Supporting hyperplane through vector projection onto convex set

Question

I am trying to proof the following property.

Let

• $F_i: \mathbb{R}^m \rightarrow \mathbb{R}$ be a $c_1$ convex function $\forall i \in I$;
• $C = \{ x \in \mathbb{R}^m : F_i(x) \leqslant 0 \quad \forall i \in I \}$ be a compact convex set;
• $P(S) = \{ x \in \mathbb{R}^m : F_i(x^{'}) + \nabla F_i(x^{'})^T (x - x^{'}) \leqslant 0 \quad \forall i \in I, x^{'} \in S\}$ be the intersection of all tangents of $F$s at the points in $S$; And
• $S^{'} = \{ x \in C : F_i(x) = 0 \quad \forall i \in I \}$ all the points at the boundary of $C$.

Then, we have that if $x \in P(S^{'})$, then $x \in C$.

To proof this, I thought in conducting a contrapositive proof, in which we

1. Consider a point $x_{*} \notin C$;
2. Obtain the boundary point through the projection of $x_{*}$ onto $C$, i.e. $x_C =$ Proj$_C(x_{*}) \in S^{'}$;
3. Calculate the supporting hyperplanes intercepting $x_C$, i.e. $F_i(x_C) + \nabla F_i(x_C)^T (x - x_C) \leqslant 0 \quad \forall i \in I$; And then
4. Show that the at least one of the recently obtained supporting hyperplane does not contain the point $x_{*}$, i.e. $\exists_{i \in I} (F_i(x_C) + \nabla F_i(x_C)^T (x_{*} - x_C) > 0)$.

However, I am struggling in this last step (Step 4). Although visually I can see this, I am not able to show the violation of $x_{*}$ in the hyperplane $F_k(x_C) + \nabla F_k(x_C)^T (x_{*} - x_C) <= 0$ such that $k \in I$.

Case there is any missing point in, either, my problem statement or proof, please, let me know.

Thank you by your attention, best regards.

Answer 1

Let some $\lambda > 0$, $-\lambda(x^*-x_C)+x_C \in C$. This is because points outside the convex region with non-empty interior when scaled and reflected about tangent hyperplane at non-corner points must lie inside $C$.

By mean value theorem, $$F_i(\lambda(x^*-x_C)+x_C) - F_i(-\lambda(x^*-x_C)+x_C) = \langle \nabla F_i(x') , 2\lambda(x^*-x_C)\rangle$$ $$\frac{F_i(\lambda(x^*-x_C)+x_C)-F_i(x_C)}{\lambda} + \frac{F_i(-\lambda(x^*-x_C)+x_C)-F_i(x_C)}{-\lambda} = \langle \nabla F_i(x') , 2(x^*-x_C)\rangle$$

$$\langle \nabla F_i(x_C) , 2(x^*-x_C)\rangle = \langle \nabla F_i(x') , 2(x^*-x_C)\rangle$$ Hence, $$ 0.5*F_i(\lambda(x^*-x_C)+x_C) +0.5* F_i(-\lambda(x^*-x_C)+x_C) = F_i(-\lambda(x^*-x_C)+x_C) + \langle \nabla F_i(x_C) , \lambda(x^*-x_C)\rangle $$ $$ 0 \leq F_i(x_C) \leq 0.5*F_i(\lambda(x^*-x_C)+x_C) +0.5* F_i(-\lambda(x^*-x_C)+x_C) = F_i(-\lambda(x^*-x_C)+x_C) + \langle \nabla F_i(x_C) , \lambda(x^*-x_C)\rangle $$

$$ 0 < - F_i(-\lambda(x^*-x_C)+x_C) \leq \langle \nabla F_i(x_C) , \lambda(x^*-x_C)\rangle $$

A contradiction to $x^* \in P(S')$ $\iff$ $F_i(-\lambda(x^*-x_C)+x_C) < 0$ or $-\lambda(x^*-x_C)+x_C \in C$

Try out cases where $-\lambda(x^*-x_C)+x_C \notin C$.