banach-hahnbanach)) extends $\mathrm{LIM}$ to $\ell^\infty_\R$ with $\mathrm{LIM} \leq p$ globally. Positivity: for $x \geq 0$, $-\mathrm{LIM}(x) =
\mathrm{LIM}(-x) \leq p(-x) = \limsup\text{avg}(-x) \leq 0$. Shift-invariance: the averages of $x - Sx$ telescope to $\frac{x_1 - x_{n+1}}n \to 0$, so $p(\pm(x - Sx)) = 0$ and $\mathrm{LIM}(x - Sx) = 0$. Bounds: $\mathrm{LIM}(x) \leq p(x)
\leq \limsup x$ (averages lag behind sups), and applying this to $-x$ gives the lower bound.

**11.** $e_n \in c_0$, so $\mathrm{LIM}(e_n) = 0$ for all $n$. If $\mathrm{LIM} = \Lambda_y$ with $y \in \ell^1$, then $y_n = \Lambda_y(e_n) = 0$ for all $n$: $\Lambda_y = 0$; but $\mathrm{LIM}(\mathbf 1) = 1$. So $\mathrm{LIM} \in
(\ell^\infty)' \setminus \{\Lambda_y : y \in \ell^1\}$: $(\ell^\infty)' \neq \ell^1$, again.

**12.** For $x = (0,1,0,1,\dots)$: $x + Sx = \mathbf 1$, so $2\,\mathrm{LIM}(x) = \mathrm{LIM}(x) + \mathrm{LIM}(Sx) =
1$: $\mathrm{LIM}(x) = \frac12$. If $\varphi$ were a shift-invariant multiplicative extension of the limit: $x\cdot
Sx = 0$ gives $\varphi(x)\varphi(Sx) = \varphi(x)^2 = 0$, so $\varphi(x) = 0$; but $\varphi(x) + \varphi(Sx) =
\varphi(\mathbf 1) = 1$ gives $2\varphi(x) = 1$: contradiction. Averaging and multiplication cannot coexist.

**13.** Let $\norm{x^{(k)}}_p \leq M$. Coordinates are bounded by $M$: a diagonal extraction gives a subsequence (still written $x^{(k)}$) with $x^{(k)}_n \to x_n$ for every $n$. Then $x \in \ell^p$: $\sum_{n\leq N}\abs{x_n}^p =
\lim_k\sum_{n\leq N}\abs{x^{(k)}_n}^p \leq M^p$ for all $N$. Weak convergence: for $y \in \ell^q$ and $N$ arbitrary,

$$
\abs{\Lambda_y(x^{(k)} - x)} \leq
\Bigl|\sum_{n \leq N}(x^{(k)}_n - x_n)y_n\Bigr|
+ 2M\Bigl(\sum_{n>N}\abs{y_n}^q\Bigr)^{1/q},
$$

where the first term tends to $0$ as $k \to \infty$ (finitely many coordinates) and the second is small for $N$ large (Hölder on the tail): $\Lambda_y(x^{(k)}) \to \Lambda_y(x)$ for every $y$ — weak convergence, since every functional is a $\Lambda_y$ (question 6). In $\ell^1$ this fails: consider $(e_n)$, bounded. Any subsequence $(e_{n_k})$ converges coordinatewise to $0$, so its only weak limit candidate is $0$; but testing against $y \in \ell^\infty$ defined by $y_{n_k} =
(-1)^k$ (and $0$ elsewhere), $\Lambda_y(e_{n_k}) = (-1)^k$ diverges. No weakly convergent subsequence: weak [sequential compactness](https://one-course.com/books/math/5/en/chapter/6-general-topology#thm-b3-topology-metriccompact) of balls characterizes the [reflexive](#rem-b3-banach-bidual) world.

**14.** For $y \in \ell^1$, $\Lambda_y(x) = \sum x_ny_n$ is defined on $c_0$ with $\abs{\Lambda_y(x)} \leq \norm
x_\infty\norm y_1$, and testing on $x^{(N)} =
(\operatorname{sign}y_1, \dots, \operatorname{sign}y_N, 0,
\dots) \in c_0$ gives $\Lambda_y(x^{(N)}) =
\sum_{n\leq N}\abs{y_n} \to \norm y_1$: $\norm{\Lambda_y} = \norm y_1$. Conversely, for $\Lambda \in
(c_0)'$ put $y_n = \Lambda(e_n)$; the same tests give $\sum_{n \leq N}\abs{y_n} = \Lambda(x^{(N)}) \leq
\norm\Lambda$, so $y \in \ell^1$, and $\Lambda = \Lambda_y$ on the dense finite sequences, hence everywhere. The chain of [duals](#def-b3-banach-operator): $(c_0)' = \ell^1$, $(\ell^1)' = \ell^\infty$ ([Exercise 8.10](#exo-b3-banach-10)), $(\ell^\infty)' \supsetneq \ell^1$ (questions 9–11): reflexivity fails at the very first step — $c_0'' = \ell^\infty \neq c_0$ — and never recovers.

**15.** $J x^{(k)} \in E'' = (E')'$ is a family of bounded functionals on the *Banach* space $E'$ ([duals](#def-b3-banach-operator) are [complete](https://one-course.com/books/math/5/en/chapter/7-complete-spaces-baire-ascoli-stoneweierstrass#def-b3-complete-complete), [Proposition 8.2](#prop-b3-banach-llcomplete)); for each $\Lambda \in E'$, the sequence $Jx^{(k)}(\Lambda) =
\Lambda(x^{(k)})$ converges, hence is bounded. Banach–Steinhaus on $E'$ gives $\sup_k\norm{Jx^{(k)}} <
\infty$, and $J$ is isometric ([Remark 8.8](#rem-b3-banach-bidual)): $\sup_k\norm{x^{(k)}} <
\infty$.

**16.** ($\Rightarrow$) Boundedness is question 15; coordinates are the functionals $\Lambda_{e_n}$. ($\Leftarrow$) Let $M = \sup_k\norm{x^{(k)}}_p$, $y \in
\ell^q$, $\varepsilon > 0$; choose $N$ with $\bigl(\sum_{n>N}\abs{y_n}^q\bigr)^{1/q} < \varepsilon$. Then

$$
\abs{\Lambda_y(x^{(k)} - x)} \leq
\sum_{n\leq N}\abs{x^{(k)}_n - x_n}\,\abs{y_n} +
\norm{x^{(k)} - x}_p\,\varepsilon
\leq \sum_{n\leq N}\abs{x^{(k)}_n - x_n}\abs{y_n} +
(M + \norm x_p)\,\varepsilon,
$$

and the finite sum tends to $0$: $\limsup \leq (M + \norm
x_p)\varepsilon$ for every $\varepsilon$. (That $x \in
\ell^p$ with $\norm x_p \leq M$ follows from Fatou-style finite-section bounds: $\sum_{n \leq N}\abs{x_n}^p =
\lim_k\sum_{n\leq N}\abs{x^{(k)}_n}^p \leq M^p$.) For $e_k$ in $\ell^2$: bounded, coordinatewise $\to 0$, so $e_k
\rightharpoonup 0$, yet $\norm{e_k} = 1$: the unit of mass escapes to infinite index, invisible to every fixed functional.

**17.** Take $\Lambda$ with $\norm\Lambda = 1$ and $\Lambda(x) = \norm x$ ([Corollary 8.7](#cor-b3-banach-hbnormed)). Then $\norm x = \lim\Lambda(x^{(k)}) \leq
\liminf\norm\Lambda\,\norm{x^{(k)}} =
\liminf\norm{x^{(k)}}$. (With $e_k \rightharpoonup 0$: $0
\leq \liminf 1$, and the inequality can be strict.)

**18.** In $\ell^2$, $\norm{x^{(k)} - x}_2^2 =
\norm{x^{(k)}}_2^2 - 2\operatorname{Re}\langle x^{(k)},
x\rangle + \norm x_2^2$ (real case: $-2\langle x^{(k)},
x\rangle$). Weak convergence applied to the functional $\Lambda_x$ gives $\langle x^{(k)}, x\rangle \to \norm
x_2^2$, and the norms converge by hypothesis: the right side tends to $\norm x^2 - 2\norm x^2 + \norm x^2 = 0$. Counterexample without norm convergence: $e_k
\rightharpoonup 0$, $\norm{e_k - 0} = 1 \not\to 0$.

**19.** Coordinate convergence: apply the functionals $\Lambda_{e_n} \in (\ell^1)' = \ell^\infty$ ($y = e_n$). Construction: having chosen $k_{j-1}, N_{j-1}$, pick $k_j >
k_{j-1}$ so large that $\sum_{n \leq
N_{j-1}}\abs{x^{(k_j)}_n} < \frac\delta{20}$ (finitely many coordinates, each $\to 0$), then $N_j > N_{j-1}$ so large that the tail satisfies $\sum_{n > N_j}\abs{x^{(k_j)}_n} <
\frac\delta{20}$ (convergence of the series defining $\norm{x^{(k_j)}}_1$). The block $B_j = \intoc{N_{j-1}}{N_j}$ then carries all but $\frac\delta{10}$ of the mass of $x^{(k_j)}$.

**20.** With $y$ as defined ($\abs{y_n} \leq 1$ everywhere):

$$
\langle x^{(k_j)}, y\rangle
= \sum_{n \in B_j}\abs{x^{(k_j)}_n}
+ \sum_{n \notin B_j}x^{(k_j)}_ny_n
\geq \Bigl(\norm{x^{(k_j)}}_1 - \frac\delta{10}\Bigr) -
\frac\delta{10}
\geq \delta - \frac{2\delta}{10} = \frac{4\delta}5 > 0,
$$

the middle inequality because the mass off the block is at most $\frac\delta{10}$ (question 19). But $y \in
\ell^\infty = (\ell^1)'$ and $x^{(k)} \rightharpoonup 0$ force $\langle x^{(k_j)}, y\rangle \to 0$: contradiction. Hence weakly null sequences of $\ell^1$ are norm-null, and by translation weakly convergent ones converge in norm: Schur’s theorem.

**21.** A weakly convergent subsequence of $(e_k)$ would have limit $0$ (coordinates), hence by Schur $\norm{e_{k_j}}_1
\to 0$ — but the norms are $1$. So no weakly convergent subsequence exists, as found by hand in question 13. No paradox: Schur says sequences cannot distinguish the weak from the norm [topology](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) in $\ell^1$ (the topologies themselves do differ — weak [neighborhoods](https://one-course.com/books/math/5/en/chapter/6-general-topology#def-b3-topology-topology) are never norm-bounded), and weak [sequential compactness](https://one-course.com/books/math/5/en/chapter/6-general-topology#thm-b3-topology-metriccompact) of the unit ball is a different, stronger property, equivalent to reflexivity (Eberlein–Šmulian, beyond our tools; the failure, at least, we have proved).

**22.** *The census.*

| $E$ | $E'$ | sep. | refl. | weak seq. cpt. balls |
| --- | --- | --- | --- | --- |
| $c_0$ | $\ell^1$ | yes | no | no ($e_1{+}\dots{+}e_k$) |
| $\ell^1$ | $\ell^\infty$ | yes | no | no ($e_k$, q. 21) |
| $\ell^p$ | $\ell^q$ | yes | yes | yes (q. 13) |
| $\ell^\infty$ | $\supsetneq\ell^1$ | no | no | no |

Signatures: $c_0$ — its bidual is $\ell^\infty$: the first [non-reflexive](#rem-b3-banach-bidual) step (question 14); $\ell^1$ — Schur’s property (question 20); $\ell^p$ — reflexivity and weak compactness (questions 7, 13); $\ell^\infty$ — non-separability and Banach limits: functionals no sequence can represent (questions 8, 10–11). One family of spaces, four different worlds.

**23.** Let $(f_k) \subseteq F$ with $\norm{x - f_k}
\to d$. Then $\norm{f_k} \leq \norm x + \sup_k\norm{x -
f_k}$: bounded, so by question 13 a subsequence $f_{k_j}
\rightharpoonup f$. If $f \notin F$, then $\delta =
\operatorname{dist}(f, F) > 0$ ($F$ closed); on $F \oplus \R
f$ the linear form $\lambda(g + tf) = t$ satisfies $\abs{\lambda(u)} \leq \norm u/\delta$ (because $\norm{g +
tf} \geq \abs t\,\delta$), and Hahn–Banach extends it to $\Lambda \in (\ell^p)'$ with $\Lambda\restriction_F = 0$, $\Lambda(f) = 1$; but then $0 = \Lambda(f_{k_j}) \to
\Lambda(f) = 1$: contradiction. So $f \in F$, and $x -
f_{k_j} \rightharpoonup x - f$ gives, by question 17,

$$
d \leq \norm{x - f} \leq \liminf_j\,\norm{x - f_{k_j}} = d :
$$

the distance is attained at $f$. In $c_0$: $\abs{\Lambda(x)}
\leq \sum_n2^{-n}\abs{x_n} < \norm x_\infty$ for every $x
\neq 0$ (a nonzero null sequence cannot satisfy $\abs{x_n} =
\norm x_\infty$ for all $n$), while the truncated ones $(1,
\dots, 1, 0, \dots)$ give $\Lambda = 1 - 2^{-N} \to 1$: so $\norm\Lambda = 1$, never attained. Distance formula: for $f
\in \ker\Lambda$, $\abs{\Lambda(x)} = \abs{\Lambda(x - f)}
\leq \norm{x - f}$, so $\operatorname{dist}(x, \ker\Lambda)
\geq \abs{\Lambda(x)}$; conversely, for $u$ in the unit ball with $\Lambda(u) \geq 1 - \varepsilon$, the vector $f = x -
\frac{\Lambda(x)}{\Lambda(u)}u$ lies in $\ker\Lambda$ with $\norm{x - f} \leq \frac{\abs{\Lambda(x)}}{1 - \varepsilon}$: equality. If some $f \in \ker\Lambda$ attained it, $z = x -
f$ would satisfy $\abs{\Lambda(z)} = \abs{\Lambda(x)} =
\norm z \neq 0$, so $\Lambda$ would attain its norm at $z/\norm z$: impossible. A closed hyperplane of $c_0$ with no nearest points anywhere — reflexivity was not decorative.

**24.** Set $y_1 = x^{(1)}$. Given $y_1, \dots, y_j$, each map $k \mapsto \langle y_i, x^{(k)}\rangle$ tends to $0$ ($y_i \in \ell^2 = (\ell^2)'$), so there is $k_{j+1}$ beyond the previous index with $\abs{\langle y_i,
x^{(k_{j+1})}\rangle} \leq \frac1{j+1}$ for $i = 1, \dots,
j$; call the choice $y_{j+1}$. Then

$$
\Bigl\lVert\sum_{j=1}^my_j\Bigr\rVert_2^2
= \sum_{j=1}^m\norm{y_j}_2^2
+ 2\sum_{j=2}^m\sum_{i<j}\langle y_i, y_j\rangle
\leq mC^2 + 2\sum_{j=2}^m\frac{j-1}j
\leq mC^2 + 2m,
$$

and dividing by $m^2$: $\norm{\frac1m\sum_jy_j}_2^2 \leq
\frac{C^2 + 2}m \to 0$. (For a weak limit $x \neq 0$, apply this to $x^{(k)} - x$.) On the orthonormal $(e_k)$ no extraction is even needed: $\norm{\frac1m(e_1 + \dots +
e_m)}_2 = \frac{\sqrt m}m = \frac1{\sqrt m}$. Averages convert weak convergence into norm convergence: the Banach–Saks property of $\ell^2$.

**25.** $A_mx = \frac1m\sum_{i=0}^{m-1}S^ix$, so linearity and shift-invariance give $L(A_mx) = L(x)$. For any bounded $u$ and $\varepsilon > 0$, pick $N$ with $u_n \leq
\limsup u + \varepsilon$ for $n \geq N$; positivity applied to $(\limsup u + \varepsilon)\mathbf 1 - S^Nu \geq 0$ and $L(S^Nu) = L(u)$ give $L(u) \leq \limsup u + \varepsilon$, and symmetrically $L(u) \geq \liminf u - \varepsilon$: hence $\liminf A_mx \leq L(x) \leq \limsup A_mx$ for every $m$. If $x$ is $T$-periodic, $A_Tx$ is the constant sequence equal to the period mean $\mu$: $L(x) = \mu$ for *every* Banach limit — $\frac13$ on $(1,0,0,\dots)$, $\frac12$ on $(0,1,0,1,\dots)$ as in question 12. For the block sequence: at $N = 3^j$ with $j$ even the last block is all ones, so the Cesàro mean is $\geq \frac{3^j - 3^{j-1}}{3^j} = \frac23$; at $N = 3^j$ with $j$ odd all the ones sit in $\intoc0
{3^{j-1}}$, so the mean is $\leq \frac13$. Hence $p(x) \geq
\frac23$ and $-p(-x) = \liminf_n\frac{x_1 + \dots + x_n}n
\leq \frac13$. On $M = c \oplus \R x$ define $\Lambda_+(y +
tx) = \lim y + t\,p(x)$. Domination by $p$: for $t > 0$, sublinearity gives $p(tx) \leq p(y + tx) + p(-y)$, i.e. $p(y + tx) \geq t\,p(x) + \lim y$ (note $p(\pm y) = \pm\lim
y$ for $y \in c$: Cesàro means of a convergent sequence converge to its limit); for $t = -s < 0$, $p(y) \leq p(y -
sx) + p(sx)$ gives $p(y - sx) \geq \lim y - s\,p(x)$; for $t
= 0$ there is equality. So $\Lambda_+ \leq p$ on $M$, and Hahn–Banach extends it to $L_+ \leq p$ on $\ell^\infty_\R$, which is a Banach limit exactly as in question 10 (domination by $p$ yields positivity, shift-invariance, and the value $\lim$ on $c$), with $L_+(x) = p(x) \geq \frac23$. The same computation with $\Lambda_-(y + tx) = \lim y - t\,p(-x)$ (using $p(-sx) \leq p(y - sx) + p(-y)$ for the case $t = -s
< 0$) yields a Banach limit $L_-$ with $L_-(x) = -p(-x) \leq
\frac13$. Two Banach limits, one sequence, two values: outside the periodic (and, more generally, *almost convergent*) world, a Banach limit is a genuine choice.