Fact-checked by Grok 2 weeks ago

Normed vector space

A normed vector space, also known as a normed linear space, is a vector space over the real or complex numbers equipped with a norm, which is a function that assigns a non-negative real number to each vector, satisfying three key axioms: positivity (the norm is zero if and only if the vector is the zero vector), absolute homogeneity (the norm of a scalar multiple is the absolute value of the scalar times the norm of the vector), and the triangle inequality (the norm of a sum is at most the sum of the norms).^[1]^[2] This structure induces a natural metric on the space, defined by the distance between two vectors as the norm of their difference, thereby turning the normed vector space into a metric space where convergence and continuity can be defined in terms of the norm.^[1] Examples include the Euclidean space \mathbb{R}^n with the p-norm \|x\|_p = \left( \sum_{i=1}^n |x_i|^p \right)^{1/p} for p \geq 1, sequence spaces like \ell^p consisting of sequences with finite p-norm, and spaces of continuous functions C[a,b] on an interval with the supremum norm \|f\|_\infty = \sup_{x \in [a,b]} |f(x)|.^[2]^[1] Normed vector spaces form the foundation of functional analysis, enabling the study of linear operators and topological properties; a normed space is called a Banach space if it is complete with respect to the metric induced by the norm, meaning every Cauchy sequence converges to an element in the space.^[1] In finite dimensions, all norms on a given vector space are equivalent, inducing the same topology, but in infinite dimensions, different norms may yield distinct topologies and completeness properties.^[2]

Foundations

Definition

A normed vector space is a vector space V over the field of real numbers \mathbb{R} or complex numbers \mathbb{C}, equipped with a norm \|\cdot\|: V \to [0, \infty) that satisfies the properties of positivity, absolute homogeneity, and the triangle inequality.^[3] Specifically, for any v \in V, the norm obeys \|v\| \geq 0, with \|v\| = 0 if and only if v = 0; for any scalar \alpha in the base field, \|\alpha v\| = |\alpha| \|v\|; and for any u, v \in V, \|u + v\| \leq \|u\| + \|v\|.^[4] The concept of normed vector spaces was introduced in the early 1920s, notably by Stefan Banach in his 1920 doctoral dissertation.^[5] Common examples include the Euclidean norm on \mathbb{R}^n, defined by \|x\|_2 = \sqrt{\sum_{i=1}^n |x_i|^2} for x = (x_1, \dots, x_n); the supremum norm on the space C[0,1] of continuous real-valued functions on [0,1], given by \|f\|_\infty = \sup_{t \in [0,1]} |f(t)|; and the \ell^p norms on sequences, such as \| (a_n) \|_p = \left( \sum_{n=1}^\infty |a_n|^p \right)^{1/p} for $1 \leq p < \infty.^[3]

Norm axioms

A norm on a vector space V over the real or complex numbers is a function \|\cdot\|: V \to [0, \infty) that satisfies three fundamental axioms for all vectors u, v \in V and scalars \alpha in the underlying field.^[6] The first axiom is positivity (or absolute homogeneity and definiteness): \|v\| \geq 0 for all v \in V, with \|v\| = 0 if and only if v = 0. This ensures the norm provides a non-negative measure of "size" that distinguishes the zero vector.^[6] The second axiom is homogeneity: \|\alpha v\| = |\alpha| \|v\| for all scalars \alpha. For \alpha = 0, this holds as \|0 \cdot v\| = \|0\| = 0 = |0| \|v\|, following directly from the positivity axiom. In the complex case, |\alpha| denotes the modulus of \alpha, ensuring the norm scales appropriately with the magnitude of the scalar while remaining real-valued.^[6]^[7] The third axiom is the triangle inequality: \|u + v\| \leq \|u\| + \|v\|. This subadditivity extends to finite sums by repeated application: for vectors v_1, \dots, v_n \in V, \left\| \sum_{i=1}^n v_i \right\| \leq \sum_{i=1}^n \|v_i\|.^[6]^[8] These axioms yield immediate corollaries. The reverse triangle inequality states that \big| \|u\| - \|v\| \big| \leq \|u - v\|, obtained by applying the triangle inequality to \|u\| = \|(u - v) + v\| \leq \|u - v\| + \|v\| and symmetrically. Additionally, \|v\| = \|-v\| follows from homogeneity with \alpha = -1, since |-1| = 1.^[9]

Properties

Algebraic properties

In a normed vector space, the norm induces several algebraic identities and inequalities that govern addition and scalar multiplication beyond the basic axioms. The triangle inequality, a core axiom, ensures subadditivity of the norm but gives rise to more refined relations, such as those involving sums and differences of vectors.^[10] A key identity is the parallelogram law, which states that for any vectors u, v in the space,

\|u + v\|^2 + \|u - v\|^2 = 2(\|u\|^2 + \|v\|^2).

This law holds in any inner product space, where the norm is derived from the inner product via \|u\|^2 = \langle u, u \rangle, as it follows directly from the bilinearity and symmetry of the inner product. Conversely, by the Jordan-von Neumann theorem, a real or complex normed space satisfies the parallelogram law for all vectors if and only if it is an inner product space (with the norm induced by that inner product).^[10] When the parallelogram law holds, the inner product can be recovered from the norm using the polarization identity. For real vector spaces, the inner product is given by

\langle u, v \rangle = \frac{1}{4} \left( \|u + v\|^2 - \|u - v\|^2 \right).

For complex vector spaces, the formula extends to account for the Hermitian property:

\langle u, v \rangle = \frac{1}{4} \left( \|u + v\|^2 - \|u - v\|^2 + i \|u + i v\|^2 - i \|u - i v\|^2 \right).

These identities allow the reconstruction of the sesquilinear form defining the space as a pre-Hilbert space.^[10] Another important algebraic feature is strict convexity, which occurs when equality in the triangle inequality \|u + v\| = \|u\| + \|v\| holds only if u and v are positively collinear, i.e., there exists \lambda \geq 0 such that v = \lambda u (assuming u, v \neq 0). Equivalently, the unit sphere contains no line segments other than single points, making the unit ball strictly convex. This property distinguishes norms with "rounded" unit balls from those with flat faces, impacting uniqueness in optimization and duality. Inner product spaces are strictly convex,^[11] but not all strictly convex spaces arise from inner products. For concrete examples, consider the sequence spaces \ell^p for $1 \leq p < \infty, equipped with the p-norm \|x\|_p = \left( \sum_{n=1}^\infty |x_n|^p \right)^{1/p}. These spaces satisfy the parallelogram law if and only if p = 2, in which case \ell^2 is a Hilbert space with the standard inner product \langle x, y \rangle = \sum_{n=1}^\infty x_n \overline{y_n}. For p \neq 2, the failure of the law highlights the absence of an underlying inner product structure.

Equivalent norms

Two norms \|\cdot\|_1 and \|\cdot\|_2 on a vector space V are said to be equivalent if there exist positive constants c and C such that

c \|v\|_1 \leq \|v\|_2 \leq C \|v\|_1

for all v \in V.^[12] This condition ensures that the norms are comparable up to scaling and thus generate the same algebraic structure in terms of boundedness and convergence properties. In finite-dimensional vector spaces over \mathbb{R} or \mathbb{C}, all norms are equivalent. To see this, fix a basis and consider the maximum norm \|\cdot\|_* = \max_i |x_i| induced by the coordinates. The unit sphere S = \{x : \|x\|_* = 1\} is compact because it is closed and bounded in the finite-dimensional Euclidean topology. The other norm \|\cdot\| is continuous with respect to \|\cdot\|_*, so by the extreme value theorem, it attains positive minimum c > 0 and maximum C < \infty on S. Extending by homogeneity yields the equivalence constants for all vectors.^[13] Equivalent norms preserve key structural features of the space. Specifically, they induce the same topology, meaning the open sets defined by balls in each norm coincide, and they identify the same bounded sets, as a set is bounded in one norm if and only if it is bounded in the other.^[12] In infinite-dimensional spaces, norms need not be equivalent. For example, consider the space c_{00} of real sequences with only finitely many non-zero terms, equipped with the \ell^1 norm \|x\|_1 = \sum_{n=1}^\infty |x_n| and the \ell^\infty norm \|x\|_\infty = \sup_n |x_n|. These norms are not equivalent because there is no constant C > 0 such that \|x\|_1 \leq C \|x\|_\infty for all x \in c_{00}. To see this, take the sequence x^{(n)} with first n components $1/n and the rest zero; then \|x^{(n)}\|_\infty = 1/n and \|x^{(n)}\|_1 = 1, so the ratio \|x^{(n)}\|_1 / \|x^{(n)}\|_\infty = n is unbounded as n \to \infty. (Note that the reverse inequality \|x\|_\infty \leq \|x\|_1 holds for all x \in c_{00}.)^[14]

Topology and Metric

Induced topology

In a normed vector space (X, \|\cdot\|), the norm induces a topology \tau on X defined by declaring a set U \subseteq X open if for every x \in U, there exists r > 0 such that the open ball B(x, r) = \{ y \in X : \|y - x\| < r \} is contained in U.^[15] This topology, known as the norm topology, ensures that the vector space operations of addition and scalar multiplication are continuous with respect to \tau.^[16] The collection of all open balls \{ B(x, r) : x \in X, r > 0 \} forms a basis for \tau, meaning every open set in \tau is a union of such balls, and the balls provide a local basis at each point x \in X.^[15] Due to the definiteness axiom of the norm (\|x\| = 0 if and only if x = 0), the induced topology is Hausdorff: for any distinct points x, y \in X with x \neq y, there exist disjoint open balls B(x, r) and B(y, s) separating them, since \|x - y\| > 0 allows choosing r, s > 0 small enough to ensure the balls do not overlap.^[15]^[16] The norm topology exhibits translation invariance, meaning that if U is open in \tau, then U + z = \{ u + z : u \in U \} is also open for every z \in X; this uniformity arises because \| (u + z) - (x + z) \| = \| u - x \|, preserving the ball structure under vector addition.^[15] Furthermore, the topology is metrizable, as the norm defines a metric d(x, y) = \|x - y\| that generates \tau, and metric spaces are first-countable, possessing a countable local basis at each point (e.g., the balls B(x, 1/n) for n \in \mathbb{N}).^[15]^[17]

Metric structure

A normed vector space (V, \|\cdot\|) naturally gives rise to a metric space structure through the induced metric defined by d(u, v) = \|u - v\| for all u, v \in V.^[18] This metric satisfies the standard axioms of a metric: non-negativity, where d(u, v) \geq 0 and d(u, v) = 0 if and only if u = v, follows directly from the positivity axiom of the norm; symmetry, d(u, v) = d(v, u), holds because \|u - v\| = \|v - u\|; and the triangle inequality, d(u, w) \leq d(u, v) + d(v, w), is inherited from the triangle inequality for the norm.^[19] These properties ensure that the induced metric defines a valid distance function on V.^[20] While every normed vector space is thus a metric space, completeness with respect to the induced metric is not assumed; normed spaces may be incomplete, meaning there exist Cauchy sequences that do not converge in V.^[18] This incompleteness distinguishes general normed spaces from Banach spaces, where the metric is complete.^[21] In the context of the induced metric, a subset B \subseteq V is bounded if its diameter \sup \{ d(u, v) : u, v \in B \} is finite.^[19] Equivalently, B is bounded if there exists M > 0 such that \|x\| \leq M for all x \in B, which aligns with the norm-based notion of boundedness.^[18] Bounded sets play a key role in analyzing geometric and analytic properties derived from the metric. The scalar multiplication operation in a normed space, viewed as a map (\lambda, x) \mapsto \lambda x from the product space \mathbb{K} \times V to V (where \mathbb{K} is the scalar field), is continuous with respect to the induced metric.^[18] Moreover, this operation is uniformly continuous when restricted to bounded subsets of V.^[18] Specifically, for any bounded set B \subseteq V and \varepsilon > 0, there exists \delta > 0 (independent of points in B) such that if |\lambda_1 - \lambda_2| < \delta and x \in B, then \|\lambda_1 x - \lambda_2 x\| < \varepsilon. This uniform continuity facilitates the study of approximations and limits within bounded regions.^[18]

Completeness

Cauchy sequences

In a normed vector space (V, \|\cdot\|), the norm induces a metric d(u, v) = \|u - v\|, which allows the study of sequences via distances between terms.^[22] A sequence \{v_n\} in V is a Cauchy sequence if, for every \epsilon > 0, there exists a positive integer N such that \|v_m - v_n\| < \epsilon whenever m, n > N.^[22] This condition ensures that the terms of the sequence become arbitrarily close to each other as the indices increase, capturing the intuitive notion of the sequence "settling down" without specifying a limit point a priori. In any metric space, including the one induced by a norm, every convergent sequence is Cauchy: if \{v_n\} converges to some v \in V, then for every \epsilon > 0, there exists N such that \|v_n - v\| < \epsilon/2 for all n > N, and by the triangle inequality, \|v_m - v_n\| < \epsilon for all m, n > N.^[23] Consequently, every convergent sequence has a Cauchy subsequence—namely, the sequence itself.^[23] A normed vector space is complete if and only if every Cauchy sequence in it converges to an element of the space. This equivalence highlights the role of Cauchy sequences in defining completeness, as the space contains all "potential limits" of such sequences.^[24] A classic example of an incomplete normed vector space is the space c_{00} of real sequences with only finitely many nonzero terms, equipped with the \ell^2 norm \|x\|_2 = \left( \sum_{i=1}^\infty |x_i|^2 \right)^{1/2}.^[21] Here, consider the sequence x = (2^{-k})_{k=1}^\infty, which is in \ell^2, and define x_n as the sequence with the first n terms of x and zeros afterward. The sequence \{x_n\} is in c_{00} and Cauchy, but it converges to x, which has infinitely many nonzero terms and thus lies outside c_{00}. This incompleteness illustrates how a dense subspace may fail to contain all limits of its Cauchy sequences.^[21]

Banach spaces

A Banach space is a normed vector space that is complete with respect to the metric induced by its norm, meaning that every Cauchy sequence in the space converges to an element within the space.^[25] This completeness property distinguishes Banach spaces from general normed spaces and forms the foundation for much of functional analysis, enabling the study of limits and convergence in infinite-dimensional settings. Every normed vector space admits a completion to a Banach space, constructed by identifying Cauchy sequences that converge to the same limit and equipping the resulting quotient space with the norm extended from the original.^[26] This completion process is unique up to isometric isomorphism, ensuring that any incomplete normed space can be embedded densely into a Banach space while preserving its algebraic and topological structure.^[27] Prominent examples of Banach spaces include the L^p spaces for $1 \leq p \leq \infty, consisting of p-integrable functions on a measure space equipped with the L^p norm \|f\|_p = \left( \int |f|^p \, d\mu \right)^{1/p} (or the essential supremum for p = \infty), which are complete and widely used in analysis and partial differential equations.^[28] Another key example is the space C[a,b] of continuous real- or complex-valued functions on the compact interval [a,b], normed by the supremum \|f\|_\infty = \sup_{x \in [a,b]} |f(x)|, which is complete due to the uniform continuity of limits of uniformly convergent sequences on compact sets.^[25] As complete metric spaces, Banach spaces satisfy the Baire category theorem, which asserts that they cannot be expressed as a countable union of nowhere dense sets, or equivalently, that the intersection of countably many dense open subsets is dense.^[29] This property underpins important results in Banach space theory, such as the open mapping and closed graph theorems, highlighting the structural robustness of complete normed spaces.^[30]

Operators and Functionals

Bounded linear operators

In a normed vector space context, a linear operator T: V \to W between normed spaces V and W (over the same field) is a map satisfying T(\alpha v + \beta u) = \alpha T v + \beta T u for all scalars \alpha, \beta and vectors u, v \in V.^[31] The operator T is bounded if there exists a constant M \geq 0 such that \|T v\|_W \leq M \|v\|_V for all v \in V, or equivalently, if \sup_{\|v\|_V \leq 1} \|T v\|_W < \infty.^[32]^[33] Boundedness is equivalent to continuity of T at the zero vector in V, since the norm topologies ensure that continuity at one point implies uniform continuity everywhere for linear maps.^[25]^[34] The operator norm of a bounded linear operator T is defined as \|T\| = \sup_{\|v\|_V = 1} \|T v\|_W, which is finite and satisfies the submultiplicative inequality \|T v\|_W \leq \|T\| \|v\|_V for all v \in V.^[31] This norm turns the space of bounded linear operators \mathcal{L}(V, W) into a normed space itself.^[35] Examples of bounded operators include integral operators on L^2 spaces; for instance, if the kernel k \in L^2(\mathbb{R}^2), the operator (T f)(x) = \int k(x, y) f(y) \, dy is bounded from L^2(\mathbb{R}) to itself.^[36] In contrast, the differentiation operator on the space of polynomials equipped with the supremum norm on [0, 1] is unbounded, as the norms of derivatives of monomials grow without bound relative to the original norms.^[37]

Dual space

The dual space of a normed vector space V over \mathbb{R} or \mathbb{C}, denoted V^*, consists of all continuous linear functionals \phi: V \to \mathbb{K}, where \mathbb{K} is the scalar field.^[38] These functionals are bounded, meaning there exists M > 0 such that |\phi(v)| \leq M \|v\| for all v \in V.^[39] The set V^* forms a vector space under pointwise addition and scalar multiplication, and it is itself a normed space with the dual norm defined by

\|\phi\| = \sup_{\|v\| \leq 1} |\phi(v)| = \sup_{\|v\| = 1} |\phi(v)|,

which is finite precisely because \phi is continuous.^[38] This norm makes V^* a Banach space, equipping it with a topology compatible with its linear structure.^[39] A key property of the dual space is the existence of extensions for functionals defined on subspaces. The Hahn-Banach extension theorem states that if M is a subspace of V and \phi: M \to \mathbb{K} is a bounded linear functional with \|\phi\| = p, then there exists an extension \tilde{\phi}: V \to \mathbb{K} that is bounded linear with \|\tilde{\phi}\| = p.^[40] This preservation of the norm ensures that the dual space captures essential information about the original space, allowing functionals to be lifted while maintaining boundedness.^[41] The theorem applies directly to normed spaces without requiring completeness, highlighting the role of the norm in controlling the growth of functionals.^[40] The bidual V^{**} is the dual of V^*, consisting of continuous linear functionals on V^*. The natural embedding J: V \to V^{**} given by J(v)(\phi) = \phi(v) for \phi \in V^* is an isometric isomorphism onto its image, but V is reflexive if and only if J is surjective, i.e., V \cong V^{**}.^[42] Hilbert spaces, equipped with their inner product-induced norm, are reflexive, as the Riesz representation theorem identifies H^* \cong H and extends to the bidual. However, not all Banach spaces are reflexive; for example, the space c_0 of sequences converging to zero under the supremum norm satisfies c_0^{**} = \ell^\infty \not\cong c_0, since \ell^\infty is non-separable while c_0 is separable.^[43] The dual space induces the weak topology on V, denoted \sigma(V, V^*), which is the initial topology making all functionals in V^* continuous.^[44] This topology is generated by subbasis sets of the form \{v \in V : |\phi(v) - \phi(v_0)| < \epsilon\} for \phi \in V^*, \epsilon > 0, and v_0 \in V, and it is coarser than the norm topology.^[45] In the weak topology, bounded sets remain bounded, but convergence is weaker: a net converges weakly if \phi(v_\alpha) \to \phi(v) for all \phi \in V^*.^[46] This structure is fundamental for studying compactness and separation in normed spaces via the dual.^[44]

Constructions

Product spaces

The direct product of two normed vector spaces V and W over the same field, denoted V \times W, is the set of ordered pairs (v, w) with v \in V and w \in W, equipped with componentwise addition and scalar multiplication: (v_1, w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2) and \alpha (v, w) = (\alpha v, \alpha w) for \alpha in the field.^[47] To make V \times W a normed space, common choices include the sum norm \|(v, w)\|_1 = \|v\| + \|w\| and the max norm \|(v, w)\|_\infty = \max\{\|v\|, \|w\|\}. Both satisfy the norm axioms: non-negativity follows from the norms on V and W; homogeneity from scalar multiplication in each component; and the triangle inequality from \|(v_1 + v_2, w_1 + w_2)\|_1 \leq \|v_1 + v_2\| + \|w_1 + w_2\| \leq (\|v_1\| + \|v_2\|) + (\|w_1\| + \|w_2\|) = \|(v_1, w_1)\|_1 + \|(v_2, w_2)\|_1, with an analogous verification for the max norm. These extend naturally to the finite product V_1 \times \cdots \times V_n by \|(v_1, \dots, v_n)\|_1 = \sum_{i=1}^n \|v_i\| and \|(v_1, \dots, v_n)\|_\infty = \max_{1 \leq i \leq n} \|v_i\|.^[47] The sum and max norms on a finite product are equivalent, since \|(v_1, \dots, v_n)\|_\infty \leq \|(v_1, \dots, v_n)\|_1 \leq n \|(v_1, \dots, v_n)\|_\infty for all (v_1, \dots, v_n) in the product space; more generally, all \ell^p-product norms for $1 \leq p < \infty are equivalent on finite products. In finite-dimensional cases, this aligns with the broader equivalence of all norms.^[47] The component projections \pi_V: V \times W \to V defined by \pi_V(v, w) = v and \pi_W: V \times W \to W defined by \pi_W(v, w) = w are bounded linear operators with operator norm 1 under either the sum or max norm, since \|\pi_V(v, w)\| = \|v\| \leq \max\{\|v\|, \|w\|\} = \|(v, w)\|_\infty and \|\pi_V(v, w)\| = \|v\| \leq \|v\| + \|w\| = \|(v, w)\|_1. The inclusions i_V: V \to V \times W given by i_V(v) = (v, 0) and i_W: W \to V \times W given by i_W(w) = (0, w) are also bounded linear operators with operator norm 1.^[47] A representative example is \mathbb{R}^n as the product of n copies of \mathbb{R} equipped with the absolute value norm | \cdot |. The Euclidean norm \|(x_1, \dots, x_n)\|_2 = \sqrt{\sum_{i=1}^n x_i^2} on \mathbb{R}^n is equivalent to the product sum norm \sum_{i=1}^n |x_i| or max norm \max_i |x_i|, as all norms on finite-dimensional spaces are equivalent.^[47]

Quotient spaces

A seminorm on a vector space V over \mathbb{R} or \mathbb{C} is a function p: V \to [0, \infty) satisfying the subadditivity condition p(v + w) \leq p(v) + p(w) for all v, w \in V and the homogeneity property p(\alpha v) = |\alpha| p(v) for all \alpha in the scalar field and v \in V. Unlike a norm, a seminorm need not be positive definite, meaning p(v) = 0 does not necessarily imply v = 0. The set \{v \in V : p(v) = 0\} forms a subspace of V, called the null space of p.^[15] Given a seminorm p on V, the quotient space V/N, where N is the null space of p, inherits a vector space structure via cosets v + N. The function \hat{p}: V/N \to [0, \infty) defined by \hat{p}(v + N) = p(v) satisfies the properties of a norm on V/N, as \hat{p}(v + N) = 0 implies v \in N, so the coset is the zero element. This construction yields a normed vector space from any seminormed space.^[15] More generally, for a normed space (V, \|\cdot\|) and a closed subspace M \subseteq V, the quotient space V/M consists of cosets v + M with the induced quotient norm \|v + M\| = \inf_{m \in M} \|v + m\|. This infimum is finite and positive for nonzero cosets because M is closed, ensuring the quotient norm separates points and satisfies the norm axioms. The natural projection \pi: V \to V/M, \pi(v) = v + M, is a continuous linear surjection with kernel M.^[15]^[48] Regarding completeness, if (V, \|\cdot\|) is a Banach space and M is a closed subspace, then (V/M, \|\cdot\|_Q) is also Banach. To see this, a Cauchy sequence in V/M lifts to a Cauchy sequence in V via the projection, which converges to some limit in V; the coset of this limit converges to the original sequence in the quotient norm.^[15] A representative example arises in sequence spaces: consider \ell^\infty, the Banach space of bounded real sequences with the supremum norm, and its closed subspace c_0 of sequences converging to zero. The quotient \ell^\infty / c_0 is a Banach space under the quotient norm \|x + c_0\| = \limsup_{n \to \infty} |x_n|, and its dual space is isometrically isomorphic to \ell^1, the space of absolutely summable sequences.^[48]

Normable topologies

A normable space is a topological vector space equipped with a topology that can be induced by a norm on the underlying vector space.^[49] In such spaces, the norm defines a metric that generates the given topology through the open balls, ensuring compatibility with the vector space operations of addition and scalar multiplication.^[50] A Hausdorff topological vector space is normable if and only if it is locally convex and has a bounded neighborhood of the origin.^[49] This characterization highlights that the existence of a convex, absorbing, and bounded set serving as a neighborhood of zero allows for the construction of a compatible norm, often via the Minkowski functional of that set.^[50] Locally boundedness ensures the topology admits a translation-invariant metric, while local convexity guarantees the norm's subadditivity and homogeneity properties hold in a way that preserves the topology.^[49] All finite-dimensional Hausdorff topological vector spaces are normable, as their topologies are equivalent to those induced by any norm, such as the Euclidean norm relative to a basis, regardless of the specific norm chosen.^[51] In contrast, the Schwartz space of rapidly decreasing smooth functions on \mathbb{R}^n, which carries a Fréchet topology defined by a countable family of seminorms, is not normable because every neighborhood of the origin is unbounded in this space.^[51] Normable spaces are always metrizable, as the inducing norm provides a translation-invariant metric compatible with the topology.^[50] However, the converse does not hold: metrizable topological vector spaces need not be normable, with the Schwartz space serving as a counterexample where the metric arises from seminorms but no single norm suffices to generate the topology.^[51] This distinction underscores that normability imposes stricter conditions, particularly the boundedness requirement absent in some infinite-dimensional metrizable settings.^[49]

References

[1]
[PDF] Chapter 3. Normed vector spaces - Lecture notes for MA2223
A normed vector space (X,|| · ||) consists of a vector space X and a norm ||x||, which is a real-valued function with specific properties.
[2]
3.6: Normed Linear Spaces - Mathematics LibreTexts
Sep 5, 2021 · By a normed linear space (briefly normed space) is meant a real or complex vector space ... Moreover, by definition,.
[3]
normed vector space - PlanetMath.org
Mar 22, 2013 · Any normed vector space (V,∥⋅∥) is a metric space under the metric d:V×V→R given by d(u,v)=∥u−v∥. This is called the metric induced by the norm ∥
[4]
Sur quelques points du calcul fonctionnel | Rendiconti del Circolo ...
Dec 23, 2008 · Sur quelques points du calcul fonctionnel. Download PDF. M. Maurice Fréchet. 1518 Accesses. 774 Citations. 23 Altmetric. Explore all metrics ...
[5]
Axioms of vector spaces
A normed real vector space is a real vector space X with an additional operation: Norm: Given an element x in X, one can form the norm ||x||, which is a non- ...
[6]
Norm Properties - Stanford CCRMA
The third property says the norm is ``absolutely homogeneous'' with respect to scalar multiplication. (The scalar $ c$ can be complex, in which case the angle ...
[7]
Math 55a: Norm basics
A normed vector space V is automatically a metric space with the distance function d(v,w):=||v-w||. (This still holds if Homogeneity is replaced by the weaker ...
[8]
[PDF] Analysis review: Norms, convergence, and continuity - MyWeb
Matrix norms. Inequalities. Reverse triangle inequality. • A related inequality: • Theorem (reverse triangle inequality): For any x,y ∈ Rd, kxk−kyk≤kx − yk.
[9]
[PDF] FROM THE PARALLELOGRAM LAW TO AN INNER PRODUCT Let (V
Let (V,|| · ||) be a normed real vector space. Suppose the norm satisfies the parallelogram law: ||v + w||2 + ||v − w||2 = 2||v||2 + 2||w||2, ∀v, w ∈ V. Define ...
[10]
[PDF] A new characterization of strict convexity on normed linear spaces
The normed space X is called strictly convex, for short (SC), if its unit sphere SX does not contain nontrivial segments, that is, for every x, y ∈ SX, x 6= y, ...Missing: source | Show results with:source
[11]
[PDF] Notes on the equivalence of norms
If we are given two norms k·ka and k·kb on some finite-dimensional vector space V over C, a very useful fact is that they are always within a constant factor ...
[12]
[PDF] Equivalence of norms Definition. Let Y be a vector space over the ...
Theorem. Let Y be finite-dimensional. Then all norms are equivalent. Proof. Let v1,...,vn be a basis of Y. We define kxk∗ := maxi=1,...,n |xi| for x = P n.
[13]
[PDF] EQUIVALENCE OF NORMS 1. Introduction Let K be a field and
Theorem 3.2. All norms on a finite-dimensional vector space over a complete valued field are equivalent. Proof. Let (K,|·|) ...
[14]
[PDF] FUNCTIONAL ANALYSIS | Second Edition Walter Rudin
Rudin, Walter, (date). Functional analysis/Walter Rudin.-2nd ed. p. em. -(international series in pure and applied mathematics).
[15]
[PDF] E.2 Topological Vector Spaces
Let X be a normed vector space. Show that the topology induced from the norm is the smallest topology with respect to which X is a topological vector space and ...<|control11|><|separator|>
[16]
[PDF] Banach Spaces I: Normed Vector Spaces - KSU Math
If (X,k.k) is a normed vector space and % denotes its norm topology, then for every linear subspace Y⊂X, the restriction of k.k to Y is a norm, and furthermore, ...
[17]
[PDF] Chapter 1: Metric and Normed Spaces - UC Davis Mathematics
¾ ...
[18]
[PDF] Norms
A norm R . R on a vector space V induces a metric d on V by d(v ,w) = R v - wR. Exercise. Show that d is a metric on V. All topological properties (e.g. ...
[19]
[PDF] Basic Properties of Metric and Normed Spaces - TTIC
A metric space (X, d) has a distance function d(x, y) satisfying non-negativity, symmetry, and triangle inequality. A normed space (V,∥·∥) has a norm ∥·∥ on a ...
[20]
[PDF] short review of metrics, norms, and convergence - Christopher Heil
A norm provides us with a notion of the length of a vector in a vector space. ... (a) Reverse Triangle Inequality: ¯. ¯kfk−kgk. ¯. ¯. ≤ kf − gk. (b) Continuity ...
[21]
[PDF] Section 1.5. Banach Spaces
Apr 19, 2019 · Definition 1.5. 1. A sequence {xn} in a normed vector space is a Cauchy sequence if for all ε > 0 there exists M ∈ N such that kxm − xnk < ε ...
[22]
[PDF] Lecture 2: Review of Metric Spaces - UW Math Department
( ∀ > 0, ∃N < ∞, such that ρ(xm,xn) < if m,n > N .) Every convergent sequence is Cauchy. The point x is a cluster point of the sequence {xn}∞ n=1 if, for.
[23]
[PDF] Sequences and Series of Functions - UC Davis Math
Definition 5.35. A normed vector space is complete if every Cauchy sequence converges. A complete normed linear space is called a Banach space.
[24]
[PDF] SAMSA Masamu Program - Auburn University
Example 3.3. An important example of a normed space not complete is the rational numbers Q endowed with the usual operations of addition and multiplication (F =.
[25]
[PDF] Banach Spaces - UC Davis Math
Definition 5.1 A Banach space is a normed linear space that is a complete metric space with respect to the metric derived from its norm.
[26]
Completion of a normed space - Branko Curgus
To prove the reverse triangle inequality let x,y \in \mathcal V be arbitrary. By the triangle inequality we have p(x) \leq p(x-y) + p(y) and hence p(x) - p(y) ...
[27]
[PDF] Funtional Analysis Lecture notes for 18.102 Richard Melrose
Feb 4, 2014 · We are particularly interested in complete, i.e. Banach, spaces and the process of completion of a normed space to a Banach space.
[28]
[PDF] Lp spaces - UC Davis Math
The Lp-spaces are perhaps the most useful and important examples of Banach spaces. 7.1. Lp spaces. For definiteness, we consider real-valued functions.
[29]
[PDF] If X is a Banach Space, because it is a complete metric space Baire ...
There are three applications we have in mind. 1. Uniform Boundedness principle. If Tn : X → Y are bounded linear maps from X to another Banach space Y and if ...
[30]
[PDF] Banach Spaces 1. Basic Definitions
Nov 13, 2017 · Abstractly, Banach spaces are less convenient than Hilbert spaces, but still sufficiently simple so many important properties hold.
[31]
[PDF] 2.4. Bounded Linear Operators
May 12, 2021 · For normed linear spaces X and Y , the set of all linear operators from. X to Y is denoted L(X, Y ). For T ∈ L(X, Y ) define the operator norm.
[32]
[PDF] 19. Normed vector spaces - People
A norm on X is a function k·k : X → K satisfying: (i) (postivity) kxk ≥ 0 for all x ∈ X, and kxk = 0 if and only if x = 0;. (ii) (homogeneity) kkxk = |k|kxk for ...
[33]
[PDF] 522 Bounded Linear Operators and the Definition of Derivatives ...
Definition. Let V , W be normed vector spaces (both over R or over C). A linear transformation or linear operator T : V → W is bounded if there is. a constant ...
[34]
[PDF] Functional Analysis and Operator Algebras - Portland State University
May 9, 2022 · ... equivalent to continuity. 3.2.1. Definition. A linear ... called a (bounded linear) operator. The family of all operators on a ...
[35]
[PDF] Norms on Operators
If V, W are vector spaces then so is the space of linear transformations from V to W denoted. _(V, W). We now consider norms on _(V, W). When V = W, _(V, V) = ...
[36]
[PDF] Metrics, Norms, Inner Products and Operator Theory Chapter 8
Oct 3, 2020 · If k ∈ L2(R2), then the integral operator Lk given by equa- tion (8.1) defines a bounded mapping of L2(R) into itself, and its operator norm ...
[37]
[PDF] Chapter 8 Several variables and partial derivatives
The functions sin(nx) have norm 1, but the derivatives have norm n. So differentiation (which is a linear operator) has unbounded norm on this space. But ...
[38]
Dual Normed Space -- from Wolfram MathWorld
If X is a normed linear space, then the set of continuous linear functionals on X is called the dual (or conjugate) space of X.
[39]
[PDF] Dual space
Mar 16, 2013 · This turns the continuous dual into a normed vector space, indeed into a Banach space so long as the underlying field is complete, which is ...
[40]
[PDF] Normed Linear Space Version of Hahn-Banach Extension Theorem
Given any closed subspace Y of a normed linear space X and x ¢ Y, there is a bounded linear functional f on X (i.e., f € X*) such that f(Y) = 0 and f(x) = 1.
[41]
[PDF] On the Hahn-Banach theorem - The Institute of Mathematical Sciences
The analytic and geometric versions of the Hahn-Banach theorem follow from a general theorem on the extension of linear functionals on a real vector space.
[42]
Reflexive Space -- from Wolfram MathWorld
X is reflexive if it coincides with its bidual, X is Banach. See also. Banach Space, Dual Vector Space, Normed Space. Portions of this entry contributed by ...
[43]
[PDF] Math 421/510 Quiz 8 Solution 1. Let c0 denote the linear space ...
Since the latter space is strictly larger than c0, this shows that c0 is not reflexive. Let Φ : `1(N) → c∗. 0 be given by. Φ(α) = Tα, where. Tα(x) = X n αnxn ...
[44]
245B, Notes 11: The strong and weak topologies - Terry Tao
Feb 21, 2009 · ... weak* topology on a dual space {V^*} , without specifying exactly what the predual space {V} is. However, in practice, the predual space is ...
[45]
[PDF] 4. The weak dual topology - KSU Math
Suppose X is a normed vector space, which is separable in the norm topology. When equipped with the w∗ topology, the compact space (X∗)1 is metrizable. Proof.<|control11|><|separator|>
[46]
[PDF] 4. Dual spaces and weak topologies Recall that if X is a Banach ...
If X is a separable Banach space, then the weak-∗ to- pology on B1(0) ⊆ X∗ (more precisely: the relative topology induced by. Tw* ) is metrizable. We don ...
[47]
https://link.springer.com/book/10.1007/978-1-4757-2828-5
[48]
None
Below is a merged response summarizing all the extracted definitions, theorems, and related information from the provided segments of Kreyszig (1989) *Introductory Functional Analysis with Applications*. To retain as much detail as possible in a dense and organized format, I will use tables where appropriate, followed by a narrative summary for additional context. The response consolidates information across all segments, noting where definitions or theorems are explicitly provided, implied, or not addressed, and includes citations, page references, and URLs where available.
[49]
[PDF] 3. Topological vector spaces
We will prove that a topological vector space is normable if and only if it is both locally convex and locally bounded. Page 6. 186. Topological vector spaces.Missing: characterization | Show results with:characterization<|control11|><|separator|>
[50]
[PDF] Functional Analysis II, Math 7321 Lecture Notes from January 24, 2017
Jan 24, 2017 · A topological vector space is normable if and only if it is locally bounded and locally convex. 1.4 Characterization (Locally compact ...
[51]
[PDF] Finite-dimensional topological vector spaces - Keith Conrad
A related theorem is that all norms on a finite-dimensional K-vector space define the same topology on the space [1, Theorem 3.2], and this has a simpler proof.