[proofplan] The proof is a direct verification of convexity from the defining equations and inequalities. Given two points in one of the sets and a parameter $t \in [0,1]$, we evaluate the [linear map](/page/Linear%20Map) $u$ on the convex combination $tx + (1-t)y$. Linearity preserves the affine equality for the hyperplane and preserves the relevant inequality for each half-space. [/proofplan] [step:Verify that the affine hyperplane contains every segment between its points] Let $x,y \in H(u,\alpha)$, and let $t \in [0,1]$. Define the point $z \in \mathbb{R}^n$ by \begin{align*} z := tx + (1-t)y. \end{align*} Since $u$ is linear and $u(x)=\alpha$, $u(y)=\alpha$, we have \begin{align*} u(z) &= u(tx + (1-t)y) \\ &= t u(x) + (1-t)u(y) \\ &= t\alpha + (1-t)\alpha \\ &= \alpha. \end{align*} Therefore $z \in H(u,\alpha)$. Since $x$, $y$, and $t$ were arbitrary, $H(u,\alpha)$ is convex. [/step] [step:Verify that the lower closed half-space contains every segment between its points] Let $x,y \in H^-(u,\alpha)$, and let $t \in [0,1]$. Define $z \in \mathbb{R}^n$ by \begin{align*} z := tx + (1-t)y. \end{align*} Because $u$ is linear, $u(x) \leq \alpha$, $u(y) \leq \alpha$, and both $t$ and $1-t$ are nonnegative, we obtain \begin{align*} u(z) &= u(tx + (1-t)y) \\ &= t u(x) + (1-t)u(y) \\ &\leq t\alpha + (1-t)\alpha \\ &= \alpha. \end{align*} Thus $z \in H^-(u,\alpha)$. Hence $H^-(u,\alpha)$ is convex. [/step] [step:Verify that the upper closed half-space contains every segment between its points] Let $x,y \in H^+(u,\alpha)$, and let $t \in [0,1]$. Define $z \in \mathbb{R}^n$ by \begin{align*} z := tx + (1-t)y. \end{align*} Since $u$ is linear, $u(x) \geq \alpha$, $u(y) \geq \alpha$, and both $t$ and $1-t$ are nonnegative, we have \begin{align*} u(z) &= u(tx + (1-t)y) \\ &= t u(x) + (1-t)u(y) \\ &\geq t\alpha + (1-t)\alpha \\ &= \alpha. \end{align*} Therefore $z \in H^+(u,\alpha)$. Hence $H^+(u,\alpha)$ is convex, completing the proof. [/step]