Characterization of Convex Absorbing Set in Vector Space

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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 if and only if:

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, $x \in t C$.

$\Box$

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:

\(\ds \frac 1 \alpha x\) \(=\) \(\ds \)
\(\ds \) \(=\) \(\ds \frac t {c \cmod \alpha} \frac 1 t x\)
\(\ds \) \(=\) \(\ds \frac t {\cmod \alpha} \frac c t x\) since $c \in \set {-1, 1}$ we have $1/c = c$

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:

\(\ds \frac 1 \alpha x\) \(=\) \(\ds \frac t \alpha \frac 1 t x\)
\(\ds \) \(=\) \(\ds \paren {c_1 \alpha_1 + i c_2 \alpha_2} \frac 1 t x\)
\(\ds \) \(=\) \(\ds \alpha_1 \frac {c_1} t x + \alpha_2 \frac {i c_2} t x\)
\(\ds \) \(=\) \(\ds \paren {\alpha_1 + \alpha_2} \paren {\frac {\alpha_1} {\alpha_1 + \alpha_2} \frac {c_1} t x + \frac {\alpha_2} {\alpha_1 + \alpha_2} \frac {i c_2} t x}\)

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.

$\blacksquare$


Sources

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