Banach space

The canonical metric ${\displaystyle d}$ of a normed space ${\displaystyle (X,\|\cdot \|)}$ induces the usual metric topology ${\displaystyle \tau _{d}}$ on ${\displaystyle X,}$ which is referred to as the canonical or norm induced topology. Every normed space is automatically assumed to carry this 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.[4] The norm ${\displaystyle \|\,\cdot \,\|:\left(X,\tau _{d}\right)\to \mathbb {R} }$ is always a continuous 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

$${\displaystyle B_{r}(x):=\{z\in X:\|z-x\|<r\}\qquad {\text{ and }}\qquad C_{r}(x):=\{z\in X:\|z-x\|\leq r\}.}$$

Any such ball is a convex and bounded subset of ${\displaystyle X,}$ but a compact ball/neighborhood exists if and only if ${\displaystyle X}$ is a finite-dimensional vector space. In particular, no infinite–dimensional normed space can be locally compact or have the Heine–Borel property. If ${\displaystyle x_{0}}$ is a vector and ${\displaystyle s\neq 0}$ is a scalar then

$${\displaystyle x_{0}+sB_{r}(x)=B_{|s|r}\left(x_{0}+sx\right)\qquad {\text{ and }}\qquad x_{0}+sC_{r}(x)=C_{|s|r}\left(x_{0}+sx\right).}$$

Using ${\displaystyle s:=1}$ shows that this norm-induced topology is translation invariant, which means that for any ${\displaystyle x\in X}$ and ${\displaystyle S\subseteq X,}$ the subset ${\displaystyle S}$ is open (respectively, closed) in ${\displaystyle X}$ if and only if this is true of its translation ${\displaystyle x+S:=\{x+s:s\in S\}.}$ Consequently, the norm induced topology is completely determined by any neighbourhood basis at the origin. Some common neighborhood bases at the origin include:

$${\displaystyle \left\{B_{r}(0):r>0\right\},\qquad \left\{C_{r}(0):r>0\right\},\qquad \left\{B_{r_{n}}(0):n\in \mathbb {N} \right\},\qquad {\text{ or }}\qquad \left\{C_{r_{n}}(0):n\in \mathbb {N} \right\}}$$

where ${\displaystyle \left(r_{n}\right)_{n=1}^{\infty }}$ is a sequence in of positive real numbers that converges to ${\displaystyle 0}$ in ${\displaystyle \mathbb {R} }$ (such as ${\displaystyle r_{n}:=1/n}$ or ${\displaystyle r_{n}:=1/2^{n}}$ for instance). So for example, every 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 every ${\displaystyle r_{x}}$ is of the form ${\displaystyle r_{x}={\tfrac {1}{n_{x}}}}$ for some integer ${\displaystyle n_{x}>0}$ (the closed ball can also be used instead of the open ball, although the indexing set ${\displaystyle I}$ and radii ${\displaystyle r_{x}}$ may need to be changed). Additionally, ${\displaystyle I}$ can always be chosen to be countable if ${\displaystyle X}$ is a separable space, which by definition means that ${\displaystyle X}$ contains some countable dense subset. The Anderson–Kadec theorem states that every infinite–dimensional separable Fréchet space is homeomorphic to the product space ${\textstyle \prod _{i\in \mathbb {N} }\mathbb {R} }$ of countably many copies of ${\displaystyle \mathbb {R} }$ (this homeomorphism need not be a linear map).[5] Since every Banach space is a Fréchet space, this is also true of all infinite–dimensional separable Banach spaces, including the separable Hilbert ${\displaystyle \ell }$² sequence space ${\displaystyle \ell ^{2}(\mathbb {N} )}$ with its usual norm ${\displaystyle \|\cdot \|_{2},}$ where (in sharp contrast to finite−dimensional spaces) ${\displaystyle \ell ^{2}(\mathbb {N} )}$ is also homeomorphic to its unit sphere ${\displaystyle \left\{x\in \ell ^{2}(\mathbb {N} ):\|x\|_{2}=1\right\}.}$

There is a compact subset ${\displaystyle S}$ of ${\displaystyle \ell ^{2}(\mathbb {N} )}$ whose convex hull ${\displaystyle \operatorname {co} (S)}$ is not closed and thus also not compact (see this footnote[note 5] for an example).[6] However, like in all Banach spaces, the closed convex hull ${\displaystyle {\overline {\operatorname {co} }}S}$ of this (and every other) compact subset will be compact.[7] But if a normed space is not complete then it is in general not guaranteed that ${\displaystyle {\overline {\operatorname {co} }}S}$ 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 ${\displaystyle \left(X,\tau _{d}\right)}$ 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 ${\displaystyle \left(X,\tau _{d}\right)}$ is only a vector space together with a certain type of topology; that is to say, when considered as a TVS, it is not associated with any particular norm or metric (both of which are "forgotten"). This Hausdorff TVS ${\displaystyle \left(X,\tau _{d}\right)}$ is even locally convex because the set of all open balls centered at the origin forms a neighbourhood basis at the origin consisting of convex balanced open sets. This TVS is also normable, which by definition refers to any TVS whose its topology is induced by some (possibly unknown) norm.

Comparison of complete metrizable vector topologies

The open mapping theorem implies that if ${\displaystyle \tau {\text{ and }}\tau _{2}}$ are topologies on ${\displaystyle X}$ that make both ${\displaystyle (X,\tau )}$ and ${\displaystyle \left(X,\tau _{2}\right)}$ into complete metrizable TVS (for example, Banach or Fréchet spaces) and if one topology is finer or coarser than the other then they must be equal (that is, if ${\displaystyle \tau \subseteq \tau _{2}{\text{ or }}\tau _{2}\subseteq \tau {\text{ then }}\tau =\tau _{2}}$).[8] So for example, if ${\displaystyle (X,p){\text{ and }}(X,q)}$ are Banach spaces with topologies ${\displaystyle \tau _{p}{\text{ and }}\tau _{q}}$ 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 ${\displaystyle p:\left(X,\tau _{q}\right)\to \mathbb {R} }$ or ${\displaystyle q:\left(X,\tau _{p}\right)\to \mathbb {R} }$ is continuous) then their topologies are identical and their norms are equivalent.

Complete norms and equivalent norms

Two norms, ${\displaystyle p}$ and ${\displaystyle q,}$ on a vector space are said to be equivalent if they induce the same topology;[9] this happens if and only if there exist positive real numbers ${\displaystyle c,C>0}$ such that ${\textstyle cq(x)\leq p(x)\leq Cq(x)}$ 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 is not equivalent to that Banach space's given norm.[note 6][9] All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space.[10]

Complete norms vs complete metrics

A metric ${\displaystyle D}$ on a vector space ${\displaystyle X}$ is induced by a norm on ${\displaystyle X}$ if and only if ${\displaystyle D}$ is translation invariant[note 3] and absolutely 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 ${\displaystyle \|x\|:=D(x,0)}$ defines a norm on ${\displaystyle X}$ and the canonical metric induced by ${\displaystyle \|\cdot \|}$ is equal to ${\displaystyle D.}$

Suppose that ${\displaystyle (X,\|\cdot \|)}$ is a normed space and that ${\displaystyle \tau }$ is the norm topology induced on ${\displaystyle X.}$ Suppose that ${\displaystyle D}$ is any metric on ${\displaystyle X}$ such that the topology that ${\displaystyle D}$ induces on ${\displaystyle X}$ is equal to ${\displaystyle \tau .}$ If ${\displaystyle D}$ is translation invariant[note 3] then ${\displaystyle (X,\|\cdot \|)}$ is a Banach space if and only if ${\displaystyle (X,D)}$ is a complete metric space.[11] If ${\displaystyle D}$ is not translation invariant, then it may be possible for ${\displaystyle (X,\|\cdot \|)}$ to be a Banach space but for ${\displaystyle (X,D)}$ to not be a complete metric space[12] (see this footnote[note 7] for an example). In contrast, a theorem of Klee,[13][14][note 8] which also applies to all metrizable topological vector spaces, implies that if there exists any[note 9] complete metric ${\displaystyle D}$ on ${\displaystyle X}$ that induces the norm topology ${\displaystyle \tau }$ on ${\displaystyle X,}$ then ${\displaystyle (X,\|\cdot \|)}$ 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 ${\textstyle \mathbb {R} ^{\mathbb {N} }=\prod _{i\in \mathbb {N} }\mathbb {R} }$ with the product topology). However, the topology of every Fréchet space is induced by some countable family of real-valued (necessarily continuous) maps called seminorms, which are generalizations of norms. It is even possible for a Fréchet space to have a topology that is induced by a countable family of norms (such norms would necessarily be continuous)[note 10][15] but to not be a Banach/normable space because its topology can not be defined by any single norm. An example of such a space is the Fréchet space ${\displaystyle C^{\infty }(K),}$ 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 spaces. Specifically, the notion of TVS-completeness uses a unique translation-invariant uniformity, called the canonical uniformity, that depends only 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 are 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 ${\displaystyle (X,\tau )}$ is a metrizable 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 a sequentially complete TVS, meaning that it is enough to check that every Cauchy sequence 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 Cauchy nets).

If ${\displaystyle (X,\tau )}$ is a topological vector space whose topology is induced by some (possibly unknown) norm (such spaces are called normable and they are characterized by being Hausdorff and having a bounded convex neighborhood of the origin), then ${\displaystyle (X,\tau )}$ is a complete topological vector space if and only if ${\displaystyle X}$ may be assigned a norm ${\displaystyle \|\cdot \|}$ that induces on ${\displaystyle X}$ the topology ${\displaystyle \tau }$ and also makes ${\displaystyle (X,\|\cdot \|)}$ into a Banach space. A Hausdorff locally convex topological vector space ${\displaystyle X}$ is normable if and only if its strong dual space ${\displaystyle X_{b}^{\prime }}$ is normable,[16] in which case ${\displaystyle X_{b}^{\prime }}$ is a Banach space (${\displaystyle X_{b}^{\prime }}$ denotes the strong dual space of ${\displaystyle X,}$ whose topology is a generalization of the dual norm-induced topology on the continuous dual space ${\displaystyle X^{\prime }}$; see this footnote[note 11] for more details). If ${\displaystyle X}$ is a metrizable locally convex TVS, then ${\displaystyle X}$ is normable if and only if ${\displaystyle X_{b}^{\prime }}$ is a Fréchet–Urysohn space.[17] This shows that in the category of locally convex TVSs, Banach spaces are exactly those complete spaces that are both metrizable and have metrizable strong dual spaces.

Every normed space can be isometrically embedded onto a dense vector subspace of some Banach space, where this Banach space is called a completion of the normed space. This Hausdorff completion is unique up to isometric isomorphism.

More precisely, for every normed space ${\displaystyle X,}$ there exist a Banach space ${\displaystyle Y}$ and a mapping ${\displaystyle T:X\to Y}$ such that ${\displaystyle T}$ is an isometric 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 ${\displaystyle Y.}$ This Banach space ${\displaystyle Y}$ is the Hausdorff completion 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 ${\displaystyle {\widehat {X}}.}$

If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed spaces over the same ground field ${\displaystyle \mathbb {K} ,}$ the set of all continuous ${\displaystyle \mathbb {K} }$-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 is bounded on the closed unit ball of ${\displaystyle X.}$ Thus, the vector space ${\displaystyle B(X,Y)}$ can be given the operator norm

$${\displaystyle \|T\|=\sup \left\{\|Tx\|_{Y}\mid x\in X,\ \|x\|_{X}\leq 1\right\}.}$$

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 the function space between two Banach spaces to only the short maps; in that case the space ${\displaystyle B(X,Y)}$ reappears as a natural bifunctor.[18]

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

If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed spaces, they are isomorphic 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 (or reflexive, separable, etc.) then so is the other space. Two normed spaces ${\displaystyle X}$ and ${\displaystyle Y}$ are isometrically isomorphic if in addition, ${\displaystyle T}$ is an isometry, that is, ${\displaystyle \|T(x)\|=\|x\|}$ for every ${\displaystyle x}$ in ${\displaystyle X.}$ The Banach–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.

Every continuous linear operator is a bounded linear operator and if dealing only with normed spaces then the converse is also true. That is, a linear operator between two normed spaces is bounded if and only if it is a continuous function. So in particular, because the scalar field (which is ${\displaystyle \mathbb {R} }$ or ${\displaystyle \mathbb {C} }$) is a normed space, a linear functional on a normed space is a bounded linear functional if and only if it is a continuous 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 a subadditive function (such as a norm, a sublinear function, or real linear functional), then[19] ${\displaystyle f}$ is continuous at the origin if and only if ${\displaystyle f}$ is uniformly continuous on all of ${\displaystyle X}$; and if in addition ${\displaystyle f(0)=0}$ then ${\displaystyle f}$ is continuous if and only if its absolute value ${\displaystyle |f|:X\to [0,\infty )}$ is continuous, which happens if and only if ${\displaystyle \{x\in X:|f(x)|<1\}}$ is an open subset of ${\displaystyle X.}$[19][note 12] And very importantly for applying the Hahn–Banach theorem, a linear functional ${\displaystyle f}$ is continuous if and only if this is true of its real part ${\displaystyle \operatorname {Re} f}$ and moreover, ${\displaystyle \|\operatorname {Re} f\|=\|f\|}$ and the real part ${\displaystyle \operatorname {Re} f}$ 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 the seminorm ${\displaystyle |f|}$ is continuous, which happens if and only if there exists a continuous seminorm ${\displaystyle p:X\to \mathbb {R} }$ such that ${\displaystyle |f|\leq p}$; 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,[20] such as

$${\displaystyle \|(x,y)\|_{1}=\|x\|+\|y\|,\qquad \|(x,y)\|_{\infty }=\max(\|x\|,\|y\|)}$$

which correspond (respectively) to the coproduct and product in the category of Banach spaces and short maps (discussed above).[18] 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 a closed linear subspace of a normed space ${\displaystyle X,}$ there is a natural norm on the quotient space ${\displaystyle X/M,}$

$${\displaystyle \|x+M\|=\inf \limits _{m\in M}\|x+m\|.}$$

The quotient ${\displaystyle X/M}$ is a Banach space when ${\displaystyle X}$ is complete.[21] 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 has 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 a complemented subspace of ${\displaystyle X}$ if ${\displaystyle M}$ is the range of a surjective bounded linear projection ${\displaystyle P:X\to M.}$ 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[21]

$${\displaystyle T=T_{1}\circ \pi ,\ \ \ T:X\ {\overset {\pi }{\longrightarrow }}\ X/ker(T)\ {\overset {T_{1}}{\longrightarrow }}\ Y}$$

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[22] of Banach spaces include: the Lp spaces ${\displaystyle L^{p}}$ and their special cases, the sequence spaces ${\displaystyle \ell ^{p}}$ that consist of scalar sequences indexed by natural numbers ${\displaystyle \mathbb {N} }$; among them, the space ${\displaystyle \ell ^{1}}$ of absolutely 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 ${\displaystyle \ell ^{\infty }}$ of bounded sequences; the space ${\displaystyle C(K)}$ of continuous scalar functions on a compact Hausdorff space ${\displaystyle K,}$ equipped with the max norm,

$${\displaystyle \|f\|_{C(K)}=\max\{|f(x)|:x\in K\},\quad f\in C(K).}$$

According to the Banach–Mazur theorem, every Banach space is isometrically isomorphic to a subspace of some ${\displaystyle C(K).}$[23] For every separable Banach space ${\displaystyle X,}$ there is a closed subspace ${\displaystyle M}$ of ${\displaystyle \ell ^{1}}$ such that ${\displaystyle X:=\ell ^{1}/M.}$[24]

Any Hilbert 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

$${\displaystyle \|x\|_{H}={\sqrt {\langle x,x\rangle }},}$$

where

$${\displaystyle \langle \cdot ,\cdot \rangle :H\times H\to \mathbb {K} }$$

is the inner product, linear in its first argument that satisfies the following:

$${\displaystyle {\begin{aligned}\langle y,x\rangle &={\overline {\langle x,y\rangle }},\quad {\text{ for all }}x,y\in H\\\langle x,x\rangle &\geq 0,\quad {\text{ for all }}x\in H\\\langle x,x\rangle =0{\text{ if and only if }}x&=0.\end{aligned}}}$$

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

The Hardy spaces, the Sobolev 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 and Partial differential equations among others.

A Banach algebra is a Banach space ${\displaystyle A}$ over ${\displaystyle \mathbb {K} =\mathbb {R} }$ or ${\displaystyle \mathbb {C} ,}$ together with a structure of algebra 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 ${\displaystyle \|ab\|\leq \|a\|\|b\|}$ for all ${\displaystyle a,b\in A.}$

• The Banach space ${\displaystyle C(K)}$ with the pointwise product, is a Banach algebra.
• The disk algebra ${\displaystyle A(\mathbf {D} )}$ consists of functions holomorphic in the open unit disk ${\displaystyle D\subseteq \mathbb {C} }$ and continuous on its closure: ${\displaystyle {\overline {\mathbf {D} }}.}$ Equipped with the max norm on ${\displaystyle {\overline {\mathbf {D} }},}$ the disk algebra ${\displaystyle A(\mathbf {D} )}$ is a closed subalgebra of ${\displaystyle C\left({\overline {\mathbf {D} }}\right).}$
• The Wiener 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 the convolution 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.
• A C*-algebra is a complex Banach algebra ${\displaystyle A}$ with an antilinear involution ${\displaystyle a\mapsto a^{*}}$ such that ${\displaystyle \left\|a^{*}a\right\|=\|a\|^{2}.}$ The space ${\displaystyle B(H)}$ of bounded linear operators on a Hilbert space ${\displaystyle H}$ is a fundamental example of C*-algebra. The Gelfand–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 ${\displaystyle f}$ its complex conjugate ${\displaystyle {\overline {f}}.}$

If ${\displaystyle X}$ is a normed space and ${\displaystyle \mathbb {K} }$ the underlying field (either the real or the complex numbers), the continuous dual space is the space of continuous linear maps from ${\displaystyle X}$ into ${\displaystyle \mathbb {K} ,}$ or continuous linear functionals. The notation for the continuous dual is ${\displaystyle X^{\prime }=B(X,\mathbb {K} )}$ in this article.[25] Since ${\displaystyle \mathbb {K} }$ is a Banach space (using the absolute value as norm), the dual ${\displaystyle X^{\prime }}$ is a Banach space, for every normed space ${\displaystyle X.}$

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

Hahn–Banach theorem — Let ${\displaystyle X}$ be a vector space over the field ${\displaystyle \mathbb {K} =\mathbb {R} ,\mathbb {C} .}$ Let further

• ${\displaystyle Y\subseteq X}$ be a linear subspace,
• ${\displaystyle p:X\to \mathbb {R} }$ be a sublinear function and
• ${\displaystyle f:Y\to \mathbb {K} }$ be a linear functional so that ${\displaystyle \operatorname {Re} (f(y))\leq p(y)}$ for all ${\displaystyle y\in Y.}$

Then, there exists a linear functional ${\displaystyle F:X\to \mathbb {K} }$ so that

$${\displaystyle F{\big \vert }_{Y}=f,\quad {\text{ and }}\quad {\text{ for all }}x\in X,\ \ \operatorname {Re} (F(x))\leq p(x).}$$

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.[26] 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

$${\displaystyle f(x)=\|x\|_{X},\quad \|f\|_{X^{\prime }}\leq 1.}$$

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

The Hahn–Banach separation theorem states that two disjoint non-empty convex sets in a real Banach space, one of them open, can be separated by a closed affine hyperplane. 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.[27]

A subset ${\displaystyle S}$ in a Banach space ${\displaystyle X}$ is total if the linear span of ${\displaystyle S}$ is dense 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 ${\displaystyle N.}$[28] If ${\displaystyle M}$ is a closed linear subspace in ${\displaystyle X,}$ one can associate the orthogonal of ${\displaystyle M}$ in the dual,

$${\displaystyle M^{bot}=\left\{x^{\prime }\in X:x^{\prime }(m)=0,\ {\text{ for all }}m\in M\right\}.}$$

The orthogonal ${\displaystyle M^{\bot }}$ is a closed linear subspace of the dual. The dual of ${\displaystyle M}$ is isometrically isomorphic to ${\displaystyle X'/M^{\bot }.}$ The dual of ${\displaystyle X/M}$ is isometrically isomorphic to ${\displaystyle M^{\bot }.}$[29]

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

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

The weak topology on a Banach space ${\displaystyle X}$ is the coarsest 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 therefore finer than the weak topology. It follows from the Hahn–Banach separation theorem that the weak topology is Hausdorff, and that a norm-closed convex subset of a Banach space is also weakly closed.[31] A norm-continuous linear map between two Banach spaces ${\displaystyle X}$ and ${\displaystyle Y}$ is also weakly continuous, that is, continuous from the weak topology of ${\displaystyle X}$ to that of ${\displaystyle Y.}$[32]

If ${\displaystyle X}$ is infinite-dimensional, there exist linear maps which are not continuous. The space ${\displaystyle X^{*}}$ of all linear maps from ${\displaystyle X}$ to the underlying field ${\displaystyle \mathbb {K} }$ (this space ${\displaystyle X^{*}}$ is called the algebraic dual space, to distinguish it from ${\displaystyle X^{\prime }}$ also induces a topology on ${\displaystyle X}$ which is finer 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 weak* 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 the Banach–Alaoglu theorem.

The Banach–Alaoglu theorem can be proved using Tychonoff's theorem about infinite products of compact Hausdorff spaces. When ${\displaystyle X}$ is separable, the unit ball ${\displaystyle B^{\prime }}$ of the dual is a metrizable compact in the weak* topology.[33]

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 ${\displaystyle y=\left\{y_{n}\right\}\in \ell ^{1}}$ such that

$${\displaystyle f(x)=\sum _{n\in \mathbb {N} }x_{n}y_{n},\qquad x=\{x_{n}\}\in c_{0},\ \ {\text{and}}\ \ \|f\|_{(c_{0})'}=\|y\|_{\ell _{1}}.}$$

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

For every vector ${\displaystyle y}$ in a Hilbert space ${\displaystyle H,}$ the mapping

$${\displaystyle x\in H\to f_{y}(x)=\langle x,y\rangle }$$

defines a continuous linear functional ${\displaystyle f_{y}}$ on ${\displaystyle H.}$The Riesz 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 an antilinear isometric bijection from ${\displaystyle H}$ onto its dual ${\displaystyle H'.}$ 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 of Radon measures in the sense of Bourbaki.[34] 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 ${\displaystyle M(K).}$ The extreme points of ${\displaystyle P(K)}$ are the Dirac measures on ${\displaystyle K.}$ The set of Dirac measures on ${\displaystyle K,}$ equipped with the w*-topology, is homeomorphic to ${\displaystyle K.}$

The result has been extended by Amir[37] and Cambern[38] to the case when the multiplicative Banach–Mazur distance between ${\displaystyle C(K)}$ and ${\displaystyle C(L)}$ is ${\displaystyle <2.}$ The theorem is no longer true when the distance is ${\displaystyle =2.}$[39]

In the commutative Banach algebra ${\displaystyle C(K),}$ the maximal ideals are precisely kernels of Dirac measures on ${\displaystyle K,}$

$${\displaystyle I_{x}=\ker \delta _{x}=\{f\in C(K):f(x)=0\},\quad x\in K.}$$

More generally, by the Gelfand–Mazur theorem, the maximal ideals of a unital commutative Banach algebra can be identified with its characters—not merely as sets but as topological spaces: the former with the hull-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 ${\displaystyle A'.}$

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 commutative C*-algebras. Gelfand's representation theorem for commutative C*-algebras states that every commutative unital C*-algebra ${\displaystyle A}$ is isometrically isomorphic to a ${\displaystyle C(K)}$ space.[40] The Hausdorff compact space ${\displaystyle K}$ here is again the maximal ideal space, also called the spectrum of ${\displaystyle A}$ in the C*-algebra context.

If ${\displaystyle X}$ is a normed space, the (continuous) dual ${\displaystyle X''}$ of the dual ${\displaystyle X'}$ is called bidual, or second dual of ${\displaystyle X.}$ For every normed space ${\displaystyle X,}$ there is a natural map,

$${\displaystyle {\begin{cases}F_{X}:X\to X''\\F_{X}(x)(f)=f(x)&{\text{ for all }}x\in X,{\text{ and for all }}f\in X'\end{cases}}}$$

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

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 ${\displaystyle \ell ^{\infty },}$ the space of bounded scalar sequences. Under these identifications, ${\displaystyle F_{X}}$ is the inclusion map from ${\displaystyle c_{0}}$ to ${\displaystyle \ell ^{\infty }.}$ It is indeed isometric, but not onto.

If ${\displaystyle F_{X}}$ is surjective, then the normed space ${\displaystyle X}$ is called reflexive (see below). 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 ${\displaystyle X^{\prime \prime }.}$ If ${\displaystyle X}$ is not reflexive, the unit ball of ${\displaystyle X}$ is a proper subset of the unit ball of ${\displaystyle X^{\prime \prime }.}$ The Goldstine 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 a net ${\displaystyle \left(x_{i}\right)_{i\in I}}$ in ${\displaystyle X}$ so that

$${\displaystyle \sup _{i\in I}\left\|x_{i}\right\|\leq \|x''\|,\ \ x''(f)=\lim _{i}f\left(x_{i}\right),\quad f\in X'.}$$

The net may be replaced by a weakly*-convergent sequence when the dual ${\displaystyle X'}$ 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 of sequences in ${\displaystyle \ell ^{1},}$ since ${\displaystyle \ell ^{1}}$ is weakly 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 the Baire category theorem. According to this theorem, a complete metric space (such as a Banach space, a Fréchet space or an F-space) cannot be equal to a union of countably many closed subsets with empty interiors. 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 countable Hamel basis is finite-dimensional.

The Banach–Steinhaus theorem is not limited to Banach spaces. It can be extended for example to the case where ${\displaystyle X}$ is a Fréchet space, provided the conclusion is modified as follows: under the same hypothesis, there exists a neighborhood ${\displaystyle U}$ of ${\displaystyle \mathbf {0} }$ in ${\displaystyle X}$ such that all ${\displaystyle T}$ in ${\displaystyle F}$ are uniformly bounded on ${\displaystyle U,}$

$${\displaystyle \sup _{T\in F}\sup _{x\in U}\;\|T(x)\|_{Y}<\infty .}$$