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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 bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

For every nonempty closed convex set $C$ in a Hilbert space and every $x$ in that space, there is a unique $p\in C$ satisfyingClosedness and completeness let a limiting approximation attain the infimum; convexity prevents two different minimizers. The proof idea is to choose a sequence approaching the infimum. The parallelogram identity and the admissible midpoint make it a Cauchy sequence; applying the same identity to two putative minimizers proves uniqueness.

For a closed subspace of a Hilbert space, this nearest-point map is an orthogonal projection and is linear. For a general closed convex set it need not be linear: projecting the real line onto $[0,1]$ clips values below zero to zero and values above one to one. The theorem is therefore useful for both linear approximation and constrained nearest-point problems.^[ref]

Closedness is essential even in familiar examples. For the open interval $(0,1)\subseteq \mathbb{R}$ and $x=0$, the distance infimum is zero but no nearest point exists. The same obstruction appears when approximating $(1/k{)}_{k\ge 1}$ by finitely supported sequences: the infimum is zero, but the limit lies outside that subspace until we take its Hilbert space completion.

The Hilbert space completion of an inner product space $V$ is a Hilbert space containing an isometric dense copy of $V$. It can be constructed from Cauchy sequences in $V$, identifying two sequences when the norm of their difference tends to zero.

The space of finitely supported sequences with the l2 sequence space norm illustrates the need for completion. Let $u^{(n)}=(1,1/2,\dots ,1/n,0,\dots )$. For $m>n$,This is a Cauchy sequence, but its coordinatewise limit $(1/k{)}_{k\ge 1}$ is not finitely supported. It belongs to the l2 sequence space, which is the completion. The same construction permits infinitely many small components while controlling their total squared size. An infinite orthonormal basis of a Hilbert space consequently permits norm-convergent series; a Hamel basis represents each vector by a finite sum, so these two uses of "basis" must be distinguished.

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.

Here “subspace” is understood to mean a closed closed linear subspace. Closedness matters: the finitely supported sequences $c_{00}$ are an infinite-dimensional algebraic subspace of ${\ell }^p$, but contain no infinite-dimensional complete subspace. Indeed, any such subspace would be the countable union of its intersections with the finite-dimensional coordinate spaces, contrary to the Baire category theorem. We prove the assertion with the intended closedness convention, for $1\le p<\infty$.

Write $E={\ell }^p$, the l-p sequence space. Choose positive numbers ${\delta }_n$ with ${\sum }_{n}2{\delta }_n<1$. Starting with $m_0=0$, choose $x_n\in X$ with $\Vert x_n{\Vert }_p=1$ and its first $m_{n-1}$ coordinates zero. This is possible because the kernel of finitely many coordinate linear functionals on an infinite-dimensional vector space is still infinite-dimensional. Choose $m_n>m_{n-1}$ so that the part of $x_n$ after coordinate $m_n$ has norm below ${\delta }_n$. Let $v_n$ be its restriction to $I_n=(m_{n-1},m_n]$, and put $u_n=v_n/\Vert v_n{\Vert }_p$. The triangle inequality givesThe vectors $u_n$ form a normalized block basic sequence. Their supports are disjoint, soConsequently $U=\overline{\mathrm{span}}\{u_n\}$ is isometric to the l-p sequence space.

For each block choose a norm-one linear functional ${\varphi }_n$ supported on $I_n$ with ${\varphi }_n(u_n)=1$. The Hahn-Banach theorem supplies such a functional, including when $p=1$. DefineThe series converges in the l-p sequence space, sinceAlso $Q{u}_n=u_n$, so $Q$ is a bounded projection onto $U$. Now define the bounded linear operatorThis series converges in operator norm and $\Vert A-I\Vert \le {\sum }_{n}2{\delta }_n<1$. The Neumann series therefore makes $A$ invertible. Disjoint support gives $A{u}_n=x_n$. Thus $Y=AU=\overline{\mathrm{span}}\{x_n\}$ lies in the closed subspace $X$, is isomorphic to the l-p sequence space, and is the range of the bounded projection $AQ{A}^{-1}$ on the whole space $E$. This proves the required copy of ${\ell }^p$ is complemented in the ambient ${\ell }^p$, not just in $X$.

Now let $Z$ be an infinite-dimensional complemented subspace of $E$. The construction supplies $Y\subset Z$ with $Y\cong E$ and a bounded projection $R:E\to Y$. Its restriction to $Z$ gives the topological direct sumIf $P:E\to Z$ is a bounded projection, then $(I-R)P$ is a bounded projection onto $W$: its range lies in $W$, and it fixes every element of $W$. Hence $W$ is a complemented subspace of $E$, so $E\cong W\oplus V$ for some closed $V$.

We give the absorption argument in the Pełczyński decomposition method explicitly. Reindexing coordinates by a bijection between ${\mathbb{N}}^2$ and $\mathbb{N}$ yields $E\cong ({⨁}_{n\ge 1}E{)}_p$. Applying the fixed Banach space isomorphism $E\cong W\oplus V$ in every coordinate givesThe extra $W$ is absorbed by shifting the countably many $W$ summands. The maps are bounded with bounded inverses because the same two-space decomposition is used uniformly in each summand; we may equip each finite direct sum with its equivalent $p$-sum norm. Since $Z\cong E\oplus W$, $Z\cong {\ell }^p$.

Read equality of the two positive series as equality of a finite total $M$, so the sequences are summable. This is the majorization of summable sequences setting. If equality is instead allowed to mean $\infty =\infty$, the assertion is false: the strictly decreasing sequences $x_i=1+1/(i+1)$ and $y_i=2+1/(i+1)$ satisfy every partial-sum inequality and have divergent equal extended totals. But a normalized nonnegative row of any doubly stochastic matrix averages numbers all at least $2$, and cannot give $x_i<2$. We prove the intended finite-total assertion constructively.

Since $y_j\to 0$ and $0<x_1\le y_1$, there is a finite $k$ such that $y_k\ge x_1>y_{k+1}$. PutThe two-coordinate stochastic averaging matrix sends $(y_k,y_{k+1})$ to $(x_1,u)$. Move the first of these values to the front by a finite permutation, leaving the other values in decreasing order. The tail iswhich is decreasing and strictly positive because $y_{k+1}\le u\le y_k$.

We need the residual majorization in order to repeat the construction. For $n<k$, the first $n$ values of $z$ are at least $x_1$, while every target-tail value is at most $x_1$. ThusFor $n\ge k$, the residual initial sum is the old initial sum with $x_1$ removed:The tail totals also agree, both equalling $M-x_1$. Therefore the same step applies to $(x_2,x_3,\dots )$ and $z$.

Iterating, let $P^{(m)}$ be the product of the first $m$ finite permutations and embedded averaging matrices. Each $P^{(m)}$ is a doubly stochastic matrix, each row is a finitely supported sequence, and the first $m$ entries of $P^{(m)}y$ are $x_1,\dots ,x_m$. Later operations act only on tail rows, so row $i$ becomes permanently fixed once $m\ge i$. Define $p_{ij}$ to be that stabilized row. ThenFor every fixed column $j$ and every $m$, its first $m$ stabilized entries sum to at most one, since they are entries of column $j$ in $P^{(m)}$. Consequently $s_j={\sum }_i{p}_{ij}\le 1$. By Tonelli's theorem,All $y_j$ are strictly positive and every summand in the last expression is nonnegative, so $s_j=1$ for every $j$. Thus both the row sums and the column sums are one, andThe finite-total argument is what prevents loss of column mass in the infinite construction. This completes the doubly stochastic realization of summable-sequence majorization.

Every finitely supported sequence $v$ lies in $Y$: if $v_n=0$ for $n>N$, set $x_n={\sum }_{k=n}^N{v}_k$ for $n\le N$ and $x_n=0$ afterwards. Then $x$ is a bounded sequence and $x-Sx=v$. Consequently $L(v)=0$. The finitely supported sequences are dense in the space of sequences converging to zero $c_0$ for the supremum norm, by truncation, so continuity of the bounded linear functional $L$ gives $L{|}_{c_0}=0$. For $x$ in the real convergent sequence space, with ordinary limit $\ell$, we have $x-\ell 1\in c_0$. Therefore $\boxed{L(x)=\ell L(1)=\underset{n\to \infty }{\lim }{x}_n}$.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.