Seminorm

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In mathematics, particularly in functional analysis, a seminorm is like a norm but need not be positive definite. Seminorms are intimately connected with convex sets: every seminorm is the Minkowski functional of some absorbing disk and, conversely, the Minkowski functional of any such set is a seminorm.

A topological vector space is locally convex if and only if its topology is induced by a family of seminorms.

Definition

Let X be a vector space over either the real numbers ℝ or the complex numbers ℂ. A real-valued function p:X→ℝ is called a seminorm if it satisfies the following two conditions:

  1. Subadditivity[1]/Triangle inequality: p(x+y)≤p(x)+p(y) for all x,y∈X.
  2. Absolute homogeneity:[1] p(sx)=|s|p(x) for all x∈X and all scalars s.

These two conditions imply that p(0)=0[proof 1] and that every seminorm p also has the following property:[proof 2]

  1. Nonnegativity:[1] p(x)≥0 for all x∈X.

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 X is a seminorm that also separates points, meaning that it has the following additional property:

  1. Positive definite/Positive[1]/Point-separating: whenever x∈X satisfies p(x)=0, then x=0.

A seminormed space is a pair (X,p) consisting of a vector space X and a seminorm p on X. If the seminorm p is also a norm then the seminormed space (X,p) is called a normed space.

Since absolute homogeneity implies positive homogeneity, every seminorm is a type of function called a sublinear function. A map p:X→ℝ 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 p:X→ℝ is a seminorm if and only if it is a sublinear and balanced function.

Examples

  • The trivial seminorm on X, which refers to the constant 0 map on X, induces the indiscrete topology on X.
  • Let μ be a measure on a space Ω. For an arbitrary constant c≥1, let X be the set of all functions f:Ω→ℝ for which ‖f‖c:=(∫Ω|f|cdμ)1/c exists and is finite. It can be shown that X is a vector space, and the functional ‖⋅‖c is a seminorm on X. However, it is not always a norm (e.g. if Ω=ℝ and μ is the Lebesgue measure) because ‖h‖c=0 does not always imply h=0. To make ‖⋅‖c a norm, quotient X by the closed subspace of functions h with ‖h‖c=0. The resulting space, Lc(μ), has a norm induced by ‖⋅‖c.
  • If f is any linear form on a vector space then its absolute value |f|, defined by x↦|f(x)|, is a seminorm.
  • A sublinear function f:X→ℝ on a real vector space X is a seminorm if and only if it is a symmetric function, meaning that f(−x)=f(x) for all x∈X.
  • Every real-valued sublinear function f:X→ℝ on a real vector space X induces a seminorm p:X→ℝ defined by p(x):=max⁡{f(x),f(−x)}.[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 p:X→ℝ and q:Y→ℝ are seminorms (respectively, norms) on X and Y then the map r:X×Y→ℝ defined by r(x,y)=p(x)+q(y) is a seminorm (respectively, a norm) on X×Y. In particular, the maps on X×Y defined by (x,y)↦p(x) and (x,y)↦q(y) are both seminorms on X×Y.
  • If p and q are seminorms on X then so are[3] (p∨q)(x)=max⁡{p(x),q(x)} and (p∧q)(x):=inf⁡{p(y)+q(z):x=y+z with y,z∈X} where p∧q≤p and p∧q≤q.[4]
  • The space of seminorms on X is generally not a distributive lattice with respect to the above operations. For example, over ℝ2, p(x,y):=max⁡(|x|,|y|),q(x,y):=2|x|,r(x,y):=2|y| are such that ((p∨q)∧(p∨r))(x,y)=inf⁡{max⁡(2|x1|,|y1|)+max⁡(|x2|,2|y2|):x=x1+x2 and y=y1+y2} while (p∨q∧r)(x,y):=max⁡(|x|,|y|)
  • If L:X→Y is a linear map and q:Y→ℝ is a seminorm on Y, then q∘L:X→ℝ is a seminorm on X. The seminorm q∘L will be a norm on X if and only if L is injective and the restriction q|L(X) is a norm on L(X).

Minkowski functionals and seminorms

Seminorms on a vector space X are intimately tied, via Minkowski functionals, to subsets of X that are convex, balanced, and absorbing. Given such a subset D of X, the Minkowski functional of D is a seminorm. Conversely, given a seminorm p on X, the sets{x∈X:p(x)<1} and {x∈X:p(x)≤1} 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 p.[5]

Algebraic properties

Every seminorm is a sublinear function, and thus satisfies all properties of a sublinear function, including convexity, p(0)=0, and for all vectors x,y∈X: the reverse triangle inequality: [2][6] |p(x)−p(y)|≤p(x−y) and also 0≤max⁡{p(x),p(−x)} and p(x)−p(y)≤p(x−y).[2][6]

For any vector x∈X and positive real r>0:[7] x+{y∈X:p(y)<r}={y∈X:p(x−y)<r} and furthermore, {x∈X:p(x)<r} is an absorbing disk in X.[3]

If p is a sublinear function on a real vector space X then there exists a linear functional f on X such that f≤p[6] and furthermore, for any linear functional g on X, g≤p on X if and only if g−1(1)∩{x∈X:p(x)<1}=∅.[6]

Other properties of seminorms

Every seminorm is a balanced function. A seminorm p is a norm on X if and only if {x∈X:p(x)<1} does not contain a non-trivial vector subspace.

If p:X→[0,∞) is a seminorm on X then ker⁡p:=p−1(0) is a vector subspace of X and for every x∈X, p is constant on the set x+ker⁡p={x+k:p(k)=0} and equal to p(x).[proof 3]

Furthermore, for any real r>0,[3] r{x∈X:p(x)<1}={x∈X:p(x)<r}={x∈X:1rp(x)<1}.

If D is a set satisfying {x∈X:p(x)<1}⊆D⊆{x∈X:p(x)≤1} then D is absorbing in X and p=pD where pD denotes the Minkowski functional associated with D (that is, the gauge of D).[5] In particular, if D is as above and q is any seminorm on X, then q=p if and only if {x∈X:q(x)<1}⊆D⊆{x∈X:q(x)≤}.[5]

If (X,‖⋅‖) is a normed space and x,y∈X then ‖x−y‖=‖x−z‖+‖z−y‖ for all z in the interval [x,y].[8]

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 p:X→ℝ be a non-negative function. The following are equivalent:

  1. p is a seminorm.
  2. p is a convex F-seminorm.
  3. p is a convex balanced G-seminorm.[9]

If any of the above conditions hold, then the following are equivalent:

  1. p is a norm;
  2. {x∈X:p(x)<1} does not contain a non-trivial vector subspace.[10]
  3. There exists a norm on X, with respect to which, {x∈X:p(x)<1} is bounded.

If p is a sublinear function on a real vector space X then the following are equivalent:[6]

  1. p is a linear functional;
  2. p(x)+p(−x)≤0 for every x∈X;
  3. p(x)+p(−x)=0 for every x∈X;

Inequalities involving seminorms

If p,q:X→[0,∞) are seminorms on X then:

  • p≤q if and only if q(x)≤1 implies p(x)≤1.[11]
  • If a>0 and b>0 are such that p(x)<a implies q(x)≤b, then aq(x)≤bp(x) for all x∈X. [12]
  • Suppose a and b are positive real numbers and q,p1,…,pn are seminorms on X such that for every x∈X, if max⁡{p1(x),…,pn(x)}<a then q(x)<b. Then aq≤b(p1+⋯+pn).[10]
  • If X is a vector space over the reals and f is a non-zero linear functional on X, then f≤p if and only if ∅=f−1(1)∩{x∈X:p(x)<1}.[11]

If p is a seminorm on X and f is a linear functional on X then:

  • |f|≤p on X if and only if Re⁡f≤p on X (see footnote for proof).[13][14]
  • f≤p on X if and only if f−1(1)∩{x∈X:p(x)<1=∅}.[6][11]
  • If a>0 and b>0 are such that p(x)<a implies f(x)≠b, then a|f(x)|≤bp(x) for all x∈X.[12]

Hahn–Banach theorem for seminorms

Seminorms offer a particularly clean formulation of the Hahn–Banach theorem:

If M is a vector subspace of a seminormed space (X,p) and if f is a continuous linear functional on M, then f may be extended to a continuous linear functional F on X that has the same norm as f.[15]

A similar extension property also holds for seminorms:

Theorem[16][12] (Extending seminorms) — If M is a vector subspace of X, p is a seminorm on M, and q is a seminorm on X such that p≤q|M, then there exists a seminorm P on X such that P|M=p and P≤q.

Proof: Let S be the convex hull of {m∈M:p(m)≤1}∪{x∈X:q(x)≤1}. Then S is an absorbing disk in X and so the Minkowski functional P of S is a seminorm on X. This seminorm satisfies p=P on M and P≤q on X. ◼

Topologies of seminormed spaces

Pseudometrics and the induced topology

A seminorm p on X induces a topology, called the seminorm-induced topology, via the canonical translation-invariant pseudometric dp:X×X→ℝ; dp(x,y):=p(x−y)=p(y−x). This topology is Hausdorff if and only if dp is a metric, which occurs if and only if p is a norm.[4] This topology makes X into a locally convex pseudometrizable topological 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: {x∈X:p(x)<r} or {x∈X:p(x)≤r} as r>0 ranges over the positive reals. Every seminormed space (X,p) 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 X with seminorm p induces a vector space quotient X/W, where W is the subspace of X consisting of all vectors x∈X with p(x)=0. Then X/W carries a norm defined by p(x+W)=p(x). The resulting topology, pulled back to X, is precisely the topology induced by p.

Any seminorm-induced topology makes X locally convex, as follows. If p is a seminorm on X and r∈ℝ, call the set {x∈X:p(x)<r} the open ball of radius r about the origin; likewise the closed ball of radius r is {x∈X:p(x)≤r}. The set of all open (resp. closed) p-balls at the origin forms a neighborhood basis of convex balanced sets that are open (resp. closed) in the p-topology on X.

Stronger, weaker, and equivalent seminorms

The notions of stronger and weaker seminorms are akin to the notions of stronger and weaker norms. If p and q are seminorms on X, then we say that q is stronger than p and that p is weaker than q if any of the following equivalent conditions holds:

  1. The topology on X induced by q is finer than the topology induced by p.
  2. If x∙=(xi)i=1∞ is a sequence in X, then q(x∙):=(q(xi))i=1∞→0 in ℝ implies p(x∙)→0 in ℝ.[4]
  3. If x∙=(xi)i∈I is a net in X, then q(x∙):=(q(xi))i∈I→0 in ℝ implies p(x∙)→0 in ℝ.
  4. p is bounded on {x∈X:q(x)<1}.[4]
  5. If inf⁡{q(x):p(x)=1,x∈X}=0 then p(x)=0 for all x∈X.[4]
  6. There exists a real K>0 such that p≤Kq on X.[4]

The seminorms p and q are called equivalent if they are both weaker (or both stronger) than each other. This happens if they satisfy any of the following conditions:

  1. The topology on X induced by q is the same as the topology induced by p.
  2. q is stronger than p and p is stronger than q.[4]
  3. If x∙=(xi)i=1∞ is a sequence in X then q(x∙):=(q(xi))i=1∞→0 if and only if p(x∙)→0.
  4. There exist positive real numbers r>0 and R>0 such that rq≤p≤Rq.

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.[17] Thus a locally convex TVS is seminormable if and only if it has a non-empty bounded open set.[18] A TVS is normable if and only if it is a T1 space and admits a bounded convex neighborhood of the origin.

If X is a Hausdorff locally convex TVS then the following are equivalent:

  1. X is normable.
  2. X is seminormable.
  3. X has a bounded neighborhood of the origin.
  4. The strong dual Xb′ of X is normable.[19]
  5. The strong dual Xb′ of X is metrizable.[19]

Furthermore, X is finite dimensional if and only if Xσ′ is normable (here Xσ′ denotes X′ 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).[18]

Topological properties

  • If X is a TVS and p is a continuous seminorm on X, then the closure of {x∈X:p(x)<r} in X is equal to {x∈X:p(x)≤r}.[3]
  • The closure of {0} in a locally convex space X whose topology is defined by a family of continuous seminorms 𝒫 is equal to ⋂p∈𝒫p−1(0).[11]
  • A subset S in a seminormed space (X,p) is bounded if and only if p(S) is bounded.[20]
  • If (X,p) is a seminormed space then the locally convex topology that p induces on X makes X into a pseudometrizable TVS with a canonical pseudometric given by d(x,y):=p(x−y) for all x,y∈X.[21]
  • 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).[18]

Continuity of seminorms

If p is a seminorm on a topological vector space X, then the following are equivalent:[5]

  1. p is continuous.
  2. p is continuous at 0;[3]
  3. {x∈X:p(x)<1} is open in X;[3]
  4. {x∈X:p(x)≤1} is closed neighborhood of 0 in X;[3]
  5. p is uniformly continuous on X;[3]
  6. There exists a continuous seminorm q on X such that p≤q.[3]

In particular, if (X,p) is a seminormed space then a seminorm q on X is continuous if and only if q is dominated by a positive scalar multiple of p.[3]

If X is a real TVS, f is a linear functional on X, and p is a continuous seminorm (or more generally, a sublinear function) on X, then f≤p on X implies that f is continuous.[6]

Continuity of linear maps

If F:(X,p)→(Y,q) is a map between seminormed spaces then let[15] ‖F‖p,q:=sup⁡{q(F(x)):p(x)≤1,x∈X}.

If F:(X,p)→(Y,q) is a linear map between seminormed spaces then the following are equivalent:

  1. F is continuous;
  2. ‖F‖p,q<∞;[15]
  3. There exists a real K≥0 such that p≤Kq;[15]
    • In this case, ‖F‖p,q≤K.

If F is continuous then q(F(x))≤‖F‖p,qp(x) for all x∈X.[15]

The space of all continuous linear maps F:(X,p)→(Y,q) between seminormed spaces is itself a seminormed space under the seminorm ‖F‖p,q. This seminorm is a norm if q is a norm.[15]

Generalizations

The concept of norm in composition algebras does not share the usual properties of a norm.

A composition algebra (A,*,N) consists of an algebra over a field A, an involution *, and a quadratic form N, which is called the "norm". In several cases N is an isotropic quadratic form so that A 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 p:X→ℝ that also satisfies p(x+y)≤max⁡{p(x),p(y)} for all x,y∈X.

Weakening subadditivity: Quasi-seminorms

A map p:X→ℝ is called a quasi-seminorm if it is (absolutely) homogeneous and there exists some b≤1 such that p(x+y)≤bp(p(x)+p(y)) for all x,y∈X. The smallest value of b for which this holds is called the multiplier of p.

A quasi-seminorm that separates points is called a quasi-norm on X.

Weakening homogeneity - k-seminorms

A map p:X→ℝ is called a k-seminorm if it is subadditive and there exists a k such that 0<k≤1 and for all x∈X and scalars s,p(sx)=|s|kp(x) A k-seminorm that separates points is called a k-norm on X.

We have the following relationship between quasi-seminorms and k-seminorms:

Suppose that q is a quasi-seminorm on a vector space X with multiplier b. If 0<k<log2b then there exists k-seminorm p on X equivalent to q.

See also

Notes

Proofs

  1. ↑ If z∈X denotes the zero vector in X while 0 denote the zero scalar, then absolute homogeneity implies that p(z)=p(0z)=|0|p(z)=0p(z)=0. ◼
  2. ↑ Suppose p:X→ℝ is a seminorm and let x∈X. Then absolute homogeneity implies p(−x)=p((−1)x)=|−1|p(x)=p(x). The triangle inequality now implies p(0)=p(x+(−x))≤p(x)+p(−x)=p(x)+p(x)=2p(x). Because x was an arbitrary vector in X, it follows that p(0)≤2p(0), which implies that 0≤p(0) (by subtracting p(0) from both sides). Thus 0≤p(0)≤2p(x) which implies 0≤p(x) (by multiplying through by 1/2). ◼
  3. ↑ Let x∈X and k∈p−1(0). It remains to show that p(x+k)=p(x). The triangle inequality implies p(x+k)≤p(x)+p(k)=p(x)+0=p(x). Since p(−k)=0, p(x)=p(x)−p(−k)≤p(x−(−k))=p(x+k), as desired. ◼

References

  1. ↑ 1.0 1.1 1.2 1.3 Kubrusly 2011, p. 200.
  2. ↑ 2.0 2.1 2.2 Narici & Beckenstein 2011, pp. 120–121.
  3. ↑ 3.00 3.01 3.02 3.03 3.04 3.05 3.06 3.07 3.08 3.09 Narici & Beckenstein 2011, pp. 116–128.
  4. ↑ 4.0 4.1 4.2 4.3 4.4 4.5 4.6 Wilansky 2013, pp. 15–21.
  5. ↑ 5.0 5.1 5.2 5.3 Schaefer & Wolff 1999, p. 40.
  6. ↑ 6.0 6.1 6.2 6.3 6.4 6.5 6.6 Narici & Beckenstein 2011, pp. 177–220.
  7. ↑ Narici & Beckenstein 2011, pp. 116−128.
  8. ↑ Narici & Beckenstein 2011, pp. 107–113.
  9. ↑ Schechter 1996, p. 691.
  10. ↑ 10.0 10.1 Narici & Beckenstein 2011, p. 149.
  11. ↑ 11.0 11.1 11.2 11.3 Narici & Beckenstein 2011, pp. 149–153.
  12. ↑ 12.0 12.1 12.2 Wilansky 2013, pp. 18–21.
  13. ↑ Obvious if X is a real vector space. For the non-trivial direction, assume that Re⁡f≤p on X and let x∈X. Let r≥0 and t be real numbers such that f(x)=reit. Then |f(x)|=r=f(e−itx)=Re⁡(f(e−itx))≤p(e−itx)=p(x).
  14. ↑ Wilansky 2013, p. 20.
  15. ↑ 15.0 15.1 15.2 15.3 15.4 15.5 Wilansky 2013, pp. 21–26.
  16. ↑ Narici & Beckenstein 2011, pp. 150.
  17. ↑ Wilansky 2013, pp. 50–51.
  18. ↑ 18.0 18.1 18.2 Narici & Beckenstein 2011, pp. 156–175.
  19. ↑ 19.0 19.1 Trèves 2006, pp. 136–149, 195–201, 240–252, 335–390, 420–433.
  20. ↑ Wilansky 2013, pp. 49–50.
  21. ↑ Narici & Beckenstein 2011, pp. 115–154.




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