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For a finite-dimensional normed vector space $E$, represent a finite family by its column operator $A:{\ell }_2^m\to E$. Its operator norm is the family's weak square norm. Since $A=AQ$ for the orthogonal projection onto $(\ker A{)}^{\perp }$, summing $\Vert A{e}_j{\Vert }^2\le \Vert A{\Vert }^2\Vert Q{e}_j{\Vert }_2^2$ proves the upper bound $\sqrt{\dim E}$. In coordinates adapted to a maximal inscribed Euclidean or Hermitian ellipsoid, the John contact decomposition gives ${\sum }_r{c}_r{u}_r{u}_r^{\ast }=I$ and ${\sum }_r{c}_r=\dim E$. The vectors $\sqrt{c_r}{u}_r$ have total squared norm $\dim E$ and weak square norm one, giving equality. This holds over both real and complex scalars.

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.

Suppose a ring has an underlying real normed vector space and its norm is multiplicative, without initially assuming real bilinearity of multiplication. For fixed $x$, distributivity makes $y\mapsto xy$ additive, while $\Vert xy-xz\Vert =\Vert x\Vert \Vert y-z\Vert$ makes it continuous. Additivity gives rational scalar compatibility, and continuity extends it to every real scalar. Apply the same argument in the other variable to obtain a bilinear map. The norm identity also excludes nonzero zero divisors.

For isomorphic finite-dimensional normed vector spaces, the Banach-Mazur distance is the infimum of $\Vert S\Vert \Vert S^{-1}\Vert$ over their linear isomorphisms. If an $N$-dimensional subspace $E$ of a Banach space of cotype 2 satisfies the summing estimate ${\pi }_2(T)\le L\Vert T\Vert$ for operators from ${\ell }_{\infty }^N$, then $d(E,{\ell }_{\infty }^N)\ge \sqrt{N}/L$: compose an isomorphism with the inclusion and use ${\pi }_2(I_E)=\sqrt{N}$.

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.

Given $f$ in a normed vector space and a vector subspace $U$, an element $p_{\ast }\in U$ is a best approximation if $\Vert f-p_{\ast }\Vert ={\inf }_{p\in U}\Vert f-p\Vert$. Existence is automatic for a finite-dimensional vector subspace: a minimizing sequence can be bounded by the triangle inequality, and a subsequence converges by compactness in finite dimension. An arbitrary infinite-dimensional vector subspace need not contain a minimizer. The norm matters: best uniform approximation uses the supremum norm, while $L^1$ approximation minimizes integrated absolute error.

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.

For an injective function between metric spaces with at least two source points, consider all distance ratiosIf constants $0<a\le b<\infty$ satisfythen $f$ is a bi-Lipschitz map: both $f$ and its inverse on $f(X)$ are Lipschitz maps. The bound $b/a$ compares the most expanded pair with the most contracted pair. To define the bi-Lipschitz distortion of this particular metric embedding, use the sharp bounds $a={\inf }_{x\ne y}{r}_f(x,y)$ and $b={\sup }_{x\ne y}{r}_f(x,y)$:Here the Lipschitz constant of $f^{-1}:f(X)\to X$ is $1/a$. If no positive finite bounds exist, the bi-Lipschitz distortion is infinite. For a finite source, every injective function has finite bi-Lipschitz distortion, but its size can depend on the source. Matoušek gives the equivalent definition using one scale factor and a multiplicative error bound. ^[ref]

Changing units is harmless: replacing $d_Y$ by $\lambda d_Y$, with $\lambda >0$, multiplies both $a$ and $b$ by $\lambda$ and leaves $b/a$ unchanged. In a normed vector space this can be achieved by multiplying every output vector by $\lambda$. Thus bi-Lipschitz distortion one means that all distances change by the same factor; an isometric embedding additionally requires that factor to be one. A stretch factor with a specified normalization is a related quantity; when comparing sources, check which overall scale has been fixed.

Take the four corners $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$ of a square, with Euclidean distance in both source and target. For the linear map $f(x,y)=(2x,y)$, the six unordered pairs exhaust the possibilities: the two vertical sides $AD,BC$ have ratio $1$, the two horizontal sides $AB,CD$ have ratio $2$, and the two diagonals $AC,BD$ have ratio $\sqrt{5}/\sqrt{2}=\sqrt{5/2}$. Consequently $a=1$, $b=2$, and $\mathrm{dist}(f)=2$. In contrast, $g(x,y)=(3x,3y)$, three times the identity function, gives ratio $3$ for every pair and has bi-Lipschitz distortion $3/3=1$.

The ratio chart makes the reason visible: a common multiplier moves every ratio together, while unequal coordinate scaling spreads the ratios apart. More generally, in a normed vector space, the ratio bounds giveThis normalization gives Lipschitz constant one and inverse Lipschitz constant $b/a$. It is only a change of scale, so it cannot improve the bi-Lipschitz distortion.

The best possible bi-Lipschitz distortion of a source into a specified target is a separate quantity:The infimum runs over all injective functions, and need not be attained. The notation $c_p(X)$ means the analogous infimum over metric embeddings into an Lp space. A calculation for one chosen $f$ gives an upper bound for $c_Y(X)$; proving optimality also requires a lower bound for every possible $f$.

For a normed vector space $X$, a family $(x_i,f_i{)}_{i\in I}$ in $X\times X^{\ast }$ is biorthogonal when the displayed Kronecker-delta equations hold. Its vectors are linearly independent: applying $f_i$ to a finite linear dependence isolates its $i$th coefficient. No inner product is required, and the family need not span a dense subspace. If its linear span is dense, each prescribed coordinate functional is unique by continuity. For a countable total family, the finite sums $P_{n}x={\sum }_{i\le n}{f}_i(x)x_i$ are bounded projection operators, since biorthogonality gives $P_n^2=P_n$.

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.

A bilinear form on a normed vector space is continuous exactly when it is bounded. For the nontrivial direction, continuity at $(0,0)$ gives $r,s>0$ such that $|B(x,y)|\le 1$ whenever $\Vert x\Vert <r$ and $\Vert y\Vert <s$; scaling nonzero $u,v$ into these balls gives $|B(u,v)|\le 4(rs{)}^{-1}\Vert u\Vert \Vert v\Vert$.

A bilinear map between normed vector spaces is jointly continuous exactly when it satisfies the displayed product bound. Continuity at zero gives a bound on a product of balls; rescale each nonzero input separately to obtain the inequality. Conversely, expand the increment as $F(x+h,y+k)-F(x,y)=F(h,y)+F(x,k)+F(h,k)$ and apply the product bound to every term. Thus continuity at zero already implies joint continuity everywhere for a bilinear map.

In a real normed vector space with a specified countable dense subset, a bounded linear functional on any vector subspace admits a norm-preserving extension without the axiom of choice. Add the dense vectors one at a time using the supremum endpoint in the one-dimensional dominated extension of a real linear functional. This makes the recursion deterministic. Extend from the resulting dense subspace by continuity, using the least index in the dense list which approximates a given vector to within $2^{-n}$.

A convex cone that is a closed set in its ambient topology. In a finite-dimensional normed vector space, this means it contains every limit of its convergent sequences. For example, the positive semidefinite cone is closed because each test $X\mapsto x^{T}Xx$ is continuous.

In a normed vector space, the closed linear span of $S$ is the closure of its linear span in the norm topology. It includes limits of finite linear combinations. In a Hilbert space, an infinite orthonormal basis spans in this closed sense, allowing norm-convergent series in addition to finite sums.

The closed unit ball of a normed vector space is the set of vectors of norm at most one. In a continuous dual space it is compact in the weak-star topology by the Banach-Alaoglu theorem.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

If a bounded linear operator from a Banach space onto a normed vector space admits the displayed lift for every $y$, then its target is complete. Given a Cauchy sequence, choose a subsequence with summable successive differences. Lift those differences to a summable sequence in the source, and use its convergent sum to produce a limit of the target subsequence. The original Cauchy sequence has the same limit.

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]

The continuous dual space $X^{\ast }$ of a topological vector space is the vector space of its continuous linear functionals. Unlike the algebraic dual, it requires that a small change in an input produce a small change in its scalar measurement. For a normed vector space, this is equivalent to a finite boundfor all vectors $x$ and some constant $C\ge 0$. The least such constant is the operator norm of $\phi$.

This is the continuity-restricted subspace of the algebraic dual space; in infinite dimension the two can differ.

For the linear functional $\phi (x,y)=2x-y$ on ${\mathbb{R}}^2$ with its Euclidean norm, the Cauchy-Schwarz inequality givesEquality holds on the unit vector $(2,-1)/\sqrt{5}$, so its operator norm is exactly $\sqrt{5}$. The same estimate controls measurement error: $|\phi (v+h)-\phi (v)|\le \sqrt{5}\Vert h{\Vert }_2$. The linear functional diagram shows which changes are entirely invisible to this measurement; its kernel of a linear map consists of the changes with zero response. All linear functionals on a finite-dimensional real or complex normed vector space are continuous, but discontinuous functional detected by vanishing sequence tails shows why the restriction matters in infinite dimension.