Properties of Convex Sets of Vectors

# Properties of Convex Sets of Vectors

Recall that if $E$ is a vector space over $\mathbf{F}$ then a subset $A$ of $E$ is said to be convex if for all $x, y \in A$ and for all $\lambda, \mu \geq 0$ with $\lambda + \mu = 1$ we have that:

(1)
\begin{align} \lambda x + \mu y \in A \end{align}

We will now look at some simple properties of convex sets.

 Proposition 1: Let $E$ be a vector space. If $A \subseteq E$ is convex then for all $\lambda, \mu \in \mathbb{R}$, $\lambda A + \mu A = (\lambda + \mu) A$.
• Proof: Note that the conclusion follows immediately if $\lambda = 0$ or $\mu = 0$. Thus assume that $\lambda, \mu > 0$. Then:
(2)
\begin{align} \quad \frac{\lambda}{\lambda + \mu} + \frac{\mu}{\lambda + \mu} = 1 \end{align}
• Since $A$ is convex we have that for all $x, y \in A$:
(3)
\begin{align} \quad \frac{\lambda}{\lambda + \mu} x + \frac{\mu}{\lambda + \mu} y \in A \end{align}
• Thus $\displaystyle{\frac{\lambda}{\lambda + \mu} A + \frac{\mu}{\lambda + \mu} A \subseteq A}$, or equivalently:
(4)
\begin{align} \lambda A + \mu A \subseteq (\lambda + \mu) A \end{align}
• On the other hand, observe that if $x \in (\lambda + \mu)A$ then $x = (\lambda + \mu) a$ for some $a \in A$, and so $x = \lambda a + \mu a$. So $x \in \lambda A + \mu A$. Thus:
(5)
\begin{align} \lambda A + \mu A \supseteq (\lambda + \mu) A \end{align}
• Hence $\lambda A + \mu A = (\lambda + \mu) A$. $\blacksquare$