Show that halfspace is not affine.

Question

Let us define half-space as $$ C = \{x\mid a^Tx\leq b\} $$ Intuitively (or geometrically), I understand why halfspace is not affine. But while I prove that half-space is convex, it seems to hold for affine case.

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Let us choose any $x_1,x_2\in C$ and $x=\theta x_1 + (1-\theta)x_2$, then $$ a^Tx = a^T(\theta x_1+(1-\theta)x_2) \leq \theta b + (1-\theta)b = b $$

As far as I know, if this inequality holds for $\theta\in\mathbb R$, $C$ is affine and if for $\theta\in[0,1]$, $C$ is convex. But in the proof, both cases seem hold. Where am I wrong?

Answer 1

The inequality $a^Tx=a^T(\theta x_1+(1-\theta)x_2)\leq \theta b+(1-\theta)b=b$ holds if and only if when $0\leq \theta \leq 1$. If suppose, $\theta <0 $ the value of $(1-\theta)$ becomes negative. We cannot really draw any conclusion about the inequality between L.H.S and R.H.S, i.e., $a^T(\theta x_1+(1-\theta)x_2)$ and $\theta b+(1-\theta)b$.

Hope this helps.

Answer 2

Suppose $a = (1, 1)$ and $b = 1$.

Take $x_1 = (1/2, 1/2)$. Clearly, $a^T x_1 = 1 \leq b$ is in the halfspace.

Take $x_2 = (0, 0)$. Clearly, $a^T x = 0 \leq b$ is also in the halfspace.

For the halfspace to be affine, all linear combinations $x = \theta x_1 + (1-\theta) x_2$ must also satisfy $a^T x \leq b$. However, for $\theta = 2$, we have $a^T x = a^T (2x_1 - 0) = 2 a^T x_1 = 2 \nleq 1$. Therefore, $a^T x \leq b$ is not affine (but it is convex).