Characterization of Convex Absorbing Set in Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$. 
Let $C$ be a convex set. 
Then $C$ is absorbing :
:for each $x \in X$ there exists $t \in \R_{> 0}$ such that $x \in t C$.
== Proof ==
=== Necessary Condition ===
Suppose that $C$ is absorbing.
Then for each $x \in X$ there exists $s \in \R_{> 0}$ such that $x \in t C$ for $s \in \C$ with $\cmod s \ge t$.
In particular, $x \in t C$. 
=== Sufficient Condition ===
Suppose that:
:for each $x \in X$ there exists $t \in \R_{> 0}$ such that $x \in t C$.
Then there exists $t \in \R_{> 0}$ such that:
:${\mathbf 0}_X \in t C$
In particular, we have:
:${\mathbf 0}_X \in C$
First suppose that $\GF = \R$. 
Let $x \in X$.
Let $t_1, t_2 > 0$ be such that: 
:$x \in t_1 C$
:$-x \in t_2 C$
Let $t = \max \set {t_1, t_2}$.
Then:
:$\ds \frac {t_i} t \in \closedint 0 1$ for each $i \in \set {1, 2}$.
From Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1, we have: 
:$\ds \frac {t_i} t C \subseteq C$
hence:
:$t_i \subseteq t C$ for each $i \in \set {1, 2}$. 
Let $\alpha \in \C$ be such that $\cmod \alpha > t$.
Let:
:$\ds c = \frac \alpha {\cmod \alpha}$ 
Then, we have:
We have:
:$\ds \frac c t x \in \set {\frac x t, -\frac x t}$
so that:
:$\ds \frac c t x \in C$
We therefore have:
:$\ds \frac c t x \in C$
so that:
:$\ds \frac 1 \alpha x \in \frac t {\cmod \alpha}$
Since:
:$\ds \frac t {\cmod \alpha} < 1$
we have:
:$\ds \frac 1 \alpha x \in C$
so:
:$x \in \alpha C$ for $\cmod \alpha > t$.
So $C$ is absorbing.
Now consider the case $\GF = \C$. 
Let $x \in X$. 
Let $t_1, t_2, t_3, t_4 \in \R_{> 0}$ be such that:
:$x \in t_1 C$
:$-x \in t_2 C$
:$i x \in t_3 C$
:$-i x \in t_4 C$
Let $t = \max \set {t_1, t_2, t_3, t_4}$. 
Then:
:$\ds \frac {t_i} t \in \closedint 0 1$ for each $i \in \set {1, 2, 3, 4}$.
As in the $\GF = \C$ case, applying Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1 we have $t_i C \subseteq t C$ for each $t \in \set {1, 2, 3, 4}$.
So we have:
:$\ds \set {\frac x t, -\frac x t, \frac {i x} t, -\frac {i x} t} \subseteq C$
Let $\alpha \in \C$ be such that $\cmod \alpha > 2 t$.
We can write: 
:$\ds \frac \alpha t = c_1 \alpha_1 + i c_2 \alpha_2$
where $\alpha_1, \alpha_2 \in \hointr 0 \infty$ and $c_1, c_2 \in \set {-1, 1}$.
We can now write:
We have:
:$\ds \frac {c_1} t x \in \set {\frac x t, -\frac x t}$
and:
:$\ds \frac {i c_2} t x \in \set {\frac {i x} t, -\frac {i x} t}$
We have:
:$\ds \frac {\alpha_1} {\alpha_1 + \alpha_2} + \frac {\alpha_2} {\alpha_1 + \alpha_2} = 1$
Since $C$ is convex, we have:
:$\ds \frac {\alpha_1} {\alpha_1 + \alpha_2} \frac {c_1} t x + \frac {\alpha_2} {\alpha_1 + \alpha_2} \frac {i c_2} t x \in C$ 
so that:
:$\ds \frac 1 \alpha x \in \paren {\alpha_1 + \alpha_2} C$
Since:
:$\ds \cmod {\frac r \alpha} < \frac 1 2$
we have that $\alpha_1 + \alpha_2 < 1$.
Hence:
:$\ds \paren {\alpha_1 + \alpha_2} C \subseteq C$
from Dilation of Convex Set containing Zero Vector by Real Number between 0 and 1.
So we have:
:$\ds \frac 1 \alpha x \in C$
and hence:
:$x \in \alpha C$ for all $\cmod \alpha > 2 r$.
So $C$ is absorbing.
== Sources ==
* 
* : $1.33$: Definitions
Category:Absorbing Sets
Category:Convex Sets (Vector Spaces)