Normed vector space ($(X,\Vert \cdot \Vert )$)

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

A total family has dense linear span; it is minimal when no vector lies in the closed span of all the others. Equivalently, no proper subfamily remains total. Such a family has unique bounded coordinate functionals making a biorthogonal system. Indeed put $M_i=\overline{\mathrm{span}}\{x_j:j\ne i\}$ and $d_i=\mathrm{dist}(x_i,M_i)>0$. The functional $m+t{x}_i\mapsto t$ on $M_i+\mathbb{R}{x}_i$ is bounded by $1/d_i$, so the Hahn-Banach theorem extends it to $X$. Its norm is exactly $1/d_i$, by the distance infimum. Uniqueness follows because its values on the total family determine it on a dense subspace. For a countable family, minimality and totality alone do not ensure norm-convergent expansions; the uniform partial-sum criterion for a Schauder basis supplies the additional condition.

In a nonzero normed vector space, a unit vector $x$ is approached from outside the unit ball by $(1+\epsilon )x$, so no boundary point is interior. It is approached from inside by $(1-\epsilon )x$, so every boundary point is in the closure of the open ball. The zero space satisfies the same identities trivially. These statements need the radial structure of a normed space; a two-point discrete metric space with interpoint distance one disproves both analogous claims for general metric balls.

A normed vector space is uniformly convex if, for each $\epsilon >0$, there is $\delta >0$ such that unit-ball vectors $x,y$ with $\Vert x-y\Vert \ge \epsilon$ satisfy $\Vert (x+y)/2\Vert \le 1-\delta$. Completeness is not part of this definition; a complete example is a uniformly convex Banach space.

A normed vector space is strictly convex when distinct unit vectors have their midpoint strictly inside the unit ball. The Lp space is strictly convex for $1<p<\infty$: strict convexity of $|z{|}^p$ gives a strict integrated midpoint inequality for distinct unit functions. The endpoint spaces $L^1$ and $L^{\infty }$ need not have this property.

If two distinct elements $u,v$ of a linear subspace both minimize distance to $f$, normalize their residuals by the common positive minimum $d$. Strictly convex normed space geometry gives $\Vert f-(u+v)/2\Vert <d$, a contradiction. If $d=0$, both elements equal $f$. This proves uniqueness when a best approximant exists, without asserting existence for a nonclosed subspace.

A strictly convex norm permits at most one best approximation from a linear subspace, but does not force the distance infimum to be attained. In the Hilbert space ${\ell }^2$, the finite-support vector subspace $c_{00}$ is dense and omits $u=(2^{-j}{)}_{j\ge 1}$. Its distance from $u$ is zero and no member attains that distance. Finite-dimensionality restores existence by compactness of bounded sets in the vector subspace. If $Y$ is a bounded projection onto a vector subspace, the estimate $\Vert u-Yu\Vert \le (1+\Vert Y\Vert ){\inf }_q\Vert u-q\Vert$ follows directly by writing $u-Yu=(I-Y)(u-q)$ for each candidate $q$, even when the infimum is not attained.

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.

For $1<p<\infty$ and $n\ge 1$, existence follows from finite-dimensional compactness and uniqueness from strictly convex normed space geometry. Translation by $2\pi /n$ and negative translation by $\pi /n$ preserve the target and the error norm. Uniqueness forces the minimizing trigonometric polynomial to be invariant under the first operation and the negative of itself under the second. Invariance removes every nonzero frequency below $n$; negative invariance removes the remaining constant.

For real functions, let $r=f-p_{\ast }$ and put $\mathrm{sign}(0)=0$. If $\int v\mathrm{sign}(r)=0$ for every $v\in U$, then $p_{\ast }$ is a best approximation in a normed space using the $L^1$ norm. The proof is the pointwise inequality $|r-v|\ge |r|-v\mathrm{sign}(r)$, followed by integration. This is a sufficient criterion even if the residual vanishes on a positive-measure set; a general necessary criterion may choose any subgradient in $[-1,1]$ on that zero set.

The volume ratio of a $d$-dimensional normed vector space is $\mathrm{vr}(E)=(\mathrm{vol}(B_E)/\mathrm{vol}(\mathcal{E}){)}^{1/d}$, where $\mathcal{E}$ is its John ellipsoid. It is unchanged by invertible linear changes of coordinates. In John position, polar integration gives $\mathrm{vr}(E{)}^d={\int }_{S^{d-1}}\Vert u{\Vert }_E^{-d}d\sigma (u)$, with normalized spherical measure. Consequently $\sigma \{\Vert u{\Vert }_E\le t\}\le (\mathrm{vr}(E)t{)}^d$.

In a $2k$-dimensional normed vector space in John position, two orthogonal $k$-dimensional subspaces admit $\Vert x{\Vert }_E\le |x|\le 32\mathrm{vr}(E{)}^2\Vert x{\Vert }_E$. To see this, take $\epsilon$-nets in two orthogonal half-dimensional unit spheres, each of size at most $(1+2/\epsilon {)}^k$, and rotate them uniformly together. The volume ratio bound makes the probability that any net point has norm below $2\epsilon$ at most $2[4\mathrm{vr}(E{)}^2({\epsilon }^2+2\epsilon ){]}^k$. With $\epsilon =(32\mathrm{vr}(E{)}^2{)}^{-1}$ this is less than $1$. Since $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric, the successful nets give the stated inequality everywhere on both spheres. A dimension-independent constant therefore requires a bound on the volume ratio; it does not hold for arbitrary normed vector spaces.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

Here is a fully computed endpoint version, now using the L2 norm on the actual continuous functions $C([0,1])$. For each integer $n\ge 1$, set $f_n(x)=\max (1-nx,0)$. Its point evaluation at zero is $E(f_n)=f_n(0)=1$, whileConsequently $f_n\to 0$ in the L2 norm but $E(f_n)\nrightarrow 0$: point evaluation is discontinuous on this normed space. Quantitatively $|E(f_n)|/\Vert f_n{\Vert }_2=\sqrt{3n}$ is unbounded. At $n=100$, the L2 norm is approximately $0.057735$, the integral is $0.005$, and the point value is still $1$. In the supremum norm, $\Vert f_n{\Vert }_{\infty }=1$ and $|E(f)|\le \Vert f{\Vert }_{\infty }$, so $E$ has operator norm one instead.

Set $g_n=1-f_n$. Then $E(g_n)=0$ and $\Vert g_n-1{\Vert }_2=1/\sqrt{3n}\to 0$, although $E(1)=1$. Thus the kernel of $E$ fails to be a closed set. On the full L2 space, a further distinction is essential: its elements are equivalence classes under equality almost everywhere, and changing a representative at zero does not change its class. Arbitrary point evaluation is therefore not even well-defined there. Restricting first to continuous functions makes the discontinuity statement precise.^[ref]

In a normed vector space, this set consists of all points at a fixed positive norm distance $r$ from $x_0$. It also makes sense in an infinite-dimensional Hilbert space, unlike a specifically Euclidean sphere. In a real Hilbert space, the sphere in a normed vector space $\Vert u{\Vert }^2=2E_0>0$ has tangent variations $v$ satisfying $(u,v)=0$; this follows by differentiating the fixed squared norm. This is the geometric constraint used in fixed-energy initial-condition optimality.

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]

For $\lambda \ge 1$, a normed vector space $X$ is lambda-injective when every bounded linear operator from a subspace $Y$ of any normed space $Z$ into $X$ extends to $Z$ with the displayed norm bound. The space of bounded scalar functions on an index set is 1-injective: extend each coordinate functional by the Hahn-Banach theorem and reassemble it using the coordinate functional representation of an operator into bounded indexed functions.

A space is a lambda-injective normed space exactly when every linear isometry $J:X\to Z$ has a bounded left inverse as displayed. Necessity extends the inverse on $J(X)$. For sufficiency, embed $X$ in bounded scalar functions on an index set, extend the composed operator coordinatewise, and compose that extension with the assumed left inverse. The associated operator $JP$ is a projection onto $J(X)$.

The norm topology is the metric topology induced by $d(x,y)=\Vert x-y\Vert$.

A sequence converges in norm when $\Vert x_n-x\Vert \to 0$.

A sequence in L2 space converges in its L2 norm when $\int |f_n-f{|}^{2}d\mu \to 0$. This concerns the integral of the squared error and need not imply pointwise convergence. In particular a complete Fourier sine basis reconstructs each odd L2 function in this norm, while pointwise claims about its Fourier series require additional hypotheses.

Every finite-dimensional vector space subspace of a normed vector space is closed. Coordinates in a finite basis identify it homeomorphically with a finite-dimensional scalar space; it is therefore complete, and every complete subspace of a metric space is closed.

An isometric isomorphism between normed spaces is a bijective linear map $T$ satisfying $\Vert Tx\Vert =\Vert x\Vert$ for every vector $x$.

A norm assigns a nonnegative length to each vector, is zero only at the zero vector, is absolutely homogeneous, and satisfies the triangle inequality.

For a real or complex polynomial $p(t)={\sum }_{k\ge 0}{a}_k{t}^k$, this finite sum defines a norm: definiteness follows from uniqueness of polynomial coefficients, absolute homogeneity of a norm follows termwise, and the triangle inequality follows by summing $|a_k+b_k|\le |a_k|+|b_k|$. On $[0,1]$, its supremum norm satisfies $\Vert p{\Vert }_{\infty }\le \Vert p{\Vert }_{\mathrm{coef},1}$, since $|t^k|\le 1$. The reverse comparison has no constant independent of the degree of a polynomial: for $p_n(t)=(1-t{)}^n$, the two norms are $1$ and $2^n$ respectively, by the binomial theorem.

A quasi-norm on a vector space is nonnegative, vanishes only at zero, is absolutely homogeneous, and satisfies a triangle inequality with one constant factor $C\ge 1$. For $0<p<1$, $(\int |f{|}^p{)}^{1/p}$ on an Lp space is a quasi-norm: $\Vert f+g{\Vert }_p^p\le \Vert f{\Vert }_p^p+\Vert g{\Vert }_p^p$ gives a quasi-triangle inequality. Measure-preserving composition preserves this quantity even though the Minkowski inequality is unavailable below exponent one.

A dual norm measures how strongly a bounded linear functional can respond to a unit-sized input. If $p$ is a bounded linear functional on a normed vector space $X$, write $⟨p,x⟩=p(x)$ and defineIt is the operator norm of $p$ when its scalar output is measured by absolute value. On ${\mathbb{R}}^n$, represent $p(x)=a^{T}x$ using a coefficient vector $a$. The unit ball is symmetric, so maximizing $|a^{T}x|$ gives the same value as maximizing $a^{T}x$: negate an input whenever its output is negative.^[ref]

For an original two-coordinate example, fix $a=(2,-1{)}^T$, so $p(x)=2x_1-x_2$. The input norm changes the allowable unit ball and hence the dual norm:Under the L1 norm, $|2x_1-x_2|\le 2(|x_1|+|x_2|)\le 2$, and $(1,0{)}^T$ attains the bound. Under the Euclidean norm, the Cauchy-Schwarz inequality gives $|a^{T}x|\le \Vert a{\Vert }_2\Vert x{\Vert }_2\le \sqrt{5}$, with equality at the displayed normalized vector. Under the maximum norm, each coordinate can reach unit absolute value, so $|2x_1-x_2|\le 2+1=3$; $(1,-1{)}^T$ attains it. Thus the dual norms of these three vector norms are respectively the maximum norm, the Euclidean norm, and the L1 norm. For $1\le p\le \infty$, Lp norms on a finite coordinate space pair with their Holder conjugate exponents.^[ref]

This geometry explains the usefulness of a dual norm: it translates a bound on inputs into the sharp bound on a linear output. The support function of the unit ball gives its value in the coefficient direction $a$. Changing the input budget changes the set of possible inputs; the three different answers do not conflict. In finite dimensions a closed unit ball is compact, so a continuous linear functional achieves its maximum absolute value. In an infinite-dimensional normed vector space, the supremum remains the definition and need not be achieved.

Suppose a perturbation $\delta x$ satisfies $\Vert \delta x\Vert \le \epsilon$, with $\epsilon \ge 0$. Absolute homogeneity of a norm and the dual norm definition giveFor nonzero $\delta x$, divide it by its norm to obtain an input in the unit ball; for zero perturbation the inequality is immediate. This is an exact error bound for a linear functional, for every perturbation in the stated input set.

For $p(x)=2x_1-x_2$ and a budget $\epsilon =0.1$, the sharp output-error bounds are $0.2$ for $\Vert \delta x{\Vert }_1\le 0.1$, $\sqrt{5}/10\simeq 0.224$ for $\Vert \delta x{\Vert }_2\le 0.1$, and $0.3$ for $\Vert \delta x{\Vert }_{\infty }\le 0.1$. Scale the maximizing vectors from the preceding example by $0.1$ to attain the positive endpoints; their negatives attain the negative endpoints. The L1 norm budget constrains a total absolute error; the maximum norm budget permits each coordinate to have that error separately. These budgets define nested sets and carry different information.

The coordinates must use the stated scale and units. Rescaling one coordinate changes both its coefficient in $a$ and the meaning of its input budget. For a nonlinear differentiable function with scalar values, its Fréchet derivative at the base point is a linear functional governing the first-order response. An exact finite-perturbation error bound additionally needs control of the Taylor remainder or of the Fréchet derivative along the path.

On the real space of sequences converging to zero $c_0$, use the supremum norm and define the linear functionalThe geometric series gives $|p(x)|\le \Vert x{\Vert }_{\infty }$, so $p$ is a bounded linear functional and $\Vert p{\Vert }_{\ast }\le 1$. The sequence with its first $N$ entries equal to one and all later entries zero belongs to $c_0$, has unit supremum norm, and gives $p(x)=1-2^{-N}$. Hence $\Vert p{\Vert }_{\ast }=1$. But if $\Vert x{\Vert }_{\infty }\le 1$ and $x_n\to 0$, not all coordinates can equal one or all equal minus one. Since every weight $2^{-n}$ is positive, $-1<p(x)<1$. The supremum is approached but never attained on the unit ball.

The block mixed norm sums the ${\ell }^q$ norm of each block, for $q\ge 1$. Its dual norm is $\Vert p{\Vert }_{q^{\prime },\infty }={\max }_i\Vert p_i{\Vert }_{q^{\prime }}$, where $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Apply Holder inequality blockwise and choose a norming block to attain equality. Absolute values are essential for signed components.

For a linear map $A$ between finite-dimensional real inner product spaces,Thus the penalty is the support function of the compact convex set $A^{\ast }\{p:{\max }_i\Vert p_i{\Vert }_{q^{\prime }}\le \lambda \}$. The Moreau decomposition makes its proximal operator the identity minus the Euclidean projection onto a convex set.

Definiteness says that a norm vanishes only at the zero vector. The box norm on a nonempty finite Cartesian product is definite because its defining sum of squares includes equal-column terms, each of which is the square of an average of squares. Omitting repeated coordinates would destroy this proof and can produce a different quantity.

Absolute homogeneity is the norm axiom $\Vert av\Vert =|a|\Vert v\Vert$ for every scalar $a$ and vector $v$. It distinguishes a norm from a merely homogeneous expression of a different degree; for example the fourth root in the box norm converts a degree-four expression into a degree-one norm.

Suppose $B$ contains zero, is convex, and meets every line $\mathbb{R}v$, $v\ne 0$, in the finite symmetric open segment $\{tv:|t|<\lambda (v)\}$ with $\lambda (v)>0$. Then its Minkowski functional is a norm, with $p_B(v)=1/\lambda (v)$ and $B=\{x:p_B(x)<1\}$. Absolute homogeneity follows by rescaling the segment. If $s>p_B(x)$ and $t>p_B(y)$, convexity puts $(x+y)/(s+t)$ in $B$, giving $p_B(x+y)<s+t$; passing to the infima proves the triangle inequality. Conversely the open unit ball uniquely determines any norm by the displayed formula. No prior topology on the vector space is needed for these radial hypotheses.

On a uniform $d$-dimensional spatial grid, the discrete L2 norm weights the squared Euclidean norm by the cell volume. It approximates the continuous L2 norm and is induced by the discrete L2 inner product. A contraction bound in this norm with a mesh-independent constant proves stability of a numerical method for the spatially discretized evolution.

For $x\in {\mathbb{R}}^d$, the L1 norm is

For $0<c\le w_k<\infty$, interpret the nonnegative sum as an extended-valued functional on the l2 sequence space. Finiteness implies $u\in {\ell }^1$, but unbounded weights can make it infinite even there. It is proper and convex. Every finite partial sum is norm-continuous, so their supremum is sequentially lower semicontinuous and also weakly lower semicontinuous. This is useful as a sparsity penalty in variational regularization.

The coordinate conditions define a subgradient precisely when $p\in {\ell }^2$ and $J_w(u)<\infty$. They follow by varying one coordinate; summing their scalar supporting inequalities proves sufficiency. Since $w_k\ge c>0$, a subgradient can exist only at finitely supported $u$. At such a base point $v$, $D_{J_w}^p(u,v)={\sum }_k(w_k|u_k|-p_k{u}_k)$, allowing the value infinity. Merely having finite penalty does not guarantee a nonempty subdifferential in the ambient l2 sequence space.

Taxicab geometry equips Euclidean coordinate space with the distance induced by the L1 norm, so the distance between two points is the sum of their coordinatewise absolute differences.

The trace norm of a finite-dimensional operator $A$ is the sum of its singular values:For a Hermitian operator $A=A_{+}-A_{-}$, it equals $\mathrm{Tr}{A}_{+}+\mathrm{Tr}{A}_{-}$.

For a finite square matrix $X$, maximize over unitary operators. Write its singular value decomposition as $X=PD{Q}^{†}$. Then $|\mathrm{Tr}(XU)|=|\mathrm{Tr}(D{Q}^{†}UP)|\le {\sum }_j{D}_{jj}$, since every diagonal entry of a unitary has modulus at most one. Choosing $U=Q{P}^{†}$ attains equality, proving the formula for the trace norm.

A Hermitian operator satisfies $\Vert X{\Vert }_1={\max }_{-I\le T\le I}\mathrm{Tr}(XT)$. Each diagonal entry of $T$ in an eigenbasis of $X$ lies between minus one and one, giving the upper bound ${\sum }_i|{\lambda }_i|$. The sign operator of $X$ attains it. The substitution $T=2E-I$ converts this formula to optimization over binary POVM effects and proves the Holevo–Helstrom theorem.

For any orthonormal basis and Hermitian operator, ${\sum }_i|⟨{\psi }_i|X|{\psi }_i⟩|\le \Vert X{\Vert }_1$. In the trace-norm variational principle for Hermitian operators, choose the diagonal $T$ whose entries are the signs of these diagonal entries. This bounds the total distinguishability visible in one fixed measurement basis.

The induced trace norm of a linear map $T$ between matrix spaces isIt measures the largest trace-norm amplification available without an auxiliary tensor factor.

The diamond norm is the stabilized induced trace normFor a map on $d$-dimensional inputs, an auxiliary space of dimension $d$ suffices. The tensor factor lets the norm detect how $T$ acts on one half of an entangled input.

The Euclidean norm of $x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n$ is

For an orthogonal matrix $Q$, the identity $Q^{T}Q=I$ gives $\Vert Qx{\Vert }_2^2=x^T{Q}^{T}Qx=x^{T}x$. Thus multiplication by $Q$ preserves the Euclidean norm. Applying this on both input and output sides proves $\Vert UA{V}^T{\Vert }_2=\Vert A{\Vert }_2$ for orthogonal matrices $U,V$ and the Euclidean operator norm. The analogous invariance of the Frobenius norm follows by summing column lengths or by a trace calculation.

For a nonzero vector $x\in {\mathbb{R}}^n$, put $r=|x|$ and $\hat{x}=x/r$. The gradient of the Euclidean norm is $\nabla r=\hat{x}$. Differentiate its components to obtain the Hessian matrixThe factor $I-\hat{x}{\hat{x}}^T$ is the orthogonal projection matrix onto the orthogonal complement of the radial direction. Thus the Hessian matrix has eigenvalue zero radially and $1/r$ on each tangential direction, explaining the convexity of the Euclidean norm. Taking the trace gives $\Delta r=(n-1)/r$ away from the origin. The Euclidean norm is not differentiable at the origin: its restriction to any line is a nonzero multiple of the absolute value.

For a real vector $v$, the square of its Euclidean norm is $\Vert v{\Vert }_2^2=v\cdot v={\sum }_j{v}_j^2$. Unlike the norm itself it is a quadratic function, so expanding sums directly exposes diagonal and cross terms. This is useful for the mean-square displacement of an isotropic planar random walk.

The Euclidean norm is self-dual:The upper bound is Cauchy-Schwarz inequality, and equality holds at $v=x/\Vert x{\Vert }_2$ when $x\ne 0$.

This is a ball for the Euclidean metric.

The closed Euclidean ball of radius $R$ about $a\in {\mathbb{R}}^n$ is $\{x:\Vert x-a{\Vert }_2\le R\}$; replacing $\le$ by $<$ gives the open ball.

The Euclidean unit ball is the closed unit ball of the Euclidean norm, centered at zero. Its radial function is identically $1$, making it a natural reference body for John ellipsoids and volume ratios.

The unit ball of a normed vector space is $\{x:\Vert x\Vert \le 1\}$ under the closed-ball convention.

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.

The open unit ball of a normed vector space $X$ is $B_X=\{x\in X:\Vert x\Vert <1\}$.

The supremum norm of a bounded real- or complex-valued function on a set $X$ isFor a continuous function on a compact space, the extreme value theorem makes this supremum a maximum.

Let $K$ be a nonempty compact space and let $\alpha :K\to \mathbb{R}$ be continuous and nowhere zero. On $C(K)$, define $\Vert f{\Vert }_{\alpha }={\sup }_{x\in K}|\alpha (x)f(x)|$. Multiplication by $\alpha$ is an isometry onto $C(K)$ with its ordinary uniform norm, because division by the continuous nonvanishing $\alpha$ provides the inverse. Thus the weighted space is a Banach space. Equivalently, the extreme value theorem gives $0<m=\min |\alpha |\le \max |\alpha |=M<\infty$, and $m\Vert f{\Vert }_{\infty }\le \Vert f{\Vert }_{\alpha }\le M\Vert f{\Vert }_{\infty }$. A weighted Cauchy sequence is uniformly Cauchy, has a continuous uniform limit, and the upper bound gives convergence also in the weighted norm.

On a nondegenerate interval, point evaluation of the derivative is unbounded for the supremum norm on smooth functions. The functions $f_N(x)=\sin (N(x-x_0))$ have supremum norm at most one but $f_N^{\prime }(x_0)=N$. Adding the supremum norm of the derivative makes this evaluation a bounded linear functional.

For a function with bounded derivatives through order $k$, the derivative supremum norm is $|u{|}_{k;U}={\sum }_{j=0}^k{\sup }_U|D^{j}u|$, with any equivalent finite-dimensional norm for each derivative tensor. This is a classical $C^k$ norm; it is different from the Lp norm with exponent $k$.

A surjective isometry $u:V\to W$ between normed vector spaces preserves every distance:

Every surjective isometry between real normed vector spaces is affine. In particular, if $u(0)=0$, then $u$ is real-linear.

For $v,w$ in a real normed space, start with the points at distance $\Vert v-w\Vert /2$ from both endpoints and repeatedly retain those within half the previous diameter of every retained point. Reflection about $(v+w)/2$ preserves every stage, the midpoint belongs to every stage, and the diameters decrease by a factor at least two. Their intersection is therefore exactly the midpoint.

Surjectivity in the Mazur-Ulam theorem is essential. The mappreserves all distances and fixes zero, but it is not linear and does not preserve every midpoint.

Two norms on one vector space are equivalent when constants $c,C>0$ satisfyfor every vector. Equivalent norms define the same convergent sequences, open sets, and continuous linear functionals.

Two norms $p,q$ on a vector space give the same topology if and only if $cp(x)\le q(x)\le Cp(x)$ for some positive constants. Indeed, continuity of the identity at zero gives $p(x)<\delta \Rightarrow q(x)<1$. For $x\ne 0$, apply this to $\delta x/(2p(x))$ to obtain $q(x)<2p(x)/\delta$. Reverse the roles of the norms for the lower bound. Conversely the two inequalities compare every translated open ball. This proof uses absolute homogeneity of a norm, not finite dimension.

Relative to a Schauder basis, take the supremum over successive finite intervals whose minima belong to a specified hereditary family $\mathcal{F}$, and signs ${\epsilon }_i\in \{-1,1\}$. If this supremum is bounded above by a multiple of the original norm and the family contains all singletons, it gives an equivalent norm. Composing two admitted signed interval projections costs at most a factor of two in this norm: split their nonempty intersections according to which interval supplies the later left endpoint. In each of the two resulting families the minima form a subset of one original list, so heredity preserves admissibility. This estimate lets bounded distortion of a Banach space give a uniform bound independent of the complexity of $\mathcal{F}$.

Bounded distortion means there is one finite bound on the ratio between the largest and smallest relative sizes of any equivalent norm after restriction to a further infinite-dimensional block subspace. The comparison is invariant under scalar multiplication of the new norm. A literal two-sided bound without an allowed rescaling would fail even for a scalar multiple of the original norm.

On $C(X)$ for compact Hausdorff $X$, any complete norm making every point evaluation functional continuous is equivalent to the uniform norm. If $f_n$ converges in the new norm to $f$ and uniformly to $g$, evaluation gives $f(x)=g(x)$ for every $x$, so the identity map between the two Banach spaces has closed graph. The closed graph theorem and open mapping theorem make the identity and its inverse bounded. Completeness of the proposed norm is essential to this argument.

Any two norms on a finite-dimensional real or complex vector space are equivalent. Relative to a fixed basis, continuity and positivity of a norm on the compact Euclidean unit sphere give uniform positive lower and finite upper bounds.

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.