Absolutely convex set

A subset S of a real or complex vector space X is called a ' and is said to be ', ', and ' if any of the following equivalent conditions is satisfied: S is a convex and balanced set. for any scalars a and b, if |a| + |b| \leq 1 then a S + b S \subseteq S. for all scalars a, b, and c, if |a| + |b| \leq |c|, then a S + b S \subseteq c S. for any scalars a_1, \ldots, a_n and c, if |a_1| + \cdots + |a_n| \leq |c| then a_1 S + \cdots + a_n S \subseteq c S. for any scalars a_1, \ldots, a_n, if |a_1| + \cdots + |a_n| \leq 1 then a_1 S + \cdots + a_n S \subseteq S. The smallest convex (respectively, balanced) subset of X containing a given set is called the convex hull (respectively, the balanced hull) of that set and is denoted by \operatorname{co} S (respectively, \operatorname{bal} S). Similarly, the ', the ', and the '''' of a set S is defined to be the smallest disk (with respect to subset inclusion) containing S. The disked hull of S will be denoted by \operatorname{disk} S or \operatorname{cobal} S and it is equal to each of the following sets: \operatorname{co} (\operatorname{bal} S), which is the convex hull of the balanced hull of S; thus, \operatorname{cobal} S = \operatorname{co} (\operatorname{bal} S). • In general, \operatorname{cobal} S \neq \operatorname{bal} (\operatorname{co} S) is possible, even in finite dimensional vector spaces. the intersection of all disks containing S. \left\{a_1 s_1 + \cdots a_n s_n ~:~ n \in \N, \, s_1, \ldots, s_n \in S, \, \text{ and } a_1, \ldots, a_n \text{ are scalars satisfying } |a_1| + \cdots + |a_n| \leq 1\right\}. ==Sufficient conditions ==

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 is a disk in X, then D is absorbing in X if and only if \operatorname{span} D = X. ==Properties==

If S is an absorbing disk in a vector space X then there exists an absorbing disk E in X such that E + E \subseteq S. If D is a disk and r and s are scalars then s D = |s| D and (r D) \cap (s D) = (\min_{} \{|r|, |s|\}) D. The absolutely convex hull of a bounded set in a locally convex topological vector space is again bounded. If D is a bounded disk in a TVS X and if x_{\bull} = \left(x_i\right)_{i=1}^{\infty} is a sequence in D, then the partial sums s_{\bull} = \left(s_n\right)_{n=1}^{\infty} are Cauchy, where for all n, s_n := \sum_{i=1}^n 2^{-i} x_i. In particular, if in addition D is a sequentially complete subset of X, then this series s_{\bull} converges in X to some point of D. The convex balanced hull of S contains both the convex hull of S and the balanced hull of S. Furthermore, it contains the balanced hull of the convex hull of S; thus \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 \subseteq X, if S \subseteq T then \operatorname{cobal} S \subseteq \operatorname{cobal} T which implies \operatorname{cobal} (\operatorname{co} S) = \operatorname{cobal} S = \operatorname{cobal} (\operatorname{bal} S). Examples Although \operatorname{cobal} S = \operatorname{co} (\operatorname{bal} S), the convex balanced hull of S is necessarily equal to the balanced hull of the convex hull of S. For an example where \operatorname{cobal} S \neq \operatorname{bal} (\operatorname{co} S) let X be the real vector space \R^2 and let S := \{(-1, 1), (1, 1)\}. Then \operatorname{bal} (\operatorname{co} S) is a strict subset of \operatorname{cobal} S that is not even convex; in particular, this example also shows that the balanced hull of a convex set is necessarily convex. The set \operatorname{cobal} S is equal to the closed and filled square in X with vertices (-1, 1), (1, 1), (-1, -1), and (1, -1) (this is because the balanced set \operatorname{cobal} S must contain both S and -S = \{(-1, -1), (1, -1)\}, where since \operatorname{cobal} S is also convex, it must consequently contain the solid square \operatorname{co} ((-S) \cup S), which for this particular example happens to also be balanced so that \operatorname{cobal} S = \operatorname{co} ((-S) \cup S)). However, \operatorname{co} (S) is equal to the horizontal closed line segment between the two points in S so that \operatorname{bal} (\operatorname{co} S) is instead a closed "hour glass shaped" subset that intersects the 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 and the other triangle whose vertices are the origin together with - S = \{(-1, -1), (1, -1)\}. This non-convex filled "hour-glass" \operatorname{bal} (\operatorname{co} S) is a proper subset of the filled square \operatorname{cobal} S = \operatorname{co} (\operatorname{bal} S). ==Generalizations==

Given a fixed real number 0 a is any subset C of a vector space X with the property that r c + s d \in C whenever c, d \in C and r, s \geq 0 are non-negative scalars satisfying r^p + s^p = 1. It is called an or a if r c + s d \in C whenever c, d \in C and r, s are scalars satisfying |r|^p + |s|^p \leq 1. A is any non-negative function q : X \to \R that satisfies the following conditions: • Subadditivity/Triangle inequality: q(x + y) \leq q(x) + q(y) for all x, y \in X. • Absolute homogeneity of degree p: q(s x) =|s|^p q(x) for all x \in X and all scalars s. This generalizes the definition of seminorms since a map is a seminorm if and only if it is a 1-seminorm (using p := 1). There exist p-seminorms that are not seminorms. For example, whenever 0 then the map q(f) = \int_{\R} |f(t)|^p d t used to define the Lp space L_p(\R) is a p-seminorm but not a seminorm. Given 0 a topological vector space is (meaning that its topology is induced by some p-seminorm) if and only if it has a bounded p-convex neighborhood of the origin. ==See also==