In mathematics, a subset C of a real or complex vector space is said to be absolutely convex or disked if it is convex and balanced (some people use the term "circled" instead of "balanced"), in which case it is called a disk. The disked hull or the absolute convex hull of a set is the intersection of all disks containing that set.
A subset S {\displaystyle S} of a real or complex vector space X {\displaystyle X} is called a disk and is said to be disked, absolutely convex, and convex balanced if any of the following equivalent conditions is satisfied:
The smallest convex (respectively, balanced) subset of X {\displaystyle X} containing a given set is called the convex hull (respectively, the balanced hull) of that set and is denoted by co S {\displaystyle \operatorname {co} S} (respectively, bal S {\displaystyle \operatorname {bal} S} ).
Similarly, the disked hull, the absolute convex hull, and the convex balanced hull of a set S {\displaystyle S} is defined to be the smallest disk (with respect to subset inclusion) containing S . {\displaystyle S.} [1] The disked hull of S {\displaystyle S} will be denoted by disk S {\displaystyle \operatorname {disk} S} or cobal S {\displaystyle \operatorname {cobal} S} and it is equal to each of the following sets:
The intersection of arbitrarily many absolutely convex sets is again absolutely convex; however, unions of absolutely convex sets need not be absolutely convex anymore.
If D {\displaystyle D} is a disk in X , {\displaystyle X,} then D {\displaystyle D} is absorbing in X {\displaystyle X} if and only if span D = X . {\displaystyle \operatorname {span} D=X.} [2]
If S {\displaystyle S} is an absorbing disk in a vector space X {\displaystyle X} then there exists an absorbing disk E {\displaystyle E} in X {\displaystyle X} such that E + E ⊆ S . {\displaystyle E+E\subseteq S.} [3] If D {\displaystyle D} is a disk and r {\displaystyle r} and s {\displaystyle s} are scalars then s D = | s | D {\displaystyle sD=|s|D} and ( r D ) ∩ ( s D ) = ( min { | r | , | s | } ) D . {\displaystyle (rD)\cap (sD)=(\min _{}\{|r|,|s|\})D.}
The absolutely convex hull of a bounded set in a locally convex topological vector space is again bounded.
If D {\displaystyle D} is a bounded disk in a TVS X {\displaystyle X} and if x ∙ = ( x i ) i = 1 ∞ {\displaystyle x_{\bullet }=\left(x_{i}\right)_{i=1}^{\infty }} is a sequence in D , {\displaystyle D,} then the partial sums s ∙ = ( s n ) n = 1 ∞ {\displaystyle s_{\bullet }=\left(s_{n}\right)_{n=1}^{\infty }} are Cauchy, where for all n , {\displaystyle n,} s n := ∑ i = 1 n 2 − i x i . {\displaystyle s_{n}:=\sum _{i=1}^{n}2^{-i}x_{i}.} [4] In particular, if in addition D {\displaystyle D} is a sequentially complete subset of X , {\displaystyle X,} then this series s ∙ {\displaystyle s_{\bullet }} converges in X {\displaystyle X} to some point of D . {\displaystyle D.}
The convex balanced hull of S {\displaystyle S} contains both the convex hull of S {\displaystyle S} and the balanced hull of S . {\displaystyle S.} Furthermore, it contains the balanced hull of the convex hull of S ; {\displaystyle S;} thus bal ( co S ) ⊆ cobal S = co ( bal S ) , {\displaystyle \operatorname {bal} (\operatorname {co} S)~\subseteq ~\operatorname {cobal} S~=~\operatorname {co} (\operatorname {bal} S),} where the example below shows that this inclusion might be strict. However, for any subsets S , T ⊆ X , {\displaystyle S,T\subseteq X,} if S ⊆ T {\displaystyle S\subseteq T} then cobal S ⊆ cobal T {\displaystyle \operatorname {cobal} S\subseteq \operatorname {cobal} T} which implies cobal ( co S ) = cobal S = cobal ( bal S ) . {\displaystyle \operatorname {cobal} (\operatorname {co} S)=\operatorname {cobal} S=\operatorname {cobal} (\operatorname {bal} S).}
Although cobal S = co ( bal S ) , {\displaystyle \operatorname {cobal} S=\operatorname {co} (\operatorname {bal} S),} the convex balanced hull of S {\displaystyle S} is not necessarily equal to the balanced hull of the convex hull of S . {\displaystyle S.} [1] For an example where cobal S ≠ bal ( co S ) {\displaystyle \operatorname {cobal} S\neq \operatorname {bal} (\operatorname {co} S)} let X {\displaystyle X} be the real vector space R 2 {\displaystyle \mathbb {R} ^{2}} and let S := { ( − 1 , 1 ) , ( 1 , 1 ) } . {\displaystyle S:=\{(-1,1),(1,1)\}.} Then bal ( co S ) {\displaystyle \operatorname {bal} (\operatorname {co} S)} is a strict subset of cobal S {\displaystyle \operatorname {cobal} S} that is not even convex; in particular, this example also shows that the balanced hull of a convex set is not necessarily convex. The set cobal S {\displaystyle \operatorname {cobal} S} is equal to the closed and filled square in X {\displaystyle X} with vertices ( − 1 , 1 ) , ( 1 , 1 ) , ( − 1 , − 1 ) , {\displaystyle (-1,1),(1,1),(-1,-1),} and ( 1 , − 1 ) {\displaystyle (1,-1)} (this is because the balanced set cobal S {\displaystyle \operatorname {cobal} S} must contain both S {\displaystyle S} and − S = { ( − 1 , − 1 ) , ( 1 , − 1 ) } , {\displaystyle -S=\{(-1,-1),(1,-1)\},} where since cobal S {\displaystyle \operatorname {cobal} S} is also convex, it must consequently contain the solid square co ( ( − S ) ∪ S ) , {\displaystyle \operatorname {co} ((-S)\cup S),} which for this particular example happens to also be balanced so that cobal S = co ( ( − S ) ∪ S ) {\displaystyle \operatorname {cobal} S=\operatorname {co} ((-S)\cup S)} ). However, co ( S ) {\displaystyle \operatorname {co} (S)} is equal to the horizontal closed line segment between the two points in S {\displaystyle S} so that bal ( co S ) {\displaystyle \operatorname {bal} (\operatorname {co} S)} is instead a closed "hour glass shaped" subset that intersects the x {\displaystyle x} -axis at exactly the origin and is the union of two closed and filled isosceles triangles: one whose vertices are the origin together with S {\displaystyle S} and the other triangle whose vertices are the origin together with − S = { ( − 1 , − 1 ) , ( 1 , − 1 ) } . {\displaystyle -S=\{(-1,-1),(1,-1)\}.} This non-convex filled "hour-glass" bal ( co S ) {\displaystyle \operatorname {bal} (\operatorname {co} S)} is a proper subset of the filled square cobal S = co ( bal S ) . {\displaystyle \operatorname {cobal} S=\operatorname {co} (\operatorname {bal} S).}
Given a fixed real number 0 < p ≤ 1 , {\displaystyle 0<p\leq 1,} a p {\displaystyle p} -convex set is any subset C {\displaystyle C} of a vector space X {\displaystyle X} with the property that r c + s d ∈ C {\displaystyle rc+sd\in C} whenever c , d ∈ C {\displaystyle c,d\in C} and r , s ≥ 0 {\displaystyle r,s\geq 0} are non-negative scalars satisfying r p + s p = 1. {\displaystyle r^{p}+s^{p}=1.} It is called an absolutely p {\displaystyle p} -convex set or a p {\displaystyle p} -disk if r c + s d ∈ C {\displaystyle rc+sd\in C} whenever c , d ∈ C {\displaystyle c,d\in C} and r , s {\displaystyle r,s} are scalars satisfying | r | p + | s | p ≤ 1. {\displaystyle |r|^{p}+|s|^{p}\leq 1.} [5]
A p {\displaystyle p} -seminorm[6] is any non-negative function q : X → R {\displaystyle q:X\to \mathbb {R} } that satisfies the following conditions:
This generalizes the definition of seminorms since a map is a seminorm if and only if it is a 1 {\displaystyle 1} -seminorm (using p := 1 {\displaystyle p:=1} ). There exist p {\displaystyle p} -seminorms that are not seminorms. For example, whenever 0 < p < 1 {\displaystyle 0<p<1} then the map q ( f ) = ∫ R | f ( t ) | p d t {\displaystyle q(f)=\int _{\mathbb {R} }|f(t)|^{p}dt} used to define the Lp space L p ( R ) {\displaystyle L_{p}(\mathbb {R} )} is a p {\displaystyle p} -seminorm but not a seminorm.[6]
Given 0 < p ≤ 1 , {\displaystyle 0<p\leq 1,} a topological vector space is p {\displaystyle p} -seminormable (meaning that its topology is induced by some p {\displaystyle p} -seminorm) if and only if it has a bounded p {\displaystyle p} -convex neighborhood of the origin.[5]