Inmathematics, more specifically infunctional analysis, aBanach space (/ˈbɑː.nʌx/,Polish pronunciation:[ˈba.nax]) is acompletenormed vector space. Thus, a Banach space is a vector space with ametric that allows the computation ofvector length and distance between vectors and is complete in the sense that aCauchy sequence of vectors always converges to a well-definedlimit that is within the space.

Banach spaces are named after the Polish mathematicianStefan Banach, who introduced this concept and studied it systematically in 1920–1922 along withHans Hahn andEduard Helly.^[1]Maurice René Fréchet was the first to use the term "Banach space" and Banach in turn then coined the term "Fréchet space".^[2]Banach spaces originally grew out of the study offunction spaces byHilbert,Fréchet, andRiesz earlier in the century. Banach spaces play a central role in functional analysis. In other areas ofanalysis, the spaces under study are often Banach spaces.

ABanach space is acompletenormed space${\displaystyle (X,\Vert \cdot \Vert ).}$ A normed space is a pair^{[note 1]}${\displaystyle (X,\Vert \cdot \Vert )}$ consisting of avector space${\displaystyle X}$ over a scalar field${\displaystyle \mathbb{K}}$ (where${\displaystyle \mathbb{K}}$ is commonly${\displaystyle \mathbb{R}}$ or${\displaystyle \mathbb{C}}$) together with a distinguished^{[note 2]}norm${\displaystyle \Vert \cdot \Vert :X\to \mathbb{R}.}$ Like all norms, this norm induces atranslation invariant^{[note 3]}distance function, called thecanonical or(norm) induced metric, defined for all vectors${\displaystyle x,y\in X}$ by^{[note 4]}$${\displaystyle d(x,y):=\Vert y-x\Vert =\Vert x-y\Vert .}$$This makes${\displaystyle X}$ into ametric space${\displaystyle (X,d).}$ A sequence${\displaystyle x_1,x_2,\dots }$ is calledCauchy in${\displaystyle (X,d)}$ or${\displaystyle d}$-Cauchy or${\displaystyle \Vert \cdot \Vert }$-Cauchy if for every real${\displaystyle r>0,}$ there exists some index${\displaystyle N}$ such thatwhenever${\displaystyle m}$ and${\displaystyle n}$ are greater than${\displaystyle N.}$ The normed space${\displaystyle (X,\Vert \cdot \Vert )}$ is called aBanach space and the canonical metric${\displaystyle d}$ is called acomplete metric if${\displaystyle (X,d)}$ is acomplete metric space, which by definition means for everyCauchy sequence${\displaystyle x_1,x_2,\dots }$ in${\displaystyle (X,d),}$ there exists some${\displaystyle x\in X}$ such that$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{x}_n=x\text{ in }(X,d),}$$where because${\displaystyle \Vert x_n-x\Vert =d(x_n,x),}$ this sequence's convergence to${\displaystyle x}$ can equivalently be expressed as$${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\Vert x_n-x\Vert =0\text{ in }\mathbb{R}.}$$

The norm${\displaystyle \Vert \cdot \Vert }$ of a normed space${\displaystyle (X,\Vert \cdot \Vert )}$ is called acomplete norm if${\displaystyle (X,\Vert \cdot \Vert )}$ is a Banach space.

For any normed space${\displaystyle (X,\Vert \cdot \Vert ),}$ there exists anL-semi-inner product${\displaystyle ⟨\cdot ,\cdot ⟩}$ on${\displaystyle X}$ such that$\Vert x\Vert =\sqrt{⟨x,x⟩}$ for all${\displaystyle x\in X.}$ In general, there may be infinitely many L-semi-inner products that satisfy this condition. L-semi-inner products are a generalization ofinner products, which are what fundamentally distinguishHilbert spaces 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.

The vector space structure allows one to relate the behavior of Cauchy sequences to that of convergingseries of vectors. A normed space${\displaystyle X}$ is a Banach space if and only if eachabsolutely convergent series in${\displaystyle X}$ converges to a value that lies within${\displaystyle X,}$^[3] symbolically

The canonical metric${\displaystyle d}$ of a normed space${\displaystyle (X,\Vert \cdot \Vert )}$ induces the usualmetric topology${\displaystyle {\tau }_d}$ on${\displaystyle X,}$ which is referred to as thecanonical ornorm inducedtopology. Every normed space is automatically assumed to carry thisHausdorff topology, unless indicated otherwise. With this topology, every Banach space is aBaire space, although there exist normed spaces that are Baire but not Banach.^[4] The norm${\displaystyle \Vert \cdot \Vert :X\to \mathbb{R}}$ is always acontinuous function with respect to the topology that it induces.

The open and closed balls of radius${\displaystyle r>0}$ centered at a point${\displaystyle x\in X}$ are, respectively, the sets Any such ball is aconvex andbounded subset of${\displaystyle X,}$ but acompact ball/neighborhood exists if and only if${\displaystyle X}$ isfinite-dimensional.In particular, no infinite–dimensional normed space can belocally compact or have theHeine–Borel property. If${\displaystyle x_0}$ is a vector and${\displaystyle s\ne 0}$ is a scalar, then$${\displaystyle x_0+s{B}_r(x)=B_{|s|r}(x_0+sx)\text{ and }{x}_0+s{C}_r(x)=C_{|s|r}(x_0+sx).}$$ Using${\displaystyle s=1}$ shows that the norm-induced topology istranslation invariant, which means that for any${\displaystyle x\in X}$ and${\displaystyle S\subseteq X,}$ the subset${\displaystyle S}$ isopen (respectively,closed) in${\displaystyle X}$ if and only if its translation${\displaystyle x+S:=\{x+s\mid s\in S\}}$ is open (respectively, closed).Consequently, the norm induced topology is completely determined by anyneighbourhood basis at the origin. Some common neighborhood bases at the origin include$${\displaystyle \{B_r(0)\mid r>0\},\{C_r(0)\mid r>0\},\{B_{r_n}(0)\mid n\in \mathbb{N}\},\text{ and }\{C_{r_n}(0)\mid n\in \mathbb{N}\},}$$where${\displaystyle r_1,r_2,\dots }$ can be any sequence of positive real numbers that converges to${\displaystyle 0}$ in${\displaystyle \mathbb{R}}$ (common choices are${\displaystyle r_n:=\frac{1}{n}}$ or${\displaystyle r_n:=1/2^n}$). So, for example, any open subset${\displaystyle U}$ of${\displaystyle X}$ can be written as a union$${\displaystyle U=\bigcup _{x\in I}{B}_{r_x}(x)=\bigcup _{x\in I}x+B_{r_x}(0)=\bigcup _{x\in I}x+r_x{B}_1(0)}$$indexed by some subset${\displaystyle I\subseteq U,}$ where each${\displaystyle r_x}$ may be chosen from the aforementioned sequence${\displaystyle r_1,r_2,\dots .}$ (The open balls can also be replaced with closed balls, although the indexing set${\displaystyle I}$ and radii${\displaystyle r_x}$ may then also need to be replaced). Additionally,${\displaystyle I}$ can always be chosen to becountable if${\displaystyle X}$ is aseparable space, which by definition means that${\displaystyle X}$ contains some countabledense subset.

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–dimensionalHilbert space is linearly isometrically isomorphic to the separable Hilbertsequence space${\displaystyle {\ell }^2(\mathbb{N})}$ with its usual norm${\displaystyle \Vert \cdot {\Vert }_2.}$

TheAnderson–Kadec theorem states that every infinite–dimensional separableFréchet space ishomeomorphic to theproduct space$\prod _{i\in \mathbb{N}}\mathbb{R}$ of countably many copies of${\displaystyle \mathbb{R}}$ (this homeomorphism need not be alinear map).^[5][6] Thus all infinite–dimensional separable Fréchet spaces are homeomorphic to each other (or said differently, their topology is uniqueup to a homeomorphism). Since every Banach space is a Fréchet space, this is also true of all infinite–dimensional separable Banach spaces, including${\displaystyle {\ell }^2(\mathbb{N}).}$ In fact,${\displaystyle {\ell }^2(\mathbb{N})}$ is evenhomeomorphic to its ownunitsphere${\displaystyle \{x\in {\ell }^2(\mathbb{N})\mid \Vert x{\Vert }_2=1\},}$ which stands in sharp contrast to finite–dimensional spaces (theEuclidean plane${\displaystyle {\mathbb{R}}^2}$ is not homeomorphic to theunit circle, for instance).

This pattern inhomeomorphism classes extends to generalizations ofmetrizable (locally Euclidean)topological manifolds known asmetricBanach manifolds, which aremetric spaces that are around every point,locally homeomorphic to some open subset of a given Banach space (metricHilbert manifolds and metricFréchet manifolds are defined similarly).^[6] For example, every open subset${\displaystyle U}$ of a Banach space${\displaystyle X}$ is canonically a metric Banach manifold modeled on${\displaystyle X}$ since theinclusion map${\displaystyle U\to X}$ is anopenlocal homeomorphism. Using Hilbert spacemicrobundles, David Henderson showed^[7] in 1969 that every metric manifold modeled on a separable infinite–dimensional Banach (orFréchet) space can betopologically embedded as anopen subset of${\displaystyle {\ell }^2(\mathbb{N})}$ and, consequently, also admits a uniquesmooth structure making it into a${\displaystyle C^{\mathrm{\infty }}}$Hilbert manifold.

There is a compact subset${\displaystyle S}$ of${\displaystyle {\ell }^2(\mathbb{N})}$ whoseconvex hull${\displaystyle \mathrm{co}(S)}$ isnot closed and thus alsonot compact.^{[note 5][8]} However, like in all Banach spaces, theclosed convex hull${\displaystyle \overline{\mathrm{co}}S}$ of this (and every other) compact subset will be compact.^[9] In a normed space that is not complete then it is in generalnot guaranteed that${\displaystyle \overline{\mathrm{co}}S}$ will be compact whenever${\displaystyle S}$ is; an example^{[note 5]} can even be found in a (non-complete)pre-Hilbert vector subspace of${\displaystyle {\ell }^2(\mathbb{N}).}$

This norm-induced topology also makes${\displaystyle (X,{\tau }_d)}$ into what is known as atopological 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${\displaystyle (X,{\tau }_d)}$ isonly a vector space together with a certain type of topology; that is to say, when considered as a TVS, it isnot associated withany particular norm or metric (both of which are "forgotten"). This Hausdorff TVS${\displaystyle (X,{\tau }_d)}$ is evenlocally convex because the set of all open balls centered at the origin forms aneighbourhood basis at the origin consisting of convexbalanced open sets. This TVS is alsonormable, which by definition refers to any TVS whose topology is induced by some (possibly unknown)norm. Normable TVSsare characterized by being Hausdorff and having aboundedconvex neighborhood of the origin. All Banach spaces arebarrelled spaces, which means that everybarrel is neighborhood of the origin (all closed balls centered at the origin are barrels, for example) and guarantees that theBanach–Steinhaus theorem holds.

Theopen mapping theorem implies that when${\displaystyle {\tau }_1}$ and${\displaystyle {\tau }_2}$ are topologies on${\displaystyle X}$ that make both${\displaystyle (X,{\tau }_1)}$ and${\displaystyle (X,{\tau }_2)}$ intocomplete metrizable TVSes (for example, Banach orFréchet spaces), if one topology isfiner or coarser than the other, then they must be equal (that is, if${\displaystyle {\tau }_1\subseteq {\tau }_2}$ or${\displaystyle {\tau }_2\subseteq {\tau }_1}$ then${\displaystyle {\tau }_1={\tau }_2}$).^[10]So, for example, if${\displaystyle (X,p)}$ and${\displaystyle (X,q)}$ are Banach spaces with topologies${\displaystyle {\tau }_p}$ and${\displaystyle {\tau }_q,}$ 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${\displaystyle p:(X,{\tau }_q)\to \mathbb{R}}$ or${\displaystyle q:(X,{\tau }_p)\to \mathbb{R}}$ is continuous), then their topologies are identical and the norms${\displaystyle p}$ and${\displaystyle q}$ areequivalent.

Two norms,${\displaystyle p}$ and${\displaystyle q,}$ on a vector space${\displaystyle X}$ are said to beequivalent if they induce the same topology;^[11] this happens if and only if there exist real numbers${\displaystyle c,C>0}$ such that$cq(x)\le p(x)\le Cq(x)$ for all${\displaystyle x\in X.}$ If${\displaystyle p}$ and${\displaystyle q}$ are two equivalent norms on a vector space${\displaystyle X}$ then${\displaystyle (X,p)}$ is a Banach space if and only if${\displaystyle (X,q)}$ is a Banach space.See this footnote for an example of a continuous norm on a Banach space that isnot equivalent to that Banach space's given norm.^{[note 6][11]} All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space.^[12]

A metric${\displaystyle D}$ on a vector space${\displaystyle X}$ is induced by a norm on${\displaystyle X}$ if and only if${\displaystyle D}$ istranslation invariant^{[note 3]} andabsolutely homogeneous, which means that${\displaystyle D(sx,sy)=|s|D(x,y)}$ for all scalars${\displaystyle s}$ and all${\displaystyle x,y\in X,}$ in which case the function${\displaystyle \Vert x\Vert :=D(x,0)}$ defines a norm on${\displaystyle X}$ and the canonical metric induced by${\displaystyle \Vert \cdot \Vert }$ is equal to${\displaystyle D.}$

Suppose that${\displaystyle (X,\Vert \cdot \Vert )}$ is a normed space and that${\displaystyle \tau }$ is the norm topology induced on${\displaystyle X.}$ Suppose that${\displaystyle D}$ isanymetric on${\displaystyle X}$ such that the topology that${\displaystyle D}$ induces on${\displaystyle X}$ is equal to${\displaystyle \tau .}$ If${\displaystyle D}$ istranslation invariant^{[note 3]} then${\displaystyle (X,\Vert \cdot \Vert )}$ is a Banach space if and only if${\displaystyle (X,D)}$ is a complete metric space.^[13] If${\displaystyle D}$ isnot translation invariant, then it may be possible for${\displaystyle (X,\Vert \cdot \Vert )}$ to be a Banach space but for${\displaystyle (X,D)}$ tonot be a complete metric space^[14] (see this footnote^{[note 7]} for an example). In contrast, a theorem of Klee,^{[15][16][note 8]} which also applies to allmetrizable topological vector spaces, implies that if there existsany^{[note 9]} complete metric${\displaystyle D}$ on${\displaystyle X}$ that induces the norm topology${\displaystyle \tau }$ on${\displaystyle X,}$ then${\displaystyle (X,\Vert \cdot \Vert )}$ is a Banach space.

AFréchet space is alocally 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 thespace of real sequences${\mathbb{R}}^{\mathbb{N}}=\prod _{i\in \mathbb{N}}\mathbb{R}$ with theproduct topology). However, the topology of every Fréchet space is induced by somecountable family of real-valued (necessarily continuous) maps calledseminorms, which are generalizations ofnorms. It is even possible for a Fréchet space to have a topology that is induced by a countable family ofnorms (such norms would necessarily be continuous)^{[note 10][17]} but to not be a Banach/normable space because its topology can not be defined by anysingle norm. An example of such a space is theFréchet space${\displaystyle C^{\mathrm{\infty }}(K),}$ whose definition can be found in the article onspaces of test functions and distributions.

There is another notion of completeness besides metric completeness and that is the notion of acomplete topological vector space (TVS) or TVS-completeness, which uses the theory ofuniform spaces. Specifically, the notion of TVS-completeness uses a unique translation-invariantuniformity, called thecanonical uniformity, that dependsonly on vector subtraction and the topology${\displaystyle \tau }$ that the vector space is endowed with, and so in particular, this notion of TVS completeness is independent of whatever norm induced the topology${\displaystyle \tau }$ (and even applies to TVSs that arenot 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${\displaystyle (X,\tau )}$ is ametrizable topological vector space (such as any norm induced topology, for example), then${\displaystyle (X,\tau )}$ is a complete TVS if and only if it is asequentially complete TVS, meaning that it is enough to check that every Cauchysequence in${\displaystyle (X,\tau )}$ converges in${\displaystyle (X,\tau )}$ to some point of${\displaystyle X}$ (that is, there is no need to consider the more general notion of arbitrary Cauchynets).

If${\displaystyle (X,\tau )}$ is a topological vector space whose topology is induced bysome (possibly unknown) norm (such spaces are callednormable), then${\displaystyle (X,\tau )}$ is a complete topological vector space if and only if${\displaystyle X}$ may be assigned anorm${\displaystyle \Vert \cdot \Vert }$ that induces on${\displaystyle X}$ the topology${\displaystyle \tau }$ and also makes${\displaystyle (X,\Vert \cdot \Vert )}$ into a Banach space. AHausdorfflocally convex topological vector space${\displaystyle X}$ isnormable if and only if itsstrong dual space${\displaystyle X_b^{\prime }}$ is normable,^[18] in which case${\displaystyle X_b^{\prime }}$ is a Banach space (${\displaystyle X_b^{\prime }}$ denotes thestrong dual space of${\displaystyle X,}$ whose topology is a generalization of thedual norm-induced topology on thecontinuous dual space${\displaystyle X^{\prime }}$; see this footnote^{[note 11]} for more details). If${\displaystyle X}$ is ametrizable locally convex TVS, then${\displaystyle X}$ is normable if and only if${\displaystyle X_b^{\prime }}$ is aFréchet–Urysohn space.^[19] This shows that in the category oflocally convex TVSs, Banach spaces are exactly those complete spaces that are bothmetrizable and have metrizablestrong dual spaces.

Every normed space can beisometrically embedded onto a dense vector subspace of a Banach space, where this Banach space is called acompletion of the normed space. This Hausdorff completion is unique up toisometric isomorphism.

More precisely, for every normed space${\displaystyle X,}$ there exists a Banach space${\displaystyle Y}$ and a mapping${\displaystyle T:X\to Y}$ such that${\displaystyle T}$ is anisometric mapping and${\displaystyle T(X)}$ is dense in${\displaystyle Y.}$ If${\displaystyle Z}$ is another Banach space such that there is an isometric isomorphism from${\displaystyle X}$ onto a dense subset of${\displaystyle Z,}$ then${\displaystyle Z}$ is isometrically isomorphic to${\displaystyle Y.}$The Banach space${\displaystyle Y}$ is the Hausdorffcompletion of the normed space${\displaystyle X.}$ The underlying metric space for${\displaystyle Y}$ is the same as the metric completion of${\displaystyle X,}$ with the vector space operations extended from${\displaystyle X}$ to${\displaystyle Y.}$ The completion of${\displaystyle X}$ is sometimes denoted by${\displaystyle \hat{X}.}$

If${\displaystyle X}$ and${\displaystyle Y}$ are normed spaces over the sameground field${\displaystyle \mathbb{K},}$ the set of allcontinuous${\displaystyle \mathbb{K}}$-linear maps${\displaystyle T:X\to Y}$ is denoted by${\displaystyle B(X,Y).}$ In infinite-dimensional spaces, not all linear maps are continuous. A linear mapping from a normed space${\displaystyle X}$ to another normed space is continuous if and only if it isbounded on the closedunit ball of${\displaystyle X.}$ Thus, the vector space${\displaystyle B(X,Y)}$ can be given theoperator norm$${\displaystyle \Vert T\Vert =sup\{\Vert Tx{\Vert }_Y\mid x\in X,\Vert x{\Vert }_X\le 1\}.}$$

For${\displaystyle Y}$ a Banach space, the space${\displaystyle B(X,Y)}$ is a Banach space with respect to this norm. In categorical contexts, it is sometimes convenient to restrict thefunction space between two Banach spaces to only theshort maps; in that case the space${\displaystyle B(X,Y)}$ reappears as a naturalbifunctor.^[20]

If${\displaystyle X}$ is a Banach space, the space${\displaystyle B(X)=B(X,X)}$ forms a unitalBanach algebra; the multiplication operation is given by the composition of linear maps.

If${\displaystyle X}$ and${\displaystyle Y}$ are normed spaces, they areisomorphic normed spaces if there exists a linear bijection${\displaystyle T:X\to Y}$ such that${\displaystyle T}$ and its inverse${\displaystyle T^{-1}}$ are continuous. If one of the two spaces${\displaystyle X}$ or${\displaystyle Y}$ is complete (orreflexive,separable, etc.) then so is the other space. Two normed spaces${\displaystyle X}$ and${\displaystyle Y}$ areisometrically isomorphic if in addition,${\displaystyle T}$ is anisometry, that is,${\displaystyle \Vert T(x)\Vert =\Vert x\Vert }$ for every${\displaystyle x}$ in${\displaystyle X.}$ TheBanach–Mazur distance${\displaystyle d(X,Y)}$ between two isomorphic but not isometric spaces${\displaystyle X}$ and${\displaystyle Y}$ gives a measure of how much the two spaces${\displaystyle X}$ and${\displaystyle Y}$ differ.

Everycontinuous linear operator is abounded linear operator and if dealing only with normed spaces then the converse is also true. That is, alinear operator between two normed spaces isbounded if and only if it is acontinuous function. So in particular, because the scalar field (which is${\displaystyle \mathbb{R}}$ or${\displaystyle \mathbb{C}}$) is a normed space, alinear functional on a normed space is abounded linear functional if and only if it is acontinuous linear functional. This allows for continuity-related results (like those below) to be applied to Banach spaces. Although boundedness is the same as continuity for linear maps between normed spaces, the term "bounded" is more commonly used when dealing primarily with Banach spaces.

If${\displaystyle f:X\to \mathbb{R}}$ is asubadditive function (such as a norm, asublinear function, or real linear functional), then^[21]${\displaystyle f}$ iscontinuous at the origin if and only if${\displaystyle f}$ isuniformly continuous on all of${\displaystyle X}$; and if in addition${\displaystyle f(0)=0}$ then${\displaystyle f}$ is continuous if and only if itsabsolute value${\displaystyle |f|:X\to [0,\mathrm{\infty })}$ is continuous, which happens if and only if is an open subset of${\displaystyle X.}$^{[21][note 12]} And very importantly for applying theHahn–Banach theorem, a linear functional${\displaystyle f}$ is continuous if and only if this is true of itsreal part${\displaystyle \mathrm{Re}f}$ and moreover,${\displaystyle \Vert \mathrm{Re}f\Vert =\Vert f\Vert }$ andthe real part${\displaystyle \mathrm{Re}f}$ completely determines${\displaystyle f,}$ which is why the Hahn–Banach theorem is often stated only for real linear functionals.Also, a linear functional${\displaystyle f}$ on${\displaystyle X}$ is continuous if and only if theseminorm${\displaystyle |f|}$ is continuous, which happens if and only if there exists a continuous seminorm${\displaystyle p:X\to \mathbb{R}}$ such that${\displaystyle |f|\le p}$; this last statement involving the linear functional${\displaystyle f}$ and seminorm${\displaystyle p}$ is encountered in many versions of the Hahn–Banach theorem.

The Cartesian product${\displaystyle X\times Y}$ of two normed spaces is not canonically equipped with a norm. However, several equivalent norms are commonly used,^[22] such as$${\displaystyle \Vert (x,y){\Vert }_1=\Vert x\Vert +\Vert y\Vert ,\Vert (x,y){\Vert }_{\mathrm{\infty }}=max(\Vert x\Vert ,\Vert y\Vert )}$$which correspond (respectively) to thecoproduct andproduct in the category of Banach spaces and short maps (discussed above).^[20] For finite (co)products, these norms give rise to isomorphic normed spaces, and the product${\displaystyle X\times Y}$ (or the direct sum${\displaystyle X\oplus Y}$) is complete if and only if the two factors are complete.

If${\displaystyle M}$ is aclosedlinear subspace of a normed space${\displaystyle X,}$ there is a natural norm on the quotient space${\displaystyle X/M,}$$${\displaystyle \Vert x+M\Vert =\underset{m\in M}{inf}\Vert x+m\Vert .}$$

The quotient${\displaystyle X/M}$ is a Banach space when${\displaystyle X}$ is complete.^[23] The quotient map from${\displaystyle X}$ onto${\displaystyle X/M,}$ sending${\displaystyle x\in X}$ to its class${\displaystyle x+M,}$ is linear, onto, and of norm${\displaystyle 1,}$ except when${\displaystyle M=X,}$ in which case the quotient is the null space.

The closed linear subspace${\displaystyle M}$ of${\displaystyle X}$ is said to be acomplemented subspace of${\displaystyle X}$ if${\displaystyle M}$ is therange of asurjective bounded linearprojection${\displaystyle P:X\to M.}$ In this case, the space${\displaystyle X}$ is isomorphic to the direct sum of${\displaystyle M}$ and${\displaystyle \ker P,}$ the kernel of the projection${\displaystyle P.}$

Suppose that${\displaystyle X}$ and${\displaystyle Y}$ are Banach spaces and that${\displaystyle T\in B(X,Y).}$ There exists a canonical factorization of${\displaystyle T}$ as^[23]$${\displaystyle T=T_1\circ \pi ,T:X\stackrel{\pi }{⟶}X/\ker T\stackrel{T_1}{⟶}Y}$$where the first map${\displaystyle \pi }$ is the quotient map, and the second map${\displaystyle T_1}$ sends every class${\displaystyle x+\ker T}$ in the quotient to the image${\displaystyle T(x)}$ in${\displaystyle Y.}$ This is well defined because all elements in the same class have the same image. The mapping${\displaystyle T_1}$ is a linear bijection from${\displaystyle X/\ker T}$ onto the range${\displaystyle T(X),}$ whose inverse need not be bounded.

Basic examples^[24] of Banach spaces include: theLp spaces${\displaystyle L^p}$ and their special cases, thesequence spaces${\displaystyle {\ell }^p}$ that consist of scalar sequences indexed bynatural numbers${\displaystyle \mathbb{N}}$; among them, the space${\displaystyle {\ell }^1}$ ofabsolutely summable sequences and the space${\displaystyle {\ell }^2}$ of square summable sequences; the space${\displaystyle c_0}$ of sequences tending to zero and the space${\displaystyle {\ell }^{\mathrm{\infty }}}$ of bounded sequences; the space${\displaystyle C(K)}$ of continuous scalar functions on a compact Hausdorff space${\displaystyle K,}$ equipped with the max norm,$${\displaystyle \Vert f{\Vert }_{C(K)}=max\{|f(x)|\mid x\in K\},f\in C(K).}$$

According to theBanach–Mazur theorem, every Banach space is isometrically isomorphic to a subspace of some${\displaystyle C(K).}$^[25] For every separable Banach space${\displaystyle X,}$ there is a closed subspace${\displaystyle M}$ of${\displaystyle {\ell }^1}$ such that${\displaystyle X:={\ell }^1/M.}$^[26]

AnyHilbert space serves as an example of a Banach space. A Hilbert space${\displaystyle H}$ on${\displaystyle \mathbb{K}=\mathbb{R},\mathbb{C}}$ is complete for a norm of the form$${\displaystyle \Vert x{\Vert }_H=\sqrt{⟨x,x⟩},}$$where$${\displaystyle ⟨\cdot ,\cdot ⟩:H\times H\to \mathbb{K}}$$is theinner product, linear in its first argument that satisfies the following:

For example, the space${\displaystyle L^2}$ is a Hilbert space.

TheHardy spaces, theSobolev spaces are examples of Banach spaces that are related to${\displaystyle L^p}$ spaces and have additional structure. They are important in different branches of analysis,Harmonic analysis andPartial differential equations among others.

ABanach algebra is a Banach space${\displaystyle A}$ over${\displaystyle \mathbb{K}=\mathbb{R}}$ or${\displaystyle \mathbb{C},}$ together with a structure ofalgebra over${\displaystyle \mathbb{K}}$, such that the product map${\displaystyle A\times A\ni (a,b)\mapsto ab\in A}$ is continuous. An equivalent norm on${\displaystyle A}$ can be found so that${\displaystyle \Vert ab\Vert \le \Vert a\Vert \Vert b\Vert }$ for all${\displaystyle a,b\in A.}$

• The Banach space${\displaystyle C(K)}$ with the pointwise product, is a Banach algebra.
• Thedisk algebra${\displaystyle A(\mathbf{D})}$ consists of functionsholomorphic in the open unit disk${\displaystyle D\subseteq \mathbb{C}}$ and continuous on itsclosure:${\displaystyle \overline{\mathbf{D}}.}$ Equipped with the max norm on${\displaystyle \overline{\mathbf{D}},}$ the disk algebra${\displaystyle A(\mathbf{D})}$ is a closed subalgebra of${\displaystyle C\left(\overline{\mathbf{D}}\right).}$
• TheWiener algebra${\displaystyle A(\mathbf{T})}$ is the algebra of functions on the unit circle${\displaystyle \mathbf{T}}$ with absolutely convergent Fourier series. Via the map associating a function on${\displaystyle \mathbf{T}}$ to the sequence of its Fourier coefficients, this algebra is isomorphic to the Banach algebra${\displaystyle {\ell }^1(Z),}$ where the product is theconvolution of sequences.
• For every Banach space${\displaystyle X,}$ the space${\displaystyle B(X)}$ of bounded linear operators on${\displaystyle X,}$ with the composition of maps as product, is a Banach algebra.
• AC*-algebra is a complex Banach algebra${\displaystyle A}$ with anantilinearinvolution${\displaystyle a\mapsto a^{\ast }}$ such that${\displaystyle \Vert a^{\ast }a\Vert =\Vert a{\Vert }^2.}$ The space${\displaystyle B(H)}$ of bounded linear operators on a Hilbert space${\displaystyle H}$ is a fundamental example of C*-algebra. TheGelfand–Naimark theorem states that every C*-algebra is isometrically isomorphic to a C*-subalgebra of some${\displaystyle B(H).}$ The space${\displaystyle C(K)}$ of complex continuous functions on a compact Hausdorff space${\displaystyle K}$ is an example of commutative C*-algebra, where the involution associates to every function${\displaystyle f}$ itscomplex conjugate${\displaystyle \overline{f}.}$

If${\displaystyle X}$ is a normed space and${\displaystyle \mathbb{K}}$ the underlyingfield (either thereals or thecomplex numbers), thecontinuous dual space is the space of continuous linear maps from${\displaystyle X}$ into${\displaystyle \mathbb{K},}$ orcontinuous linear functionals.The notation for the continuous dual is${\displaystyle X^{\prime }=B(X,\mathbb{K})}$ in this article.^[27] Since${\displaystyle \mathbb{K}}$ is a Banach space (using theabsolute value as norm), the dual${\displaystyle X^{\prime }}$ is a Banach space, for every normed space${\displaystyle X.}$ TheDixmier–Ng theorem characterizes the dual spaces of Banach spaces.

The main tool for proving the existence of continuous linear functionals is theHahn–Banach theorem.

In particular, every continuous linear functional on a subspace of a normed space can be continuously extended to the whole space, without increasing the norm of the functional.^[28] An important special case is the following: for every vector${\displaystyle x}$ in a normed space${\displaystyle X,}$ there exists a continuous linear functional${\displaystyle f}$ on${\displaystyle X}$ such that$${\displaystyle f(x)=\Vert x{\Vert }_X,\Vert f{\Vert }_{X^{\prime }}\le 1.}$$

When${\displaystyle x}$ is not equal to the${\displaystyle \mathbf{0}}$ vector, the functional${\displaystyle f}$ must have norm one, and is called anorming functional for${\displaystyle x.}$

TheHahn–Banach separation theorem states that two disjoint non-emptyconvex sets in a real Banach space, one of them open, can be separated by a closedaffinehyperplane. The open convex set lies strictly on one side of the hyperplane, the second convex set lies on the other side but may touch the hyperplane.^[29]

A subset${\displaystyle S}$ in a Banach space${\displaystyle X}$ istotal if thelinear span of${\displaystyle S}$ isdense in${\displaystyle X.}$ The subset${\displaystyle S}$ is total in${\displaystyle X}$ if and only if the only continuous linear functional that vanishes on${\displaystyle S}$ is the${\displaystyle \mathbf{0}}$ functional: this equivalence follows from the Hahn–Banach theorem.

If${\displaystyle X}$ is the direct sum of two closed linear subspaces${\displaystyle M}$ and${\displaystyle N,}$ then the dual${\displaystyle X^{\prime }}$ of${\displaystyle X}$ is isomorphic to the direct sum of the duals of${\displaystyle M}$ and${\displaystyle N.}$^[30] If${\displaystyle M}$ is a closed linear subspace in${\displaystyle X,}$ one can associate theorthogonal of${\displaystyle M}$ in the dual,$${\displaystyle M^{\mathrm{\perp }}=\{x^{\prime }\in X\mid x^{\prime }(m)=0\text{ for all }m\in M\}.}$$

The orthogonal${\displaystyle M^{\mathrm{\perp }}}$ is a closed linear subspace of the dual. The dual of${\displaystyle M}$ is isometrically isomorphic to${\displaystyle X^{\prime }/M^{\mathrm{\perp }}.}$ The dual of${\displaystyle X/M}$ is isometrically isomorphic to${\displaystyle M^{\mathrm{\perp }}.}$^[31]

The dual of a separable Banach space need not be separable, but:

When${\displaystyle X^{\prime }}$ is separable, the above criterion for totality can be used for proving the existence of a countable total subset in${\displaystyle X.}$

Theweak topology on a Banach space${\displaystyle X}$ is thecoarsest topology on${\displaystyle X}$ for which all elements${\displaystyle x^{\prime }}$ in the continuous dual space${\displaystyle X^{\prime }}$ are continuous. The norm topology is thereforefiner than the weak topology. It follows from the Hahn–Banach separation theorem that the weak topology isHausdorff, and that a norm-closedconvex subset of a Banach space is also weakly closed.^[33] A norm-continuous linear map between two Banach spaces${\displaystyle X}$ and${\displaystyle Y}$ is alsoweakly continuous, that is, continuous from the weak topology of${\displaystyle X}$ to that of${\displaystyle Y.}$^[34]

If${\displaystyle X}$ is infinite-dimensional, there exist linear maps which are not continuous. The space${\displaystyle X^{\ast }}$ of all linear maps from${\displaystyle X}$ to the underlying field${\displaystyle \mathbb{K}}$ (this space${\displaystyle X^{\ast }}$ is called thealgebraic dual space, to distinguish it from${\displaystyle X^{\prime }}$ also induces a topology on${\displaystyle X}$ which isfiner than the weak topology, and much less used in functional analysis.

On a dual space${\displaystyle X^{\prime },}$ there is a topology weaker than the weak topology of${\displaystyle X^{\prime },}$ called theweak* topology. It is the coarsest topology on${\displaystyle X^{\prime }}$ for which all evaluation maps${\displaystyle x^{\prime }\in X^{\prime }\mapsto x^{\prime }(x),}$ where${\displaystyle x}$ ranges over${\displaystyle X,}$ are continuous. Its importance comes from theBanach–Alaoglu theorem.

The Banach–Alaoglu theorem can be proved usingTychonoff's theorem about infinite products of compact Hausdorff spaces. When${\displaystyle X}$ is separable, the unit ball${\displaystyle B^{\prime }}$ of the dual is ametrizable compact in the weak* topology.^[35]

The dual of${\displaystyle c_0}$ is isometrically isomorphic to${\displaystyle {\ell }^1}$: for every bounded linear functional${\displaystyle f}$ on${\displaystyle c_0,}$ there is a unique element${\displaystyle y=\{y_n\}\in {\ell }^1}$ such that$${\displaystyle f(x)=\sum _{n\in \mathbb{N}}{x}_n{y}_n,x=\{x_n\}\in c_0,\text{and}\Vert f{\Vert }_{(c_0{)}^{\prime }}=\Vert y{\Vert }_{{\ell }_1}.}$$

The dual of${\displaystyle {\ell }^1}$ is isometrically isomorphic to${\displaystyle {\ell }^{\mathrm{\infty }}}$. The dual ofLebesgue space${\displaystyle L^p([0,1])}$ is isometrically isomorphic to${\displaystyle L^q([0,1])}$ when and${\displaystyle \frac{1}{p}+\frac{1}{q}=1.}$

For every vector${\displaystyle y}$ in a Hilbert space${\displaystyle H,}$ the mapping$${\displaystyle x\in H\to f_y(x)=⟨x,y⟩}$$

defines a continuous linear functional${\displaystyle f_y}$ on${\displaystyle H.}$TheRiesz representation theorem states that every continuous linear functional on${\displaystyle H}$ is of the form${\displaystyle f_y}$ for a uniquely defined vector${\displaystyle y}$ in${\displaystyle H.}$The mapping${\displaystyle y\in H\to f_y}$ is anantilinear isometric bijection from${\displaystyle H}$ onto its dual${\displaystyle H^{\prime }.}$ When the scalars are real, this map is an isometric isomorphism.

When${\displaystyle K}$ is a compact Hausdorff topological space, the dual${\displaystyle M(K)}$ of${\displaystyle C(K)}$ is the space ofRadon measures in the sense of Bourbaki.^[36] The subset${\displaystyle P(K)}$ of${\displaystyle M(K)}$ consisting of non-negative measures of mass 1 (probability measures) is a convex w*-closed subset of the unit ball of${\displaystyle M(K).}$ Theextreme points of${\displaystyle P(K)}$ are theDirac measures on${\displaystyle K.}$ The set of Dirac measures on${\displaystyle K,}$ equipped with the w*-topology, ishomeomorphic to${\displaystyle K.}$

The result has been extended by Amir^[39] and Cambern^[40] to the case when the multiplicativeBanach–Mazur distance between${\displaystyle C(K)}$ and${\displaystyle C(L)}$ is The theorem is no longer true when the distance is${\displaystyle =2.}$^[41]

In the commutativeBanach algebra${\displaystyle C(K),}$ themaximal ideals are precisely kernels of Dirac measures on${\displaystyle K,}$$${\displaystyle I_x=\ker {\delta }_x=\{f\in C(K)\mid f(x)=0\},x\in K.}$$

More generally, by theGelfand–Mazur theorem, the maximal ideals of a unital commutative Banach algebra can be identified with itscharacters—not merely as sets but as topological spaces: the former with thehull-kernel topology and the latter with the w*-topology. In this identification, the maximal ideal space can be viewed as a w*-compact subset of the unit ball in the dual${\displaystyle A^{\prime }.}$

Not every unital commutative Banach algebra is of the form${\displaystyle C(K)}$ for some compact Hausdorff space${\displaystyle K.}$ However, this statement holds if one places${\displaystyle C(K)}$ in the smaller category of commutativeC*-algebras.Gelfand'srepresentation theorem for commutative C*-algebras states that every commutative unitalC*-algebra${\displaystyle A}$ is isometrically isomorphic to a${\displaystyle C(K)}$ space.^[42] The Hausdorff compact space${\displaystyle K}$ here is again the maximal ideal space, also called thespectrum of${\displaystyle A}$ in the C*-algebra context.

If${\displaystyle X}$ is a normed space, the (continuous) dual${\displaystyle X^{″}}$ of the dual${\displaystyle X^{\prime }}$ is called thebidual orsecond dual of${\displaystyle X.}$ For every normed space${\displaystyle X,}$ there is a natural map,

This defines${\displaystyle F_X(x)}$ as a continuous linear functional on${\displaystyle X^{\prime },}$ that is, an element of${\displaystyle X^{″}.}$ The map${\displaystyle F_X:x\to F_X(x)}$ is a linear map from${\displaystyle X}$ to${\displaystyle X^{″}.}$ As a consequence of the existence of anorming functional${\displaystyle f}$ for every${\displaystyle x\in X,}$ this map${\displaystyle F_X}$ is isometric, thusinjective.

For example, the dual of${\displaystyle X=c_0}$ is identified with${\displaystyle {\ell }^1,}$ and the dual of${\displaystyle {\ell }^1}$ is identified with${\displaystyle {\ell }^{\mathrm{\infty }},}$ the space of bounded scalar sequences. Under these identifications,${\displaystyle F_X}$ is the inclusion map from${\displaystyle c_0}$ to${\displaystyle {\ell }^{\mathrm{\infty }}.}$ It is indeed isometric, but not onto.

If${\displaystyle F_X}$ issurjective, then the normed space${\displaystyle X}$ is calledreflexive (seebelow). Being the dual of a normed space, the bidual${\displaystyle X^{″}}$ is complete, therefore, every reflexive normed space is a Banach space.

Using the isometric embedding${\displaystyle F_X,}$ it is customary to consider a normed space${\displaystyle X}$ as a subset of its bidual. When${\displaystyle X}$ is a Banach space, it is viewed as a closed linear subspace of${\displaystyle X^{″}.}$ If${\displaystyle X}$ is not reflexive, the unit ball of${\displaystyle X}$ is a proper subset of the unit ball of${\displaystyle X^{″}.}$ TheGoldstine theorem states that the unit ball of a normed space is weakly*-dense in the unit ball of the bidual. In other words, for every${\displaystyle x^{″}}$ in the bidual, there exists anet${\displaystyle (x_i{)}_{i\in I}}$ in${\displaystyle X}$ so that$${\displaystyle \underset{i\in I}{sup}\Vert x_i\Vert \le \Vert x^{″}\Vert ,x^{″}(f)=\underset{i}{lim}f(x_i),f\in X^{\prime }.}$$

The net may be replaced by a weakly*-convergent sequence when the dual${\displaystyle X^{\prime }}$ is separable. On the other hand, elements of the bidual of${\displaystyle {\ell }^1}$ that are not in${\displaystyle {\ell }^1}$ cannot be weak*-limit ofsequences in${\displaystyle {\ell }^1,}$ since${\displaystyle {\ell }^1}$ isweakly sequentially complete.

Here are the main general results about Banach spaces that go back to the time of Banach's book (Banach (1932)) and are related to theBaire category theorem. According to this theorem, a complete metric space (such as a Banach space, aFréchet space or anF-space) cannot be equal to a union of countably many closed subsets with emptyinteriors. Therefore, a Banach space cannot be the union of countably many closed subspaces, unless it is already equal to one of them; a Banach space with a countableHamel basis is finite-dimensional.