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I understand from Show that halfspace is not affine. that halfspace is not an affine set.

Would like to ask is half-space a convex cone? How do you prove it mathematically? I am unsure if the following is right:

$a^Tx=\theta_1a^Tx_1+\theta_2a^Tx_2$ where $\theta_1\ge0$ and $\theta_2\ge0$.

Since $a^Tx_1\le b$ and $a^Tx_2\le b$, we could not conclude that $a^Tx\le b$ as the $\theta$s does not have an upper-bound.

However, I read from How is a halfspace an affine convex cone? that "An (affine) half-space is an affine convex cone". I am confused as I thought isn't half-space not an affine set. What is an affine half-space then?

user1018172
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  • Hello, please add the definition of "half space" you are working with. If the half-space is affine, then technically it is not a cone in the usual sense. However, after shifting to the origin, it is a cone (perhaps pending some technicalities about closure) – Zim Jan 24 '22 at 17:26
  • The definition of a (closed) halfspace is the set of the form: ${x | a^T x \le b}$ where $a \neq 0$. A hyperplane divides $R^n$ into two halfspaces. – user1018172 Jan 25 '22 at 01:02

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A halfspace is not always a cone.

A cone $C$ satisfies $(\forall \lambda\in\mathbb{R}_{++}) (\forall x\in C)\quad \lambda x\in C$.

Consider the halfspace $H = \{x\in \mathbb{R}\,|\, x\leq-1\}$ (i.e., $a=1, b=-1$) and set $\lambda=0.5$ and $x=-1$. Since $x\in H$ and $\lambda x=-0.5\not\in H$, $H$ is not a cone.

Zim
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