Editing Banach space (section)

===Completeness===

====Complete norms and equivalent norms====

Two norms, <math>p</math> and <math>q,</math> on a vector space <math>X</math> are said to be ''[[Equivalent norms|equivalent]]'' if they induce the same topology;<ref name="Conrad Equiv norms">{{cite web|url=https://kconrad.math.uconn.edu/blurbs/gradnumthy/equivnorms.pdf |archive-url=https://ghostarchive.org/archive/20221009/https://kconrad.math.uconn.edu/blurbs/gradnumthy/equivnorms.pdf |archive-date=2022-10-09 |url-status=live|title=Equivalence of norms|last=Conrad|first=Keith|website=kconrad.math.uconn.edu|access-date=September 7, 2020}}</ref> this happens if and only if there exist real numbers <math>c,C > 0</math> such that <math display=inline>c\,q(x) \leq p(x) \leq C\,q(x)</math> for all <math>x \in X.</math> If <math>p</math> and <math>q</math> are two equivalent norms on a vector space <math>X</math> then <math>(X, p)</math> is a Banach space if and only if <math>(X, q)</math> is a Banach space.
See this footnote for an example of a continuous norm on a Banach space that is {{em|not}} equivalent to that Banach space's given norm.<ref group=note>Let <math>(C([0, 1]), |{\cdot}\|_{\infty})</math> denote the [[Continuous functions on a compact Hausdorff space|Banach space of continuous functions]] with the supremum norm and let <math>\tau_{\infty}</math> denote the topology on <math>C([0, 1])</math> induced by <math>\|{\cdot}\|_{\infty}.</math> The vector space <math>C([0, 1])</math> can be identified (via the [[inclusion map]]) as a proper [[Dense set|dense]] vector subspace <math>X</math> of the [[Lp-space|<math>L^1</math> space]] <math>(L^1([0, 1]), \|{\cdot}\|_1),</math> which satisfies <math>\|f\|_1 \leq \|f\|_{\infty}</math> for all <math>f \in X.</math> Let <math>p</math> denote the restriction of <math>\|{\cdot}\|_1</math> to <math>X,</math> which makes this map <math>p : X \to \R</math> a norm on <math>X</math> (in general, the restriction of any norm to any vector subspace will necessarily again be a norm). The normed space <math>(X, p)</math> is {{em|not}} a Banach space since its completion is the proper superset <math>(L^1([0, 1]), \|{\cdot}\|_1).</math> Because <math>p \leq \|{\cdot}\|_{\infty}</math> holds on <math>X,</math> the map <math>p : (X, \tau_{\infty}) \to \R</math> is continuous. Despite this, the norm <math>p</math> is {{em|not}} equivalent to the norm <math>\|{\cdot}\|_{\infty}</math> (because <math>(X, \|{\cdot}\|_{\infty})</math> is complete but <math>(X, p)</math> is not).</ref><ref name="Conrad Equiv norms"/> 
All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space.<ref>see Corollary&nbsp;1.4.18, p.&nbsp;32 in {{harvtxt|Megginson|1998}}.</ref>

====Complete norms vs complete metrics====

A metric <math>D</math> on a vector space <math>X</math> is induced by a norm on <math>X</math> if and only if <math>D</math> is [[translation invariant]]<ref group=note name="translation invariant metric"/> and ''absolutely homogeneous'', which means that <math>D(sx, sy) = |s| D(x, y)</math> for all scalars <math>s</math> and all <math>x, y \in X,</math> in which case the function <math>\|x\| := D(x, 0)</math> defines a norm on <math>X</math> and the canonical metric induced by <math>\|{\cdot}\|</math> is equal to <math>D.</math>

Suppose that <math>(X, \|{\cdot}\|)</math> is a normed space and that <math>\tau</math> is the norm topology induced on <math>X.</math> Suppose that <math>D</math> is {{em|any}} [[Metric (mathematics)|metric]] on <math>X</math> such that the topology that <math>D</math> induces on <math>X</math> is equal to <math>\tau.</math> If <math>D</math> is [[translation invariant]]<ref group=note name="translation invariant metric"/> then <math>(X, \|{\cdot}\|)</math> is a Banach space if and only if <math>(X, D)</math> is a complete metric space.{{sfn|Narici|Beckenstein|2011|pp=47-66}} 
If <math>D</math> is {{em|not}} translation invariant, then it may be possible for <math>(X, \|{\cdot}\|)</math> to be a Banach space but for <math>(X, D)</math> to {{em|not}} be a complete metric space{{sfn|Narici|Beckenstein|2011|pp=47-51}} (see this footnote<ref group=note>The [[normed space]] <math>(\R,|\cdot |)</math> is a Banach space where the absolute value is a [[Norm (mathematics)|norm]] on the real line <math>\R</math> that induces the usual [[Euclidean topology]] on <math>\R.</math> Define a metric <math>D : \R \times \R \to \R</math> on <math>\R</math> by <math>D(x, y) =|\arctan(x) - \arctan(y)|</math> for all <math>x, y \in \R.</math> Just like {{nowrap|<math>|\cdot|</math>{{hsp}}'s}} induced metric, the metric <math>D</math> also induces the usual Euclidean topology on <math>\R.</math> However, <math>D</math> is not a complete metric because the sequence <math>x_{\bull} = (x_i)_{i=1}^{\infty}</math> defined by <math>x_i := i</math> is a [[Cauchy sequence|{{nowrap|<math>D</math>-Cauchy}} sequence]] but it does not converge to any point of <math>\R.</math> As a consequence of not converging, this {{nowrap|<math>D</math>-Cauchy}} sequence cannot be a Cauchy sequence in <math>(\R,|\cdot |)</math> (that is, it is not a Cauchy sequence with respect to the norm <math>|\cdot|</math>) because if it was {{nowrap|<math>|\cdot|</math>-Cauchy,}} then the fact that <math>(\R,|\cdot |)</math> is a Banach space would imply that it converges (a contradiction).{{harvnb|Narici|Beckenstein|2011|pp=47–51}}</ref> for an example). In contrast, a theorem of Klee,{{sfn|Schaefer|Wolff|1999|p=35}}<ref name="Klee Inv metrics">{{Cite journal|last1=Klee|first1=V. L.|title=Invariant metrics in groups (solution of a problem of Banach)|year=1952|journal=Proc. Amer. Math. Soc.|volume=3|issue=3|pages=484–487|url=https://www.ams.org/journals/proc/1952-003-03/S0002-9939-1952-0047250-4/S0002-9939-1952-0047250-4.pdf |archive-url=https://ghostarchive.org/archive/20221009/https://www.ams.org/journals/proc/1952-003-03/S0002-9939-1952-0047250-4/S0002-9939-1952-0047250-4.pdf |archive-date=2022-10-09 |url-status=live|doi=10.1090/s0002-9939-1952-0047250-4|doi-access=free}}</ref><ref group=note>The statement of the theorem is: Let <math>d</math> be {{em|any}} metric on a vector space <math>X</math> such that the topology <math>\tau</math> induced by <math>d</math> on <math>X</math> makes <math>(X, \tau)</math> into a topological vector space. If <math>(X, d)</math> is a [[complete metric space]] then <math>(X, \tau)</math> is a [[complete topological vector space]].</ref> which also applies to all [[metrizable topological vector space]]s, implies that if there exists {{em|any}}<ref group=note>This metric <math>D</math> is {{em|not}} assumed to be translation-invariant. So in particular, this metric <math>D</math> does {{em|not}} even have to be induced by a norm.</ref> complete metric <math>D</math> on <math>X</math> that induces the norm topology <math>\tau</math> on <math>X,</math> then <math>(X, \|{\cdot}\|)</math> is a Banach space.

A [[Fréchet space]] is a [[locally convex topological vector space]] whose topology is induced by some translation-invariant complete metric. 
Every Banach space is a Fréchet space but not conversely; indeed, there even exist Fréchet spaces on which no norm is a continuous function (such as the [[space of real sequences]] <math display=inline>\R^{\N} = \prod_{i \in \N} \R</math> with the [[product topology]]). 
However, the topology of every Fréchet space is induced by some [[Countable set|countable]] family of real-valued (necessarily continuous) maps called [[seminorm]]s, which are generalizations of [[Norm (mathematics)|norm]]s. 
It is even possible for a Fréchet space to have a topology that is induced by a countable family of {{em|norms}} (such norms would necessarily be continuous)<ref group=note name=CharacterizationOfContinuityOfANorm>A norm (or [[seminorm]]) <math>p</math> on a topological vector space <math>(X, \tau)</math> is continuous if and only if the topology <math>\tau_p</math> that <math>p</math> induces on <math>X</math> is [[Comparison of topologies|coarser]] than <math>\tau</math> (meaning, <math>\tau_p \subseteq \tau</math>), which happens if and only if there exists some open ball <math>B</math> in <math>(X, p)</math> (such as maybe <math>\{x \in X \mid p(x) < 1\}</math> for example) that is open in <math>(X, \tau).</math></ref>{{sfn|Trèves|2006|pp=57–69}} 
but to not be a Banach/[[normable space]] because its topology can not be defined by any {{em|single}} norm. 
An example of such a space is the [[Fréchet space]] <math>C^{\infty}(K),</math> whose definition can be found in the article on [[spaces of test functions and distributions]].

====Complete norms vs complete topological vector spaces====

There is another notion of completeness besides metric completeness and that is the notion of a [[complete topological vector space]] (TVS) or TVS-completeness, which uses the theory of [[uniform space]]s. 
Specifically, the notion of TVS-completeness uses a unique translation-invariant [[Uniformity (topology)|uniformity]], called the [[Complete topological vector space#Canonical uniformity|canonical uniformity]], that depends {{em|only}} on vector subtraction and the topology <math>\tau</math> that the vector space is endowed with, and so in particular, this notion of TVS completeness is independent of whatever norm induced the topology <math>\tau</math> (and even applies to TVSs that are {{em|not}} even metrizable). 
Every Banach space is a complete TVS. Moreover, a normed space is a Banach space (that is, its norm-induced metric is complete) if and only if it is complete as a topological vector space. 
If <math>(X, \tau)</math> is a [[metrizable topological vector space]] (such as any norm induced topology, for example), then <math>(X, \tau)</math> is a complete TVS if and only if it is a {{em|sequentially}} complete TVS, meaning that it is enough to check that every Cauchy {{em|sequence}} in <math>(X, \tau)</math> converges in <math>(X, \tau)</math> to some point of <math>X</math> (that is, there is no need to consider the more general notion of arbitrary Cauchy [[Net (mathematics)|nets]]).

If <math>(X, \tau)</math> is a topological vector space whose topology is induced by {{em|some}} (possibly unknown) norm (such spaces are called {{em|[[Normable space|normable]]}}), then <math>(X, \tau)</math> is a complete topological vector space if and only if <math>X</math> may be assigned a [[Norm (mathematics)|norm]] <math>\|{\cdot}\|</math> that induces on <math>X</math> the topology <math>\tau</math> and also makes <math>(X, \|{\cdot}\|)</math> into a Banach space. 
A [[Hausdorff space|Hausdorff]] [[locally convex topological vector space]] <math>X</math> is [[Normable space|normable]] if and only if its [[strong dual space]] <math>X'_b</math> is normable,{{sfn|Trèves|2006|p=201}} in which case <math>X'_b</math> is a Banach space (<math>X'_b</math> denotes the [[strong dual space]] of <math>X,</math> whose topology is a generalization of the [[dual norm]]-induced topology on the [[continuous dual space]] <math>X'</math>; see this footnote<ref group=note><math>X'</math> denotes the [[continuous dual space]] of <math>X.</math> When <math>X'</math> is endowed with the [[Strong topology (polar topology)|strong dual space topology]], also called the [[topology of uniform convergence]] on [[Bounded set (functional analysis)|bounded subsets]] of <math>X,</math> then this is indicated by writing <math>X'_b</math> (sometimes, the subscript <math>\beta</math> is used instead of <math>b</math>). When <math>X</math> is a normed space with norm <math>\|{\cdot}\|</math> then this topology is equal to the topology on <math>X'</math> induced by the [[dual norm]]. In this way, the [[Strong topology (polar topology)|strong topology]] is a generalization of the usual dual norm-induced topology on <math>X'.</math></ref> for more details). 
If <math>X</math> is a [[Metrizable topological vector space|metrizable]] locally convex TVS, then <math>X</math> is normable if and only if <math>X'_b</math> is a [[Fréchet–Urysohn space]].<ref name="Gabriyelyan 2014">Gabriyelyan, S.S. [https://arxiv.org/pdf/1412.1497.pdf "On topological spaces and topological groups with certain local countable networks] (2014)</ref> 
This shows that in the category of [[Locally convex topological vector space|locally convex TVSs]], Banach spaces are exactly those complete spaces that are both [[Metrizable topological vector space|metrizable]] and have metrizable [[strong dual space]]s.

====Completions====

Every normed space can be [[isometry|isometrically]] embedded onto a dense vector subspace of a Banach space, where this Banach space is called a ''[[Completion (metric space)|completion]]'' of the normed space. This Hausdorff completion is unique up to [[Isometry|isometric]] isomorphism.

More precisely, for every normed space <math>X,</math> there exists a Banach space <math>Y</math> and a mapping <math>T : X \to Y</math> such that <math>T</math> is an [[Isometry|isometric mapping]] and <math>T(X)</math> is dense in <math>Y.</math> If <math>Z</math> is another Banach space such that there is an isometric isomorphism from <math>X</math> onto a dense subset of <math>Z,</math> then <math>Z</math> is isometrically isomorphic to <math>Y.</math>
The Banach space <math>Y</math> is the Hausdorff ''[[Complete metric space#Completion|completion]]'' of the normed space <math>X.</math> The underlying metric space for <math>Y</math> is the same as the metric completion of <math>X,</math> with the vector space operations extended from <math>X</math> to <math>Y.</math> The completion of <math>X</math> is sometimes denoted by <math>\widehat{X}.</math>