# Absolutely Convex Sets

Definition: Let $X$ be a linear space and let $E \subseteq X$. Then $E$ is said to be Absolutely Convex if whenever $x, y \in E$ and $a, b \in \mathbf{F}$ are such that $|a| + |b| \leq 1$ then $(ax + by) \in E$. |

Proposition 1: Let $X$ be a linear space and let $E \subseteq X$. If $E$ is absolutely convex then $E$ is convex. |

*The converse of proposition 1 is not true in general. See example 1.*

**Proof:**Suppose that $E$ is absolutely convex. Then whenever $x, y \in E$ and $a, b \in \mathbf{F}$ are such that $|a| + |b| \leq 1$ then $ax + by \in E$. In particular, when $a \in [0, 1]$, and $b = 1 - a$ then $|a| + |b| = a + (1 - a) = 1$, and so $ax + (1 - a)y \in E$ for all $a \in [0, 1]$. Therefore $E$ is convex.

Proposition 2: Let $X$ be a seminormed linear space. Then the open unit ball $B(0, 1) = \{ x \in X : p(x) < 1 \} $] and the closed unit ball [[$ \bar{B}(0, 1) = \{ x \in X : p(x) \leq 1 \}$ are both absolutely convex sets. |

**Proof:**Let $x, y \in B(0, 1)$ so that $p(x) < 1$ and $p(y) < 1$. Let $a, b \in \mathbf{F}$ be such that $|a| + |b| \leq 1$. Then $|b| \leq 1 - |a|$. Using this and the properties of seminorms and we get:

- This shows that $ax + by \in B(0, 1)$. A similar argument shows that $\bar{B}(0, 1)$ is absolutely convex. $\blacksquare$

Let's look at some examples.

## Example 1

Consider the linear space $\mathbb{C}$ with the usual norm let $A \subset \mathbb{C}$ be defined by:

(2)Observe that $A$ is simply the closed interval $[a, b]$ embedded into $\mathbb{C}$. The set $A$ is clearly convex, however, it is not absolutely convex. To see this, observe that if $a = i$ and $b =0$ then $a$ and $b$ are such that $|a| + |b| \leq 1$ (the sum actually equals $1$). Take the points $x = 1$ and $y = 0$, both of which are in $A$. Then:

(3)Visually, multiplication by $i$ rotates the line segment $A$ counterclockwise $90$ degrees about the origin, and clearly this rotation is not in $A$. This shows that there exists some convex sets that are not absolutely convex, and hence, the converse of proposition 1 is not true in general.