Editing Banach space (section)

==Definition==

A '''Banach space''' is a [[Complete metric space|complete]] [[normed space]] <math>(X, \|{\cdot}\|).</math> 
A normed space is a pair<ref group=note>It is common to read {{nowrap|"<math>X</math> is a normed space"}} instead of the more technically correct but (usually) pedantic {{nowrap|"<math>(X, \|{\cdot}\|)</math> is a normed space",}} especially if the norm is well known (for example, such as with [[Lp space|<math>\mathcal{L}^p</math> spaces]]) or when there is no particular need to choose any one (equivalent) norm over any other (especially in the more abstract theory of [[topological vector space]]s), in which case this norm (if needed) is often automatically assumed to be denoted by <math>\|{\cdot}\|.</math> However, in situations where emphasis is placed on the norm, it is common to see <math>(X, \|{\cdot}\|)</math> written instead of <math>X.</math> The technically correct definition of normed spaces as pairs <math>(X, \|{\cdot}\|)</math> may also become important in the context of [[category theory]] where the distinction between the categories of normed spaces, [[normable space]]s, [[metric space]]s, [[topological vector space|TVS]]s, [[topological space]]s, etc. is usually important.</ref> 
<math>(X, \|{\cdot}\|)</math> consisting of a [[vector space]] <math>X</math> over a scalar field <math>\mathbb{K}</math> (where <math>\mathbb{K}</math> is commonly <math>\Reals</math> or <math>\Complex</math>) together with a distinguished<ref group=note>This means that if the norm <math>\|{\cdot}\|</math> is replaced with a different norm <math>\|{\cdot}\|'</math> on <math>X,</math> then <math>(X, \|{\cdot}\|)</math> is {{em|not}} the same normed space as <math>(X, \|{\cdot}\|'),</math> not even if the norms are equivalent. However, equivalence of norms on a given vector space does form an [[equivalence relation]].</ref> 
[[Norm (mathematics)|norm]] <math>\|{\cdot}\| : X \to \Reals.</math> Like all norms, this norm induces a [[translation invariant]]<ref group=note name="translation invariant metric">A metric <math>D</math> on a vector space <math>X</math> is said to be ''translation invariant'' if <math>D(x, y) = D(x + z, y + z)</math> for all vectors <math>x, y, z \in X.</math> This happens if and only if <math>D(x, y) = D(x - y, 0)</math> for all vectors <math>x, y \in X.</math> A metric that is induced by a norm is always translation invariant.</ref> 
[[Metric (mathematics)|distance function]], called the ''canonical'' or [[Norm induced metric|''(norm) induced metric'']], defined for all vectors <math>x, y \in X</math> by<ref group=note>Because <math>\|{-z}\| = \|z\|</math> for all <math>z \in X,</math> it is always true that <math>d(x, y) := \|y - x\| = \|x - y\|</math> for all <math>x, y \in X.</math> So the order of <math>x</math> and <math>y</math> in this definition does not matter.</ref>
<math display=block>d(x, y) := \|y - x\| = \|x - y\|.</math>
This makes <math>X</math> into a [[metric space]] <math>(X, d).</math> 
A sequence <math>x_1, x_2, \ldots</math> is called {{nobr|''Cauchy in <math>(X, d)</math>''}} or {{nowrap|''[[Cauchy sequence|<math>d</math>-Cauchy]]''}} or {{nowrap|''<math>\|{\cdot}\|</math>-Cauchy''}} if for every real <math>r > 0,</math> there exists some index <math>N</math> such that
<math display=block>d(x_n, x_m) = \|x_n - x_m\| < r</math>
whenever <math>m</math> and <math>n</math> are greater than <math>N.</math> 
The normed space <math>(X, \|{\cdot}\|)</math> is called a '''Banach space''' and the canonical metric <math>d</math> is called a ''complete metric'' if <math>(X, d)</math> is a [[complete metric space]], which by definition means for every [[Cauchy sequence]] <math>x_1, x_2, \ldots</math> in <math>(X, d),</math> there exists some <math>x \in X</math> such that
<math display=block>\lim_{n \to \infty} x_n = x \; \text{ in } (X, d),</math>
where because <math>\|x_n - x\| = d(x_n, x),</math> this sequence's convergence to <math>x</math> can equivalently be expressed as
<math display=block>\lim_{n \to \infty} \|x_n - x\| = 0 \; \text{ in } \Reals.</math>

The norm <math>\|{\cdot}\|</math> of a normed space <math>(X, \|{\cdot}\|)</math> is called a '''{{visible anchor|complete norm|Complete norm}}''' if <math>(X, \|{\cdot}\|)</math> is a Banach space.

=== L-semi-inner product ===

For any normed space <math>(X, \|{\cdot}\|),</math> there exists an [[L-semi-inner product]] <math>\langle\cdot, \cdot\rangle</math> on <math>X</math> such that <math display=inline>\|x\| = \sqrt{\langle x, x \rangle}</math> for all <math>x \in X.</math><ref name="Lumer_1961">{{cite journal |last1=Lumer |first1=G. |title=Semi-inner-product spaces |journal=Transactions of the American Mathematical Society |date=1961 |volume=100 |issue=1 |pages=29–43 |doi=10.1090/S0002-9947-1961-0133024-2|doi-access=free}}</ref> In general, there may be infinitely many L-semi-inner products that satisfy this condition and the proof of the existence of L-semi-inner products relies on the non-constructive [[Hahn–Banach theorem]]<ref name="Lumer_1961" />. L-semi-inner products are a generalization of [[inner product]]s, which are what fundamentally distinguish [[Hilbert space]]s from all other Banach spaces. This shows that all normed spaces (and hence all Banach spaces) can be considered as being generalizations of (pre-)Hilbert spaces.

=== Characterization in terms of series ===

The vector space structure allows one to relate the behavior of Cauchy sequences to that of converging [[Series (mathematics)#Generalizations|series of vectors]]. 
A normed space <math>X</math> is a Banach space if and only if each [[absolute convergence|absolutely convergent]] series in <math>X</math> converges to a value that lies within <math>X,</math><ref>see Theorem&nbsp;1.3.9, p.&nbsp;20 in {{harvtxt|Megginson|1998}}.</ref> symbolically
<math display=block>\sum_{n=1}^{\infty} \|v_n\| < \infty \implies \sum_{n=1}^{\infty} v_n\text{ converges in } X.</math>

===Topology===

The canonical metric <math>d</math> of a normed space <math>(X, \|{\cdot}\|)</math> induces the usual [[metric topology]] <math>\tau_d</math> on <math>X,</math> which is referred to as the ''canonical'' or ''norm induced [[topology]]''. 
Every normed space is automatically assumed to carry this [[Hausdorff space|Hausdorff]] topology, unless indicated otherwise. 
With this topology, every Banach space is a [[Baire space]], although there exist normed spaces that are Baire but not Banach.{{sfn|Wilansky|2013|p=29}} The norm <math>\|{\cdot}\| : X \to \Reals</math> is always a [[continuous function]] with respect to the topology that it induces.

The open and closed balls of radius <math>r > 0</math> centered at a point <math>x \in X</math> are, respectively, the sets 
<math display=block>B_r(x) := \{z \in X \mid \|z - x\| < r\} \qquad \text{ and } \qquad C_r(x) := \{z \in X \mid \|z - x\| \leq r\}.</math> 
Any such ball is a [[Convex set|convex]] and [[Bounded set (topological vector space)|bounded subset]] of <math>X,</math> but a [[Compact space|compact]] ball/[[Neighbourhood (topology)|neighborhood]] exists if and only if <math>X</math> is [[finite-dimensional]].
In particular, no infinite–dimensional normed space can be [[Locally compact space|locally compact]] or have the [[Montel space|Heine–Borel property]]. 
If <math>x_0</math> is a vector and <math>s \neq 0</math> is a scalar, then 
<math display=block>x_0 + s\,B_r(x) = B_{|s| r}(x_0 + s x) \qquad \text{ and } \qquad x_0 + s\,C_r(x) = C_{|s| r}(x_0 + s x).</math> 
Using <math>s = 1</math> shows that the norm-induced topology is [[Translation invariant topology|translation invariant]], which means that for any <math>x \in X</math> and <math>S \subseteq X,</math> the subset <math>S</math> is [[Open set|open]] (respectively, [[Closed set|closed]]) in <math>X</math> if and only if its translation <math>x + S := \{x + s \mid s \in S\}</math> is open (respectively, closed).
Consequently, the norm induced topology is completely determined by any [[Neighbourhood system|neighbourhood basis]] at the origin. Some common neighborhood bases at the origin include
<math display=block>\{B_r(0) \mid r > 0\}, \qquad \{C_r(0) \mid r > 0\}, \qquad \{B_{r_n}(0) \mid n \in \N\}, \qquad \text{ and }  \qquad \{C_{r_n}(0) \mid n \in \N\},</math>
where <math>r_1, r_2, \ldots</math> can be any sequence of positive real numbers that converges to <math>0</math> in <math>\R</math> (common choices are <math>r_n := \tfrac{1}{n}</math> or <math>r_n := 1/2^n</math>). 
So, for example, any open subset <math>U</math> of <math>X</math> can be written as a union 
<math display=block>U = \bigcup_{x \in I} B_{r_x}(x) = \bigcup_{x \in I} x + B_{r_x}(0) = \bigcup_{x \in I} x + r_x\,B_1(0)</math>
indexed by some subset <math>I \subseteq U,</math> where each <math>r_x</math> may be chosen from the aforementioned sequence <math>r_1, r_2, \ldots.</math> (The open balls can also be replaced with closed balls, although the indexing set <math>I</math> and radii <math>r_x</math> may then also need to be replaced). 
Additionally, <math>I</math> can always be chosen to be [[Countable set|countable]] if <math>X</math> is a {{em|[[separable space]]}}, which by definition means that <math>X</math> contains some countable [[Dense set|dense subset]]. 

====Homeomorphism classes of separable Banach spaces====

All finite–dimensional normed spaces are separable Banach spaces and any two Banach spaces of the same finite dimension are linearly homeomorphic. 
Every separable infinite–dimensional [[Hilbert space]] is linearly isometrically isomorphic to the separable Hilbert [[ℓ2 space|sequence space <math>\ell^2(\N)</math>]] with its usual norm <math>\|{\cdot}\|_2.</math> 

The [[Anderson–Kadec theorem]] states that every infinite–dimensional separable [[Fréchet space]] is [[Homeomorphism|homeomorphic]] to the [[product space]] <math display=inline>\prod_{i \in \N} \Reals</math> of countably many copies of <math>\Reals</math> (this homeomorphism need not be a [[linear map]]).<ref>{{harvnb|Bessaga|Pełczyński|1975|p=189}}</ref>{{sfn|Anderson|Schori|1969|p=315}} 
Thus all infinite–dimensional separable Fréchet spaces are homeomorphic to each other (or said differently, their topology is unique [[up to]] a homeomorphism). <!-- and so as with finite–dimensional spaces, any two separable Fréchet spaces (of any dimensions) are homeomorphic if and only if they have the same dimension.<ref group=note>This means that their dimensions are either both finite and equal or else both infinite.</ref>-->
Since every Banach space is a Fréchet space, this is also true of all infinite–dimensional separable Banach spaces, including <math>\ell^2(\N).</math> 
In fact, <math>\ell^2(\N)</math> is even [[Homeomorphism|homeomorphic]] to its own [[Unit sphere|unit {{em|sphere}}]] <math>\{x \in \ell^2(\N) \mid \|x\|_2 = 1\},</math> which stands in sharp contrast to finite–dimensional spaces (the [[Euclidean plane]] <math>\Reals^2</math> is not homeomorphic to the [[unit circle]], for instance). 

This pattern in [[homeomorphism class]]es extends to generalizations of [[Metrizable topological space|metrizable]] ([[locally Euclidean]]) [[topological manifold]]s known as {{em|metric [[Banach manifold]]s}}, which are [[metric space]]s that are around every point, [[locally homeomorphic]] to some open subset of a given Banach space (metric [[Hilbert manifold]]s and metric [[Fréchet manifold]]s are defined similarly).{{sfn|Anderson|Schori|1969|p=315}} 
For example, every open subset <math>U</math> of a Banach space <math>X</math> is canonically a metric Banach manifold modeled on <math>X</math> since the [[inclusion map]] <math>U \to X</math> is an [[Open map|open]] [[local homeomorphism]]. 
Using Hilbert space [[microbundle]]s, David Henderson showed{{sfn|Henderson|1969|p=}} in 1969 that every metric manifold modeled on a separable infinite–dimensional Banach (or [[Fréchet space|Fréchet]]) space can be [[Topological embedding|topologically embedded]] as an [[Open set|{{em|open}} subset]] of <math>\ell^2(\N)</math> and, consequently, also admits a unique [[smooth structure]] making it into a <math>C^\infty</math> [[Hilbert manifold]]. 

====Compact and convex subsets====

There is a compact subset <math>S</math> of <math>\ell^2(\N)</math> whose [[convex hull]] <math>\operatorname{co}(S)</math> is {{em|not}} closed and thus also {{em|not}} compact.<ref group=note name=ExampleCompactButHullIsNotCompact>Let <math>H</math> be the separable [[Hilbert space]] [[ℓ2 space|<math>\ell^2(\N)</math>]] of square-summable sequences with the usual norm <math>\|{\cdot}\|_2,</math> and let <math>e_n = (0, \ldots, 0, 1, 0, \ldots, 0)</math> be the standard [[orthonormal basis]] (that is, each <math>e_n</math> has zeros in every position except for a <math>1</math> in the <math>n</math><sup>th</sup>-position). The closed set <math>S = \{0\} \cup \{\tfrac{1}{n} e_n \mid n = 1, 2, \ldots\}</math> is compact (because it is [[Sequentially compact space|sequentially compact]]) but its convex hull <math>\operatorname{co} S</math> is {{em|not}} a closed set because the point <math display=inline>h := \sum_{n=1}^{\infty} \tfrac{1}{2^n} \tfrac{1}{n} e_n</math> belongs to the closure of <math>\operatorname{co} S</math> in <math>H</math> but <math>h \not\in\operatorname{co} S</math> (since every point <math>z=(z_1,z_2,\ldots) \in \operatorname{co} S</math> is a finite [[convex combination]] of elements of <math>S</math> and so <math>z_n = 0</math> for all but finitely many coordinates, which is not true of <math>h</math>). However, like in all [[Complete topological vector space|complete]] Hausdorff locally convex spaces, the {{em|closed}} convex hull <math>K := \overline{\operatorname{co}} S</math> of this compact subset is compact. The vector subspace <math>X := \operatorname{span} S = \operatorname{span} \{e_1, e_2, \ldots\}</math> is a [[pre-Hilbert space]] when endowed with the substructure that the Hilbert space <math>H</math> induces on it, but <math>X</math> is not complete and <math>h \not\in C := K \cap X</math> (since <math>h \not\in X</math>). The closed convex hull of <math>S</math> in <math>X</math> (here, "closed" means with respect to <math>X,</math> and not to <math>H</math> as before) is equal to <math>K \cap X,</math> which is not compact (because it is not a complete subset). This shows that in a Hausdorff locally convex space that is not complete, the closed convex hull of a compact subset might {{em|fail}} to be compact (although it will be [[Totally bounded space|precompact/totally bounded]]).</ref>{{sfn|Aliprantis|Border|2006|p=185}} 
However, like in all Banach spaces, the [[Closed convex hull|{{em|closed}} convex hull]] <math>\overline{\operatorname{co}} S</math> of this (and every other) compact subset will be compact.{{sfn|Trèves|2006|p=145}} In a normed space that is not complete then it is in general {{em|not}} guaranteed that <math>\overline{\operatorname{co}} S</math> will be compact whenever <math>S</math> is; an example<ref group=note name=ExampleCompactButHullIsNotCompact /> can even be found in a (non-complete) [[pre-Hilbert space|pre-Hilbert]] vector subspace of <math>\ell^2(\N).</math>

====As a topological vector space====

This norm-induced topology also makes <math>(X, \tau_d)</math> into what is known as a [[topological vector space]] (TVS), which by definition is a vector space endowed with a topology making the operations of addition and scalar multiplication continuous. It is emphasized that the TVS <math>(X, \tau_d)</math> is {{em|only}} a vector space together with a certain type of topology; that is to say, when considered as a TVS, it is {{em|not}} associated with {{em|any}} particular norm or metric (both of which are "[[Forgetful functor|forgotten]]"). This Hausdorff TVS <math>(X, \tau_d)</math> is even [[Locally convex topological vector space|locally convex]] because the set of all open balls centered at the origin forms a [[neighbourhood basis]] at the origin consisting of convex [[Balanced set|balanced]] open sets. This TVS is also {{em|[[Normable space|normable]]}}, which by definition refers to any TVS whose topology is induced by some (possibly unknown) [[Norm (mathematics)|norm]]. Normable TVSs [[Kolmogorov's normability criterion|are characterized by]] being Hausdorff and having a [[Bounded set (topological vector space)|bounded]] [[Convex set|convex]] neighborhood of the origin. 
All Banach spaces are [[barrelled space]]s, which means that every [[Barrelled set|barrel]] is neighborhood of the origin (all closed balls centered at the origin are barrels, for example) and guarantees that the [[Uniform boundedness principle|Banach–Steinhaus theorem]] holds. 

====Comparison of complete metrizable vector topologies====

The [[Open mapping theorem (functional analysis)|open mapping theorem]] implies that when <math>\tau_1</math> and <math>\tau_2</math> are topologies on <math>X</math> that make both <math>(X, \tau_1)</math> and <math>(X, \tau_2)</math> into [[F-space|complete metrizable TVS]]es (for example, Banach or [[Fréchet space]]s), if one topology is [[Comparison of topologies|finer or coarser]] than the other, then they must be equal (that is, if <math>\tau_1 \subseteq \tau_2</math> or <math>\tau_2 \subseteq \tau_1</math> then <math>\tau_1 = \tau_2</math>).{{sfn|Trèves|2006|pp=166–173}}
So, for example, if <math>(X, p)</math> and <math>(X, q)</math> are Banach spaces with topologies <math>\tau_p</math> and <math>\tau_q,</math> and if one of these spaces has some open ball that is also an open subset of the other space (or, equivalently, if one of <math>p : (X, \tau_q) \to \Reals</math> or <math>q : (X, \tau_p) \to \Reals</math> is continuous), then their topologies are identical and the norms <math>p</math> and <math>q</math> are [[Equivalent norm|equivalent]].

===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>