# Supporting hyperplane for Bayes boundary of a convex set

I was reading the book 'Optimal Statistical Decisions' by DeGroot. I came across the following claim, without proof:

G is a convex set in $$R^k$$ that is the convex hull of a finite number of points. A point y ∈ G belongs to the Bayes boundary of G if there is no point x ∈ G such that for every i = 1, . . ., k, $$x_i$$ < $$y_i$$. Then if y is on the Bayes boundary of G, then there exists a supporting hyperplane ⟨a, x⟩ = c to G at y such that a ≥ 0. (Assuming a is such that for every z ∈ G, ⟨a, z⟩ ≥ c.)

I tried proving it, and while it seems obvious to realise, I am not able to prove it formally. Could someone help me with it?

Suppose $$y$$ is on the Bayes boundary of $$G$$. The set $$L = \left\lbrace x \colon x_{i} < y_{i} \forall i \right\rbrace$$ is convex and disjoint from $$G$$. The Separating Hyperplane Theorem shows that there is a vector $$p$$ such that $$p^{T}x \leq p^{T} z$$ whenever $$x \in L$$ and $$z \in G$$. The point $$y$$ is in the closure of $$L$$ and $$G$$, and so we infer that $$z \mapsto p^{T}z$$ is minimized over $$G$$; that is, the hyperplane $$\lbrace x \in \mathbb{R}^{k} \colon p^{T}x = p^{T}y\rbrace$$ supports $$G$$ at $$y$$. The only missing part is to show that $$p \geq 0$$. If $$p \geq 0$$ were false, then we could define $$y^{\prime}$$ by choosing $$y_{i}^{\prime} = \begin{cases} y_{i} - 1 \quad \text{if p_{i} < 0}, \\ y_{i} - \varepsilon \quad \text{if p_{i} \geq 0}, \end{cases}$$ for all $$i$$. If $$\varepsilon$$ is sufficiently small, then $$p^{T}y^{\prime} > p^{T}y$$. However, by construction, $$y_{i}^{\prime} < y_{i}$$ for every entry $$i$$, contradicting the fact that $$L$$ and $$G$$ are separated by $$p$$.