University Mathematics — Year 3 · Bachelor Year 3

8Banach 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, via Baire’s theorem, forces strong uniformity: pointwise bounded families of operators are norm-bounded (Banach–Steinhaus), continuous bijections have continuous inverses (open mapping), and graphs detect continuity (closed graph). The other pillar, Hahn–Banach, needs no completeness at all — only Zorn’s lemma — and guarantees that dual spaces 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 computations, and on a genuinely surprising application: there exist continuous $2\pi$ -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 normed space.

8.1 Bounded operators; sequence spaces

Definition 8.1

$\mathcal L(E, F)$ denotes the space of bounded (= continuous, 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, 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 (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).

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. — 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 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). Consequently the set of invertible operators is open, and inversion is continuous on it.

Proof. The series converges absolutely ( $\vertiii{T^n} \leq \vertiii T^n$ , geometric) in the Banach $\mathcal L(E)$ (Proposition 8.2; Exercise 7.1(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 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 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(a) exploited exactly this), which is why initial value problems never suffer the resonance phenomena of boundary value problems (Chapter 15).

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 $\C$ ).

Every $f \in F'$ ( $F$ a subspace) extends to $\tilde f \in E'$ with $\norm{\tilde f}_{E'} = \norm f_{F'}$ .
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 $\norm x = \sup_{\norm f \leq 1}\abs{f(x)}$ .
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 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): 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 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) 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 $2\pi$ -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 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, $\norm f_\infty \leq 1$ , 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 $f$ 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 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 by the above applied in any ball, is 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) gives interior to $\overline{T(B)}$ : some $B_F(y_0, 4c) \subseteq \overline{T(B)}$ . Re-center 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 (here completeness 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: $B_F(0, c) \subseteq T(B)$ . Openness of $T$ on arbitrary opens follows by translation and scaling; for the corollary, openness of $T$ means $T^{-1}$ is continuous. ∎

Corollary 8.13 (Equivalent norms)

If a vector space is 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): $\norm{Tx} \leq \norm{(x, Tx)} \leq C\norm x$ . ∎

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 needed. Banach–Steinhaus: to convert pointwise information into uniform bounds — typically to show a limit operation is continuous, or (contrapositive) to prove divergence for some element, as for Fourier series. Open mapping / closed graph: to get continuity for free from algebraic bijectivity or from a closure property of the graph — typical use: comparing two complete norms, or proving automatic continuity. All three Baire theorems require completeness of the source; counterexamples otherwise (Exercise 8.7).

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. 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 functionals agreeing on a dense set 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: (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

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, $\abs f \leq 1$ ) $f \equiv 1$ 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 with $T$ invertible, a sufficiently small perturbation of $T$ keeps it uniquely solvable, with a quantitative bound on the change of solution.

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, $S$ 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, 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 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

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 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, 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

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(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 (finite sequences are dense in $\ell^1$ ). Where does the argument break for $(\ell^\infty)'$ ?

Solution

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).

Exercise 8.6 ★★

Let $E, F, G$ be normed with $E$ Banach, and $B \colon E \times F \to G$ bilinear, continuous in each variable separately. Show that $B$ is (jointly) continuous: $\norm{B(x,y)} \leq C\norm x\norm y$ . (Apply Banach–Steinhaus to the family $(B(\cdot, y))_{\norm y \leq 1}$ .)

Solution

Solution of Exercise 8.6.

For each fixed $x$ , $y \mapsto B(x, y)$ is continuous 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, by continuity in $x$ ) is pointwise bounded on the Banach space $E$ : Banach–Steinhaus (Theorem 8.9) 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 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

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: $(\mathcal C(\intcc01), \norm\cdot_1)$ is not complete (Exercise 7.1); the corollary requires completeness 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 — and $\varphi$ has no continuous extension to $\ell^2$ , illustrating that density without uniform continuity is powerless (Theorem 7.2).

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

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 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). 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 $f$ 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

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) 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); 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(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(3).)

Solution

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(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(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.

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. (Use the series criterion of Exercise 7.1(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

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(b)), and continuity of $\pi$ gives $\sum_k\bar x_k = \bar s$ : every absolutely convergent series of $E/F$ converges, which is equivalent to completeness (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 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 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), are the exception, not the rule.

Solution

Solution of Exercise 8.12.

(a) $W = \ker P$ is closed (preimage of $0$ under a continuous map); $V = \operatorname{im}P = \ker(I - P)$ (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).

(c) (a) and (b) together are the equivalence. In a Hilbert space every closed $V$ has the closed complement $V^\perp$ (Chapter 13): 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.

(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 $t \mapsto \frac{t^p}p + \frac1q - t$ .
(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.
(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.)
Prove that $\ell^p$ is complete and that the finite sequences are dense in it.

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

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 $\norm{\Lambda_y} = \norm y_q$ .
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 $\ell^q \to (\ell^p)'$ .
Deduce that $\ell^p$ is reflexive 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.

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 $1$ .)
Deduce from Exercise 8.10 that $(\ell^1)' \cong \ell^\infty$ but $(\ell^\infty)' \not\cong \ell^1$ : $\ell^1$ is not reflexive. (If $(\ell^\infty)'$ were $\ell^1$ , it would be separable, forcing $\ell^\infty$ separable.)
(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 $\liminf x \leq \mathrm{LIM}(x)\leq \limsup x$ .
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 $1$ .)
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.

Using Corollary 8.10 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.

Part V — The weak 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 $\Lambda \in E'$ .

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 and mark where reflexivity fails.
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) — on which Banach space, and why is completeness available there?
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.
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).
(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.
(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 $B_j = \intoc{N_{j-1}}{N_j}$ :

$\sum_{n \in B_j}\bigl|x^{(k_j)}_n\bigr| \geq \norm{x^{(k_j)}}_1 - \frac\delta{10} .$
(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.
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 — no contradiction, only the failure of reflexivity.
(Synthesis table) For $E \in \{c_0,\ \ell^1,\ \ell^p\ (1{<}p{<}\infty),\ \ell^\infty\}$ , tabulate: the dual; 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.

(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 $\ker\Lambda$ .
(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.
(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

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: as for $\ell^1$ (Exercise 8.3), 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.

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 must meet each: no countable dense set exists.

9. If $\ell^1$ were reflexive, then $(\ell^\infty)' \cong ((\ell^1)')' = J(\ell^1)$ would be separable (isometric image of the separable $\ell^1$ ); by Exercise 8.10, the separability of the dual $(\ell^\infty)'$ would force $\ell^\infty$ separable — contradicting question 8. So $\ell^1$ is not reflexive (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) 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 of balls characterizes the reflexive 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: $(c_0)' = \ell^1$ , $(\ell^1)' = \ell^\infty$ (Exercise 8.10), $(\ell^\infty)' \supsetneq \ell^1$ (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 are complete, Proposition 8.2); 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): $\sup_k\norm{x^{(k)}} < \infty$ .

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). 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 in $\ell^1$ (the topologies themselves do differ — weak neighborhoods are never norm-bounded), and weak sequential compactness 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 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.

Terms defined in this chapter

See all 395 terms in the glossary

8.1 Bounded operators; sequence spaces

8.2 Hahn–Banach

8.3 The Baire trilogy

8.4 Dual spaces, concretely

8.5 Exercises

8.6 Problem: the duality of the ℓp\ell^pℓp spaces

Terms defined in this chapter

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