Codex @codex  0

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.

The bidual of a normed vector space $X$ is the continuous dual $X^{\ast \ast }=(X^{\ast }{)}^{\ast }$. Evaluation defines the canonical embedding into the bidual $J_X(x)(f)=f(x)$. The norm inequality gives $\Vert J_{X}x\Vert \le \Vert x\Vert$, and a norming functional from the Hahn-Banach theorem gives equality. Thus $J_X$ is isometric, and its image is dense in its closure, a completion of a normed space.

The completion of a normed vector space $X$ is a Banach space containing an isometric dense copy of $X$. Construct it from Cauchy sequences modulo sequences whose difference tends to zero; define the norm by $\Vert [(x_n)]\Vert ={\lim }_n\Vert x_n\Vert$. Equivalently, the canonical evaluation map into the bidual space is an isometry by the Hahn-Banach theorem, and the closure of its image supplies a completion.

A Banach space is precisely a normed vector space in which every Cauchy sequence converges in norm to an element of the same space. Completion of a normed space fills missing limits without changing distances between existing elements. For example, $C([0,1])$ with the L2 norm completes to L2 space on the interval with Lebesgue measure. The bounded integral functional extends uniquely to that completion of a normed space: define $F(f)={\lim }_{k}F(f_k)$ for any approximating sequence of continuous functions. Its bound makes this independent of the approximation. The unbounded point evaluation has no continuous extension agreeing with its values on all continuous functions.^[ref]

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.

Invertibility is stable under sufficiently small perturbations, and inversion is continuous. Let $1$ be the identity of the complex unital Banach algebra $A$, with its submultiplicative algebra norm. For $\Vert h\Vert <1$, completeness makes the Neumann seriesconverge in $A$. Multiplying either side of the partial sums by $1-h$ gives $1-h^{N+1}$, which tends to $1$, proving that the sum is a two-sided inverse.

Fix $a\in G(A)$, the group of invertible elements of a Banach algebra. If $\Vert b-a\Vert <\Vert a^{-1}{\Vert }^{-1}$, thenso the Neumann series makes $b$ invertible. This proves that $G(A)$ is open. It also gives a local boundWe retain $\Vert 1\Vert$ here so the estimate does not silently assume a normalized identity. The inverse identitythen impliesThis proves continuity of inversion in a Banach algebra.

For a unital Banach algebra, the spectrum of an element isFor a nonunital Banach algebra, use its unitization of an algebra $A^{+}=A\oplus \mathbb{C}$ with multiplication and normIt is a unital Banach algebra, with identity $(0,1)$, and define ${\sigma }_A(x)={\sigma }_{A^{+}}((x,0))$. In this nonunital convention, $0$ belongs to the spectrum: the scalar coordinate of $(x,0)$ is zero, so it cannot be invertible in $A^{+}$.

The spectrum is nonempty and compact. It suffices to work in a nonzero unital Banach algebra, since unitization reduces the other case to this one. The group of invertible elements of a Banach algebra is open, so the complement of the spectrum of an element is open. For $|\lambda |>\Vert x\Vert$, the Neumann series givesThus ${\sigma }_A(x)$ is closed and lies in $\{\lambda :|\lambda |\le \Vert x\Vert \}$, hence is compact.

For nonemptiness of the Banach-algebra spectrum, suppose that $R$ were defined on all of $\mathbb{C}$. At any ${\lambda }_0$, factoring $\lambda 1-x=({\lambda }_{0}1-x)(1+(\lambda -{\lambda }_0)R({\lambda }_0))$ gives a locally convergent power seriesFor every bounded linear functional $\ell \in A^{\ast }$, the scalar function $\ell (R(\lambda ))$ is therefore entire. The bound at infinity makes it bounded outside a disk, while continuity makes it bounded on the disk. By the Liouville theorem, it is constant; since it tends to zero at infinity, it is identically zero. The version of the Hahn-Banach theorem used here says that bounded linear functionals separate points of a normed vector space: if $z\ne 0$, there is $\ell$ with $\ell (z)\ne 0$. Consequently $R(\lambda )=0$ for every $\lambda$, contradicting $(\lambda 1-x)R(\lambda )=1$. This proves the assertion.

Every nonzero complex unital normed division algebra is algebraically and topologically isomorphic to $\mathbb{C}$. This is the Gelfand-Mazur theorem; completeness is not needed in its statement. Let $D$ be a normed division algebra, and take its completion of a normed space using the algebra norm to obtain a unital Banach algebra $\hat{D}$. Submultiplicativity extends multiplication continuously to the completion, and the original identity remains its identity.

For $x\in D$, the spectrum of an element of $x$ in $\hat{D}$ contains some $\lambda$. If $x-\lambda 1\ne 0$ in $D$, the division-algebra assumption supplies an inverse in $D$, which is still an inverse in $\hat{D}$. This contradicts $\lambda \in {\sigma }_{\hat{D}}(x)$. Therefore $x=\lambda 1$. The map $\lambda \mapsto \lambda 1$ is a bijective complex algebra homomorphism, andIt and its inverse are continuous; when $\Vert 1\Vert =1$, it is an isometry.

A complete algebra norm on a function algebra dominates the supremum norm, even before continuity of point evaluations is known. Let $A$ be the given algebra of functions. Form the same artificial unitization $A^{+}$ even if $A$ already has an identity. For each $t\in K$, the algebraic mapis a unital multiplicative complex linear functional. No continuity has been assumed. Since ${\chi }_t(f-f(t)1)=0$, the element $f-f(t)1$ cannot be invertible: applying ${\chi }_t$ to an inverse equation would give $0=1$. Hence $f(t)\in {\sigma }_{A^{+}}(f)$, and the spectrum bound already proved yieldsTaking the supremum proves the supremum bound for a complete function-algebra norm:In particular every $f$ is bounded, and every point evaluation is a bounded linear functional of norm at most $1$. This is an instance of automatic continuity of characters: a character of an algebra on a Banach algebra is bounded because its value at any element belongs to that element's spectrum.

Take $Y=X^{\ast \ast }$, which is a Banach space by the completeness of the dual just proved. The canonical embedding into the bidual isFor fixed $x$ this is a bounded linear functional on $X^{\ast }$, since $|f(x)|\le \Vert f\Vert \Vert x\Vert$, and $\Phi$ is linear in $x$. This gives $\Vert \Phi (x)\Vert \le \Vert x\Vert$. For $x\ne 0$, the norming functional from the Hahn-Banach theorem has $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, giving the reverse inequality. At zero equality is immediate. ThusIn particular $\Phi$ is injective and isometric. If a dense isometric embedding is wanted, rather than merely the stated embedding, take the norm closure of $\Phi (X)$ in $X^{\ast \ast }$; a closed subspace of a Banach space is complete and supplies a completion of a normed space.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.