recession cone

Let $C$ be a convex set in ${ℝ}^n$. If $C$ is bounded, then for any $x\in C$, any ray emanating from $x$ will eventually “exit” $C$ (that is, there is a point $z$ on the ray such that $z\notin C$). If $C$ is unbounded, however, then there exists a point $x\in C$, and a ray $\rho$ emanating from $x$ such that $\rho \subseteq C$. A direction $d$ in $C$ is a point in ${ℝ}^n$ such that for any $x\in C$, the ray $\{x+rd\mid r\ge 0\}$ is also in $C$ (a subset of $C$).

The recession cone of $C$ is the set of all directions in $C$, and is denoted by denoted by $0^{+}C$. In other words,

$$0^{+}C=\{d\mid x+rd\in C,\forall x\in C,\forall r\ge 0\}.$$

If a convex set $C$ is bounded, then the recession cone of $C$ is pretty useless; it is $\{0\}$. The converse is not true, as illustrated by the convex set

Clearly, $C$ is not bounded but $0^{+}C=\{0\}$. However, if the additional condition that $C$ is closed is imposed, then we recover the converse.

Here are some other examples of recession cones of unbounded convex sets:

• If $C=\{(x,y)\mid |x|\le y\}$, then $0^{+}C=C$.

• If , then $0^{+}C=\overline{C}$, the closure of $C$.

• If $C=\{(x,y)\mid {|x|}^n\le y,n>1\}$, then $0^{+}C=\{(0,y)\mid y\ge 0\}$.

Remark. The recession cone of a convex set is convex, and, if the convex set is closed, its recession cone is closed as well.