Convexity in R N Supplemental Notes 1

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1 John Nachbar Washington University November 1, 2014 Convexity in R N Supplemental Notes 1 1 Introduction. These notes provide exact characterizations of support and separation in R N. The statement of the Supporting Hyperplane Theorem, in particular, is stronger than that in the main notes on convexity, at the cost of a more complicated proof. See the main notes on convexity for basic definitions (e.g., the definition of convexity) and other related material (e.g., the Basic Separation Theorem). 1.1 Convex Hulls. Definition 1. Given a set A R N, the convex hull of A, written co(a), is the intersection of all convex sets containing A. Informally, co(a) is the smallest convex subset containing A. An equivalent characterization is the following. Given a finite set of points {x 1,..., x L } in R N, x is a convex combination of these points iff there are weights θ 1,..., θ L in R, θ l 0, L l=1 θ l = 1, such that L x = θ l x l. l=1 The following result shows that co(a) equals the set of all convex combinations of all finite subsets of A. Theorem 1. A point x is in co(a) iff x is a convex combination of a finite subset of A. Proof. Induction on L establishes that if x is a convex combination of a finite number points in A then x is in any convex set containing A, hence x co(a). Let V be the set formed by taking taking all convex combinations of all finite subsets of A. Then, by the preceding argument, V co(a). On the other hand, again by the induction argument, V is convex. Therefore, since A V, co(a) V. Hence co(a) = V. 1 cbna. This work is licensed under the Creative Commons Attribution-NonCommercial- ShareAlike 4.0 License. 1

2 2 Supporting Hyperplane Theorem. Recall the following definition. Definition 2. Let A R N and x R N. A is supported at x iff there is a v R N, v 0, such that for all a A, v a v x. Theorem 2 (Supporting Hyperplane Theorem). Let A R N and x R N. Then A is supported at x iff x is not interior to co(a). Proof. If A = then the theorem holds trivially. Henceforth assume A is not empty. If A is supported at x then there is a v 0 such that v a v x for all a A. Theorem 1 then implies that v x v x for all x co(a): if x co(a) then x = l θ la l for some finite set of a l A; but then v x = θ l (v a l ) θ l (v x ) = v x. Therefore, for any ε > 0, the set N ε (x ) {x R N : v x < v x }, which is half of N ε (x ), is not contained in co(a), hence x is not interior to co(a). I need to show that there is a v R N, v 0, such that for any a A, v a v x. To simplify notation, I assume that A is convex. If A is not convex, substitute co(a) for A in the argument below, which then shows that co(a) is supported at x. Since A co(a), if co(a) is supported at x, then so is A. I claim that if x is not interior to A then x is also not interior to A. This claim is not true in general. For example, if A = [0, 1) (1, 2] then 1 is not interior to A (since 1 / A) but 1 is interior to A. Notice that in this example, A is not convex. I need to argue, therefore, that the claim is true provided both that x is not interior to A and that A is convex. For the moment, suppose that I have established the claim. Then the remainder of the proof is exactly as in the main convexity notes. More explicitly, the claim implies that for any ε > 0, N ε (x ) A c, which in turn implies that there is a sequence of points (x t ) in A c that converges to x. For each x t, one can show that C t = A {x t } is closed, convex and does not contain the origin. Apply the Basic Separation Theorem of the main notes to C t to get a v t C t, hence v t 0, such that for any a A, v t a > v t x. Finally, argue that this implies that there is a v R N, v 0, such that for any a A, and hence for any a A, v a v x. See the main convexity notes for details. To establish the claim that if x is not interior to A then it is not interior to A, fix any ε > 0. Suppose that I can find an open set U (which may depend on ε) with the property that U A c and N ε (x ) U. Since U is an open subset of A c, any element of U is interior to A c, hence is not in A, hence is in A c. Thus, U A c, which implies N ε (x ) A c. Since ε was arbitrary, this implies that x is not interior to A. It remains to show the existence of U. The following argument is not elegant, but it works. 2

3 Fix any w N ε (x ) A c. For each n, let e n be the unit vector for coordinate n. Let U 1 + be the half-line from w (but not including w) in the direction e1 : U 1 + = {x R N : γ > 0 such that x = w + γe 1 }. Let U1 be half-line from w in the direction e 1 : U1 = {x R N : γ > 0 such that x = w γe 1 }. I claim that either U 1 + or U 1 (or both) are subsets of Ac. Suppose to the contrary that there are γ 1, γ 2 > 0 such that x 1 = w + γ 1 e 1 A and x 2 = w γ 2 e 1 A. Then, setting θ = γ 2 /(γ 1 + γ 2 ), x = θ 1 x 1 + (1 θ 2 )x 2 = w. But since w / A, this implies A is not convex. By contraposition, then, since A is convex, either U 1 + or U 1 (or both) are subsets of A c. If U 1 + Ac, let U 1 = U 1 +. Otherwise, let U 1 = U1. Similarly, for each z U 1, let U 2 + (z) be the half-line from z (but not including z) in the direction e 2 : U 2 + (z) = {x RN : γ > 0 such that x = z + γe 2 }. Let U2 (z) be the open half-line from z in the direction e2 : U2 (z) = {x RN : γ > 0 such that x = z γe 2 }. As in the argument for U 1, since z A c, convexity of A implies that either U 1 + (z) or U 1 (z) (or both) are subsets of Ac I claim that, in fact, either U 2 + (z) Ac for all z U 1 or U2 (z) Ac for all z U 1 (or both). Suppose that there are z 1, z 2 U 1 and γ 1, γ 2 > 0 such that x 1 = z 1 + γ 1 e 2 U 2 + (z 1) A and x 2 = z 2 γ 2 e 2 U2 (ẑ) A. Then, setting θ = γ 2 /(γ 1 + γ 2 ), z = θ 1 x 1 + (1 θ 2 )x 2 = θ 1 z 1 + (1 θ 2 )z 2, which is in U 1, since U 1 is convex. Since U 1 A c, this implies A is not convex. By contraposition, then, since A is convex, either U 2 + (z) Ac for all z U 1 or U2 (z) Ac for all z U 1 (or both). If U 2 + (z) Ac for all z U 1, let U 2 = z U 1 U 2 + (z). Otherwise, let U 2 = z U 1 U2 (z). Proceeding in this way, I construct an open set U = U N. U is a subset of A c and has w as a limit point. Since w is a limit point of U, N ε (x ) U. Then as argued above, x is not interior to A, as was to be shown. 3 Separating Hyperplane Theorem. Definition 3. Let A, B R N. 1. A and B can be separated iff there is a v R N, v 0, and an r R such that for all a A, b B, v a r v b. 2. A and B can be strictly separated iff there is a v R N, v 0, and an r R such that for all a A, b B, v a > r > v b. 3. A and B can be separated with a gap iff there is a v R N, v 0, and r, r R such that for all a A, b B, v a r > r v b. Some texts call separation with a gap strong separation. Separation with a gap implies strict separation, but not conversely, as the next example shows. 3

4 Example 1. Let A = {x R 2 : x 1 > 0 and x 2 1/x 1 }. Let B = {x R 2 : x 1 > 0 and x 2 1/x 1 }. Then A and B can be strictly separated, with v = (0, 1) and r = 0, but they cannot be separated with a gap. In the main notes on convexity, both the Basic Separation Theorem and the Separating Hyperplane Theorem establish separation with a gap rather than merely strict separation. Theorem 3 (Separating Hyperplane Theorem). Let A, B R N. Let C = A B. 1. A and B can be separated iff the origin is not interior to co(c). 2. If A and B can be strictly separated then the origin is not contained in co(c). The converse direction, however, is false. 3. A and B can be separated with a gap iff the origin is not contained in co(c). Proof. If either A or B is empty then the result holds trivially. Henceforth, assume A and B are not empty. 1.. If A and B can be separated then there is a v 0 such that for any a A and any b B, v a v b, hence v (a b) 0. Hence for any c C, v c 0. By Theorem 1 (as in the proof of the direction of Theorem 2), this implies that the origin is not interior to co(c).. To simply notation, I assume in the following that C is convex; if C is not convex, simply substitute co(c) for C in the argument below, which then shows that co(c) is supported at 0, which implies that C is supported at 0, which implies that A and B are separated. Since 0 is not interior to C, Theorem 2 implies that there is a v R N, v 0, such that for any c C, v c v 0 = 0. Therefore, for any a A, b B, letting c = a b, v (a b) 0, hence v a v b. Let r = inf v a, a A r = sup v b. b B r and r are well defined because, for example, the set of v a is bounded below by any v b and hence the inf exists. Then r r. Choose r [r, r] and the result follows. 2.. If A and B can be strictly separated then there is a v 0 such that for any a A and any b B, v a > v b, hence v (a b) > 0. Hence for any c C, v c > 0. By Theorem 1 (as in the proof of the direction of Theorem 2), this implies that for any x co(c), v x > 0. Since v 0 = 0, this implies that 0 / co(c). 4

5 . It suffices to give a counterexample. Let A = {x R 2 : x 1 > 0 and x 2 1/x 1 } and B = {x R 2 : x 2 = 0} (that is, B is the x 1 axis). Then C = {x R 2 : x 2 > 0} (that is, C is the open half-space above the x 1 axis). C is convex and does not contain the origin, but while A and B can be separated, with v = (0, 1) and r = 0, they cannot be not strictly separated. 3. If A and B can be separated with a gap there is a v 0 and r > r such that for any a A, b B, v a r > r v b. Let q = r r > 0. Then for any a A, b B, v (a b) q. Hence for any c C, v c q. By Theorem 1 (as in the proof of the direction of Theorem 2), this implies v x q for any x co(c). By continuity, for any x co(c), v x q. Since q > 0, but v 0 = 0, this implies that 0 / co(c).. The proof is essentially the same as the proof in the main convexity notes (and not very different from the proof of the direction of part 1 above) and hence I omit it. 5