Let [math]\displaystyle{ X }[/math] be a vector space over either the real numbers[math]\displaystyle{ \R }[/math] or the complex numbers [math]\displaystyle{ \Complex. }[/math]
A real-valued function[math]\displaystyle{ p : X \to \R }[/math] is called a seminorm if it satisfies the following two conditions:
Subadditivity[1]/Triangle inequality: [math]\displaystyle{ p(x + y) \leq p(x) + p(y) }[/math] for all [math]\displaystyle{ x, y \in X. }[/math]
Absolute homogeneity:[1][math]\displaystyle{ p(s x) =|s|p(x) }[/math] for all [math]\displaystyle{ x \in X }[/math] and all scalars [math]\displaystyle{ s. }[/math]
These two conditions imply that [math]\displaystyle{ p(0) = 0 }[/math][proof 1] and that every seminorm [math]\displaystyle{ p }[/math] also has the following property:[proof 2]
Nonnegativity:[1][math]\displaystyle{ p(x) \geq 0 }[/math] for all [math]\displaystyle{ x \in X. }[/math]
Some authors include non-negativity as part of the definition of "seminorm" (and also sometimes of "norm"), although this is not necessary since it follows from the other two properties.
By definition, a norm on [math]\displaystyle{ X }[/math] is a seminorm that also separates points, meaning that it has the following additional property:
Positive definite/Positive[1]/Point-separating: whenever [math]\displaystyle{ x \in X }[/math] satisfies [math]\displaystyle{ p(x) = 0, }[/math] then [math]\displaystyle{ x = 0. }[/math]
A seminormed space is a pair [math]\displaystyle{ (X, p) }[/math] consisting of a vector space [math]\displaystyle{ X }[/math] and a seminorm [math]\displaystyle{ p }[/math] on [math]\displaystyle{ X. }[/math] If the seminorm [math]\displaystyle{ p }[/math] is also a norm then the seminormed space [math]\displaystyle{ (X, p) }[/math] is called a normed space.
Since absolute homogeneity implies positive homogeneity, every seminorm is a type of function called a sublinear function. A map [math]\displaystyle{ p : X \to \R }[/math] is called a sublinear function if it is subadditive and positive homogeneous. Unlike a seminorm, a sublinear function is not necessarily nonnegative. Sublinear functions are often encountered in the context of the Hahn–Banach theorem.
A real-valued function [math]\displaystyle{ p : X \to \R }[/math] is a seminorm if and only if it is a sublinear and balanced function.
Examples
The trivial seminorm on [math]\displaystyle{ X, }[/math] which refers to the constant [math]\displaystyle{ 0 }[/math] map on [math]\displaystyle{ X, }[/math] induces the indiscrete topology on [math]\displaystyle{ X. }[/math]
Let [math]\displaystyle{ \mu }[/math] be a measure on a space [math]\displaystyle{ \Omega }[/math]. For an arbitrary constant [math]\displaystyle{ c \geq 1 }[/math], let [math]\displaystyle{ X }[/math] be the set of all functions [math]\displaystyle{ f: \Omega \rightarrow \mathbb{R} }[/math] for which
[math]\displaystyle{ \lVert f \rVert_c := \left( \int_{\Omega}| f |^c \, d\mu \right)^{1/c} }[/math]
exists and is finite. It can be shown that [math]\displaystyle{ X }[/math] is a vector space, and the functional [math]\displaystyle{ \lVert \cdot \rVert_c }[/math] is a seminorm on [math]\displaystyle{ X }[/math]. However, it is not always a norm (e.g. if [math]\displaystyle{ \Omega = \mathbb{R} }[/math] and [math]\displaystyle{ \mu }[/math] is the Lebesgue measure) because [math]\displaystyle{ \lVert h \rVert_c = 0 }[/math] does not always imply [math]\displaystyle{ h = 0 }[/math]. To make [math]\displaystyle{ \lVert \cdot \rVert_c }[/math] a norm, quotient [math]\displaystyle{ X }[/math] by the closed subspace of functions [math]\displaystyle{ h }[/math] with [math]\displaystyle{ \lVert h \rVert_c = 0 }[/math]. The resulting space, [math]\displaystyle{ L^c(\mu) }[/math], has a norm induced by [math]\displaystyle{ \lVert \cdot \rVert_c }[/math].
If [math]\displaystyle{ f }[/math] is any linear form on a vector space then its absolute value[math]\displaystyle{ |f|, }[/math] defined by [math]\displaystyle{ x \mapsto |f(x)|, }[/math] is a seminorm.
A sublinear function[math]\displaystyle{ f : X \to \R }[/math] on a real vector space [math]\displaystyle{ X }[/math] is a seminorm if and only if it is a symmetric function, meaning that [math]\displaystyle{ f(-x) = f(x) }[/math] for all [math]\displaystyle{ x \in X. }[/math]
Every real-valued sublinear function[math]\displaystyle{ f : X \to \R }[/math] on a real vector space [math]\displaystyle{ X }[/math] induces a seminorm [math]\displaystyle{ p : X \to \R }[/math] defined by [math]\displaystyle{ p(x) := \max \{f(x), f(-x)\}. }[/math][2]
Any finite sum of seminorms is a seminorm. The restriction of a seminorm (respectively, norm) to a vector subspace is once again a seminorm (respectively, norm).
If [math]\displaystyle{ p : X \to \R }[/math] and [math]\displaystyle{ q : Y \to \R }[/math] are seminorms (respectively, norms) on [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y }[/math] then the map [math]\displaystyle{ r : X \times Y \to \R }[/math] defined by [math]\displaystyle{ r(x, y) = p(x) + q(y) }[/math] is a seminorm (respectively, a norm) on [math]\displaystyle{ X \times Y. }[/math] In particular, the maps on [math]\displaystyle{ X \times Y }[/math] defined by [math]\displaystyle{ (x, y) \mapsto p(x) }[/math] and [math]\displaystyle{ (x, y) \mapsto q(y) }[/math] are both seminorms on [math]\displaystyle{ X \times Y. }[/math]
If [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are seminorms on [math]\displaystyle{ X }[/math] then so are[3][math]\displaystyle{ (p \vee q)(x) = \max \{p(x), q(x)\} }[/math] and [math]\displaystyle{ (p \wedge q)(x) := \inf \{p(y) + q(z) : x = y + z \text{ with } y, z \in X\} }[/math]
where [math]\displaystyle{ p \wedge q \leq p }[/math] and [math]\displaystyle{ p \wedge q \leq q. }[/math][4]
The space of seminorms on [math]\displaystyle{ X }[/math] is generally not a distributive lattice with respect to the above operations. For example, over [math]\displaystyle{ \R^2 }[/math], [math]\displaystyle{ p(x, y) := \max(|x|, |y|), q(x, y) := 2|x|, r(x, y) := 2|y| }[/math] are such that
[math]\displaystyle{ ((p \vee q) \wedge (p \vee r)) (x, y) = \inf \{\max(2|x_1|, |y_1|) + \max(|x_2|, 2|y_2|) : x = x_1 + x_2 \text{ and } y = y_1 + y_2\} }[/math] while [math]\displaystyle{ (p \vee q \wedge r) (x, y) := \max(|x|, |y|) }[/math]
If [math]\displaystyle{ L : X \to Y }[/math] is a linear map and [math]\displaystyle{ q : Y \to \R }[/math] is a seminorm on [math]\displaystyle{ Y, }[/math] then [math]\displaystyle{ q \circ L : X \to \R }[/math] is a seminorm on [math]\displaystyle{ X. }[/math] The seminorm [math]\displaystyle{ q \circ L }[/math] will be a norm on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ L }[/math] is injective and the restriction [math]\displaystyle{ q\big\vert_{L(X)} }[/math] is a norm on [math]\displaystyle{ L(X). }[/math]
Seminorms on a vector space [math]\displaystyle{ X }[/math] are intimately tied, via Minkowski functionals, to subsets of [math]\displaystyle{ X }[/math] that are convex, balanced, and absorbing. Given such a subset [math]\displaystyle{ D }[/math] of [math]\displaystyle{ X, }[/math] the Minkowski functional of [math]\displaystyle{ D }[/math] is a seminorm. Conversely, given a seminorm [math]\displaystyle{ p }[/math] on [math]\displaystyle{ X, }[/math] the sets[math]\displaystyle{ \{x \in X : p(x) \lt 1\} }[/math] and [math]\displaystyle{ \{x \in X : p(x) \leq 1\} }[/math] are convex, balanced, and absorbing and furthermore, the Minkowski functional of these two sets (as well as of any set lying "in between them") is [math]\displaystyle{ p. }[/math][5]
Algebraic properties
Every seminorm is a sublinear function, and thus satisfies all properties of a sublinear function, including convexity, [math]\displaystyle{ p(0) = 0, }[/math] and for all vectors [math]\displaystyle{ x, y \in X }[/math]:
the reverse triangle inequality: [6][7][math]\displaystyle{ |p(x) - p(y)| \leq p(x - y) }[/math]
and also
[math]\displaystyle{ 0 \leq \max \{p(x), p(-x)\} }[/math] and [math]\displaystyle{ p(x) - p(y) \leq p(x - y). }[/math][6][7]
For any vector [math]\displaystyle{ x \in X }[/math] and positive real [math]\displaystyle{ r \gt 0: }[/math][8][math]\displaystyle{ x + \{y \in X : p(y) \lt r\} = \{y \in X : p(x - y) \lt r\} }[/math]
and furthermore, [math]\displaystyle{ \{x \in X : p(x) \lt r\} }[/math] is an absorbingdisk in [math]\displaystyle{ X. }[/math][3]
If [math]\displaystyle{ p }[/math] is a sublinear function on a real vector space [math]\displaystyle{ X }[/math] then there exists a linear functional [math]\displaystyle{ f }[/math] on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ f \leq p }[/math][7] and furthermore, for any linear functional [math]\displaystyle{ g }[/math] on [math]\displaystyle{ X, }[/math][math]\displaystyle{ g \leq p }[/math] on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ g^{-1}(1) \cap \{x \in X : p(x) \lt 1 = \varnothing\}. }[/math][7]
Other properties of seminorms
Every seminorm is a balanced function.
A seminorm [math]\displaystyle{ p }[/math] is a norm on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ \{x \in X : p(x) \lt 1\} }[/math] does not contain a non-trivial vector subspace.
If [math]\displaystyle{ p : X \to [0, \infty) }[/math] is a seminorm on [math]\displaystyle{ X }[/math] then [math]\displaystyle{ \ker p := p^{-1}(0) }[/math] is a vector subspace of [math]\displaystyle{ X }[/math] and for every [math]\displaystyle{ x \in X, }[/math][math]\displaystyle{ p }[/math] is constant on the set [math]\displaystyle{ x + \ker p = \{x + k : p(k) = 0\} }[/math] and equal to [math]\displaystyle{ p(x). }[/math][proof 3]
Furthermore, for any real [math]\displaystyle{ r \gt 0, }[/math][3][math]\displaystyle{ r \{x \in X : p(x) \lt 1\} = \{x \in X : p(x) \lt r\} = \left\{x \in X : \tfrac{1}{r} p(x) \lt 1 \right\}. }[/math]
If [math]\displaystyle{ D }[/math] is a set satisfying [math]\displaystyle{ \{x \in X : p(x) \lt 1\} \subseteq D \subseteq \{x \in X : p(x) \leq 1\} }[/math] then [math]\displaystyle{ D }[/math] is absorbing in [math]\displaystyle{ X }[/math] and [math]\displaystyle{ p = p_D }[/math] where [math]\displaystyle{ p_D }[/math] denotes the Minkowski functional associated with [math]\displaystyle{ D }[/math] (that is, the gauge of [math]\displaystyle{ D }[/math]).[5] In particular, if [math]\displaystyle{ D }[/math] is as above and [math]\displaystyle{ q }[/math] is any seminorm on [math]\displaystyle{ X, }[/math] then [math]\displaystyle{ q = p }[/math] if and only if [math]\displaystyle{ \{x \in X : q(x) \lt 1\} \subseteq D \subseteq \{x \in X : q(x) \leq\}. }[/math][5]
If [math]\displaystyle{ (X, \|\,\cdot\,\|) }[/math] is a normed space and [math]\displaystyle{ x, y \in X }[/math] then [math]\displaystyle{ \|x - y\| = \|x - z\| + \|z - y\| }[/math] for all [math]\displaystyle{ z }[/math] in the interval [math]\displaystyle{ [x, y]. }[/math][9]
Every norm is a convex function and consequently, finding a global maximum of a norm-based objective function is sometimes tractable.
Relationship to other norm-like concepts
Let [math]\displaystyle{ p : X \to \R }[/math] be a non-negative function. The following are equivalent:
[math]\displaystyle{ p }[/math] is a seminorm.
[math]\displaystyle{ p }[/math] is a convex[math]\displaystyle{ F }[/math]-seminorm.
[math]\displaystyle{ p }[/math] is a convex balanced G-seminorm.[10]
If any of the above conditions hold, then the following are equivalent:
[math]\displaystyle{ p }[/math] is a norm;
[math]\displaystyle{ \{x \in X : p(x) \lt 1\} }[/math] does not contain a non-trivial vector subspace.[11]
There exists a norm on [math]\displaystyle{ X, }[/math] with respect to which, [math]\displaystyle{ \{x \in X : p(x) \lt 1\} }[/math] is bounded.
If [math]\displaystyle{ p }[/math] is a sublinear function on a real vector space [math]\displaystyle{ X }[/math] then the following are equivalent:[7]
[math]\displaystyle{ p }[/math] is a linear functional;
[math]\displaystyle{ p(x) + p(-x) \leq 0 \text{ for every } x \in X }[/math];
[math]\displaystyle{ p(x) + p(-x) = 0 \text{ for every } x \in X }[/math];
Inequalities involving seminorms
If [math]\displaystyle{ p, q : X \to [0, \infty) }[/math] are seminorms on [math]\displaystyle{ X }[/math] then:
[math]\displaystyle{ p \leq q }[/math] if and only if [math]\displaystyle{ q(x) \leq 1 }[/math] implies [math]\displaystyle{ p(x) \leq 1. }[/math][12]
If [math]\displaystyle{ a \gt 0 }[/math] and [math]\displaystyle{ b \gt 0 }[/math] are such that [math]\displaystyle{ p(x) \lt a }[/math] implies [math]\displaystyle{ q(x) \leq b, }[/math] then [math]\displaystyle{ a q(x) \leq b p(x) }[/math] for all [math]\displaystyle{ x \in X. }[/math][13]
Suppose [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] are positive real numbers and [math]\displaystyle{ q, p_1, \ldots, p_n }[/math] are seminorms on [math]\displaystyle{ X }[/math] such that for every [math]\displaystyle{ x \in X, }[/math] if [math]\displaystyle{ \max \{p_1(x), \ldots, p_n(x)\} \lt a }[/math] then [math]\displaystyle{ q(x) \lt b. }[/math] Then [math]\displaystyle{ a q \leq b \left(p_1 + \cdots + p_n\right). }[/math][11]
If [math]\displaystyle{ X }[/math] is a vector space over the reals and [math]\displaystyle{ f }[/math] is a non-zero linear functional on [math]\displaystyle{ X, }[/math] then [math]\displaystyle{ f \leq p }[/math] if and only if [math]\displaystyle{ \varnothing = f^{-1}(1) \cap \{x \in X : p(x) \lt 1\}. }[/math][12]
If [math]\displaystyle{ p }[/math] is a seminorm on [math]\displaystyle{ X }[/math] and [math]\displaystyle{ f }[/math] is a linear functional on [math]\displaystyle{ X }[/math] then:
[math]\displaystyle{ |f| \leq p }[/math] on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ \operatorname{Re} f \leq p }[/math] on [math]\displaystyle{ X }[/math] (see footnote for proof).[14][15]
[math]\displaystyle{ f \leq p }[/math] on [math]\displaystyle{ X }[/math] if and only if [math]\displaystyle{ f^{-1}(1) \cap \{x \in X : p(x) \lt 1 = \varnothing\}. }[/math][7][12]
If [math]\displaystyle{ a \gt 0 }[/math] and [math]\displaystyle{ b \gt 0 }[/math] are such that [math]\displaystyle{ p(x) \lt a }[/math] implies [math]\displaystyle{ f(x) \neq b, }[/math] then [math]\displaystyle{ a |f(x)| \leq b p(x) }[/math] for all [math]\displaystyle{ x \in X. }[/math][13]
Hahn–Banach theorem for seminorms
Seminorms offer a particularly clean formulation of the Hahn–Banach theorem:
If [math]\displaystyle{ M }[/math] is a vector subspace of a seminormed space [math]\displaystyle{ (X, p) }[/math] and if [math]\displaystyle{ f }[/math] is a continuous linear functional on [math]\displaystyle{ M, }[/math] then [math]\displaystyle{ f }[/math] may be extended to a continuous linear functional [math]\displaystyle{ F }[/math] on [math]\displaystyle{ X }[/math] that has the same norm as [math]\displaystyle{ f. }[/math][16]
A similar extension property also holds for seminorms:
Theorem[17][13](Extending seminorms) — If [math]\displaystyle{ M }[/math] is a vector subspace of [math]\displaystyle{ X, }[/math][math]\displaystyle{ p }[/math] is a seminorm on [math]\displaystyle{ M, }[/math] and [math]\displaystyle{ q }[/math] is a seminorm on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ p \leq q\big\vert_M, }[/math] then there exists a seminorm [math]\displaystyle{ P }[/math] on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ P\big\vert_M = p }[/math] and [math]\displaystyle{ P \leq q. }[/math]
Proof: Let [math]\displaystyle{ S }[/math] be the convex hull of [math]\displaystyle{ \{m \in M : p(m) \leq 1\} \cup \{x \in X : q(x) \leq 1\}. }[/math] Then [math]\displaystyle{ S }[/math] is an absorbingdisk in [math]\displaystyle{ X }[/math] and so the Minkowski functional[math]\displaystyle{ P }[/math] of [math]\displaystyle{ S }[/math] is a seminorm on [math]\displaystyle{ X. }[/math] This seminorm satisfies [math]\displaystyle{ p = P }[/math] on [math]\displaystyle{ M }[/math] and [math]\displaystyle{ P \leq q }[/math] on [math]\displaystyle{ X. }[/math][math]\displaystyle{ \blacksquare }[/math]
Topologies of seminormed spaces
Pseudometrics and the induced topology
A seminorm [math]\displaystyle{ p }[/math] on [math]\displaystyle{ X }[/math] induces a topology, called the seminorm-induced topology, via the canonical translation-invariant pseudometric[math]\displaystyle{ d_p : X \times X \to \R }[/math]; [math]\displaystyle{ d_p(x, y) := p(x - y) = p(y - x). }[/math]
This topology is Hausdorff if and only if [math]\displaystyle{ d_p }[/math] is a metric, which occurs if and only if [math]\displaystyle{ p }[/math] is a norm.[4]
This topology makes [math]\displaystyle{ X }[/math] into a locally convexpseudometrizabletopological vector space that has a bounded neighborhood of the origin and a neighborhood basis at the origin consisting of the following open balls (or the closed balls) centered at the origin:
[math]\displaystyle{ \{x \in X : p(x) \lt r\} \quad \text{ or } \quad \{x \in X : p(x) \leq r\} }[/math]
as [math]\displaystyle{ r \gt 0 }[/math] ranges over the positive reals.
Every seminormed space [math]\displaystyle{ (X, p) }[/math] should be assumed to be endowed with this topology unless indicated otherwise. A topological vector space whose topology is induced by some seminorm is called seminormable.
Equivalently, every vector space [math]\displaystyle{ X }[/math] with seminorm [math]\displaystyle{ p }[/math] induces a vector space quotient[math]\displaystyle{ X / W, }[/math] where [math]\displaystyle{ W }[/math] is the subspace of [math]\displaystyle{ X }[/math] consisting of all vectors [math]\displaystyle{ x \in X }[/math] with [math]\displaystyle{ p(x) = 0. }[/math] Then [math]\displaystyle{ X / W }[/math] carries a norm defined by [math]\displaystyle{ p(x + W) = p(v). }[/math] The resulting topology, pulled back to [math]\displaystyle{ X, }[/math] is precisely the topology induced by [math]\displaystyle{ p. }[/math]
Any seminorm-induced topology makes [math]\displaystyle{ X }[/math]locally convex, as follows. If [math]\displaystyle{ p }[/math] is a seminorm on [math]\displaystyle{ X }[/math] and [math]\displaystyle{ r \in \R, }[/math] call the set [math]\displaystyle{ \{x \in X : p(x) \lt r\} }[/math] the open ball of radius [math]\displaystyle{ r }[/math] about the origin; likewise the closed ball of radius [math]\displaystyle{ r }[/math] is [math]\displaystyle{ \{x \in X : p(x) \leq r\}. }[/math] The set of all open (resp. closed) [math]\displaystyle{ p }[/math]-balls at the origin forms a neighborhood basis of convexbalanced sets that are open (resp. closed) in the [math]\displaystyle{ p }[/math]-topology on [math]\displaystyle{ X. }[/math]
Stronger, weaker, and equivalent seminorms
The notions of stronger and weaker seminorms are akin to the notions of stronger and weaker norms. If [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are seminorms on [math]\displaystyle{ X, }[/math] then we say that [math]\displaystyle{ q }[/math] is stronger than [math]\displaystyle{ p }[/math] and that [math]\displaystyle{ p }[/math] is weaker than [math]\displaystyle{ q }[/math] if any of the following equivalent conditions holds:
The topology on [math]\displaystyle{ X }[/math] induced by [math]\displaystyle{ q }[/math] is finer than the topology induced by [math]\displaystyle{ p. }[/math]
If [math]\displaystyle{ x_{\bull} = \left(x_i\right)_{i=1}^{\infty} }[/math] is a sequence in [math]\displaystyle{ X, }[/math] then [math]\displaystyle{ q\left(x_{\bull}\right) := \left(q\left(x_i\right)\right)_{i=1}^{\infty} \to 0 }[/math] in [math]\displaystyle{ \R }[/math] implies [math]\displaystyle{ p\left(x_{\bull}\right) \to 0 }[/math] in [math]\displaystyle{ \R. }[/math][4]
If [math]\displaystyle{ x_{\bull} = \left(x_i\right)_{i \in I} }[/math] is a net in [math]\displaystyle{ X, }[/math] then [math]\displaystyle{ q\left(x_{\bull}\right) := \left(q\left(x_i\right)\right)_{i \in I} \to 0 }[/math] in [math]\displaystyle{ \R }[/math] implies [math]\displaystyle{ p\left(x_{\bull}\right) \to 0 }[/math] in [math]\displaystyle{ \R. }[/math]
[math]\displaystyle{ p }[/math] is bounded on [math]\displaystyle{ \{x \in X : q(x) \lt 1\}. }[/math][4]
If [math]\displaystyle{ \inf{} \{q(x) : p(x) = 1, x \in X\} = 0 }[/math] then [math]\displaystyle{ p(x) = 0 }[/math] for all [math]\displaystyle{ x \in X. }[/math][4]
There exists a real [math]\displaystyle{ K \gt 0 }[/math] such that [math]\displaystyle{ p \leq K q }[/math] on [math]\displaystyle{ X. }[/math][4]
The seminorms [math]\displaystyle{ p }[/math] and [math]\displaystyle{ q }[/math] are called equivalent if they are both weaker (or both stronger) than each other. This happens if they satisfy any of the following conditions:
The topology on [math]\displaystyle{ X }[/math] induced by [math]\displaystyle{ q }[/math] is the same as the topology induced by [math]\displaystyle{ p. }[/math]
[math]\displaystyle{ q }[/math] is stronger than [math]\displaystyle{ p }[/math] and [math]\displaystyle{ p }[/math] is stronger than [math]\displaystyle{ q. }[/math][4]
If [math]\displaystyle{ x_{\bull} = \left(x_i\right)_{i=1}^{\infty} }[/math] is a sequence in [math]\displaystyle{ X }[/math] then [math]\displaystyle{ q\left(x_{\bull}\right) := \left(q\left(x_i\right)\right)_{i=1}^{\infty} \to 0 }[/math] if and only if [math]\displaystyle{ p\left(x_{\bull}\right) \to 0. }[/math]
There exist positive real numbers [math]\displaystyle{ r \gt 0 }[/math] and [math]\displaystyle{ R \gt 0 }[/math] such that [math]\displaystyle{ r q \leq p \leq R q. }[/math]
Normability and seminormability
A topological vector space (TVS) is said to be a seminormable space (respectively, a normable space) if its topology is induced by a single seminorm (resp. a single norm).
A TVS is normable if and only if it is seminormable and Hausdorff or equivalently, if and only if it is seminormable and T1 (because a TVS is Hausdorff if and only if it is a T1 space).
A locally bounded topological vector space is a topological vector space that possesses a bounded neighborhood of the origin.
Normability of topological vector spaces is characterized by Kolmogorov's normability criterion.
A TVS is seminormable if and only if it has a convex bounded neighborhood of the origin.[18]
Thus a locally convex TVS is seminormable if and only if it has a non-empty bounded open set.[19]
A TVS is normable if and only if it is a T1 space and admits a bounded convex neighborhood of the origin.
If [math]\displaystyle{ X }[/math] is a Hausdorff locally convex TVS then the following are equivalent:
[math]\displaystyle{ X }[/math] is normable.
[math]\displaystyle{ X }[/math] is seminormable.
[math]\displaystyle{ X }[/math] has a bounded neighborhood of the origin.
The strong dual [math]\displaystyle{ X^{\prime}_b }[/math] of [math]\displaystyle{ X }[/math] is normable.[20]
The strong dual [math]\displaystyle{ X^{\prime}_b }[/math] of [math]\displaystyle{ X }[/math] is metrizable.[20]
Furthermore, [math]\displaystyle{ X }[/math] is finite dimensional if and only if [math]\displaystyle{ X^{\prime}_{\sigma} }[/math] is normable (here [math]\displaystyle{ X^{\prime}_{\sigma} }[/math] denotes [math]\displaystyle{ X^{\prime} }[/math] endowed with the weak-* topology).
The product of infinitely many seminormable space is again seminormable if and only if all but finitely many of these spaces trivial (that is, 0-dimensional).[21]
Topological properties
If [math]\displaystyle{ X }[/math] is a TVS and [math]\displaystyle{ p }[/math] is a continuous seminorm on [math]\displaystyle{ X, }[/math] then the closure of [math]\displaystyle{ \{x \in X : p(x) \lt r\} }[/math] in [math]\displaystyle{ X }[/math] is equal to [math]\displaystyle{ \{x \in X : p(x) \leq r\}. }[/math][3]
The closure of [math]\displaystyle{ \{0\} }[/math] in a locally convex space [math]\displaystyle{ X }[/math] whose topology is defined by a family of continuous seminorms [math]\displaystyle{ \mathcal{P} }[/math] is equal to [math]\displaystyle{ \bigcap_{p \in \mathcal{P}} p^{-1}(0). }[/math][22]
A subset [math]\displaystyle{ S }[/math] in a seminormed space [math]\displaystyle{ (X, p) }[/math] is bounded if and only if [math]\displaystyle{ p(S) }[/math] is bounded.[23]
If [math]\displaystyle{ (X, p) }[/math] is a seminormed space then the locally convex topology that [math]\displaystyle{ p }[/math] induces on [math]\displaystyle{ X }[/math] makes [math]\displaystyle{ X }[/math] into a pseudometrizable TVS with a canonical pseudometric given by [math]\displaystyle{ d(x, y) := p(x - y) }[/math] for all [math]\displaystyle{ x, y \in X. }[/math][24]
The product of infinitely many seminormable spaces is again seminormable if and only if all but finitely many of these spaces are trivial (that is, 0-dimensional).[21]
Continuity of seminorms
If [math]\displaystyle{ p }[/math] is a seminorm on a topological vector space [math]\displaystyle{ X, }[/math] then the following are equivalent:[5]
[math]\displaystyle{ p }[/math] is continuous.
[math]\displaystyle{ p }[/math] is continuous at 0;[3]
[math]\displaystyle{ \{x \in X : p(x) \lt 1\} }[/math] is open in [math]\displaystyle{ X }[/math];[3]
[math]\displaystyle{ \{x \in X : p(x) \leq 1\} }[/math] is closed neighborhood of 0 in [math]\displaystyle{ X }[/math];[3]
[math]\displaystyle{ p }[/math] is uniformly continuous on [math]\displaystyle{ X }[/math];[3]
There exists a continuous seminorm [math]\displaystyle{ q }[/math] on [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ p \leq q. }[/math][3]
In particular, if [math]\displaystyle{ (X, p) }[/math] is a seminormed space then a seminorm [math]\displaystyle{ q }[/math] on [math]\displaystyle{ X }[/math] is continuous if and only if [math]\displaystyle{ q }[/math] is dominated by a positive scalar multiple of [math]\displaystyle{ p. }[/math][3]
If [math]\displaystyle{ X }[/math] is a real TVS, [math]\displaystyle{ f }[/math] is a linear functional on [math]\displaystyle{ X, }[/math] and [math]\displaystyle{ p }[/math] is a continuous seminorm (or more generally, a sublinear function) on [math]\displaystyle{ X, }[/math] then [math]\displaystyle{ f \leq p }[/math] on [math]\displaystyle{ X }[/math] implies that [math]\displaystyle{ f }[/math] is continuous.[7]
Continuity of linear maps
If [math]\displaystyle{ F : (X, p) \to (Y, q) }[/math] is a map between seminormed spaces then let[16][math]\displaystyle{ \|F\|_{p,q} := \sup \{q(F(x)) : p(x) \leq 1, x \in X\}. }[/math]
If [math]\displaystyle{ F : (X, p) \to (Y, q) }[/math] is a linear map between seminormed spaces then the following are equivalent:
There exists a real [math]\displaystyle{ K \geq 0 }[/math] such that [math]\displaystyle{ p \leq K q }[/math];[16]
In this case, [math]\displaystyle{ \|F\|_{p,q} \leq K. }[/math]
If [math]\displaystyle{ F }[/math] is continuous then [math]\displaystyle{ q(F(x)) \leq \|F\|_{p,q} p(x) }[/math] for all [math]\displaystyle{ x \in X. }[/math][16]
The space of all continuous linear maps [math]\displaystyle{ F : (X, p) \to (Y, q) }[/math] between seminormed spaces is itself a seminormed space under the seminorm [math]\displaystyle{ \|F\|_{p,q}. }[/math]
This seminorm is a norm if [math]\displaystyle{ q }[/math] is a norm.[16]
Generalizations
The concept of norm in composition algebras does not share the usual properties of a norm.
A composition algebra [math]\displaystyle{ (A, *, N) }[/math] consists of an algebra over a field[math]\displaystyle{ A, }[/math] an involution[math]\displaystyle{ \,*, }[/math] and a quadratic form[math]\displaystyle{ N, }[/math] which is called the "norm". In several cases [math]\displaystyle{ N }[/math] is an isotropic quadratic form so that [math]\displaystyle{ A }[/math] has at least one null vector, contrary to the separation of points required for the usual norm discussed in this article.
An ultraseminorm or a non-Archimedean seminorm is a seminorm [math]\displaystyle{ p : X \to \R }[/math] that also satisfies [math]\displaystyle{ p(x + y) \leq \max \{p(x), p(y)\} \text{ for all } x, y \in X. }[/math]
Weakening subadditivity: Quasi-seminorms
A map [math]\displaystyle{ p : X \to \R }[/math] is called a quasi-seminorm if it is (absolutely) homogeneous and there exists some [math]\displaystyle{ b \leq 1 }[/math] such that [math]\displaystyle{ p(x + y) \leq b p(p(x) + p(y)) \text{ for all } x, y \in X. }[/math]
The smallest value of [math]\displaystyle{ b }[/math] for which this holds is called the multiplier of [math]\displaystyle{ p. }[/math]
A quasi-seminorm that separates points is called a quasi-norm on [math]\displaystyle{ X. }[/math]
Weakening homogeneity - [math]\displaystyle{ k }[/math]-seminorms
A map [math]\displaystyle{ p : X \to \R }[/math] is called a [math]\displaystyle{ k }[/math]-seminorm if it is subadditive and there exists a [math]\displaystyle{ k }[/math] such that [math]\displaystyle{ 0 \lt k \leq 1 }[/math] and for all [math]\displaystyle{ x \in X }[/math] and scalars [math]\displaystyle{ s, }[/math][math]\displaystyle{ p(s x) = |s|^k p(x) }[/math] A [math]\displaystyle{ k }[/math]-seminorm that separates points is called a [math]\displaystyle{ k }[/math]-norm on [math]\displaystyle{ X. }[/math]
We have the following relationship between quasi-seminorms and [math]\displaystyle{ k }[/math]-seminorms:
Suppose that [math]\displaystyle{ q }[/math] is a quasi-seminorm on a vector space [math]\displaystyle{ X }[/math] with multiplier [math]\displaystyle{ b. }[/math] If [math]\displaystyle{ 0 \lt \sqrt{k} \lt \log_2 b }[/math] then there exists [math]\displaystyle{ k }[/math]-seminorm [math]\displaystyle{ p }[/math] on [math]\displaystyle{ X }[/math] equivalent to [math]\displaystyle{ q. }[/math]
↑If [math]\displaystyle{ z \in X }[/math] denotes the zero vector in [math]\displaystyle{ X }[/math] while [math]\displaystyle{ 0 }[/math] denote the zero scalar, then absolute homogeneity implies that [math]\displaystyle{ p(0) = p(0 z) = |0|p(z) = 0 p(z) = 0. }[/math][math]\displaystyle{ \blacksquare }[/math]
↑Suppose [math]\displaystyle{ p : X \to \R }[/math] is a seminorm and let [math]\displaystyle{ x \in X. }[/math] Then absolute homogeneity implies [math]\displaystyle{ p(-x) = p((-1) x) =|-1|p(x) = p(x). }[/math] The triangle inequality now implies [math]\displaystyle{ p(0) = p(x + (- x)) \leq p(x) + p(-x) = p(x) + p(x) = 2 p(x). }[/math] Because [math]\displaystyle{ x }[/math] was an arbitrary vector in [math]\displaystyle{ X, }[/math] it follows that [math]\displaystyle{ p(0) \leq 2 p(0), }[/math] which implies that [math]\displaystyle{ 0 \leq p(0) }[/math] (by subtracting [math]\displaystyle{ p(0) }[/math] from both sides). Thus [math]\displaystyle{ 0 \leq p(0) \leq 2 p(x) }[/math] which implies [math]\displaystyle{ 0 \leq p(x) }[/math] (by multiplying thru by [math]\displaystyle{ 1/2 }[/math]).
↑Let [math]\displaystyle{ x \in X }[/math] and [math]\displaystyle{ k \in p^{-1}(0). }[/math] It remains to show that [math]\displaystyle{ p(x + k) = p(x). }[/math] The triangle inequality implies [math]\displaystyle{ p(x + k) \leq p(x) + p(k) = p(x) + 0 = p(x). }[/math] Since [math]\displaystyle{ p(-k) = 0, }[/math][math]\displaystyle{ p(x) = p(x) - p(-k) \leq p(x - (-k)) = p(x + k), }[/math] as desired. [math]\displaystyle{ \blacksquare }[/math]
↑Obvious if [math]\displaystyle{ X }[/math] is a real vector space. For the non-trivial direction, assume that [math]\displaystyle{ \operatorname{Re} f \leq p }[/math] on [math]\displaystyle{ X }[/math] and let [math]\displaystyle{ x \in X. }[/math] Let [math]\displaystyle{ r \geq 0 }[/math] and [math]\displaystyle{ t }[/math] be real numbers such that [math]\displaystyle{ f(x) = r e^{i t}. }[/math] Then [math]\displaystyle{ |f(x)|= r = f\left(e^{-it} x\right) = \operatorname{Re}\left(f\left(e^{-it} x\right)\right) \leq p\left(e^{-it} x\right) = p(x). }[/math]
Adasch, Norbert; Ernst, Bruno; Keim, Dieter (1978). Topological Vector Spaces: The Theory Without Convexity Conditions. Lecture Notes in Mathematics. {3834. Berlin New York: Springer-Verlag. ISBN978-3-540-08662-8. OCLC297140003.
Wilansky, Albert (2013). Modern Methods in Topological Vector Spaces. Mineola, New York: Dover Publications, Inc. ISBN978-0-486-49353-4. OCLC849801114.