Banach space

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

A normed vector space is a Banach space exactly when every absolutely convergent series converges in the space. Completeness implies the assertion because the triangle inequality makes the partial sums a Cauchy sequence. Conversely, select from any Cauchy sequence a subsequence whose consecutive differences have norm at most $2^{-k}$. The series of these differences converges by hypothesis, so the subsequence converges. The original Cauchy sequence then has the same limit. This provides a useful way to demonstrate incompleteness by a summable-norm series whose candidate sum lies outside the space.

For a closed linear subspace $Y$ of a Banach space $X$, finite codimension means the vector quotient $X/Y$ has finite dimension. This definition does not subtract two infinite dimensions. A local stable manifold has the codimension of its tangent subspace. If the linearized phase space has a direct sum decomposition $X=E^s\oplus E^u$ with $\dim E^u=1$, then $X/E^s$ is isomorphic to $E^u$, so a local stable manifold tangent to $E^s$ has codimension one.

A Banach space has the Banach-Saks property if every bounded sequence contains a subsequence whose arithmetic averages converge in norm. The weak Banach-Saks property requires this only for weakly convergent sequences. The weak Banach–Saks theorem in a Hilbert space establishes that weak version in Hilbert spaces.

A Banach space $X$ has cotype $q$, $2\le q<\infty$, if a constant $C$ independent of the number of vectors satisfies $({\sum }_j\Vert x_j{\Vert }^q{)}^{1/q}\le C(\mathbb{E}\Vert {\sum }_j{\epsilon }_j{x}_j{\Vert }^2{)}^{1/2}$ for independent symmetric Rademacher random variables. It prevents many large vectors from having a small random sum.

This is cotype of a Banach space with $q=2$. The Lp space $L^1$ has cotype 2 with constant at most $\sqrt{2}$: the triangle inequality for the Euclidean norm gives $(\sum \Vert f_j{\Vert }_1^2{)}^{1/2}\le \int (\sum |f_j(t){|}^2{)}^{1/2}dt$, and the sharp Rademacher second-moment inequality, followed by Fubini theorem, bounds this by $\sqrt{2}\mathbb{E}\Vert \sum {\epsilon }_j{f}_j{\Vert }_1$.

If a Banach space $Y$ has cotype 2 constant $C$, every bounded linear operator $T:{\ell }_{\infty }^N\to Y$ satisfies ${\pi }_2(T)\le 3^{1/4}C{\pi }_4(T)$ and ${\pi }_2(T)\le 4\sqrt{3}{C}^2\Vert T\Vert$, uniformly in $N$. The first bound follows from Pietsch factorization theorem and the fourth moment bound for Rademacher sums. For the second, 2-summing domination gives a weighted coordinate $L^2$ norm; clipping coordinates at a variable threshold yields ${\pi }_4(T)\le 2\sqrt{\Vert T\Vert {\pi }_2(T)}$, which absorbs the first bound.

A closed vector subspace $Y$ of a Banach space $X$ is complemented if a bounded linear operator $P:X\to X$ satisfies $P^2=P$ and $P(X)=Y$. Equivalently, $X$ is the topological direct sum of $Y$ and a closed subspace: a projection gives $X=Y\oplus \ker P$, and the coordinate projection of such a decomposition is bounded.

For a Banach space $E$ satisfying $E\cong ({⨁}_{n\ge 1}E{)}_p$, a complemented subspace $W$ is absorbed by $E$: write $E\cong W\oplus V$, expand the countable sum, and absorb one additional copy of $W$ into the countably many $W$ summands. Hence $E\oplus W\cong E$. If $Z$ is complemented in $E$ and contains a complemented copy of $E$, write $Z\cong E\oplus W$ with $W$ complemented in $E$; the preceding argument gives $Z\cong E$. The decompositions are Banach space isomorphisms, not merely algebraic decompositions.

An infinite-dimensional Banach space is hereditarily indecomposable if no closed infinite-dimensional subspace is a topological direct sum of two infinite-dimensional closed subspaces. Equivalently, for any two infinite-dimensional closed subspaces $U,V$ and any $\epsilon >0$, some unit vectors $u\in U,v\in V$ satisfy $\Vert u-v\Vert <\epsilon$. A positive lower bound on such distances would make the addition map from $U\oplus V$ an isomorphism onto a closed subspace.

Every infinite-dimensional Banach space contains either an unconditional basic sequence or a hereditarily indecomposable Banach space. For a space with a basis and no unconditional basic sequence, apply the block Ramsey theorem to finite sequences whose odd and even spans contain nearby unit vectors. A diagonal sequence of successively sharper games gives a subspace in which every pair of infinite-dimensional subspaces has angle zero.

For a closed vector subspace $Y$ of a Banach space $X$, the quotient vector space $X/Y$ is complete with its quotient norm. Closedness makes the quotient seminorm a norm. Completeness follows by choosing representatives of a rapidly Cauchy subsequence whose successive differences have summable norms in $X$.

If a Banach space $X$ has separable dual and $q:X\to X/Y$ is a quotient by a closed subspace, every weakly null sequence in the quotient's closed unit ball has a subsequence with approximate weakly null lifts of norm at most $3$. Choose representatives of norm below $3/2$, pass to a common bidual weak-star limit, and subtract convex blocks whose quotient images tend to zero in norm. Convex-block cancellation of a weak-star limit gives weak nullity and the triangle inequality gives the bound $3$.

The quotient norm measures the smallest norm among representatives of a coset. Its infimum need not be attained. Nevertheless, for every $\epsilon >0$, an element $z\in X/Y$ has a representative $v\in X$ with $\Vert v\Vert <\Vert z\Vert +\epsilon$. The quotient map is contractive.

A convex block of a sequence in a Banach space is a sequence of convex combinations supported on successively disjoint finite intervals, with $p_n<q_n<p_{n+1}$, $a_i\ge 0$, and ${\sum }_{i=p_n}^{q_n}{a}_i=1$. Zero coefficients can fill gaps and ensure strict endpoint inequalities. By the Mazur theorem, every weakly null sequence has convex blocks tending to zero in norm.

If a bounded sequence $(y_n)$ in $X$ converges weak-star to $\varphi \in X^{\ast \ast }$ under the canonical embedding into the bidual, then any convex blocks $(u_n)$ have the same scalar limit on every $f\in X^{\ast }$. Thus $y_n-u_n$ is a weakly null sequence, even when $\varphi$ does not belong to the embedded space $X$.

The duality mapping takes a vector to its supporting dual functionals for half its squared norm. Equivalently $p\in \mathcal{J}(x)$ when $\Vert p{\Vert }_{\ast }=\Vert x\Vert$ and $⟨x,p⟩=\Vert x{\Vert }^2$. This follows from the dual-norm inequality and equality in the conjugate relation for squared norms. In a Hilbert space, the Riesz representation theorem identifies it with the identity; in a general Banach space it can be set-valued.

For an arbitrary set $\Gamma$ and scalar field $\mathbb{F}$, this is the space of bounded functions $u:\Gamma \to \mathbb{F}$ with the supremum norm $\Vert u\Vert ={\sup }_{\gamma }|u(\gamma )|$. Uniform limits of Cauchy sequences prove it is a Banach space. For $\Gamma =\mathbb{N}$ it is the l-infinity sequence space; for the empty index set it is the zero space, with norm zero.

A bounded linear operator $T:X\to {\ell }_{\infty }(\Gamma )$ is exactly a uniformly bounded family of bounded linear functionals $f_{\gamma }\in X^{\ast }$ through $(Tx)(\gamma )=f_{\gamma }(x)$. Evaluation shows $\Vert f_{\gamma }\Vert \le \Vert T\Vert$; conversely the family bound gives $\Vert Tx\Vert \le (\sup \Vert f_{\gamma }\Vert )\Vert x\Vert$, proving equality. Taking all functionals in the dual unit ball gives an isometric embedding by the Hahn-Banach theorem.

A Banach space is separable when it has a countable dense subset for its norm topology. It then has a countable base of norm balls and a countable norming family. Its continuous dual space need not be norm separable.

For $1\le p<\infty$, the space ${\ell }^p$ consists of scalar sequences $x=(x_n)$ satisfyingIt is a Banach space, and ${\ell }^q\subset {\ell }^p$ when $1\le q<p$.

As a Lebesgue space it uses counting measure; as a space of sequences it is a sequence space.

The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

For example, $x=(2^{-1},2^{-2},\dots )$ belongs to the absolutely summable sequence space and has norm one, by the geometric series formula. Merely tending to zero is insufficient: $(1/n)$ tends to zero but its series of absolute values diverges. Thus this sequence space records total absolute size, rather than just the eventual size of individual coordinates. Its completeness ensures that a Cauchy sequence in this total-error norm has an absolutely summable limit.^[ref]

Suppose the distinct complex numbers ${\beta }_k$ satisfy $|{\beta }_k|\le \delta <1$. The geometric sequences in the displayed formula belong to the absolutely summable sequence space and have dense linear span. Otherwise the Hahn-Banach theorem gives a nonzero annihilating functional, represented by a bounded sequence $b_j$. Its power series $F(z)={\sum }_{j\ge 1}{b}_j{z}^{j-1}$ is holomorphic in the unit disk and vanishes at every ${\beta }_k$. Compactness supplies an accumulation point inside that disk. The identity theorem forces $F=0$ and hence all its coefficients zero, a contradiction. This is a useful form of density by annihilators that converts a Banach-space problem into uniqueness of an analytic function.

For positive weights $w_j$, the sequence space with norm $\Vert \alpha \Vert ={\sum }_j{w}_j|{\alpha }_j|$ is isometrically isomorphic to ${\ell }_1$: multiplication by $w_j$ is a linear bijection onto the absolutely summable sequence space, with inverse multiplication by $w_j^{-1}$. Thus it is a separable Banach space. Density can also be seen directly by truncating a weighted summable sequence and then approximating its finitely many coordinates by rational numbers.

The space ${\ell }^2$ consists of scalar sequences $x=(x_n)$ satisfying ${\sum }_n|x_n{|}^2<\infty$. With inner product $⟨x,y⟩={\sum }_n{x}_n\overline{y_n}$, it is a Hilbert space.

The l2 sequence space is a uniformly convex Banach space. If $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$, the parallelogram law givesThus $\Vert x-y{\Vert }_2\ge \epsilon$ permits the modulus $\delta (\epsilon )=1-\sqrt{1-{\epsilon }^2/4}$.

A finitely supported sequence has only finitely many nonzero coordinates. Their vector space is commonly denoted $c_{00}$.

Equip the vector space $c_{00}$ of finitely supported sequences with the norm inherited from the absolutely summable sequence space. Consider the truncationsFor $M>N$, the geometric series givesTherefore $(x^{(N)})$ is a Cauchy sequence. Any possible norm limit must have coordinate $k$ equal to $2^{-k}$, since $|x_k-y_k|\le \Vert x-y{\Vert }_1$. That forces infinitely many nonzero coordinates, so no limit lies in $c_{00}$. There is nothing wrong with the approximations: the space is missing their limit.

Inside the absolutely summable sequence space, the limit is $x=(2^{-k}{)}_{k\ge 1}$ and its exact truncation error is $2^{-N}$. Four retained coordinates leave error $1/16$; ten leave $1/1024$. More generally, truncating any element of the absolutely summable sequence space leaves a tail whose norm tends to zero. Together with completeness, this identifies that Banach space as the completion of a normed space $c_{00}$ with this norm. Finite storage and an infinite mathematical object fit together through a controlled error.

For $1\le p<\infty$, coordinate truncations converge in ${\ell }^p$, so $c_{00}$ is dense. It is not dense in ${\ell }^{\infty }$: every finitely supported sequence is at distance at least one from the constant-one sequence.

The space ${\ell }^{\infty }$ consists of bounded scalar sequences with norm $\Vert x{\Vert }_{\infty }={\sup }_n|x_n|$.

It is the bounded-sequence example of a sequence space.

A generalized limit on real bounded sequences is a positive linear functional of operator norm one that extends the ordinary limit of a sequence. Such a functional can be chosen by the Hahn-Banach theorem: extend the limit on the convergent sequence space while dominating it by the sublinear functional $p(x)={\mathrm{limsup}}_n{x}_n$. Applying the same domination to $-x$ gives the displayed lower bound. A Banach limit additionally has invariance under the left shift on bounded sequences.

A Banach limit is a positive linear functional on the real l-infinity sequence space that extends the ordinary limit on the convergent sequence space, has operator norm one, and is invariant under the left shift on bounded sequences. The Hahn-Banach theorem constructs one by separating the constant sequence $1$ from the subspace $(I-S){\ell }^{\infty }$, whose distance from $1$ is one. The functional is generally not unique.

Suppose $q$ is a sublinear functional on the bounded real sequences, dominated by the limit superior and nonpositive on every shift difference $x-Sx$. The Hahn-Banach theorem then yields a Banach limit. Indeed $0\le q(x)+q(-x)$ implies $q(x)\ge -q(-x)\ge \mathrm{liminf}{x}_n$, so $q$ agrees with the ordinary limit on convergent sequences. Extend this linear limit functional by Hahn-Banach to a linear $\psi \le q$. Domination on $x-Sx$ and on its negative forces $\psi (x)=\psi (Sx)$. The two-sided limit bounds give positivity, norm one and agreement with ordinary limits. In finite-shift averaging constructions, the needed bound on shift differences follows by telescoping consecutive averages to $(x_{n+1}-x_{n+k+1})/k$.

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

A nonprincipal ultrafilter on the natural numbers gives a Banach limit by taking the ultralimit of the Cesàro means of each bounded sequence. Compactness of a closed bounded interval, or disk for complex sequences, supplies the limit. Finite intersections in the filter give linearity; nonnegative averages give positivity; constants give norm one. The Cesàro theorem for convergent sequences gives agreement with ordinary limits because the ultrafilter contains every cofinite set. For the left shift on bounded sequences, the two means differ by $(x_{n+1}-x_1)/n$, which tends to zero, proving shift invariance.

The left shift is a bounded linear operator of operator norm one on the l-infinity sequence space. For $y=x-Sx$, telescoping gives $N^{-1}{\sum }_{n=1}^N{y}_n=(x_1-x_{N+1})/N\to 0$. This identity is useful in constructing a Banach limit.

If $Y$ is a proper closed subspace of a normed space $X$ and $0<\alpha <1$, there is a unit vector $x\in X$ whose distance from $Y$ exceeds $\alpha$.

A normed vector space has compact closed unit ball if and only if it is finite-dimensional. In an infinite-dimensional space, repeated application of Riesz lemma produces unit vectors separated pairwise by a fixed positive distance, contradicting sequential compactness.

A Banach space $X$ is uniformly convex when, for every $\epsilon >0$, some $\delta >0$ satisfies

This is a complete uniformly convex space.

Every nonempty closed convex subset of a uniformly convex Banach space has a unique nearest point to each vector. A minimizing sequence is Cauchy: otherwise uniform convexity would put some midpoint strictly closer than the infimum. Completeness and closedness give its limiting point. In ${\ell }^p$, $p\ge 2$, the Clarkson inequality makes this quantitative: the distance between two minimizing vectors tends to zero after subtracting the midpoint's lower distance bound. Nonemptiness is required.

For $2\le p<\infty$, the Clarkson inequality isFor $1<p\le 2$ and $q=p/(p-1)$, the other Clarkson inequality isTogether these inequalities prove uniform convexity of $L^p$ for $1<p<\infty$.

For $p\ge 2$ and $p^{\prime }=p/(p-1)$, interpolate the two-coordinate map $(x,y)\mapsto (x+y,x-y)$ between ${\ell }^1\to {\ell }^{\infty }$ norm one and ${\ell }^2\to {\ell }^2$ norm $\sqrt{2}$. This proves the scalar inequality with factor $2^{1/p}$. Integrating that estimate and applying Minkowski inequality to $|f{|}^{p^{\prime }}+|g{|}^{p^{\prime }}$ in $L^{p/p^{\prime }}$ gives the displayed function inequality.

If a family of bounded linear operators from a Banach space to a normed space is pointwise bounded, then their operator norms are uniformly bounded.

For operators on a complex Hilbert space, pointwise boundedness of $⟨Tx,x⟩$ over a family implies uniform boundedness of their operator norms. The complex polarization identity recovers each $⟨Tx,y⟩$ from four diagonal values. Apply the Uniform boundedness principle first to the functionals in $y$ for fixed $x$, then to the operators in $x$. On a real Hilbert space the claim fails: multiples of a nonzero skew-symmetric operator have zero diagonal forms but unbounded norms. For example, rotations through a quarter turn on ${\mathbb{R}}^2$ multiplied by integers suffice.

Let $X$ be a normed space, $Y$ a Banach space, and $A_n:X\to Y$ be bounded linear operators with $C={\sup }_n\Vert A_n\Vert <\infty$. Convergence of $A_{n}d$ for every $d$ in a dense subset $D$ implies convergence of $A_{n}x$ for every $x\in X$ to a bounded linear operator $A$, with $\Vert A\Vert \le {\mathrm{liminf}}_n\Vert A_n\Vert$. Indeed, $\Vert A_{n}x-A_{m}x\Vert \le 2C\Vert x-d\Vert +\Vert A_{n}d-A_{m}d\Vert$ makes $(A_{n}x)$ a Cauchy sequence; completeness gives its limit, and linearity and the bound pass to that limit. If $X$ is a Banach space, pointwise convergence conversely implies the uniform bound by the Uniform boundedness principle. If the limit bounded linear operator is specified in advance, completeness of $Y$ is unnecessary.

A subset of a normed space is norm bounded if it is bounded under every continuous linear functional. Apply the Uniform boundedness principle on the complete dual space to the evaluation family $f\mapsto f(x)$. Hahn-Banach theorem identifies the evaluation norm with $\Vert x\Vert$, even when the original normed space is not complete. As a consequence, two norms with exactly the same continuous dual as sets must be equivalent: apply this criterion to either norm's unit ball in the other norm.

Let $(x_n)$ lie in a normed space $X$ and suppose ${\sum }_n|f(x_n)|<\infty$ for every $f\in X^{\ast }$. Then some $C$ satisfiesApply the uniform boundedness principle on the Banach space $X^{\ast }$ to the finite signed sums $f\mapsto {\sum }_n{a}_{n}f(x_n)$ with $|a_n|\le 1$.