A Set is Convex iff it contains all of its Convex Combinations

Last updated Nov 1, 2022

Let $V$ be a Vector Space over $\mathbb{R}$ and $S \subset V$. $S$ is a Convex Set If and Only If $\forall n \in \mathbb{N}$, $x_{1}, \dots, x_{n} \in S$, all Convex Combinations (of $x_{1}, \cdots, x_{n}$) $\lambda_{1} x_{1} + \cdots + \lambda_{n} x_{n} \in S$.

$(\Leftarrow)$: Let $u, v \in S$. Note that for $\lambda \in [0,1]$, $\lambda u + (1 - \lambda) v = w$ is a Convex Combination of $u,v$ and, thus, $w \in S$. Thus, by definition, $S$ is a Convex Set. $\checkmark$

$(\Rightarrow)$: We proceed by Induction. We skip the trivial case where $n=1$.

Base Case ($n = 2$): Let $x_{1}, x_{2} \in S$ and let $\lambda_{1}, \lambda_{2} \in [0,1]$ so that $\lambda_{1} + \lambda_{2} = 1$. Then $\lambda_{2} = (1- \lambda_{1})$. Since $S$ is a Convex Set, we have that $$S \ni \lambda_{1}x_{1} + (1 - \lambda_{1})x_{2} = \lambda_{1}x_{1} + \lambda_{2}x_{2}$$ establishing the case when $n = 2$.

Inductive Step: Let $n \in \mathbb{N}$ so that $n > 2$, and assume the result holds true for $k \in [n-1]$. Let $\lambda_{1}, \dots, \lambda_{n} \in [0,1]$ so that $\sum\limits_{i=1}^{n} \lambda_{i} = 1$. If $\lambda_{1} = 1$, then we can drop $\lambda_{i} = 0$ for $i \in {2, \dots, n}$ and our Convex Combination is just $x_{1} \in S$ $\checkmark$. Otherwise, assuming $\lambda_{1} \neq 1$. Then $\sum\limits_{i=2}^{n} \lambda_{i} = 1 - \lambda_{1}$ and $\sum\limits_{i=2}^{n} \frac{\lambda_{i}}{1- \lambda_{1}} = 1$. By Induction we know that $$\frac{\lambda_{2}}{1-\lambda_{1}}x_{2} + \cdots + \frac{\lambda_{n}}{1-\lambda_{1}}x_{n} \in S.$$ Because $S$ is a Convex Set $$\begin{align*} S &\ni \lambda_{1} x_{1} + (1 - \lambda_{1}) \left(\frac{\lambda_{2}}{1-\lambda_{1}}x_{2} + \cdots + \frac{\lambda_{n}}{1-\lambda_{1}}x_{n}\right)\\ &=\lambda_{1}x_{1} + \cdots + \lambda_{n} x_{n} \end{align*}$$ so $S$ is closed under Convex Combinations. $\checkmark \blacksquare$