I am currently studying Convex Optimization by Boyd and Vandenberghe. Chapter 2.2.4 Polyhedra gives the following description of a nonnegative orthant:
Example 2.4 The nonnegative orthant is the set of points with nonnegative components, i.e.,
$$\mathbf{R}^n_+ = \{ x \in \mathbf{R}^n \mid x_i \ge 0, i = 1, \dots, n \} = \{ x \in \mathbf{R}^n \mid x \succeq 0 \}.$$
(Here $\mathbf{R}_+$ denotes the set of nonnegative numbers: $\mathbf{R}_+ = \{ x \in \mathbf{R} \mid x \ge 0 \}$.) The nonnegative orthant is a polyhedron and a cone (and therefore called a polyhedral cone).
Chapter 2.1.5 Cones gives the following description of a cone and convex cone:
A set $C$ is called a cone, or nonnegative homogeneous, if for every $x \in C$ and $\theta \ge 0$ we have $\theta x \in C$. A set $C$ is a convex cone if it is convex and a cone, which means that for any $x_1, x_2 \in C$ and $\theta_1, \theta_2 \ge 0$, we have
$$\theta_1 x_1 + \theta_2 x_2 \in C$$
It seems to me that, not only is the nonnegative orthant a polyhedron and a cone, but it is also a convex cone? So it would be a "convex polyhedral cone"?
Best Answer
There exist convex cones that are not polyhedral (since they are not $\it{finite} $ intersection of hyperplanes). For instance, the cone $\{(x,y,z)\ \mid \ x^2 + y^2 \le z^2 \}$. Another very important example is the (convex) cone of positive semidefinite matrices.