Absolutely convex set

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 ${\displaystyle D}$ is a disk in ${\displaystyle X,}$ then ${\displaystyle D}$ is absorbing in ${\displaystyle X}$ if and only if ${\displaystyle \mathrm{span}D=X.}$

Properties

If ${\displaystyle S}$ is an absorbing disk in a vector space ${\displaystyle X}$ then there exists an absorbing disk ${\displaystyle E}$ in ${\displaystyle X}$ such that ${\displaystyle E+E\subseteq S.}$ If ${\displaystyle D}$ is a disk and ${\displaystyle r}$ and ${\displaystyle s}$ are scalars then ${\displaystyle sD=|s|D}$ and ${\displaystyle (rD)\cap (sD)=(\underset{}{min}\{|r|,|s|\})D.}$

The absolutely convex hull of a bounded set in a locally convex topological vector space is again bounded.

If ${\displaystyle D}$ is a bounded disk in a TVS ${\displaystyle X}$ and if ${\displaystyle x_{\bullet }={\left(x_i\right)}_{i=1}^{\mathrm{\infty }}}$ is a sequence in ${\displaystyle D,}$ then the partial sums ${\displaystyle s_{\bullet }={\left(s_n\right)}_{n=1}^{\mathrm{\infty }}}$ are Cauchy, where for all ${\displaystyle n,}$ ${\displaystyle s_n:=\sum _{i=1}^n{2}^{-i}{x}_i.}$ In particular, if in addition ${\displaystyle D}$ is a sequentially complete subset of ${\displaystyle X,}$ then this series ${\displaystyle s_{\bullet }}$ converges in ${\displaystyle X}$ to some point of ${\displaystyle D.}$

The convex balanced hull of ${\displaystyle S}$ contains both the convex hull of ${\displaystyle S}$ and the balanced hull of ${\displaystyle S.}$ Furthermore, it contains the balanced hull of the convex hull of ${\displaystyle S;}$ thus $${\displaystyle \mathrm{bal}(\mathrm{co}S)\subseteq \mathrm{cobal}S=\mathrm{co}(\mathrm{bal}S),}$$ where the example below shows that this inclusion might be strict. However, for any subsets ${\displaystyle S,T\subseteq X,}$ if ${\displaystyle S\subseteq T}$ then ${\displaystyle \mathrm{cobal}S\subseteq \mathrm{cobal}T}$ which implies ${\displaystyle \mathrm{cobal}(\mathrm{co}S)=\mathrm{cobal}S=\mathrm{cobal}(\mathrm{bal}S).}$

Examples

Although ${\displaystyle \mathrm{cobal}S=\mathrm{co}(\mathrm{bal}S),}$ the convex balanced hull of ${\displaystyle S}$ is not necessarily equal to the balanced hull of the convex hull of ${\displaystyle S.}$ For an example where ${\displaystyle \mathrm{cobal}S\ne \mathrm{bal}(\mathrm{co}S)}$ let ${\displaystyle X}$ be the real vector space ${\displaystyle {\mathbb{R}}^2}$ and let ${\displaystyle S:=\{(-1,1),(1,1)\}.}$ Then ${\displaystyle \mathrm{bal}(\mathrm{co}S)}$ is a strict subset of ${\displaystyle \mathrm{cobal}S}$ 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 ${\displaystyle \mathrm{cobal}S}$ is equal to the closed and filled square in ${\displaystyle X}$ with vertices ${\displaystyle (-1,1),(1,1),(-1,-1),}$ and ${\displaystyle (1,-1)}$ (this is because the balanced set ${\displaystyle \mathrm{cobal}S}$ must contain both ${\displaystyle S}$ and ${\displaystyle -S=\{(-1,-1),(1,-1)\},}$ where since ${\displaystyle \mathrm{cobal}S}$ is also convex, it must consequently contain the solid square ${\displaystyle \mathrm{co}((-S)\cup S),}$ which for this particular example happens to also be balanced so that ${\displaystyle \mathrm{cobal}S=\mathrm{co}((-S)\cup S)}$). However, ${\displaystyle \mathrm{co}(S)}$ is equal to the horizontal closed line segment between the two points in ${\displaystyle S}$ so that ${\displaystyle \mathrm{bal}(\mathrm{co}S)}$ is instead a closed "hour glass shaped" subset that intersects the ${\displaystyle x}$-axis at exactly the origin and is the union of two closed and filled isosceles triangles: one whose vertices are the origin together with ${\displaystyle S}$ and the other triangle whose vertices are the origin together with ${\displaystyle -S=\{(-1,-1),(1,-1)\}.}$ This non-convex filled "hour-glass" ${\displaystyle \mathrm{bal}(\mathrm{co}S)}$ is a proper subset of the filled square ${\displaystyle \mathrm{cobal}S=\mathrm{co}(\mathrm{bal}S).}$

Generalizations

Given a fixed real number ${\displaystyle 0<p\le 1,}$ a ${\displaystyle p}$-convex set is any subset ${\displaystyle C}$ of a vector space ${\displaystyle X}$ with the property that ${\displaystyle rc+sd\in C}$ whenever ${\displaystyle c,d\in C}$ and ${\displaystyle r,s\ge 0}$ are non-negative scalars satisfying ${\displaystyle r^p+s^p=1.}$ It is called an absolutely ${\displaystyle p}$-convex set or a ${\displaystyle p}$-disk if ${\displaystyle rc+sd\in C}$ whenever ${\displaystyle c,d\in C}$ and ${\displaystyle r,s}$ are scalars satisfying ${\displaystyle |r{|}^p+|s{|}^p\le 1.}$

A ${\displaystyle p}$-seminorm is any non-negative function ${\displaystyle q:X\to \mathbb{R}}$ that satisfies the following conditions:

1. Subadditivity/Triangle inequality: ${\displaystyle q(x+y)\le q(x)+q(y)}$ for all ${\displaystyle x,y\in X.}$
2. Absolute homogeneity of degree ${\displaystyle p}$: ${\displaystyle q(sx)=|s{|}^{p}q(x)}$ for all ${\displaystyle x\in X}$ and all scalars ${\displaystyle s.}$

This generalizes the definition of seminorms since a map is a seminorm if and only if it is a ${\displaystyle 1}$-seminorm (using ${\displaystyle p:=1}$). There exist ${\displaystyle p}$-seminorms that are not seminorms. For example, whenever ${\displaystyle 0<p<1}$ then the map ${\displaystyle q(f)={\int }_{\mathbb{R}}|f(t){|}^{p}dt}$ used to define the Lp space ${\displaystyle L_p(\mathbb{R})}$ is a ${\displaystyle p}$-seminorm but not a seminorm.

Given ${\displaystyle 0<p\le 1,}$ a topological vector space is ${\displaystyle p}$-seminormable (meaning that its topology is induced by some ${\displaystyle p}$-seminorm) if and only if it has a bounded ${\displaystyle p}$-convex neighborhood of the origin.