Convex Subsets of Vector Spaces

# Convex Subsets of Vector Spaces

 Definition: Let $X$ be a vector space. A Convex Subset of $X$ is a subset $K \subseteq X$ such that for every pair of points $x, y \in K$ and for every $0 \leq t \leq 1$ we have that $tx + (1 - t)y \in K$. Elements of the form $tx + (1 - t)y$ are called Convex Combinations of $x$ and $y$.

Geometrically, a subset $K$ of $X$ is convex if for every pair of points $x, y \in K$, the set $K$ contains the "line segment" connecting $x$ and $y$. For example, let $Y \subseteq X$ be a subspace of $X$. Then $Y$ is a convex subset of $X$. This is because $Y$ is closed under addition and scalar multiplication and so if $x, y \in Y$ then for $0 \leq t \leq 1$ we immediately get that:

(1)
\begin{align} \quad tx + (1 - t)y \in K \end{align}

For another example, let $X$ be a normed linear space. Then for every $x_0 \in X$ and for every $r > 0$, the open and closed balls centered at $x_0$ with radius $r > 0$ are convex subsets of $X$. For a more concrete example, consider the closed unit ball centered at the origin:

(2)
\begin{align} \quad B(0, 1) = \{ z \in X : \| z \| \leq 1 \} \end{align}

Let $x, y \in B(0, 1)$ and let $0 \leq t \leq 1$. Then $\| x \| \leq 1$ and $\| y \| \leq 1$, and:

(3)
\begin{align} \quad \| tx + (1 - t)y \| \leq \| tx \| + \| (1 - t) y \| = t \| x \| + (1 - t) \| y \| \leq t \max \{ \| x \|, \| y \| \} + (1 - t) \max \{ \| x \|, \| y \| \} \leq \max \{ \| x \|, \| y \| \} \leq 1 \end{align}

Therefore $tx + (1 - t)y \in B(0, 1)$, so $B(0, 1)$ is a convex set.