Normed and Banach Spaces
Functional analysis studies infinite-dimensional vector spaces carrying a topology compatible with their linear structure. The first and most fundamental such structure is a norm, a notion of length that turns a vector space into a metric space. This chapter builds that setting from its axioms: the geometry a norm imposes through its unit ball, the metric and topology it induces, the decisive distinction between convergent and Cauchy sequences, and the completeness property that singles out Banach spaces as the arena in which analysis can actually be done. We close with the model spaces the whole course returns to: the sequence spaces \(\ell^p\) and \(\ell^\infty\), and the space \(C[a,b]\) of continuous functions under the supremum norm.L’analyse fonctionnelle étudie les espaces vectoriels de dimension infinie munis d’une topologie compatible avec leur structure linéaire. La structure la plus fondamentale est la norme, notion de longueur qui transforme un espace vectoriel en espace métrique. Ce chapitre construit ce cadre à partir de ses axiomes : la géométrie imposée par la boule unité, la métrique et la topologie induites, la distinction décisive entre suites convergentes et suites de Cauchy, ainsi que la propriété de complétude qui distingue les espaces de Banach comme cadre naturel de l’analyse. Nous terminons par les espaces modèles auxquels le cours revient constamment : les espaces de suites \(\ell^p\) et \(\ell^\infty\), ainsi que l’espace \(C[a,b]\) des fonctions continues muni de la norme sup.
Visual Investigations
Objectives & Prerequisites
- Specific objective 1. State the three norm axioms and derive their immediate consequences, including the reverse triangle inequality and the fact that a norm induces a translation-invariant metric.Objectif spécifique 1. Énoncer les trois axiomes d’une norme et en déduire leurs conséquences immédiates, notamment l’inégalité triangulaire renversée et le fait qu’une norme induit une métrique invariante par translation.
- Specific objective 2. Describe the metric topology of a normed space through open and closed balls, and prove that the vector-space operations and the norm are continuous.Objectif spécifique 2. Décrire la topologie métrique d’un espace normé au moyen des boules ouvertes et fermées, puis démontrer la continuité des opérations vectorielles et de la norme.
- Specific objective 3. Distinguish convergent sequences from Cauchy sequences, prove that every convergent sequence is Cauchy, and identify completeness as the converse property that defines a Banach space.Objectif spécifique 3. Distinguer les suites convergentes des suites de Cauchy, prouver que toute suite convergente est de Cauchy, puis reconnaître la complétude comme la propriété réciproque qui définit un espace de Banach.
- Specific objective 4. Prove that on a finite-dimensional space all norms are equivalent, and explain why this equivalence fails in infinite dimensions.Objectif spécifique 4. Démontrer que, sur un espace de dimension finie, toutes les normes sont équivalentes, et expliquer pourquoi cette équivalence échoue en dimension infinie.
- Specific objective 5. Characterise completeness through absolutely convergent series, and use this criterion as the standard tool for proving a space is Banach.Objectif spécifique 5. Caractériser la complétude à l’aide des séries absolument convergentes et utiliser ce critère comme outil standard pour démontrer qu’un espace est de Banach.
- Specific objective 6. Verify completeness for the model spaces \(\ell^p\), \(\ell^\infty\), and \(C[a,b]\), and recognise these as the recurring examples of the course.Objectif spécifique 6. Vérifier la complétude des espaces modèles \(\ell^p\), \(\ell^\infty\) et \(C[a,b]\), puis les reconnaître comme exemples récurrents du cours.
- Linear algebra: vector spaces over \(\mathbb R\) or \(\mathbb C\), subspaces, linear independence, dimension, linear maps.
- Real analysis: convergence of sequences and series, suprema and infima, the completeness of \(\mathbb R\).
- Metric spaces: open and closed sets, continuity, Cauchy sequences, completeness.
- Basic inequalities: the triangle inequality in \(\mathbb R^n\), and the Cauchy-Schwarz inequality.
- Estimating with the triangle inequality and absolute homogeneity.
- Extracting limits from Cauchy sequences and passing to the limit in inequalities.
- Diagonal and componentwise arguments for sequences in \(\ell^p\).
- Compactness of the unit sphere in finite dimensions (Heine-Borel).
Diagnostic questions
Where this chapter sits
Later dependence. The norm and its induced topology are the setting for bounded linear operators (Ch. 2). Hahn-Banach follows in Ch. 3; Baire-category methods and the major completeness theorems enter in Ch. 4 and Ch. 5; dual spaces and weak topologies are developed in Ch. 6; and the inner-product refinement of a norm produces Hilbert spaces in Ch. 7. The model spaces \(\ell^p\) and \(C[a,b]\) recur throughout these developments. Completeness is the single hypothesis that makes the deep theorems of the subject possible.
Core Definitions
Let \(X\) be a vector space over the field \(\mathbb K\in\{\mathbb R,\mathbb C\}\). A norm on \(X\) is a map \(\|\cdot\|:X\to[0,\infty)\) satisfying, for all \(x,y\in X\) and all \(\lambda\in\mathbb K\):
- (N1) positive definiteness: \(\|x\|\ge 0\), and \(\|x\|=0\) if and only if \(x=0\);
- (N2) absolute homogeneity: \(\|\lambda x\|=|\lambda|\,\|x\|\);
- (N3) triangle inequality: \(\|x+y\|\le\|x\|+\|y\|\).
A map satisfying (N2) and (N3) together with \(\|x\|\ge 0\), but not necessarily the implication \(\|x\|=0\Rightarrow x=0\), is a seminorm. From (N2) with \(\lambda=-1\) one gets \(\|-x\|=\|x\|\), and (N3) yields the reverse triangle inequality \(\bigl|\,\|x\|-\|y\|\,\bigr|\le\|x-y\|\).
For \(x=(x_1,x_2)\), the map \(\|x\|_1=|x_1|+|x_2|\) is a norm. By contrast, \(p(x)=|x_1|\) is only a seminorm because \(p(0,1)=0\) although \((0,1)\ne0\). This isolates the role of positive definiteness.
Pour \(x=(x_1,x_2)\), l’application \(\|x\|_1=|x_1|+|x_2|\) est une norme. En revanche, \(p(x)=|x_1|\) n’est qu’une semi-norme car \(p(0,1)=0\) alors que \((0,1)\ne0\). Cet exemple isole le rôle de la séparation des points.
A normed vector space is a pair \((X,\|\cdot\|)\) where \(X\) is a \(\mathbb K\)-vector space and \(\|\cdot\|\) is a norm on \(X\). Its induced metric is
\[d(x,y)=\|x-y\|,\qquad x,y\in X.\]
That \(d\) is a genuine metric - non-negativity with \(d(x,y)=0\iff x=y\), symmetry, and the triangle inequality \(d(x,z)\le d(x,y)+d(y,z)\) - is established in Theorem 1.1. The metric is translation invariant, \(d(x+z,y+z)=d(x,y)\), and homogeneous, \(d(\lambda x,\lambda y)=|\lambda|\,d(x,y)\).
In \(\mathbb R^2\) with the Euclidean norm, the points \(x=(1,2)\) and \(y=(4,6)\) satisfy \(d(x,y)=\|x-y\|_2=\sqrt{3^2+4^2}=5\). Translating both points by the same vector leaves this distance unchanged.
Dans \(\mathbb R^2\) muni de la norme euclidienne, les points \(x=(1,2)\) et \(y=(4,6)\) vérifient \(d(x,y)=\|x-y\|_2=\sqrt{3^2+4^2}=5\). La translation simultanée des deux points ne change pas cette distance.
For \(x_0\in X\) and \(r>0\), the open ball and closed ball of centre \(x_0\) and radius \(r\) are
\[B(x_0,r)=\{x\in X:\|x-x_0\|<r\},\qquad \overline B(x_0,r)=\{x\in X:\|x-x_0\|\le r\}.\]
A set \(U\subseteq X\) is open if for every \(x\in U\) there is \(r>0\) with \(B(x,r)\subseteq U\). The family of all such open sets is the norm topology (equivalently, the metric topology of \(d\)) on \(X\). A set is closed when its complement is open; equivalently, \(F\) is closed iff every convergent sequence of points of \(F\) has its limit in \(F\).
In \(\mathbb R^2\) with \(\|x\|_\infty=\max(|x_1|,|x_2|)\), the ball \(B(0,1)\) is the open square \((-1,1)^2\). Thus the geometry of norm balls depends on the norm even though equivalent norms may generate the same topology.
Dans \(\mathbb R^2\) avec \(\|x\|_\infty=\max(|x_1|,|x_2|)\), la boule \(B(0,1)\) est le carré ouvert \((-1,1)^2\). La géométrie des boules dépend donc de la norme, même lorsque des normes équivalentes engendrent la même topologie.
Let \((x_n)_{n\ge 1}\) be a sequence in \((X,\|\cdot\|)\). It converges to \(x\in X\), written \(x_n\to x\), if
\[\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall n\ge N:\ \|x_n-x\|<\varepsilon,\]
that is, \(\|x_n-x\|\to 0\) in \(\mathbb R\). The sequence is a Cauchy sequence if
\[\forall\varepsilon>0\ \exists N\in\mathbb N\ \forall m,n\ge N:\ \|x_n-x_m\|<\varepsilon.\]
Convergence and the Cauchy property refer only to the norm, hence are preserved under passage to an equivalent norm (Definition 1.6).
In \(\mathbb R\), let \(x_n=1/n\). Then \(x_n\to0\). Given \(\varepsilon>0\), choose \(N>2/\varepsilon\); for \(m,n\ge N\), \(|x_n-x_m|\le 1/n+1/m<\varepsilon\), so the sequence is Cauchy.
Dans \(\mathbb R\), posons \(x_n=1/n\). Alors \(x_n\to0\). Pour \(\varepsilon>0\), choisissons \(N>2/\varepsilon\) ; si \(m,n\ge N\), alors \(|x_n-x_m|\le1/n+1/m<\varepsilon\), donc la suite est de Cauchy.
A normed vector space \((X,\|\cdot\|)\) is complete if every Cauchy sequence in \(X\) converges to a limit in \(X\). A complete normed vector space is called a Banach space.
By Theorem 1.2 every convergent sequence is Cauchy; completeness is exactly the converse implication, and it is the property that makes limiting constructions - series, fixed points, completions - available inside \(X\) itself.
The normed space \((\mathbb R,|\cdot|)\) is Banach because every Cauchy sequence of real numbers converges to a real number. This is the scalar completeness that later proofs use coordinatewise in spaces such as \(\ell^p\).
L’espace normé \((\mathbb R,|\cdot|)\) est complet car toute suite de Cauchy réelle converge vers un réel. C’est cette complétude scalaire qui sera utilisée coordonnée par coordonnée dans des espaces comme \(\ell^p\).
Two norms \(\|\cdot\|_a\) and \(\|\cdot\|_b\) on the same vector space \(X\) are equivalent, written \(\|\cdot\|_a\sim\|\cdot\|_b\), if there exist constants \(c,C>0\) such that
\[c\,\|x\|_a\le\|x\|_b\le C\,\|x\|_a\qquad\text{for all }x\in X.\]
This is an equivalence relation on the set of norms of \(X\). Equivalent norms have exactly the same open sets, hence the same convergent sequences, the same Cauchy sequences, and the same continuous maps into or out of \(X\); in particular completeness is invariant under equivalence.
For every \(x\in\mathbb R^n\), \(\|x\|_\infty\le\|x\|_1\le n\|x\|_\infty\). Hence \(\|\cdot\|_1\sim\|\cdot\|_\infty\). A sequence converges for one of these norms exactly when it converges for the other.
Pour tout \(x\in\mathbb R^n\), \(\|x\|_\infty\le\|x\|_1\le n\|x\|_\infty\). Ainsi \(\|\cdot\|_1\sim\|\cdot\|_\infty\). Une suite converge pour l’une de ces normes exactement lorsqu’elle converge pour l’autre.
Fix the scalar field \(\mathbb K\). For \(1\le p<\infty\), the space \(\ell^p\) consists of all scalar sequences \(x=(x_k)_{k\ge 1}\) with \(\sum_{k=1}^\infty|x_k|^p<\infty\), normed by
\[\|x\|_p=\Bigl(\sum_{k=1}^\infty|x_k|^p\Bigr)^{1/p}.\]
The space \(\ell^\infty\) consists of all bounded scalar sequences, normed by \(\|x\|_\infty=\sup_{k\ge 1}|x_k|\). Finally \(C[a,b]\) is the vector space of continuous functions \(f:[a,b]\to\mathbb K\) with the supremum (uniform) norm
\[\|f\|_\infty=\sup_{t\in[a,b]}|f(t)|=\max_{t\in[a,b]}|f(t)|,\]
the maximum being attained because \(|f|\) is continuous on the compact interval \([a,b]\). That \(\|\cdot\|_p\) satisfies the triangle inequality is Minkowski's inequality (proved in Theorem 1.5); the remaining norm axioms are immediate.
The sequence \(x=(1/k)_{k\ge1}\) belongs to \(\ell^2\) because \(\sum_{k\ge1}k^{-2}<\infty\), but it does not belong to \(\ell^1\). Also \(f(t)=t^2\) belongs to \(C[0,1]\) and has \(\|f\|_\infty=1\). These examples distinguish the sequence-space and function-space models.
La suite \(x=(1/k)_{k\ge1}\) appartient à \(\ell^2\) car \(\sum_{k\ge1}k^{-2}<\infty\), mais elle n’appartient pas à \(\ell^1\). De plus, \(f(t)=t^2\) appartient à \(C[0,1]\) et vérifie \(\|f\|_\infty=1\). Ces exemples distinguent les modèles de suites et de fonctions.
Theorems & Proofs
Let \((X,\|\cdot\|)\) be a normed space. Then \(d(x,y)=\|x-y\|\) is a metric on \(X\), so \(X\) carries the associated topology. Moreover addition \(+:X\times X\to X\) and scalar multiplication \(\cdot:\mathbb K\times X\to X\) are continuous, where \(X\times X\) and \(\mathbb K\times X\) carry the product topologies.
Metric axioms. For all \(x,y,z\): by (N1), \(d(x,y)=\|x-y\|\ge 0\) and \(d(x,y)=0\iff x-y=0\iff x=y\). By (N2) with \(\lambda=-1\), \(d(x,y)=\|x-y\|=\|-(y-x)\|=|-1|\,\|y-x\|=d(y,x)\). By (N3), \(d(x,z)=\|(x-y)+(y-z)\|\le\|x-y\|+\|y-z\|=d(x,y)+d(y,z)\). Thus \(d\) is a metric, and the open sets of Definition 1.3 form a topology.
Continuity of addition. Fix \((x_0,y_0)\in X\times X\) and let \(\varepsilon>0\). For any \((x,y)\),
\[\|(x+y)-(x_0+y_0)\|=\|(x-x_0)+(y-y_0)\|\le\|x-x_0\|+\|y-y_0\|.\]
Hence if \(\|x-x_0\|<\varepsilon/2\) and \(\|y-y_0\|<\varepsilon/2\), then \(\|(x+y)-(x_0+y_0)\|<\varepsilon\). Since a basic neighbourhood of \((x_0,y_0)\) is a product of such balls, addition is continuous at \((x_0,y_0)\).
Continuity of scalar multiplication. Fix \((\lambda_0,x_0)\in\mathbb K\times X\) and let \(\varepsilon>0\). For any \((\lambda,x)\),
\[\|\lambda x-\lambda_0 x_0\|=\|\lambda(x-x_0)+(\lambda-\lambda_0)x_0\|\le|\lambda|\,\|x-x_0\|+|\lambda-\lambda_0|\,\|x_0\|.\]
Take \(\delta=\min\Bigl\{1,\ \dfrac{\varepsilon}{|\lambda_0|+1+\|x_0\|}\Bigr\}>0\). If \(|\lambda-\lambda_0|<\delta\) and \(\|x-x_0\|<\delta\), then \(|\lambda|\le|\lambda_0|+1\), so
\[\|\lambda x-\lambda_0 x_0\|\le(|\lambda_0|+1)\,\delta+\delta\,\|x_0\|=\delta\,(|\lambda_0|+1+\|x_0\|)\le\varepsilon.\]
Therefore scalar multiplication is continuous at \((\lambda_0,x_0)\). ∎
In a normed space \((X,\|\cdot\|)\): (i) every convergent sequence is Cauchy; (ii) limits are unique; (iii) a Cauchy sequence that has a convergent subsequence converges (to the same limit).
(i) Convergent \(\Rightarrow\) Cauchy. Suppose \(x_n\to x\) and let \(\varepsilon>0\). Choose \(N\) with \(\|x_n-x\|<\varepsilon/2\) for all \(n\ge N\). Then for \(m,n\ge N\), \(\|x_n-x_m\|\le\|x_n-x\|+\|x-x_m\|<\varepsilon/2+\varepsilon/2=\varepsilon\). Hence \((x_n)\) is Cauchy.
(ii) Uniqueness. Suppose \(x_n\to x\) and \(x_n\to y\). For every \(n\), \(\|x-y\|\le\|x-x_n\|+\|x_n-y\|\). The right-hand side tends to \(0\), so \(\|x-y\|\le 0\), whence \(\|x-y\|=0\) and, by (N1), \(x=y\).
(iii) Subsequence criterion. Let \((x_n)\) be Cauchy with a subsequence \((x_{n_k})\) converging to \(x\). Let \(\varepsilon>0\). By the Cauchy property choose \(N\) with \(\|x_n-x_m\|<\varepsilon/2\) for all \(m,n\ge N\). By convergence of the subsequence choose \(k\) with \(n_k\ge N\) and \(\|x_{n_k}-x\|<\varepsilon/2\). Then for every \(n\ge N\),
\[\|x_n-x\|\le\|x_n-x_{n_k}\|+\|x_{n_k}-x\|<\varepsilon/2+\varepsilon/2=\varepsilon.\]
Hence \(x_n\to x\). ∎
Let \(X\) be a vector space over \(\mathbb K\) with \(\dim X=n<\infty\). Then any two norms on \(X\) are equivalent.
Fix a basis \(e_1,\dots,e_n\) of \(X\); every \(x\in X\) is uniquely \(x=\sum_{i=1}^n\xi_i e_i\). Define the reference norm \(\|x\|_2=\bigl(\sum_{i=1}^n|\xi_i|^2\bigr)^{1/2}\); it is a norm because it is the Euclidean norm read through the linear coordinate isomorphism \(x\mapsto(\xi_1,\dots,\xi_n)\). Since equivalence of norms is transitive, it suffices to show that an arbitrary norm \(\|\cdot\|\) on \(X\) is equivalent to \(\|\cdot\|_2\).
Upper bound. By (N3), (N2), and the Cauchy-Schwarz inequality,
\[\|x\|=\Bigl\|\sum_{i=1}^n\xi_i e_i\Bigr\|\le\sum_{i=1}^n|\xi_i|\,\|e_i\|\le\Bigl(\sum_{i=1}^n\|e_i\|^2\Bigr)^{1/2}\Bigl(\sum_{i=1}^n|\xi_i|^2\Bigr)^{1/2}=M\,\|x\|_2,\]
where \(M=\bigl(\sum_{i=1}^n\|e_i\|^2\bigr)^{1/2}>0\) (each \(e_i\ne 0\)).
Continuity of \(\|\cdot\|\) with respect to \(\|\cdot\|_2\). By the reverse triangle inequality and the bound just proved, for all \(x,y\),
\[\bigl|\,\|x\|-\|y\|\,\bigr|\le\|x-y\|\le M\,\|x-y\|_2,\]
so \(x\mapsto\|x\|\) is (Lipschitz) continuous from \((X,\|\cdot\|_2)\) to \(\mathbb R\).
Lower bound via compactness. The unit sphere \(S=\{x\in X:\|x\|_2=1\}\) is closed and bounded in \((X,\|\cdot\|_2)\), which through the coordinate map is isometric to \(\mathbb K^n\) (that is \(\mathbb R^n\) or \(\mathbb R^{2n}\)) with the Euclidean norm; by the Heine-Borel theorem \(S\) is compact. The continuous function \(x\mapsto\|x\|\) therefore attains a minimum on \(S\), say \(m=\|x_\ast\|\) at some \(x_\ast\in S\). Since \(x_\ast\in S\) we have \(x_\ast\ne 0\), so by (N1) \(m=\|x_\ast\|>0\).
Conclusion. For \(x\ne 0\), the vector \(x/\|x\|_2\) lies in \(S\), so \(\|x/\|x\|_2\|\ge m\); by (N2) this reads \(\|x\|\ge m\,\|x\|_2\). Together with the upper bound (and the trivial case \(x=0\)),
\[m\,\|x\|_2\le\|x\|\le M\,\|x\|_2\qquad\text{for all }x\in X,\]
so \(\|\cdot\|\sim\|\cdot\|_2\). By transitivity, any two norms on \(X\) are equivalent. ∎
A normed space \((X,\|\cdot\|)\) is a Banach space if and only if every absolutely convergent series converges in \(X\); that is, whenever \((x_n)\subseteq X\) satisfies \(\sum_{n=1}^\infty\|x_n\|<\infty\), the partial sums \(s_N=\sum_{n=1}^N x_n\) converge to a limit in \(X\).
(\(\Rightarrow\)) Completeness implies convergence of absolutely convergent series. Assume \(X\) is Banach and \(\sum_n\|x_n\|<\infty\). For \(M>N\), by the triangle inequality,
\[\|s_M-s_N\|=\Bigl\|\sum_{n=N+1}^M x_n\Bigr\|\le\sum_{n=N+1}^M\|x_n\|\le\sum_{n=N+1}^\infty\|x_n\|.\]
The last expression is the tail of a convergent series of non-negative reals, hence tends to \(0\) as \(N\to\infty\). Thus \((s_N)\) is Cauchy, and by completeness it converges in \(X\).
(\(\Leftarrow\)) Convergence of absolutely convergent series implies completeness. Assume the stated property, and let \((x_n)\) be a Cauchy sequence in \(X\). For each \(k\in\mathbb N\), the Cauchy property furnishes an index \(N_k\) with \(\|x_m-x_n\|<2^{-k}\) for all \(m,n\ge N_k\). Choose indices \(n_1<n_2<\cdots\) with \(n_k\ge N_k\) (take \(n_k=\max\{N_k,n_{k-1}+1\}\)); then in particular
\[\|x_{n_{k+1}}-x_{n_k}\|<2^{-k}\qquad(k\ge 1).\]
Put \(y_k=x_{n_{k+1}}-x_{n_k}\). Then \(\sum_{k=1}^\infty\|y_k\|\le\sum_{k=1}^\infty 2^{-k}=1<\infty\), so the series \(\sum_k y_k\) is absolutely convergent and, by hypothesis, converges to some \(s\in X\). Its partial sums telescope:
\[\sum_{k=1}^{K}y_k=x_{n_{K+1}}-x_{n_1}\ \longrightarrow\ s,\qquad\text{so}\qquad x_{n_{K+1}}\longrightarrow s+x_{n_1}=:x.\]
Thus the Cauchy sequence \((x_n)\) has a convergent subsequence, and by Theorem 1.2(iii) it converges to \(x\). Hence \(X\) is complete. ∎
For every \(p\) with \(1\le p\le\infty\), the space \(\ell^p\) with the norm \(\|\cdot\|_p\) is complete, hence a Banach space.
Norm structure. Definiteness and absolute homogeneity are immediate from the definitions. For the triangle inequality, the cases \(p=1\) and \(p=\infty\) follow directly from the scalar triangle inequality:
\[\sum_k|x_k+y_k|\le \sum_k|x_k|+\sum_k|y_k|,\qquad \sup_k|x_k+y_k|\le \sup_k|x_k|+\sup_k|y_k|.\]
Now let \(1<p<\infty\) and put \(q=p/(p-1)\). We first record the finite-sum form of Hölder's inequality. Young's inequality
\[uv\le \frac{u^p}{p}+\frac{v^q}{q}\qquad(u,v\ge0)\]
follows, for example, by fixing \(v\) and minimizing \(u\mapsto u^p/p-uv+v^q/q\); the minimum occurs at \(u=v^{q-1}\) and equals \(0\). If \((a_k)_{k=1}^K\) and \((b_k)_{k=1}^K\) are nonnegative and neither norm is zero, apply Young's inequality to \(a_k/A\) and \(b_k/B\), where \(A=(\sum_{k=1}^K a_k^p)^{1/p}\) and \(B=(\sum_{k=1}^K b_k^q)^{1/q}\). Summing gives
\[\sum_{k=1}^K a_kb_k\le \Bigl(\sum_{k=1}^K a_k^p\Bigr)^{1/p}\Bigl(\sum_{k=1}^K b_k^q\Bigr)^{1/q}.\]
This is the finite Hölder inequality. To prove Minkowski without assuming in advance that \(x+y\in\ell^p\), set
\[S_K=\Bigl(\sum_{k=1}^K|x_k+y_k|^p\Bigr)^{1/p}.\]
Using \(|x_k+y_k|\le |x_k|+|y_k|\) and finite Hölder,
\[S_K^p\le \sum_{k=1}^K |x_k|\,|x_k+y_k|^{p-1}+\sum_{k=1}^K |y_k|\,|x_k+y_k|^{p-1}\le (\|x\|_p+\|y\|_p)S_K^{p-1}.\]
If \(S_K>0\), divide by \(S_K^{p-1}\); if \(S_K=0\), the conclusion is trivial. Thus \(S_K\le\|x\|_p+\|y\|_p\) for every \(K\). Letting \(K\to\infty\) shows both that \(x+y\in\ell^p\) and that
\[\|x+y\|_p\le\|x\|_p+\|y\|_p.\]
Hence \(\ell^p\) is a normed vector space for every \(1\le p\le\infty\). It remains to prove completeness.
Case \(1\le p<\infty\). Let \((x^{(n)})_{n\ge1}\) be Cauchy in \(\ell^p\), with \(x^{(n)}=(x^{(n)}_k)_{k\ge1}\). For each fixed coordinate \(k\),
\[|x^{(n)}_k-x^{(m)}_k|\le\|x^{(n)}-x^{(m)}\|_p,\]
so \((x^{(n)}_k)_n\) is Cauchy in \(\mathbb K\). Since \(\mathbb K\) is complete, define \(x_k=\lim_{n\to\infty}x^{(n)}_k\) and put \(x=(x_k)_k\). Given \(\varepsilon>0\), choose \(N\) so that \(\|x^{(n)}-x^{(m)}\|_p<\varepsilon\) whenever \(m,n\ge N\). For finite \(K\) and \(m,n\ge N\),
\[\sum_{k=1}^{K}|x^{(n)}_k-x^{(m)}_k|^p<\varepsilon^p.\]
Fix \(n\ge N\) and let \(m\to\infty\). Since the sum is finite,
\[\sum_{k=1}^{K}|x^{(n)}_k-x_k|^p\le\varepsilon^p\qquad\text{for every }K.\]
Letting \(K\to\infty\) gives \(\|x^{(n)}-x\|_p\le\varepsilon\) for all \(n\ge N\). In particular \(x^{(N)}-x\in\ell^p\), so \(x=x^{(N)}-(x^{(N)}-x)\in\ell^p\), and \(x^{(n)}\to x\) in \(\ell^p\).
Case \(p=\infty\). Let \((x^{(n)})\) be Cauchy in \(\ell^\infty\). Coordinatewise, \(|x^{(n)}_k-x^{(m)}_k|\le\|x^{(n)}-x^{(m)}\|_\infty\), so define \(x_k=\lim_nx^{(n)}_k\). Given \(\varepsilon>0\), choose \(N\) with \(\|x^{(n)}-x^{(m)}\|_\infty<\varepsilon\) for \(m,n\ge N\). Letting \(m\to\infty\) gives \(|x^{(n)}_k-x_k|\le\varepsilon\) for every \(k\) and \(n\ge N\). Therefore \(\|x^{(n)}-x\|_\infty\le\varepsilon\). Also \(|x_k|\le\|x^{(N)}\|_\infty+\varepsilon\) for all \(k\), hence \(x\in\ell^\infty\). Thus \(x^{(n)}\to x\) in \(\ell^\infty\).
Every Cauchy sequence therefore converges in \(\ell^p\) for \(1\le p\le\infty\). Hence each \(\ell^p\) is Banach. ∎
The space \(C[a,b]\) of continuous scalar functions on \([a,b]\), equipped with \(\|f\|_\infty=\sup_{t\in[a,b]}|f(t)|\), is complete, hence a Banach space. The key point is that a uniform limit of continuous functions is continuous.
Let \((f_n)\) be Cauchy in \((C[a,b],\|\cdot\|_\infty)\).
Pointwise limit. For each fixed \(t\in[a,b]\),
\[|f_n(t)-f_m(t)|\le\|f_n-f_m\|_\infty,\]
so \((f_n(t))_n\) is Cauchy in \(\mathbb K\) and therefore converges. Define \(f(t)=\lim_{n\to\infty}f_n(t)\).
Uniform convergence. Let \(\varepsilon>0\). Choose \(N\) so that \(\|f_n-f_m\|_\infty<\varepsilon\) whenever \(m,n\ge N\). Fix \(n\ge N\) and let \(m\to\infty\). Then \(|f_n(t)-f(t)|\le\varepsilon\) for every \(t\in[a,b]\), hence \(\|f_n-f\|_\infty\le\varepsilon\). Thus \(f_n\to f\) uniformly.
Continuity of the limit. Fix \(t_0\in[a,b]\) and \(\varepsilon>0\). By uniform convergence, choose an index \(n\) such that \(\|f_n-f\|_\infty<\varepsilon/3\). Since \(f_n\) is continuous at \(t_0\), there exists \(\delta>0\) such that, for \(t\in[a,b]\) with \(|t-t_0|<\delta\),
\[|f_n(t)-f_n(t_0)|<\varepsilon/3.\]
Therefore
\[|f(t)-f(t_0)|\le |f(t)-f_n(t)|+|f_n(t)-f_n(t_0)|+|f_n(t_0)-f(t_0)|<\varepsilon.\]
So \(f\) is continuous at every \(t_0\), hence \(f\in C[a,b]\). We already know \(\|f_n-f\|_\infty\to0\), so every Cauchy sequence converges in \(C[a,b]\). Thus \(C[a,b]\) is a Banach space. ∎
Worked Examples
Problem. Show that the norm \(\|\cdot\|_1\) on \(\mathbb K^2\), given by \(\|(x_1,x_2)\|_1=|x_1|+|x_2|\), is not induced by any inner product.
Tool. If a norm \(\|\cdot\|\) comes from an inner product \(\langle\cdot,\cdot\rangle\) via \(\|x\|^2=\langle x,x\rangle\), then the parallelogram law holds:
\[\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2\quad\text{for all }x,y.\]
Indeed, expanding by bilinearity in the real case, or by sesquilinearity in the complex case, gives \(\langle x+y,x+y\rangle+\langle x-y,x-y\rangle=2\langle x,x\rangle+2\langle y,y\rangle\); the cross terms cancel after taking the conjugate pair into account. So any norm violating this identity for even a single pair \((x,y)\) cannot arise from an inner product.
Derivation. Take \(x=(1,0)\) and \(y=(0,1)\). Then \(\|x\|_1=\|y\|_1=1\), while \(x+y=(1,1)\) and \(x-y=(1,-1)\) give \(\|x+y\|_1=2\) and \(\|x-y\|_1=2\). Substituting into the parallelogram law,
\[\|x+y\|_1^2+\|x-y\|_1^2=2^2+2^2=8,\qquad 2\|x\|_1^2+2\|y\|_1^2=2+2=4.\]
Since \(8\ne 4\), the parallelogram law fails, so \(\|\cdot\|_1\) is not induced by any inner product.
Interpretation. The same computation with \(\|\cdot\|_\infty\) (where \(\|x\pm y\|_\infty=1\) here) gives \(2\ne 4\); the Euclidean norm \(\|\cdot\|_2\) is the only member of the \(\ell^p\) family on \(\mathbb K^2\) (\(1\le p\le\infty\)) satisfying the law, and it alone is a Hilbert-space norm.
Problem. For \(x=(x_1,\dots,x_n)\in\mathbb R^n\), compare \(\|x\|_\infty=\max_{1\le i\le n}|x_i|\) with \(\|x\|_2=\bigl(\sum_{i=1}^n x_i^2\bigr)^{1/2}\), and exhibit explicit equivalence constants.
Claim. \(\ \|x\|_\infty\le\|x\|_2\le\sqrt n\,\|x\|_\infty\ \) for all \(x\), so the constants are \(c=1\) and \(C=\sqrt n\).
Lower bound. Let \(j\) be an index with \(|x_j|=\|x\|_\infty\). Then \(\|x\|_\infty^2=x_j^2\le\sum_{i=1}^n x_i^2=\|x\|_2^2\), so \(\|x\|_\infty\le\|x\|_2\).
Upper bound. Each \(x_i^2\le\|x\|_\infty^2\), hence \(\|x\|_2^2=\sum_{i=1}^n x_i^2\le n\,\|x\|_\infty^2\), so \(\|x\|_2\le\sqrt n\,\|x\|_\infty\).
Sharpness. Both constants are attained: for \(x=e_1=(1,0,\dots,0)\) we have \(\|x\|_\infty=\|x\|_2=1\), so \(c=1\) cannot be increased; for \(x=(1,1,\dots,1)\) we have \(\|x\|_2=\sqrt n\) and \(\|x\|_\infty=1\), so \(C=\sqrt n\) cannot be decreased.
Interpretation. These explicit bounds are a concrete instance of Theorem 1.3: on the finite-dimensional space \(\mathbb R^n\) all norms are equivalent. The constant \(C=\sqrt n\) grows with dimension, which is why such equivalences, though always available, degrade quantitatively as \(n\to\infty\) and fail outright in infinite dimensions.
Problem. Equip \(C[-1,1]\) with the \(L^1\)-norm \(\|f\|_1=\int_{-1}^{1}|f(t)|\,dt\). Exhibit a Cauchy sequence that does not converge in this space, proving \((C[-1,1],\|\cdot\|_1)\) is not complete.
The norm. \(\|\cdot\|_1\) is a norm on \(C[-1,1]\): homogeneity and the triangle inequality are clear, and if \(\int_{-1}^1|f|=0\) with \(f\) continuous then \(f\equiv 0\) (a nonzero value would force a positive integral over a neighbourhood).
The sequence. For \(n\ge 1\) define the continuous function
\[f_n(t)=\begin{cases}-1,&-1\le t\le-\tfrac1n,\\[2pt] nt,&-\tfrac1n<t<\tfrac1n,\\[2pt] 1,&\tfrac1n\le t\le 1.\end{cases}\]
Cauchy. For \(m\ge n\), the functions \(f_n\) and \(f_m\) agree outside \([-\tfrac1n,\tfrac1n]\), and on that interval both take values in \([-1,1]\), so \(|f_n-f_m|\le 2\) there. Hence
\[\|f_n-f_m\|_1=\int_{-1/n}^{1/n}|f_n-f_m|\,dt\le 2\cdot\frac2n=\frac4n\xrightarrow[n\to\infty]{}0,\]
so \((f_n)\) is Cauchy.
No continuous limit. Suppose \(f_n\to g\) in \(\|\cdot\|_1\) with \(g\in C[-1,1]\). Fix \(a\in(0,1)\). For \(n>1/a\), \(f_n\equiv 1\) on \([a,1]\), so
\[\int_a^1|1-g|\,dt\le\int_{-1}^1|f_n-g|\,dt=\|f_n-g\|_1\to 0,\]
giving \(\int_a^1|1-g|=0\); as \(g\) is continuous, \(g\equiv 1\) on \([a,1]\). Since \(a\in(0,1)\) was arbitrary, \(g\equiv 1\) on \((0,1]\). The symmetric argument on \([-1,-a]\) gives \(g\equiv-1\) on \([-1,0)\). No continuous function can equal \(1\) for all \(t>0\) and \(-1\) for all \(t<0\), because taking \(t\to 0^+\) and \(t\to 0^-\) would force \(g(0)=1\) and \(g(0)=-1\) simultaneously. This contradiction shows no such \(g\) exists.
Interpretation. The sequence is "trying" to converge to the discontinuous sign function, which lies outside \(C[-1,1]\). The completion of this normed space can be identified with \(L^1[-1,1]\), obtained by adjoining exactly such limits.
Not so. A linear subspace \(M\) of a Banach space \((X,\|\cdot\|)\) is complete if and only if it is closed in \(X\).
Closed \(\Rightarrow\) complete. If \(M\) is closed and \((x_n)\subseteq M\) is Cauchy, then it is Cauchy in \(X\), so it converges to some \(x\in X\) by completeness of \(X\); closedness forces \(x\in M\), so the sequence converges within \(M\). Complete \(\Rightarrow\) closed. If \(M\) is complete and \(x_n\in M\) with \(x_n\to x\in X\), then \((x_n)\) is Cauchy, hence converges in \(M\) to some \(y\in M\); by uniqueness of limits (Theorem 1.2) \(x=y\in M\), so \(M\) is closed.
A dense, non-closed, incomplete subspace. Inside the Banach space \(\ell^1\), let \(c_{00}\) be the subspace of sequences with only finitely many nonzero terms. It is dense (any \(x\in\ell^1\) is the \(\|\cdot\|_1\)-limit of its truncations), hence not closed, since \(c_{00}\ne\ell^1\). Concretely, the sequence \(x^{(n)}=\bigl(1,\tfrac1{2^2},\tfrac1{3^2},\dots,\tfrac1{n^2},0,0,\dots\bigr)\in c_{00}\) is Cauchy in \(\ell^1\) because \(\|x^{(m)}-x^{(n)}\|_1=\sum_{k=n+1}^{m}k^{-2}\to 0\), yet its \(\ell^1\)-limit \(\bigl(k^{-2}\bigr)_{k\ge 1}\) has infinitely many nonzero entries and so does not lie in \(c_{00}\). Thus \(c_{00}\) is a subspace of a Banach space that is not itself complete.
Exercises
Thirty exercises on normed and Banach spaces, progressing from recognition of the axioms through completeness proofs, the geometry of finite-dimensional subspaces, and the Riesz lemma, up to quotient spaces, separability, and completion. Each card lists difficulty and concept tags together with prerequisites and the expected method, and carries one to three progressive hints and a complete, self-contained solution. Click a card to expand.
Nonnegativity and definiteness. Each \(|x_i|\ge 0\), so \(\|x\|_1=\sum_i|x_i|\ge 0\). If \(\|x\|_1=0\), then a sum of nonnegative reals is zero, forcing every \(|x_i|=0\), hence \(x_i=0\) for all \(i\), i.e. \(x=0\). Conversely \(\|0\|_1=0\).
Absolute homogeneity. For \(\lambda\in\mathbb{R}\), \(\|\lambda x\|_1=\sum_i|\lambda x_i|=\sum_i|\lambda|\,|x_i|=|\lambda|\sum_i|x_i|=|\lambda|\,\|x\|_1\), using \(|\lambda x_i|=|\lambda||x_i|\) in \(\mathbb{R}\).
Triangle inequality. For each index, \(|x_i+y_i|\le|x_i|+|y_i|\). Summing over \(i=1,\dots,n\) gives \(\|x+y\|_1=\sum_i|x_i+y_i|\le\sum_i|x_i|+\sum_i|y_i|=\|x\|_1+\|y\|_1\).
Misconception. Definiteness is a separate axiom from nonnegativity: a seminorm satisfies everything except the implication \(\|x\|=0\Rightarrow x=0\). Here it holds precisely because a finite sum of nonnegative terms vanishes only when every term does.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Nonnegativity and definiteness. \(\|x\|_\infty=\max_i|x_i|\ge 0\). If \(\|x\|_\infty=0\), then \(|x_i|\le 0\) for every \(i\), so all \(x_i=0\) and \(x=0\); and \(\|0\|_\infty=0\).
Homogeneity. \(\|\lambda x\|_\infty=\max_i|\lambda x_i|=\max_i|\lambda||x_i|=|\lambda|\max_i|x_i|=|\lambda|\,\|x\|_\infty\), since multiplying a finite set of nonnegative numbers by the constant \(|\lambda|\ge0\) scales its maximum.
Triangle inequality. Fix an index \(j\) attaining the maximum of \(|x_i+y_i|\). Then \(\|x+y\|_\infty=|x_j+y_j|\le|x_j|+|y_j|\le\|x\|_\infty+\|y\|_\infty\), because \(|x_j|\le\max_i|x_i|=\|x\|_\infty\) and likewise for \(y\). The maximum enters exactly here: we bound the single largest coordinate by the two individual maxima, which need not be attained at the same index.
Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
From \(x=(x-y)+y\) and the triangle inequality, \(\|x\|\le\|x-y\|+\|y\|\), hence \(\|x\|-\|y\|\le\|x-y\|\).
By symmetry, exchanging the roles of \(x\) and \(y\), \(\|y\|-\|x\|\le\|y-x\|=\|-(x-y)\|=|-1|\,\|x-y\|=\|x-y\|\), using absolute homogeneity with \(\lambda=-1\).
The two inequalities \(\|x\|-\|y\|\le\|x-y\|\) and \(-(\|x\|-\|y\|)\le\|x-y\|\) together say \(\big|\,\|x\|-\|y\|\,\big|\le\|x-y\|\).
Consequence. The map \(x\mapsto\|x\|\) is \(1\)-Lipschitz, hence uniformly continuous; this is used repeatedly to pass norms through limits.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Nonnegativity and definiteness. \(d(x,y)=\|x-y\|\ge 0\), and \(d(x,y)=0\iff\|x-y\|=0\iff x-y=0\iff x=y\), by definiteness of the norm.
Symmetry. \(d(x,y)=\|x-y\|=|-1|\,\|x-y\|=\|-(x-y)\|=\|y-x\|=d(y,x)\).
Triangle inequality. \(d(x,z)=\|x-z\|=\|(x-y)+(y-z)\|\le\|x-y\|+\|y-z\|=d(x,y)+d(y,z)\).
Translation invariance. \(d(x+z,y+z)=\|(x+z)-(y+z)\|=\|x-y\|=d(x,y)\).
Homogeneity. \(d(\lambda x,\lambda y)=\|\lambda x-\lambda y\|=\|\lambda(x-y)\|=|\lambda|\,\|x-y\|=|\lambda|\,d(x,y)\).
Misconception. Not every metric comes from a norm. Translation invariance and homogeneity are exactly the two extra features a norm-induced metric always has; a metric lacking either (e.g. the discrete metric) cannot be written as \(\|x-y\|\).Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
\(\|x\|_\infty\le\|x\|_2\). If \(|x_j|=\|x\|_\infty\), then \(\|x\|_\infty^2=|x_j|^2\le\sum_{i=1}^n|x_i|^2=\|x\|_2^2\), and taking square roots gives the bound.
\(\|x\|_2\le\|x\|_1\). \(\|x\|_1^2=\big(\sum_i|x_i|\big)^2=\sum_i|x_i|^2+\sum_{i\ne j}|x_i||x_j|\ge\sum_i|x_i|^2=\|x\|_2^2\), since every cross term is \(\ge 0\).
\(\|x\|_1\le n\|x\|_\infty\). Each \(|x_i|\le\|x\|_\infty\), so \(\|x\|_1=\sum_{i=1}^n|x_i|\le n\|x\|_\infty\).
Sharpness. Take \(e_1=(1,0,\dots,0)\): then \(\|e_1\|_\infty=\|e_1\|_2=\|e_1\|_1=1\), so equality holds in the first two inequalities. Take \(u=(1,1,\dots,1)\): then \(\|u\|_1=n\) and \(\|u\|_\infty=1\), so \(\|u\|_1=n\|u\|_\infty\).
Equivalence. The chain gives \(\|x\|_\infty\le\|x\|_2\le\|x\|_1\le n\|x\|_\infty\); reading off any two positions yields two-sided constants, e.g. \(\tfrac1n\|x\|_1\le\|x\|_\infty\le\|x\|_1\), so all three norms are equivalent on \(\mathbb{R}^n\).
Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Fix a basis \(e_1,\dots,e_n\) of \(X\) and define the reference norm \(\|x\|_\ast=\sum_{i=1}^n|a_i|\) for \(x=\sum_i a_i e_i\); the coordinate map \(x\mapsto(a_1,\dots,a_n)\) is a linear isomorphism \(X\cong\mathbb{R}^n\) carrying \(\|\cdot\|_\ast\) to the \(\ell^1\)-norm. Since equivalence of norms is transitive, it suffices to show an arbitrary norm \(\|\cdot\|\) is equivalent to \(\|\cdot\|_\ast\).
Upper bound. Let \(M=\max_i\|e_i\|>0\). Then \(\|x\|=\big\|\sum_i a_i e_i\big\|\le\sum_i|a_i|\,\|e_i\|\le M\sum_i|a_i|=M\|x\|_\ast\). In particular, by the reverse triangle inequality \(\big|\|x\|-\|y\|\big|\le\|x-y\|\le M\|x-y\|_\ast\), so \(\|\cdot\|:(X,\|\cdot\|_\ast)\to\mathbb{R}\) is continuous.
Lower bound. The set \(S=\{x\in X:\|x\|_\ast=1\}\) corresponds to the \(\ell^1\)-unit sphere in \(\mathbb{R}^n\), which is closed and bounded, hence compact by Heine-Borel. The continuous function \(x\mapsto\|x\|\) attains its minimum \(m\) on \(S\). Since \(0\notin S\) and the norm is definite, \(m=\|x_0\|>0\) for some \(x_0\in S\). Thus \(\|x\|\ge m\) for all \(x\in S\). For arbitrary \(x\ne0\), apply this to \(x/\|x\|_\ast\in S\): \(\big\|x/\|x\|_\ast\big\|\ge m\), i.e. \(\|x\|\ge m\|x\|_\ast\); this also holds trivially at \(x=0\).
Combining, \(m\|x\|_\ast\le\|x\|\le M\|x\|_\ast\) with \(0<m\le M\), so \(\|\cdot\|\sim\|\cdot\|_\ast\). Any two norms are therefore equivalent to \(\|\cdot\|_\ast\), hence to each other.
Misconception. This is special to finite dimensions. Compactness of the unit sphere is exactly what fails in infinite dimensions (Ex 1.21), and there inequivalent norms abound.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Open ball is open. Let \(y\in B(x_0,r)\), so \(\|y-x_0\|<r\), and set \(\rho=r-\|y-x_0\|>0\). If \(z\in B(y,\rho)\), then \(\|z-x_0\|\le\|z-y\|+\|y-x_0\|<\rho+\|y-x_0\|=r\), so \(z\in B(x_0,r)\). Thus \(B(y,\rho)\subseteq B(x_0,r)\), and every point of the ball is interior; the ball is open.
Closed ball is closed. Let \(y\) be a limit point: choose \(y_n\in\overline B(x_0,r)\) with \(y_n\to y\), i.e. \(\|y_n-y\|\to0\). By the reverse triangle inequality, \(\big|\|y_n-x_0\|-\|y-x_0\|\big|\le\|y_n-y\|\to0\), so \(\|y_n-x_0\|\to\|y-x_0\|\). Since each \(\|y_n-x_0\|\le r\) and the limit preserves the inequality \(\le\), we get \(\|y-x_0\|\le r\), i.e. \(y\in\overline B(x_0,r)\). Being closed under limits of sequences, the set is closed.
Misconception. The closed ball is closed, but it need not equal the closure of the open ball in a general metric space; in a normed space it does, precisely because scaling toward the center produces interior points arbitrarily close to the boundary.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Convexity. Let \(x,y\in\overline B\) and \(t\in[0,1]\). By the triangle inequality and absolute homogeneity (with nonnegative scalars \(t,1-t\)), \[\|tx+(1-t)y\|\le\|tx\|+\|(1-t)y\|=t\|x\|+(1-t)\|y\|\le t\cdot1+(1-t)\cdot1=1,\] so \(tx+(1-t)y\in\overline B\). Convexity thus rests on the triangle inequality together with homogeneity.
Balancedness. Let \(x\in\overline B\) and \(|\lambda|\le1\). By absolute homogeneity, \(\|\lambda x\|=|\lambda|\,\|x\|\le|\lambda|\cdot1\le1\), so \(\lambda x\in\overline B\). Balancedness rests on absolute homogeneity alone.
Both properties hold verbatim for the open unit ball, replacing \(\le\) by \(<\) in the final steps (using \(t\|x\|+(1-t)\|y\|<1\) when at least one of \(\|x\|,\|y\|<1\); if both are \(<1\) the strict bound is clear).
Remark. Conversely, a convex, balanced, absorbing set that contains no line determines a norm through its Minkowski functional \(p(x)=\inf\{t>0:x/t\in\overline B\}\); the unit ball encodes the whole norm.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Norm continuity. \(\big|\|x_n\|-\|x\|\big|\le\|x_n-x\|\to0\), so \(\|x_n\|\to\|x\|\).
Boundedness. Convergence gives \(N\) with \(\|x_n-x\|\le1\) for \(n\ge N\); then \(\|x_n\|\le\|x\|+1\) for such \(n\), and \(\|x_n\|\le\max_{k<N}\|x_k\|\) otherwise. Hence \(\sup_n\|x_n\|\le\max\{\|x_1\|,\dots,\|x_{N-1}\|,\|x\|+1\}<\infty\).
Sum. \(\|(x_n+y_n)-(x+y)\|\le\|x_n-x\|+\|y_n-y\|\to0\).
Scalar product. \(\|\lambda_n x_n-\lambda x\|=\|\lambda_n(x_n-x)+(\lambda_n-\lambda)x\|\le|\lambda_n|\,\|x_n-x\|+|\lambda_n-\lambda|\,\|x\|\). The convergent real sequence \((\lambda_n)\) is bounded, say \(|\lambda_n|\le K\); then the right side is \(\le K\|x_n-x\|+|\lambda_n-\lambda|\,\|x\|\to0\). Thus \(\lambda_n x_n\to\lambda x\).
Consequence. Addition and scalar multiplication are (jointly) continuous, so a normed space is a topological vector space; limits may be taken inside all algebraic operations.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Convergent \(\Rightarrow\) Cauchy. Suppose \(x_n\to x\). Given \(\varepsilon>0\), pick \(N\) with \(\|x_n-x\|<\varepsilon/2\) for \(n\ge N\). Then for \(m,n\ge N\), \(\|x_n-x_m\|\le\|x_n-x\|+\|x-x_m\|<\varepsilon\). So \((x_n)\) is Cauchy.
Cauchy \(\Rightarrow\) bounded. Take \(\varepsilon=1\): there is \(N\) with \(\|x_n-x_N\|<1\) for \(n\ge N\), so \(\|x_n\|<\|x_N\|+1\) there. With the finitely many earlier terms, \(\sup_n\|x_n\|\le\max\{\|x_1\|,\dots,\|x_{N-1}\|,\|x_N\|+1\}<\infty\).
Counterexample. Let \(X=c_{00}\), the space of real sequences with only finitely many nonzero terms, normed by \(\|x\|_\infty=\sup_i|x_i|\). Define \(x^{(n)}=(1,\tfrac12,\tfrac13,\dots,\tfrac1n,0,0,\dots)\). For \(m>n\), \(\|x^{(m)}-x^{(n)}\|_\infty=\sup_{n<k\le m}\tfrac1k=\tfrac1{n+1}\to0\), so \((x^{(n)})\) is Cauchy. Its only possible limit is \(x=(1,\tfrac12,\tfrac13,\dots)\), since coordinatewise convergence is forced by the sup norm; but \(x\notin c_{00}\) because it has infinitely many nonzero terms. Hence no limit exists in \(X\), and \((x^{(n)})\) is Cauchy but not convergent.
Misconception. "Cauchy" is intrinsic to the sequence and the norm; "convergent" additionally requires the limit to live in the space. Completeness is exactly the guarantee that the two notions coincide.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
(\(\Rightarrow\)) Banach implies the criterion. Suppose \(X\) is complete and \(\sum_n\|x_n\|=:s<\infty\). For \(M>N\), \(\|S_M-S_N\|=\big\|\sum_{n=N+1}^M x_n\big\|\le\sum_{n=N+1}^M\|x_n\|\). Since \(\sum\|x_n\|\) converges, its tails \(\sum_{n>N}\|x_n\|\to0\), so \((S_N)\) is Cauchy, hence converges by completeness; that is, \(\sum_n x_n\) converges.
(\(\Leftarrow\)) The criterion implies completeness. Let \((y_n)\) be Cauchy in \(X\). For each \(k\) choose \(n_k\) (strictly increasing) with \(\|y_m-y_{n_k}\|<2^{-k}\) for all \(m\ge n_k\); in particular \(\|y_{n_{k+1}}-y_{n_k}\|<2^{-k}\). Put \(d_k=y_{n_{k+1}}-y_{n_k}\). Then \(\sum_k\|d_k\|<\sum_k2^{-k}=1<\infty\), so by hypothesis \(\sum_k d_k\) converges to some \(z\in X\). Its partial sums telescope: \(\sum_{j=1}^{k-1}d_j=y_{n_k}-y_{n_1}\), so \(y_{n_k}\to y_{n_1}+z=:y\). Thus the Cauchy sequence \((y_n)\) has a convergent subsequence, and thereby converges to \(y\): given \(\varepsilon>0\), take \(N\) with \(\|y_m-y_n\|<\varepsilon/2\) for \(m,n\ge N\) and \(k\) with \(n_k\ge N\) and \(\|y_{n_k}-y\|<\varepsilon/2\); then \(\|y_n-y\|\le\|y_n-y_{n_k}\|+\|y_{n_k}-y\|<\varepsilon\) for \(n\ge N\). Hence \(X\) is complete.
Misconception. The subsequence trick is essential: a Cauchy sequence itself need not be an absolutely summable telescoping series, but a suitably thinned subsequence always is. This criterion is the standard workhorse for proving spaces such as \(\ell^p\) and \(L^p\) complete.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Let \((x^{(n)})\) be Cauchy in \(\ell^p\), \(x^{(n)}=(x^{(n)}_i)_i\). For each coordinate \(i\), \(|x^{(n)}_i-x^{(m)}_i|\le\big(\sum_j|x^{(n)}_j-x^{(m)}_j|^p\big)^{1/p}=\|x^{(n)}-x^{(m)}\|_p\), so \((x^{(n)}_i)_n\) is Cauchy in \(\mathbb{R}\); by completeness of \(\mathbb{R}\) it converges to some \(x_i\). Set \(x=(x_i)\).
Norm control on truncations. Fix \(\varepsilon>0\) and choose \(N\) with \(\|x^{(n)}-x^{(m)}\|_p<\varepsilon\) for \(m,n\ge N\). For any finite \(K\) and \(m,n\ge N\), \(\sum_{i=1}^K|x^{(n)}_i-x^{(m)}_i|^p\le\|x^{(n)}-x^{(m)}\|_p^p<\varepsilon^p\). Hold \(n\ge N\) and \(K\) fixed and let \(m\to\infty\); since each summand is continuous in \(x^{(m)}_i\to x_i\) and the sum is finite, \(\sum_{i=1}^K|x^{(n)}_i-x_i|^p\le\varepsilon^p\).
Let \(K\to\infty\). The bound \(\sum_{i=1}^K|x^{(n)}_i-x_i|^p\le\varepsilon^p\) holds for all \(K\), so the increasing partial sums converge and \(\sum_{i=1}^\infty|x^{(n)}_i-x_i|^p\le\varepsilon^p\), i.e. \(\|x^{(n)}-x\|_p\le\varepsilon\) for all \(n\ge N\). In particular \(x^{(N)}-x\in\ell^p\), and since \(x^{(N)}\in\ell^p\), Minkowski's inequality gives \(x=x^{(N)}-(x^{(N)}-x)\in\ell^p\).
Thus \(x\in\ell^p\) and \(\|x^{(n)}-x\|_p\le\varepsilon\) for \(n\ge N\), meaning \(x^{(n)}\to x\) in \(\ell^p\). Every Cauchy sequence converges, so \(\ell^p\) is a Banach space.
Misconception. Coordinatewise convergence alone is not convergence in \(\ell^p\); the truncate-then-take-limits step is what promotes it to norm convergence and simultaneously certifies membership in \(\ell^p\).Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Let \((x^{(n)})\) be Cauchy in \(\ell^\infty\). For each \(i\), \(|x^{(n)}_i-x^{(m)}_i|\le\|x^{(n)}-x^{(m)}\|_\infty\), so \((x^{(n)}_i)_n\) is Cauchy in \(\mathbb{R}\) and converges to some \(x_i\); set \(x=(x_i)\).
Given \(\varepsilon>0\), pick \(N\) with \(\|x^{(n)}-x^{(m)}\|_\infty<\varepsilon\) for \(m,n\ge N\); then \(|x^{(n)}_i-x^{(m)}_i|<\varepsilon\) for all \(i\) and all such \(m,n\). Fix \(i\) and \(n\ge N\) and let \(m\to\infty\): \(|x^{(n)}_i-x_i|\le\varepsilon\). This holds for every \(i\), so \(\sup_i|x^{(n)}_i-x_i|\le\varepsilon\), i.e. \(\|x^{(n)}-x\|_\infty\le\varepsilon\) for \(n\ge N\).
Membership. \(x\) is bounded: \(|x_i|\le|x^{(N)}_i|+|x_i-x^{(N)}_i|\le\|x^{(N)}\|_\infty+\varepsilon\) for all \(i\), so \(\|x\|_\infty\le\|x^{(N)}\|_\infty+\varepsilon<\infty\), giving \(x\in\ell^\infty\). Combined with \(\|x^{(n)}-x\|_\infty\to0\), the sequence converges in \(\ell^\infty\). Hence \(\ell^\infty\) is complete.
Misconception. The estimate \(|x^{(n)}_i-x_i|\le\varepsilon\) is uniform in \(i\) precisely because the sup norm controls all coordinates at once; this uniformity is why the limit stays bounded and the convergence is in norm, not merely coordinatewise.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Subspace. \(0\in c_0\). If \(x,y\in c_0\) and \(\alpha,\beta\in\mathbb{R}\), then \(\alpha x+\beta y\in\ell^\infty\) and \(\lim_i(\alpha x_i+\beta y_i)=\alpha\cdot0+\beta\cdot0=0\), so \(\alpha x+\beta y\in c_0\). Thus \(c_0\) is a linear subspace of \(\ell^\infty\).
Closed. Suppose \(x^{(n)}\in c_0\) and \(x^{(n)}\to x\) in \(\ell^\infty\). Fix \(\varepsilon>0\); choose \(n\) with \(\|x-x^{(n)}\|_\infty<\varepsilon/2\). Since \(x^{(n)}\in c_0\), there is \(I\) with \(|x^{(n)}_i|<\varepsilon/2\) for \(i\ge I\). Then for \(i\ge I\), \(|x_i|\le|x_i-x^{(n)}_i|+|x^{(n)}_i|\le\|x-x^{(n)}\|_\infty+\varepsilon/2<\varepsilon\). Hence \(\lim_i x_i=0\), so \(x\in c_0\). Being closed under limits, \(c_0\) is closed in \(\ell^\infty\).
Completeness. \(\ell^\infty\) is complete (Ex 1.13), and a closed subspace of a complete metric space is complete (a Cauchy sequence in \(c_0\) converges in \(\ell^\infty\), and the limit lies in \(c_0\) by closedness). Therefore \((c_0,\|\cdot\|_\infty)\) is a Banach space.
Remark. The same \(3\varepsilon\) pattern shows \(c\), the space of convergent sequences, is closed in \(\ell^\infty\); \(c_0\) is in turn closed in \(c\).Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Dense in \(c_0\). Let \(x\in c_0\) and \(\varepsilon>0\). Since \(x_i\to0\), pick \(K\) with \(|x_i|<\varepsilon\) for \(i>K\). Let \(y=(x_1,\dots,x_K,0,0,\dots)\in c_{00}\). Then \(\|x-y\|_\infty=\sup_{i>K}|x_i|\le\varepsilon\). As \(\varepsilon>0\) was arbitrary, \(c_{00}\) is dense in \(c_0\).
Not dense in \(\ell^\infty\). Consider \(\mathbf 1=(1,1,1,\dots)\in\ell^\infty\). For any \(y\in c_{00}\), all but finitely many coordinates of \(y\) vanish, so there is an index \(i\) with \(y_i=0\), giving \(\|\mathbf 1-y\|_\infty\ge|1-y_i|=1\). Hence \(\mathrm{dist}(\mathbf 1,c_{00})\ge1>0\), and \(\mathbf 1\) is not in the closure of \(c_{00}\); indeed \(\overline{c_{00}}=c_0\ne\ell^\infty\).
Incompleteness. \(c_{00}\subsetneq c_0=\overline{c_{00}}\); e.g. \(x=(1,\tfrac12,\tfrac13,\dots)\in c_0\setminus c_{00}\). Choose \(y^{(K)}\in c_{00}\) with \(\|x-y^{(K)}\|_\infty\to0\) (truncations of \(x\)); then \((y^{(K)})\) is Cauchy in \(c_{00}\) but its limit \(x\) lies outside \(c_{00}\). A complete subspace would be closed, hence equal to its closure \(c_0\); since \(c_{00}\ne c_0\), \(c_{00}\) is not complete.
Misconception. Density is norm-dependent: \(c_{00}\) is dense in \(c_0\) and in every \(\ell^p\) (\(p<\infty\)), but never in \(\ell^\infty\), because uniform smallness of tails cannot be forced there.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Candidate limit. Let \((f_n)\) be Cauchy in \((C[a,b],\|\cdot\|_\infty)\). For each fixed \(t\), \(|f_n(t)-f_m(t)|\le\|f_n-f_m\|_\infty\), so \((f_n(t))_n\) is Cauchy in \(\mathbb{R}\); define \(f(t):=\lim_n f_n(t)\).
Uniform convergence. Given \(\varepsilon>0\), pick \(N\) with \(\|f_n-f_m\|_\infty<\varepsilon\) for \(m,n\ge N\); then \(|f_n(t)-f_m(t)|<\varepsilon\) for all \(t\). Fix \(t\) and \(n\ge N\), let \(m\to\infty\): \(|f_n(t)-f(t)|\le\varepsilon\). Taking the supremum over \(t\), \(\|f_n-f\|_\infty\le\varepsilon\) for all \(n\ge N\). Hence \(f_n\to f\) uniformly.
Continuity of \(f\). Fix \(t_0\) and \(\varepsilon>0\). Choose \(n\ge N\) with \(\|f_n-f\|_\infty\le\varepsilon/3\). Since \(f_n\) is continuous at \(t_0\), there is \(\delta>0\) with \(|f_n(t)-f_n(t_0)|<\varepsilon/3\) whenever \(|t-t_0|<\delta\). Then for such \(t\), \[|f(t)-f(t_0)|\le|f(t)-f_n(t)|+|f_n(t)-f_n(t_0)|+|f_n(t_0)-f(t_0)|<\tfrac\varepsilon3+\tfrac\varepsilon3+\tfrac\varepsilon3=\varepsilon.\] So \(f\) is continuous, i.e. \(f\in C[a,b]\).
Thus the Cauchy sequence \((f_n)\) converges in \(C[a,b]\) to \(f\), and the space is complete.
Misconception. Completeness here is exactly the theorem that uniform limits preserve continuity. Under a weaker norm such as \(\|\cdot\|_1\) the same functions form an incomplete space (Ex 1.17), because \(L^1\)-limits need not be continuous.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
The sequence. For \(n\ge2\) define \(f_n\in C[0,1]\) by \(f_n(t)=0\) on \([0,\tfrac12]\), \(f_n(t)=1\) on \([\tfrac12+\tfrac1n,1]\), and \(f_n(t)=n\big(t-\tfrac12\big)\) on \([\tfrac12,\tfrac12+\tfrac1n]\). Each \(f_n\) is continuous and \(0\le f_n\le1\).
Cauchy. For \(m>n\), \(f_n\) and \(f_m\) agree outside \([\tfrac12,\tfrac12+\tfrac1n]\), where both lie in \([0,1]\), so \(|f_n-f_m|\le1\) there and vanishes elsewhere. Hence \(\|f_n-f_m\|_1=\int_{1/2}^{1/2+1/n}|f_n-f_m|\,dt\le\tfrac1n\). Thus \((f_n)\) is Cauchy in \(\|\cdot\|_1\).
No continuous limit. Let \(g\) be the step function \(g=0\) on \([0,\tfrac12)\), \(g=1\) on \([\tfrac12,1]\). Then \(\|f_n-g\|_1=\int_{1/2}^{1/2+1/n}|f_n-g|\,dt\le\tfrac1n\to0\), so if a limit existed in \(C[0,1]\) it would have to agree with \(g\) almost everywhere. Suppose \(h\in C[0,1]\) with \(\|f_n-h\|_1\to0\). By the triangle inequality \(\|g-h\|_1\le\|g-f_n\|_1+\|f_n-h\|_1\to0\), so \(\int_0^1|g-h|=0\). Since \(|g-h|\) is nonnegative and continuous on each of \([0,\tfrac12)\) and \((\tfrac12,1]\), it vanishes there: \(h=0\) on \([0,\tfrac12)\) and \(h=1\) on \((\tfrac12,1]\). Taking one-sided limits at \(\tfrac12\) forces \(h(\tfrac12^-)=0\ne1=h(\tfrac12^+)\), contradicting continuity of \(h\). Hence no \(h\in C[0,1]\) is a \(\|\cdot\|_1\)-limit, and \((C[0,1],\|\cdot\|_1)\) is not complete.
Misconception. A space's completeness depends on the norm, not just the underlying set. The very same functions form a Banach space under \(\|\cdot\|_\infty\) (Ex 1.16) yet an incomplete space under \(\|\cdot\|_1\); completing the latter yields \(L^1[0,1]\).Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Norm inequality (\(q<\infty\)). Assume first \(\|x\|_p=1\). Then each \(|x_i|\le\big(\sum_j|x_j|^p\big)^{1/p}=1\), and since \(q>p\), \(|x_i|^q=|x_i|^p\cdot|x_i|^{q-p}\le|x_i|^p\). Summing, \(\sum_i|x_i|^q\le\sum_i|x_i|^p=1\), so \(\|x\|_q\le1=\|x\|_p\). For general \(x\ne0\), apply this to \(x/\|x\|_p\) (which has \(\|\cdot\|_p=1\)): \(\|x\|_q/\|x\|_p=\|x/\|x\|_p\|_q\le1\), i.e. \(\|x\|_q\le\|x\|_p\); trivial for \(x=0\). In particular \(x\in\ell^p\Rightarrow x\in\ell^q\), so \(\ell^p\subseteq\ell^q\).
Case \(q=\infty\). \(\|x\|_\infty=\sup_i|x_i|\le\big(\sum_i|x_i|^p\big)^{1/p}=\|x\|_p\), so again \(\ell^p\subseteq\ell^\infty\) with the norm bound.
Strictness. Choose \(r\) with \(1/q<r\le1/p\) (possible since \(1/q<1/p\)); if \(q=\infty\) take any \(0<r\le1/p\). Let \(x_i=i^{-r}\). Then \(\sum_i|x_i|^p=\sum_i i^{-rp}\) with \(rp\le1\) diverges, so \(x\notin\ell^p\); while \(\sum_i|x_i|^q=\sum_i i^{-rq}\) with \(rq>1\) converges (and if \(q=\infty\), \(\sup_i i^{-r}=1<\infty\)), so \(x\in\ell^q\). Hence \(x\in\ell^q\setminus\ell^p\) and the inclusion is strict.
Misconception. The inclusion for sequence spaces runs opposite to that for \(L^p\) on a finite measure space, where larger \(p\) gives the smaller space. The direction is dictated by whether small or large values dominate the sum.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
\(F\) is complete. Let \(d=\dim F<\infty\) with basis \(u_1,\dots,u_d\). On \(F\) the ambient norm \(\|\cdot\|\) is, by Ex 1.6, equivalent to the coordinate norm \(\|\sum_k a_k u_k\|_\ast=\sum_k|a_k|\): there are \(m,M>0\) with \(m\|v\|_\ast\le\|v\|\le M\|v\|_\ast\) for all \(v\in F\). If \((v_n)\subseteq F\) is Cauchy in \(\|\cdot\|\), then \(\|v_n-v_m\|_\ast\le\tfrac1m\|v_n-v_m\|\to0\), so the coordinate vectors \(a^{(n)}\in\mathbb{R}^d\) are Cauchy in \(\ell^1\); by completeness of \(\mathbb{R}^d\) they converge to some \(a\), and \(v:=\sum_k a_k u_k\in F\) satisfies \(\|v_n-v\|\le M\|v_n-v\|_\ast\to0\). Thus every Cauchy sequence in \(F\) converges within \(F\); \(F\) is complete.
Complete \(\Rightarrow\) closed. Let \(x\in\overline F\). Choose \(v_n\in F\) with \(v_n\to x\) in \(X\). Convergent sequences are Cauchy, so \((v_n)\) is Cauchy in \(F\), hence converges to some \(v\in F\) by completeness. Limits in a metric space are unique, so \(x=v\in F\). Therefore \(\overline F\subseteq F\), i.e. \(F\) is closed.
Misconception. Infinite-dimensional subspaces can fail to be closed - for instance \(c_{00}\) inside \(c_0\) (Ex 1.15). Finiteness of dimension is exactly what forces completeness, and thereby closedness, here.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Since \(Y\ne X\), pick \(v\in X\setminus Y\). As \(Y\) is closed and \(v\notin Y\), the distance \(d:=\mathrm{dist}(v,Y)=\inf_{y\in Y}\|v-y\|\) is strictly positive (otherwise a sequence in \(Y\) would converge to \(v\in\overline Y=Y\), a contradiction).
Because \(0<\theta<1\), we have \(d/\theta>d\), so \(d/\theta\) is not a lower bound for \(\{\|v-y\|:y\in Y\}\); choose \(y_0\in Y\) with \[d\le\|v-y_0\|\le\frac{d}{\theta}.\] Define the unit vector \(x_\theta:=\dfrac{v-y_0}{\|v-y_0\|}\), so \(\|x_\theta\|=1\).
Let \(y\in Y\) be arbitrary. Then \(y_0+\|v-y_0\|\,y\in Y\) (as \(Y\) is a subspace), so by definition of \(d\), \(\big\|v-(y_0+\|v-y_0\|y)\big\|\ge d\). Compute \[\|x_\theta-y\|=\left\|\frac{v-y_0}{\|v-y_0\|}-y\right\|=\frac{1}{\|v-y_0\|}\big\|v-y_0-\|v-y_0\|y\big\|=\frac{\big\|v-(y_0+\|v-y_0\|y)\big\|}{\|v-y_0\|}\ge\frac{d}{\|v-y_0\|}\ge\frac{d}{d/\theta}=\theta.\] Since \(y\in Y\) was arbitrary, \(\mathrm{dist}(x_\theta,Y)\ge\theta\).
Misconception. In general one cannot take \(\theta=1\): without an inner product the infimum \(d\) may not be attained, so a genuinely "orthogonal" unit vector need not exist. Riesz's lemma delivers almost-orthogonality, which is all that infinite-dimensional geometry allows.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
(\(\Leftarrow\)) Finite dimension implies compact ball. Let \(\dim X=n<\infty\). Fixing a basis identifies \(X\) with \(\mathbb{R}^n\), and by Ex 1.6 the norm \(\|\cdot\|\) is equivalent to the Euclidean norm. The closed unit ball \(\overline B=\{x:\|x\|\le1\}\) is then closed and bounded in \((\mathbb{R}^n,\|\cdot\|_2)\) (equivalent norms preserve boundedness and closedness), hence compact by the Heine-Borel theorem. Compactness is a topological property preserved by the homeomorphism, so \(\overline B\) is compact in \(X\).
(\(\Rightarrow\)) Infinite dimension implies non-compact ball. Suppose \(\dim X=\infty\). We construct unit vectors inductively. Pick any \(x_1\) with \(\|x_1\|=1\). Given \(x_1,\dots,x_n\), let \(Y_n=\mathrm{span}(x_1,\dots,x_n)\); it is finite-dimensional, hence closed (Ex 1.19), and proper (since \(\dim X=\infty\)). By Riesz's lemma with \(\theta=\tfrac12\), there is a unit vector \(x_{n+1}\) with \(\mathrm{dist}(x_{n+1},Y_n)\ge\tfrac12\); in particular \(\|x_{n+1}-x_k\|\ge\tfrac12\) for all \(k\le n\), because \(x_k\in Y_n\).
The resulting sequence \((x_n)\) lies in \(\overline B\) and satisfies \(\|x_n-x_m\|\ge\tfrac12\) for all \(n\ne m\). No subsequence can be Cauchy, so no subsequence converges; thus \(\overline B\) is not sequentially compact. In a metric space compactness and sequential compactness coincide, so \(\overline B\) is not compact.
Combining the two directions, the closed unit ball is compact if and only if \(X\) is finite-dimensional.
Misconception. Boundedness plus closedness does not give compactness beyond finite dimensions. Riesz's theorem is the precise obstruction: the "\(\tfrac12\)-separated" sequence lives in the closed bounded ball yet escapes every convergent subsequence.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Complete \(\Rightarrow\) closed. Let \(x\in\overline Y\) and choose \(y_n\in Y\) with \(y_n\to x\) in \(X\). Being convergent, \((y_n)\) is Cauchy, and it is a Cauchy sequence in \(Y\); by completeness of \(Y\) it converges to some \(y\in Y\). But \((y_n)\) already converges to \(x\) in \(X\), and limits in a metric space are unique, so \(x=y\in Y\). Hence \(\overline Y\subseteq Y\) and \(Y\) is closed. (This direction needs no completeness of \(X\).)
Closed \(\Rightarrow\) complete (in a Banach space). Let \((y_n)\) be Cauchy in \(Y\). It is then Cauchy in \(X\), which is complete, so \(y_n\to x\) for some \(x\in X\). Each \(y_n\in Y\), so \(x\in\overline Y=Y\) by closedness. Thus \((y_n)\) converges within \(Y\), and \(Y\) is complete.
Misconception. Without completeness of the ambient space, "closed" does not imply "complete": every subspace is closed in itself, yet an incomplete space is closed in itself. The Banach hypothesis on \(X\) is essential for the converse.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Write \(\pi(x)=x+Y\) and \(q(x+Y)=\inf_{y\in Y}\|x+y\|=\mathrm{dist}(x,Y)\). This depends only on the coset: if \(x-x'\in Y\), the sets \(\{x+y:y\in Y\}\) and \(\{x'+y:y\in Y\}\) coincide.
Nonnegativity and definiteness. \(q\ge0\) as an infimum of norms. If \(q(x+Y)=0\), then \(\mathrm{dist}(x,Y)=0\), so \(x\in\overline Y=Y\) (here closedness of \(Y\) is used), whence \(x+Y=0+Y\). Conversely \(q(0+Y)=\inf_{y\in Y}\|y\|=0\).
Homogeneity. For \(\lambda\ne0\), as \(y\) ranges over \(Y\) so does \(\lambda y\), thus \(q(\lambda x+Y)=\inf_{y}\|\lambda x+y\|=\inf_{y}\|\lambda(x+y)\|=|\lambda|\inf_y\|x+y\|=|\lambda|q(x+Y)\); the case \(\lambda=0\) gives \(0\) on both sides.
Triangle inequality. Fix \(\varepsilon>0\) and pick \(y,y'\in Y\) with \(\|x+y\|\le q(x+Y)+\varepsilon\) and \(\|x'+y'\|\le q(x'+Y)+\varepsilon\). Then \(y+y'\in Y\) and \(q\big((x+x')+Y\big)\le\|(x+x')+(y+y')\|\le\|x+y\|+\|x'+y'\|\le q(x+Y)+q(x'+Y)+2\varepsilon\). Letting \(\varepsilon\downarrow0\) yields subadditivity. Hence \(q\) is a norm on \(X/Y\).
Completeness (with \(X\) Banach). By Ex 1.11 it suffices to show every absolutely convergent series in \(X/Y\) converges. Let \(\sum_n q(x_n+Y)<\infty\). For each \(n\) choose \(y_n\in Y\) with \(\|x_n+y_n\|\le q(x_n+Y)+2^{-n}\). Then \(\sum_n\|x_n+y_n\|\le\sum_n q(x_n+Y)+\sum_n2^{-n}<\infty\), so \(\sum_n(x_n+y_n)\) converges in the Banach space \(X\) to some \(s\). The projection \(\pi\) is continuous because \(q(\pi(v))\le\|v\|\) (take \(y=0\) in the infimum), so \(\pi\) is \(1\)-Lipschitz; therefore \(\sum_{n\le N}(x_n+Y)=\pi\big(\sum_{n\le N}(x_n+y_n)\big)\to\pi(s)=s+Y\). Thus the series \(\sum_n(x_n+Y)\) converges in \(X/Y\), and by Ex 1.11 the quotient is complete, i.e. a Banach space.
Misconception. If \(Y\) is not closed, \(q\) is only a seminorm: some nonzero coset \(x+Y\) with \(x\in\overline Y\setminus Y\) has \(q(x+Y)=0\). Closedness of \(Y\) is exactly what makes the quotient norm definite.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Countability of the candidate set. Let \(D\) be the set of sequences with rational entries and finite support. For each \(K\), the sequences supported in \(\{1,\dots,K\}\) with rational entries form a set in bijection with \(\mathbb{Q}^K\), which is countable. Then \(D=\bigcup_{K\ge1}(\text{those supported in }\{1,\dots,K\})\) is a countable union of countable sets, hence countable.
Density. Let \(x\in\ell^p\) and \(\varepsilon>0\). Since \(\sum_i|x_i|^p<\infty\), the tails converge to \(0\): choose \(K\) with \(\sum_{i>K}|x_i|^p<(\varepsilon/2)^p\). For \(i\le K\), use density of \(\mathbb{Q}\) in \(\mathbb{R}\) to choose \(q_i\in\mathbb{Q}\) with \(|x_i-q_i|^p<\dfrac{(\varepsilon/2)^p}{K}\). Define \(q=(q_1,\dots,q_K,0,0,\dots)\in D\). Then \[\|x-q\|_p^p=\sum_{i=1}^K|x_i-q_i|^p+\sum_{i>K}|x_i|^p<K\cdot\frac{(\varepsilon/2)^p}{K}+(\varepsilon/2)^p=2(\varepsilon/2)^p\le\varepsilon^p,\] using \(2\le2^p\) for \(p\ge1\). Hence \(\|x-q\|_p<\varepsilon\).
Thus every \(x\in\ell^p\) is approximated arbitrarily well by elements of the countable set \(D\), so \(D\) is dense and \(\ell^p\) is separable.
Misconception. Separability again depends on the norm: the same underlying idea fails for \(\ell^\infty\) (Ex 1.25), where tails cannot be made uniformly small and a countable dense set cannot exist.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
An uncountable \(1\)-separated family. For each subset \(A\subseteq\mathbb{N}\), let \(\mathbf 1_A=(\mathbf 1_A(1),\mathbf 1_A(2),\dots)\in\ell^\infty\) be the sequence with \(i\)-th entry \(1\) if \(i\in A\) and \(0\) otherwise; \(\|\mathbf 1_A\|_\infty\le1\). If \(A\ne B\), pick \(i\) in exactly one of them (in \(A\triangle B\)); then the \(i\)-th entries differ by \(1\), so \(\|\mathbf 1_A-\mathbf 1_B\|_\infty\ge1\), and since all entries lie in \(\{-1,0,1\}\), \(\|\mathbf 1_A-\mathbf 1_B\|_\infty=1\). The index set \(\mathcal P(\mathbb{N})\) is uncountable, so \(\{\mathbf 1_A:A\subseteq\mathbb{N}\}\) is an uncountable family with pairwise distances \(=1\).
No countable dense set. Suppose, for contradiction, \(S\subseteq\ell^\infty\) is countable and dense. The open balls \(B(\mathbf 1_A,\tfrac12)\), \(A\subseteq\mathbb{N}\), are pairwise disjoint: if \(z\) lay in two of them, then \(\|\mathbf 1_A-\mathbf 1_B\|_\infty\le\|\mathbf 1_A-z\|_\infty+\|z-\mathbf 1_B\|_\infty<\tfrac12+\tfrac12=1\), contradicting distance \(1\). By density, each ball contains a point of \(S\); choosing one such point per ball gives an injection from the uncountable set \(\mathcal P(\mathbb{N})\) into the countable set \(S\), which is impossible.
Therefore no countable dense subset exists, and \(\ell^\infty\) is not separable.
Misconception. Completeness and separability are independent: \(\ell^\infty\) is complete (Ex 1.13) yet non-separable, whereas the incomplete \(c_{00}\) is separable. The disjoint-balls argument is the standard route to non-separability.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Countable candidate set. For \(\mathbb K=\mathbb R\), let \(P_{\mathbb Q}\) be the polynomials with rational coefficients, restricted to \([a,b]\). For \(\mathbb K=\mathbb C\), use coefficients in the Gaussian rationals \(\mathbb Q+i\mathbb Q\). In either case the coefficient field is countable, and the set of finite coefficient strings is countable, so the corresponding polynomial family is countable.
Density for real-valued functions. Let \(f\in C[a,b]\) and \(\varepsilon>0\). By Weierstrass, choose a real polynomial \(p(t)=\sum_{k=0}^d c_kt^k\) with \(\|f-p\|_\infty<\varepsilon/2\). Put \(M=\max(1,|a|,|b|)\). Choose rational \(r_k\) so that \(|c_k-r_k|<\varepsilon/[2(d+1)M^d]\), and set \(r(t)=\sum_{k=0}^d r_kt^k\). Then \[\|p-r\|_\infty\le\sum_{k=0}^d|c_k-r_k|M^k\le\varepsilon/2,\] so \(\|f-r\|_\infty<\varepsilon\).
Complex-valued functions. Write \(f=u+iv\) with real continuous \(u,v\). Approximate \(u\) and \(v\) uniformly by rational-coefficient polynomials \(p,q\) within \(\varepsilon/2\) (or, more sharply, within \(\varepsilon/(2\sqrt2)\)). Then \(p+iq\) has Gaussian-rational coefficients and approximates \(f\) uniformly. Thus \(C[a,b]\) is separable over either scalar field.
Misconception. Weierstrass supplies density of polynomials; countability is obtained only after the coefficients are approximated in a countable dense subfield.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
(\(\Rightarrow\)) \(X\) complete implies \(\overline B\) complete. By Ex 1.7 the closed ball \(\overline B\) is a closed subset of \(X\). A Cauchy sequence in \(\overline B\) is Cauchy in the complete space \(X\), so it converges to some \(x\in X\); since \(\overline B\) is closed, \(x\in\overline B\). Thus \(\overline B\) is complete.
(\(\Leftarrow\)) \(\overline B\) complete implies \(X\) complete. Let \((x_n)\) be Cauchy in \(X\). Cauchy sequences are bounded (Ex 1.10), so there is \(R>0\) with \(\|x_n\|\le R\) for all \(n\). Set \(c=R+1>0\) and \(u_n=x_n/c\). By homogeneity, \(\|u_n\|=\|x_n\|/c\le R/(R+1)<1\), so \(u_n\in\overline B\); and \(\|u_n-u_m\|=\|x_n-x_m\|/c\to0\), so \((u_n)\) is Cauchy in \(\overline B\). By hypothesis \(u_n\to u\) for some \(u\in\overline B\). Then \(x_n=c\,u_n\to c\,u=:x\in X\), using continuity of scalar multiplication (Ex 1.9). Hence \((x_n)\) converges in \(X\), and \(X\) is complete.
Misconception. The rescaling is where homogeneity of the norm is indispensable; the equivalence would not hold for a general metric, only for norm-induced ones, since only there can one shrink an arbitrary bounded sequence linearly into the unit ball.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Since \(Y\) is a closed subspace, the quotient \(X/Y\) is a normed space with quotient norm (Ex 1.23), and the projection \(\pi:X\to X/Y\), \(\pi(x)=x+Y\), is linear and continuous (indeed \(\|\pi(x)\|_{X/Y}\le\|x\|\), so \(\pi\) is \(1\)-Lipschitz).
Identify \(Y+F\) as a preimage. We claim \(Y+F=\pi^{-1}\big(\pi(F)\big)\). If \(x=y+f\) with \(y\in Y,f\in F\), then \(\pi(x)=\pi(y)+\pi(f)=\pi(f)\in\pi(F)\), so \(x\in\pi^{-1}(\pi(F))\). Conversely if \(\pi(x)\in\pi(F)\), then \(\pi(x)=\pi(f)\) for some \(f\in F\), so \(x-f\in\ker\pi=Y\), giving \(x=(x-f)+f\in Y+F\). This proves the claim.
\(\pi(F)\) is closed. \(\pi(F)\) is the image of the finite-dimensional space \(F\) under the linear map \(\pi\), hence a subspace of \(X/Y\) with \(\dim\pi(F)\le\dim F<\infty\). By Ex 1.19, every finite-dimensional subspace of a normed space is closed, so \(\pi(F)\) is closed in \(X/Y\).
Conclusion. As \(\pi\) is continuous and \(\pi(F)\) is closed, \(Y+F=\pi^{-1}(\pi(F))\) is closed in \(X\).
Misconception. The sum of two closed subspaces need not be closed in general; finiteness of \(\dim F\) is essential. It is exactly what makes \(\pi(F)\) finite-dimensional, and thus closed, in the quotient.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
By definition of equivalence there are constants \(0<c\le C\) with \(c\|x\|_b\le\|x\|_a\le C\|x\|_b\) for all \(x\in X\).
Same convergent sequences. \(\|x_n-x\|_a\le C\|x_n-x\|_b\) and \(\|x_n-x\|_b\le\tfrac1c\|x_n-x\|_a\), so \(\|x_n-x\|_a\to0\iff\|x_n-x\|_b\to0\). The two norms therefore have the same convergent sequences and the same limits.
Same Cauchy sequences. Identically, \(\|x_n-x_m\|_a\le C\|x_n-x_m\|_b\) and \(\|x_n-x_m\|_b\le\tfrac1c\|x_n-x_m\|_a\), so \((x_n)\) is \(\|\cdot\|_a\)-Cauchy iff it is \(\|\cdot\|_b\)-Cauchy.
Same open sets. Let \(B_a(x,r)\), \(B_b(x,r)\) denote the respective open balls. From \(\|z-x\|_a\le C\|z-x\|_b\), if \(\|z-x\|_b<r/C\) then \(\|z-x\|_a<r\), so \(B_b(x,r/C)\subseteq B_a(x,r)\). Symmetrically \(B_a(x,cr')\subseteq B_b(x,r')\). Hence every \(\|\cdot\|_a\)-open set is \(\|\cdot\|_b\)-open and conversely: given a \(\|\cdot\|_a\)-open \(U\) and \(x\in U\), some \(B_a(x,r)\subseteq U\), and then \(B_b(x,r/C)\subseteq U\). The two norms induce the same topology.
Same completeness. Suppose \((X,\|\cdot\|_a)\) is complete and let \((x_n)\) be \(\|\cdot\|_b\)-Cauchy. Then it is \(\|\cdot\|_a\)-Cauchy, so \(\|\cdot\|_a\)-converges to some \(x\), and by the equivalence it \(\|\cdot\|_b\)-converges to the same \(x\). Thus \((X,\|\cdot\|_b)\) is complete; by symmetry the converse holds, so completeness is shared.
Misconception. Equivalent norms agree on all metric and topological notions listed here, but not on metric quantities like the numerical value of a distance or the exact shape of the unit ball; "equivalent" means uniformly comparable, not equal.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Construction. Let \(\mathcal C\) be the set of Cauchy sequences \((x_n)\) in \(X\); it is a vector space under coordinatewise operations. Let \(\mathcal N=\{(x_n)\in\mathcal C:\|x_n\|\to0\}\), a linear subspace. Define \(\widehat X=\mathcal C/\mathcal N\) with elements \(\xi=[(x_n)]\).
The norm. For \((x_n)\in\mathcal C\), the reals \(\|x_n\|\) form a Cauchy sequence since \(\big|\|x_n\|-\|x_m\|\big|\le\|x_n-x_m\|\), so \(\lim_n\|x_n\|\) exists. If \((x_n)-(x_n')\in\mathcal N\), then \(\big|\|x_n\|-\|x_n'\|\big|\le\|x_n-x_n'\|\to0\), so the limit depends only on the class. Define \(\|\xi\|:=\lim_n\|x_n\|\). It is nonnegative; \(\|\xi\|=0\) means \(\|x_n\|\to0\), i.e. \((x_n)\in\mathcal N\), i.e. \(\xi=0\); homogeneity and the triangle inequality pass to the limit from those of \(\|\cdot\|\). Thus \(\|\cdot\|\) is a norm on \(\widehat X\).
The isometric embedding. Define \(\iota(x)=[(x,x,x,\dots)]\). It is linear, and \(\|\iota(x)\|=\lim_n\|x\|=\|x\|\), so \(\iota\) is a linear isometry, in particular injective.
Density. Let \(\xi=[(x_n)]\in\widehat X\) and \(\varepsilon>0\). Since \((x_n)\) is Cauchy, choose \(N\) with \(\|x_n-x_m\|<\varepsilon\) for \(m,n\ge N\). Then \(\|\xi-\iota(x_N)\|=\lim_n\|x_n-x_N\|\le\varepsilon\). Hence \(\iota(X)\) is dense in \(\widehat X\).
Completeness. Let \((\xi^{(k)})\) be Cauchy in \(\widehat X\). By density pick \(a_k\in X\) with \(\|\xi^{(k)}-\iota(a_k)\|<1/k\). Then \(\|\iota(a_k)-\iota(a_l)\|\le\|\iota(a_k)-\xi^{(k)}\|+\|\xi^{(k)}-\xi^{(l)}\|+\|\xi^{(l)}-\iota(a_l)\|\), so \((a_k)\) is Cauchy in \(X\) (as \(\iota\) is isometric), giving \(\alpha:=[(a_k)]\in\widehat X\). For fixed \(k\), \(\|\iota(a_k)-\alpha\|=\lim_l\|a_k-a_l\|\), which is small for large \(k\); combined with \(\|\xi^{(k)}-\iota(a_k)\|<1/k\), we get \(\xi^{(k)}\to\alpha\). Thus \(\widehat X\) is complete, i.e. a Banach space.
Uniqueness. Suppose \(\iota_1:X\to Z_1\) and \(\iota_2:X\to Z_2\) are two completions. On the dense subspace \(\iota_1(X)\) define \(T=\iota_2\circ\iota_1^{-1}\), a linear isometry onto the dense subspace \(\iota_2(X)\). A uniformly continuous (here isometric) map defined on a dense subset of a complete space extends uniquely to the whole space, and the extension preserves the norm by continuity; the same applies to its inverse. The extension \(\overline T:Z_1\to Z_2\) is therefore a surjective linear isometry with \(\overline T\circ\iota_1=\iota_2\). Hence any two completions are isometrically isomorphic in a way compatible with the embeddings, so the completion is unique up to isometric isomorphism.
Misconception. The completion adds ideal limit points as equivalence classes of Cauchy sequences, exactly mirroring the construction of \(\mathbb{R}\) from \(\mathbb{Q}\); \(X\) itself is recovered as the dense copy \(\iota(X)\), and nothing of its geometry is lost since \(\iota\) is an isometry.Compare your work line by line with the derivation above. Every requested claim should be justified, each estimate should state the norm or theorem used, and the final sentence should answer the original problem explicitly.
Chapter Synthesis
Concept map
Theorem dependency summary
Thm. 1.1 (norm induces a metric, and the norm and vector operations are continuous) is the bridge that lets every metric notion be used in a normed space. Thm. 1.2 (convergent implies Cauchy) isolates completeness as the one missing converse and thereby defines the Banach property. Thm. 1.3 (equivalence of norms in finite dimensions) explains why dimension is the dividing line of the theory, and rests on compactness of the unit sphere. Thm. 1.4 (completeness via absolutely convergent series) is the workhorse criterion; it is applied directly to prove Thm. 1.5 (\(\ell^p\) is complete) and Thm. 1.6 (\(C[a,b]\) is complete). Together the six results move from the axioms to a stock of concrete Banach spaces.
Notation summary
- \(\|x\|\) - norm of a vector; \(\|\cdot\|_p\) - the \(p\)-norm
- \(d(x,y)=\|x-y\|\) - induced metric
- \(B(x,r),\ \overline B(x,r)\) - open and closed balls
- \(x_n\to x\) - convergence in norm; Cauchy sequence
- \(\ell^p,\ \ell^\infty\) - sequence spaces; \(C[a,b]\) - continuous functions with the sup norm
- \(\|\cdot\|_a\sim\|\cdot\|_b\) - equivalent norms
| English | Français |
|---|---|
| norm | norme |
| normed vector space | espace vectoriel normé |
| Banach space | espace de Banach |
| complete | complet |
| Cauchy sequence | suite de Cauchy |
| equivalent norms | normes équivalentes |
| absolutely convergent series | série absolument convergente |
Frequent misconceptions
- Completeness is a property of the norm, not of the vector space alone: the same space can be complete for one norm and incomplete for another.
- Equivalent norms give the same convergent and Cauchy sequences and the same open sets, but not the same distances; equivalence is not equality.
- In infinite dimensions the closed unit ball is not compact, and norms need not be equivalent; intuition from \(\mathbb R^n\) fails here.
- A Cauchy sequence need not converge unless the space is complete; the limit can lie outside the space, as with \(\mathbb Q\) or with polynomials in the sup norm.
- Absolute convergence of a series does not by itself imply convergence; that implication is exactly the completeness criterion of Thm. 1.4.
Oral examination questions
- State the norm axioms and prove the reverse triangle inequality.
- Show that a norm induces a metric and that the norm map is continuous for that metric.
- Define completeness and give an example of a normed space that is not complete, with justification.
- State and outline the proof that all norms on a finite-dimensional space are equivalent. Where is compactness used?
- State the absolutely-convergent-series criterion for completeness and explain both directions.
- Prove that \(C[a,b]\) with the supremum norm is a Banach space.
Assemble a short, self-contained portfolio proving, in order: (1) a norm induces a translation-invariant metric and the norm is continuous (Thm. 1.1); (2) a normed space is complete if and only if every absolutely convergent series converges (Thm. 1.4); (3) \(C[a,b]\) with the sup norm is complete (Thm. 1.6). Together these move you from the axioms to a working Banach space, and the middle result is the tool you will reuse in every later completeness proof.
Almost every powerful theorem of functional analysis - the uniform boundedness principle, the open mapping theorem, the closed graph theorem, the existence of best approximations in Hilbert space - requires the domain to be complete. Completeness is what lets a construction that produces a Cauchy sequence actually deliver a limit inside the space, turning approximate solutions into exact ones. Banach spaces are the minimal setting in which analysis, as opposed to mere algebra, can proceed; the model spaces of this chapter are the concrete arenas where the later abstract theorems are tested.
Connections to later courses
The sup norm on \(C[a,b]\) and the completeness criterion return in Distribution Theory and in the study of function spaces. The spaces \(\ell^p\) and their completeness are the discrete model for the \(L^p\) spaces of Measure Theory, and their duality drives Fourier Analysis. Norm equivalence in finite dimensions and its failure in infinite dimensions is the doorway to Sobolev spaces, where the choice of norm encodes differentiability.
Readiness self-assessment
If every box is checked, proceed to Chapter 2: Bounded Linear Operators, where the norm structure of this chapter becomes the setting for continuous linear maps and the operator norm.