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 $t > 0$ such that $x \in t C$.

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.