A problem regarding a convex closed set

Question

When studying about convex sets, I encountered this problem:

Given a convex closed set $C \subset \mathbb{R}^n$. Prove that there exists a family of hyperplanes $\{ H_ i \}$ ($i \in I$) with

$$H_i = \{ x \in \mathbb{R}^n, \langle a_i,x \rangle = \alpha_i \} $$ so that $$C = \{x \in \mathbb{R}^n: \langle a_i,x \rangle \le \alpha_i , i \in I \}. $$

Right now, I have trouble imagining how it is possible for such thing to happen. For some particular cases, like when $C$ is a hypersphere, I can tell that the desired family of hyperplanes is just simply all the tangent hyperplanes. However, when it comes to the general case (a convex closed set), I really have no idea how to describe the desired family. Still, I think that tangent hyperplanes may be a good place to start with. But I don't know how to proceed.

Please help me. Any help is greatly appreciated.

Thank you.

Answer 1

Maybe this form makes things more clear \begin{align*} C = \bigcap_{i \in I}\{ x \in \mathbb R^n : \langle a_i, x \rangle \le \alpha_i\}. \end{align*} We want to prove above equality. Note for each $i \in I$, $C$ is contained in the closed half space $\{x \in \mathbb R^n : \langle a_i, x \rangle \le \alpha_i\}$. So if we denote the right hand-side as $D$, it is clear $C \subseteq D$. It suffices to prove $D \subset C$. This comes from the separation of convex sets. If $y \notin C$, we can find a hyperplane such that $\langle y, a \rangle > \langle x, a \rangle $ for every $x \in C$. Let $ \alpha = \sup_{x \in C} \langle x, a \rangle$. So this shows the hyperplane $(a, \alpha)$ is in the index family and it follows $x \notin D$ since $D$ is the intersection. This implies $C^c \subseteq D^c$ and consequently $D \subseteq C$.

Answer 2

Take any unit vector $v$. Roughly speaking, if $C$ is not unbounded in the direction defined by $v$, then $v$ will define a tangent hyperplane.

Consider $C_v = \{\langle x, v \rangle : x \in C\}$. This is the image of a closed set under a linear function it is closed as well, and thus either has a maximum or is unbounded above.

If $C_v$ is unbounded above then there will be no tangent hyperplane of $C$ with normal $v$ (there might be one with $-v$ though). If $C_v$ is bounded above, it has a maximum, $\alpha_v$.

Let $I = \{v : C_v \;\text{is bounded above}\}$. Define $H_v = \{x \in \mathbb{R}^n, \langle v, x \rangle = \alpha_v\}$ for all $v \in I$.

To see that $C$ lies on one side of $H_v$, suppose that there is a point $x \in C$ such that $\langle x, v \rangle > \alpha_v$. This violates the definition of $\alpha_v$ as the maximum of $C_v$, so $C \subseteq H_v$ for all $v$ and thus $C \subseteq \bigcap\limits_{v \in I} H_v$.

Consider any point $y \notin C$. Then, by the separating hyperplane theorem, there is a $u \in \mathbb{R}^n$, $t \in \mathbb{R}$ such that $C \subseteq \{x : \langle u, x \rangle \le t\}$ and $y \in \{x: \langle u, x \rangle > t\}$. This shows that $C_u$ is bounded above, so $u \in I$, and $\alpha_u \le t$. This shows that $y$ is not on the same side of $H_u$ as $C$, so $\bigcap\limits_{v \in I} H_v \subseteq C$.