{p(x)<1} is a Convex Set

This theorem can be considered a converse of a previous theorem.

Theorem: Let $p$ denote a positive homogenous, subadditive function defined on a linear space $X$ over the reals.

(i) The set of points $x$ satisfying $p(x)<1$ is a convex subset of $X$ , and 0 is an interior point of it.

(ii) The set of points $x$ satisfying $p(x)\leq 1$ is a convex subset of $X$ .

Proof: (i) Let $K=\{x\in X\mid p(x)<1\}$ . Let $x_1,x_2\in K$ . For $0\leq\alpha\leq 1$ ,

$\begin{aligned}p(\alpha x_1+(1-\alpha)x_2)&\leq \alpha p(x_1)+(1-\alpha)p(x_2)\\ &<\alpha+(1-\alpha)\\ &=1 \end{aligned}$

Therefore $K$ is convex. We also have $p(0)=0\in K$ .

The proof of (ii) is similar.

Author: mathtuition88