Theorem 6.34.Let C be a nonempty convex set in a vector spaceE. If M is an affine set such that rai(C)∩M =∅, then there exists a hyperplane H extendingM such that rai(C)∩H =∅.
Proof. The proof of the theorem is the same as the proof of Theorem 6.17 except that we replace ri(C) in that proof by rai(C) and invoke Theorem 6.33
instead of Theorem 6.15. ut
6.7 Separation of Several Convex Sets 163 Proof. Both sets of conditions cannot hold simultaneously, because ifxsatis- fies (6.10), then we have the contradiction
0<
k
X
i=1
λi(αi−`i(x)) =
k
X
i=1
λiαi≤0.
Suppose that (6.10) is inconsistent, and define the sets
L:={u∈Rk :∃x∈E, ui=`i(x)−αi, i= 1, . . . , k}, K:={u∈Rk :u <0};
Lis affine, andK is an open convex set. Note thatK∩L=∅; it follows from Theorem 6.34 (in fact Theorem 6.17 suffices) that there exists a hyperplane H =H(λ,γ)⊂Rk such that L⊆H and K⊆H−. On the one hand, the fact K⊆H− gives
hλ, vi< γ for all v∈Rk, v <0,
which implies thatλ≥0 andγ≥0; on the other hand, the factL⊆H means that
Xk
i=1
λi`i
(x)−
k
X
i=1
λiαi=
k
X
i=1
λi(`i(x)−αi) =γ for all x∈E,
which implies thatPk
i=1λi`i= 0 andPk
i=1λiαi=−γ≤0. ut An independent proof of the lemma can be found in Appendix A.
Theorem 6.37 gives both analytic and geometric characterizations of the disjointness condition ∩k1raiCi = ∅ of k proper nonempty sets {Ci}k1 under the mild conditions raiCi 6=∅ for alli= 1, . . . , k, which are always satisfied in the finite-dimensional case.
Theorem 6.37.Let{Ci}k1,k >1, be proper convex sets in a vector spaceE such thatraiCi6=∅ for alli= 1, . . . , k.
The following conditions are equivalent:
(a)∩k1raiCi=∅.
(b) There exist linear functionals {`i}k1 on E, not all identically zero, and scalars{αi}k1 such that
`i(xi)≤αi for all xi ∈Ci, i= 1, . . . , k,
k
X
1
`i = 0,
k
X
1
αi≤0, and there exists a point xbelonging to a setCi with corresponding`i6= 0 such that `i(xi)< αi.
(c) The sets {Ci}k1 are properly separable.
Proof. We first prove the equivalence of parts (a) and (b). We start by proving (a) implies (b), the longest part of the whole theorem. Suppose that (a) is true.
We prove (b) by induction onk. If k = 2, Theorem 6.33 implies that there exists a hyperplaneH=H(`,α)such that`(x1)≤αfor allx1∈C1,`(x2)≥α for all x2 ∈ C2, and C1∪C2 6⊆ H; (b) is clearly satisfied with the choices
`1=`,`2=−`,α1=α, andα2=−α.
Assume thatk >2, and that we have proved (a) implies (b) for all integers smaller thank. If the relative algebraic interiors ofk−1 of the sets have empty intersection, say∩k2raiCi =∅, then by the induction hypothesis, there exist {(`i, αi)}k2 satisfying (b). If we define`1= 0 andα1= 0, then (b) is satisfied with{(`i, αi)}k1.
Thus, we may assume that ∩k2raiCi6=∅. Define the sets K1:={(x1, x1, . . . , x1) :x1∈C1}, K2:=C2× · · · ×Ck.
We have, by elementary arguments,
raiK1={(x1, x1, . . . , x1) :x1∈raiC1} 6=∅, raiK2= raiC2× · · · ×raiCk6=∅.
Since ∩k1raiCi = ∅, we see that raiK1∩raiK2 = ∅. Theorem 6.33 implies that there exists (`, α) = ((`2, . . . , `k), α),`6= 0, such that
`2(x2)+· · ·+`k(xk)≤α≤Xk
2
`i
(x1) for all xi ∈Ci, i= 1, . . . , k, (6.11) and strict inequality holds in one of the inequalities above for some choice of x1, . . . , xk.
Defineαi:= supxi∈Ci`i(xi) fori= 2, . . . , k, and`1:=−Pk
2`i,α1:=−α.
It follows from (6.11) that α2+· · ·+αk ≤ α. Since we have `i 6= 0 for some 2≤i≤k, (b) holds except possibly when `i(xi) =αi for all xi ∈Ci, i= 2, . . . , k, andPk
2αi=α. However, in this last case we haveα < `1(¯x1) for some ¯x1∈C1 andα≤`1(x1) for allx1∈C1. We must have`1 6= 0, because otherwise lettingxi=x∈ ∩k2rai(Ci) gives
0 = (
k
X
1
`i)(x) =
k
X
2
`i(x) =α < `1(¯x1) = 0, a contradiction. Thus, (b) holds in this case as well.
Conversely, let us prove that (b) implies (a). Suppose that (b) is true but (a) is false, that is, there exists a pointx∈ ∩k1raiCi6=∅. If the setCiis such that `i 6= 0 andCi does not lie on the hyperplane Hi := H(`i,αi), we have rai(Ci)⊆Hi−; otherwise, there existw∈rai(Ci)∩Hi and u∈Ci such that w∈(xi, u), and this gives a contradiction becauseu∈C∩Hi+=∅. Therefore, we have
6.7 Separation of Several Convex Sets 165
0>
k
X
i=1
(`i(x)−αi) =−
k
X
i=1
αi≥0, a contradiction, which proves that (a) must be true.
It remains to demonstrate that parts (b) and (c) are equivalent. It follows immediately from Lemma 6.36 that (c) implies (b). Conversely, suppose that (b) is true. DefineI:={i: 1≤i≤k, `i6= 0}andHi :=H(`i,αi)fori∈I. We have Ci ⊆H¯i−. The open half-spaces {Hi−}i∈I must already have an empty intersection, becausex∈ ∩i∈IHi−6=∅gives the contradiction
0>
k
X
i=1
(`i(x)−αi) =−
k
X
i=1
αi≥0.
It remains to show that every Ci, i /∈ I, is contained in a half-space. Pick a point x /∈ Ci and invoke Theorem 6.34 to obtain a hyperplane Hi such that rai(Ci)⊆Hi−. The hyperplanes{Hi}k1 properly separate the sets{Ci}k1,
proving (c). ut
The next theorem is a vast generalization of the Dubovitskii–Milyutin theorem, whose finite-dimensional versions were given in Lemma 6.22 and Theorem 6.23. The cone version of the theorem, with a different proof, is given in [82].
Theorem 6.38. (Dubovitskii–Milyutin) Let {Ci}k1, k >1, be nonempty convex sets in a vector spaceE, such that{Ci}k−11 are algebraically open, that is,ai(Ci) =Ci,i= 1, . . . , k−1.
The following conditions are equivalent:
(a)∩k1Ci =∅.
(b) There exist linear functionals {`i}k1 on E, not all identically zero, and scalars{αi}k1 such that
`i(xi)≤αi for all xi ∈Ci, i= 1, . . . , k,
k
X
1
`i = 0,
k
X
1
αi≤0.
Proof. We first prove that (a) implies (b). Define the sets K1:={(xk, xk, . . . , xk) :xk∈Ck}, K2:=C1× · · · ×Ck−1.
We have, by elementary arguments,
aiK2= aiC1× · · · ×aiCk−1=C1× · · · ×Ck−1=K2,
so that K2 is algebraically open. Clearly,K1∩K2 =∅, so by Theorem 6.32 there exists (`, α) = ((`1, . . . , `k−1), α),`6= 0, such that
`1(x1) +· · ·+`k−1(xk−1)< α≤k−1X
1
`i
(xk) for all xi∈Ci, i= 1, . . . , k.
Defineαi:= supxi∈Ci`i(xi) fori= 1, . . . , k−1, so thatα1+· · ·+αk−1≤α, and`k :=−Pk−1
1 `i,αk :=−α; this proves (b).
Conversely, let us assume (b) and prove (a). If (a) is false, then there exists a pointx∈ ∩k1Ci6=∅that satisfies (b). On the one hand, we have
0 =
k
X
1
`i(x)≤
k
X
1
αi≤0,
so that`i(x) =αifor alli= 1, . . . , k, andPk
1αi= 0. On the other hand, ifz∈ Eis an arbitrary point, then there exists >0 such that [x−z, x+z]⊆Ci
fori= 1, . . . , k−1, becauseCiis algebraically open. This gives`i(x∓z)≤αi, so that
αi∓`i(z) =`i(x)∓`i(z) =`i(x∓z)≤αi,
which means that`i(z) = 0, that is, the linear functions{`i}k−11 are identically zero, and hence all the{`i}k1 are identically zero, which contradicts (b). This
proves that (a) must be true. ut
The topological version of Theorem 6.38 is now easy to establish. The following proof should serve as a model for obtaining a topological separation theorem from an algebraic one.
Theorem 6.39. (Dubovitskii–Milyutin) Let {Ci}k1, k >1, be nonempty convex sets in a topological vector spaceE, such that{Ci}k−11 are open, that is,int(Ci) =Ci,i= 1, . . . , k−1.
The following conditions are equivalent:
(a)∩k1Ci =∅.
(b) There exist continuous linear functionals {`i}k1 onE, not all identically zero, and scalars{αi}k1 such that
`i(xi)≤αi for all xi ∈Ci, i= 1, . . . , k,
k
X
1
`i = 0,
k
X
1
αi≤0.
Proof. By virtue of Theorem 5.20, int(Ci) = ai(Ci) fori= 1, . . . , k−1, so it follows from Theorem 6.38 that we need to prove only that if (a) is true, then the linear functionals{`i}k1in (b) are continuous; in fact, since`k =−Pk−1
1 `i, it suffices to prove the continuity of`:= (`1, . . . , `k−1).
Suppose that (a) holds. As shown in the proof of Theorem 6.38, there is a hyperplaneH :=H(`,α) such that the open convex setC1× · · · ×Ck−1 lies in the algebraically open half-spaceH−;`is continuous by Corollary 6.28. ut