Banach space

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 sequence space ${\displaystyle \ell ^{2}(\mathbb {N} )}$ with its usual norm ${\displaystyle \|{\cdot }\|_{2}.}$

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).^[6][7] Thus all infinite–dimensional separable Fréchet spaces are homeomorphic to each other (or said differently, their topology is unique up 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 even homeomorphic to its own unit sphere ${\displaystyle \{x\in \ell ^{2}(\mathbb {N} )\mid \|x\|_{2}=1\},}$ which stands in sharp contrast to finite–dimensional spaces (the Euclidean plane ${\displaystyle \mathbb {R} ^{2}}$ is not homeomorphic to the unit circle, for instance).

This pattern in homeomorphism classes extends to generalizations of metrizable (locally Euclidean) topological manifolds known as metric Banach manifolds, which are metric spaces that are around every point, locally homeomorphic to some open subset of a given Banach space (metric Hilbert manifolds and metric Fréchet manifolds are defined similarly).^[7] For example, every open subset ${\displaystyle U}$ of a Banach space ${\displaystyle X}$ is canonically a metric Banach manifold modeled on ${\displaystyle X}$ since the inclusion map ${\displaystyle U\to X}$ is an open local homeomorphism. Using Hilbert space microbundles, David Henderson showed^[8] in 1969 that every metric manifold modeled on a separable infinite–dimensional Banach (or Fréchet) space can be topologically embedded as an open subset of ${\displaystyle \ell ^{2}(\mathbb {N} )}$ and, consequently, also admits a unique smooth structure making it into a ${\displaystyle C^{\infty }}$ Hilbert manifold.

Compact and convex subsets

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.^{[note 5][9]} 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.^[10] In a normed space that 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 ${\displaystyle \ell ^{2}(\mathbb {N} ).}$

As a topological vector space

This norm-induced topology also makes ${\displaystyle (X,\tau _{d})}$ 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 ${\displaystyle (X,\tau _{d})}$ 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 ${\displaystyle (X,\tau _{d})}$ 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 topology is induced by some (possibly unknown) norm. Normable TVSs are characterized by being Hausdorff and having a bounded convex neighborhood of the origin. All Banach spaces are barrelled spaces, which means that every barrel is neighborhood of the origin (all closed balls centered at the origin are barrels, for example) and guarantees that the Banach–Steinhaus theorem holds.

Comparison of complete metrizable vector topologies

The open 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 ${\displaystyle (X,\tau _{2})}$ into complete metrizable TVSes (for example, Banach or Fréchet spaces), if one topology is finer 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 ${\displaystyle \tau _{1}=\tau _{2}}$).^[11] 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 ${\displaystyle q}$ are equivalent.

Complete norms and equivalent norms

Two norms, ${\displaystyle p}$ and ${\displaystyle q,}$ on a vector space ${\displaystyle X}$ are said to be equivalent if they induce the same topology;^[12] this happens if and only if there exist real numbers ${\displaystyle c,C>0}$ such that ${\textstyle c\,q(x)\leq p(x)\leq C\,q(x)}$ 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][12]} All norms on a finite-dimensional vector space are equivalent and every finite-dimensional normed space is a Banach space.^[13]

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.^[14] 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^[15] (see this footnote^{[note 7]} for an example). In contrast, a theorem of Klee,^{[16][17][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][18]} 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), 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 ${\displaystyle X'_{b}}$ is normable,^[19] in which case ${\displaystyle X'_{b}}$ is a Banach space (${\displaystyle X'_{b}}$ denotes the strong dual space of ${\displaystyle X,}$ whose topology is a generalization of the dual norm-induced topology on the continuous dual space ${\displaystyle X'}$; 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 ${\displaystyle X'_{b}}$ is a Fréchet–Urysohn space.^[20] 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.

Completions

Every normed space can be isometrically embedded onto a dense vector subspace of a 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 exists 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 ${\displaystyle Y.}$ The 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 ${\displaystyle {\widehat {X}}.}$

Continuous and bounded linear functions and seminorms

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^[22] ${\displaystyle f}$ 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 ${\displaystyle \{x\in X\mid |f(x)|<1\}}$ is an open subset of ${\displaystyle X.}$^{[22][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.

Examples

• 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 ${\displaystyle \|a^{*}a\|=\|a\|^{2}.}$ 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 ${\displaystyle {\overline {f}}.}$

Weak topologies

The weak topology on a Banach space ${\displaystyle X}$ is the coarsest topology on ${\displaystyle X}$ for which all elements ${\displaystyle x'}$ in the continuous dual space ${\displaystyle X'}$ 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.^[34] 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 ${\displaystyle Y.}$^[35]

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 ${\displaystyle X'}$ 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 ${\displaystyle X',}$ there is a topology weaker than the weak topology of ${\displaystyle X',}$ called the weak* topology. It is the coarsest topology on ${\displaystyle X'}$ for which all evaluation maps ${\displaystyle x'\in X'\mapsto x'(x),}$ 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 ${\displaystyle B'}$ of the dual is a metrizable compact in the weak* topology.^[36]

Examples of dual spaces

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

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.^[37] 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^[40] and Cambern^[41] 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 ${\displaystyle =2.}$^[42]

In the commutative Banach algebra ${\displaystyle C(K),}$ the maximal ideals are precisely kernels of Dirac measures on ${\displaystyle K,}$ $${\displaystyle I_{x}=\ker \delta _{x}=\{f\in C(K)\mid f(x)=0\},\quad x\in K.}$$

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.^[43] 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.

Bidual

If ${\displaystyle X}$ is a normed space, the (continuous) dual ${\displaystyle X''}$ of the dual ${\displaystyle X'}$ is called the bidual or second dual of ${\displaystyle X.}$ For every normed space ${\displaystyle X,}$ there is a natural map, $${\displaystyle {\begin{cases}F_{X}\colon X\to X''\\F_{X}(x)(f)=f(x)&{\text{ for all }}x\in X,{\text{ and for all }}f\in X'\end{cases}}}$$

This defines ${\displaystyle F_{X}(x)}$ as a continuous linear functional on ${\displaystyle X',}$ that is, an element of ${\displaystyle X''.}$ The map ${\displaystyle F_{X}\colon x\to F_{X}(x)}$ is a linear map from ${\displaystyle X}$ to ${\displaystyle X''.}$ 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 ${\displaystyle X''.}$ If ${\displaystyle X}$ is not reflexive, the unit ball of ${\displaystyle X}$ is a proper subset of the unit ball of ${\displaystyle X''.}$ 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 ${\displaystyle (x_{i})_{i\in I}}$ in ${\displaystyle X}$ so that $${\displaystyle \sup _{i\in I}\|x_{i}\|\leq \|x''\|,\ \ x''(f)=\lim _{i}f(x_{i}),\quad f\in X'.}$$

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.