Absolutely Convex Sets of Vectors

# Absolutely Convex Sets of Vectors

 Definition: Let $E$ be a vector space. A subset $A \subseteq E$ is Absolutely Convex if it is both convex and balanced.
 Proposition 1: Let $E$ be a vector space and let $A, B \subseteq E$. (1) If $A$ and $B$ are absolutely convex then $A + B$ is absolutely convex. (2) If $A$ is absolutely convex then $\lambda A$ is absolutely convex for all $\lambda \in \mathbf{F}$.
• Proof of (1): Let $A$ and $B$ be absolutely convex. Then $A$ and $B$ are both convex and balanced. Let $z, w \in A + B$ and let $\lambda, \mu \in \mathbf{F}$ with $\lambda + \mu = 1$. Write $z = x_1 + y_1$ and $w = x_2 + y_2$ with $x_1, x_2 \in A$ and $y_1, y_2 \in B$. Since $A$ and $B$ are convex we have that:
(1)
\begin{align} \quad \lambda x_1 + \mu x_2 \in A \quad \mathrm{and} \quad \lambda y_1 + \mu y_2 \in B \end{align}
• Summing the two elements above yield:
(2)
\begin{align} \quad \lambda (x_1 + y_2) + \mu (x_2 + y_2) \in A + B \end{align}
• Or equivalently, $\lambda z + \mu w \in A + B$, so $A + B$ is convex.
• Now let $\lambda \in \mathbf{F}$ be such that $|\lambda| \leq 1$. Then since $A$ and $B$ are balanced, $\lambda x_1 \in A$ and $\lambda y_1 \in B$. Summing the two elements above yield:
(3)
\begin{align} \quad \lambda (x_1 + y_1) \in A + B \end{align}
• Or equivalently, $\lambda z \in A + B$. So $A + B$ is balanced. Thus $A + B$ is absolutely convex. $\blacksquare$
• Proof of (2): Let $A$ be absolutely convex. Then $A$ is convex and balanced. Let $\lambda \in \mathbf{F}$. If $\lambda = 0$ then trivially $\lambda A = 0 A$ is absolutely convex. So assume that $\lambda \neq 0$.
• Let $x, y \in \lambda A$. Then $x = \lambda x_1$ and $y = \lambda y_1$ with $x_1, y_1 \in A$. Let $\mu, \delta \in \mathbf{F}$ with $\mu + \delta = 1$. Then, since $A$ is convex, we have that $\mu x_1 + \delta y_1 \in A$. Therefore:
(4)
\begin{align} \quad \lambda [\mu x_1 + \delta y_1] \in \lambda A \end{align}
• Or equivalently, $\mu x + \delta y \in \lambda A$, and thus $\lambda A$ is convex.
• Now let $\mu \in \mathbf{F}$ be such that $|\mu| \leq 1$. Since $A$ is balanced, we have that $\mu x_1 \in A$. Thus:
(5)
\begin{align} \quad \lambda [\mu x_1] \in \lambda A \end{align}
• Or equivalently, $\mu x \in \lambda A$, and thus $\lambda A$ is balanced. So $\lambda A$ is absolutely convex. $\blacksquare$
 Proposition 2: Let $E$ be a vector space and let $A \subseteq E$ be a nonempty absolutely convex set. Then: (1) $o \in A$. (2) If $\lambda, \mu \in \mathbf{F}$ are such that $|\lambda| \leq |\mu|$ then $\lambda A \subseteq \mu A$.
• Proof of (1): Since $A$ is absolutely convex, it is by definition, balanced, and so $o \in A$. $\blacksquare$
• Proof of (2): Let $\lambda, \mu \in \mathbf{F}$ be such that $|\lambda| \leq |\mu|$. There are two main cases to consider.
• Case 1: Suppose that $\mu = 0$. Then $\lambda = 0$. We see that $\lambda A = 0 A = \{ o \}$ and $\mu A = 0 A = \{ o \}$, so trivially, $\lambda A \subseteq \mu A$.
• Case 2: Suppose that $\mu \neq 0$. Since $|\lambda| \leq |\mu|$, we have that $\frac{|\lambda|}{|\mu|} \leq 1$. Since $A$ is absolutely convex, it is balanced, and so:
(6)
\begin{align} \quad \frac{\lambda}{\mu} A \subseteq A \end{align}
• Consequently, $\lambda A \subseteq \mu A$. $\blacksquare$