orthogonal to u = x − y passing through y strictly separates x and C , namely: Proof. Since C is closed and convex, the projection y = P$_{C}$ ( x ) of x onto C is unique. Define, for any given z ∈ C, a functional ϕ$_{z}$ : [0 , 1] −→ R by: As y is closest to x, ϕ$_{z}$ ( t ) achieves its minimum on [0, 1] at t = 0, thus ϕ $^{′}$$_{z}$(0) ≥ 0. Since ϕ $^{′}$$_{z}$( t ) = 2 t ‖ y − z ‖ 2 + 2 〈 x − y, y − z 〉 , it follows ϕ $^{′}$$_{z}$(0) = 2 〈 x − y, y − z 〉 = 2 〈 u, y − z 〉 ≥ 0 , i.e., 〈 u, z 〉 ≤ 〈 u, y 〉 . On the other hand, 0 < ‖ x − y ‖ 2 = 〈 u, x − y 〉 = 〈 u, x 〉 − 〈 u, y 〉 . Thus, 〈 u, z 〉 ≤ 〈 u, y 〉 < 〈 u, x 〉 . Proposition 2 Let K and C be disjoint convex subsets of R with K compact and C closed. Then, C and K are strictly separated by a hyperplane H, i.e., there exists u ∈ R $^{n}$, u = 0 , with sup x ∈ C 〈 u, x 〉 < min x ′ ∈ $_{K}$〈 u, x $^{′}$〉 . Proof. Since C is closed and K is compact, the set C − K := { y ∈ R n : y = x − x $^{′}$, x ∈ C, x ′ ∈ K } is also closed. It is, moreover, convex as the difference of convex sets. Since C ∩ K =  , then 0 / ∈ C − K. Lemma 1 applies, yielding for u = 0 − P$_{C}$$_{−}$$_{K}$ (0) , the inequalities: 〈 u, z 〉 ≤ 〈 u, − u 〉 < 〈 u, 0 〉 = 0 , ∀ z ∈ C − K. Thus, as z = x − x $^{′}$, x ∈ C, x ′ ∈ K, 〈 u, x 〉 ≤ 〈 u, x $^{′}$〉 − ‖ u ‖ 2 < 〈 u, x $^{′}$〉 , ∀ x ∈ C, ∀ x ′ ∈ K. ϕ$_{z}$ ( t ) := ‖ x − ( tz + (1 − t ) y ‖ $^{2}$. 〈 u, z 〉 ≤ 〈 u, y 〉 < 〈 u, x 〉 , ∀ z ∈ C 4