7. Separating Hyperplane Theorems I Daisuke Oyama Mathematics II - PowerPoint PPT Presentation

7. Separating Hyperplane Theorems I Daisuke Oyama Mathematics II May 1, 2020

Separating Hyperplane Theorem Proposition 7.1 (Separating Hyperplane Theorem) Suppose that B ⊂ R N , B ̸ = ∅ , is convex and closed, and that b / ∈ B . Then there exist p ∈ R N with p ̸ = 0 and c ∈ R such that p · y ≤ c < p · b for all y ∈ B. 1 / 28

Lemma 7.2 Suppose that B ⊂ R N , B ̸ = ∅ , is closed, and that b / ∈ B . Let δ = inf {∥ z − b ∥ | z ∈ B } . Then δ > 0 , and there exists y ∗ ∈ B such that δ = ∥ y ∗ − b ∥ . Lemma 7.3 Suppose that B ⊂ R N , B ̸ = ∅ , is closed and convex, and that b / ∈ B . Then there exists a unique y ∗ ∈ B such that ∥ y ∗ − b ∥ = min {∥ z − b ∥ | z ∈ B } . Lemma 7.4 Suppose that B ⊂ R N , B ̸ = ∅ , is closed and convex, and ∈ B . that b / Let y ∗ ∈ B be as in Lemma 7.3. Then ( b − y ∗ ) · ( z − y ∗ ) ≤ 0 for all z ∈ B. 2 / 28

Proof of Lemma 7.4 ▶ Let y ∗ ∈ B be such that ∥ b − y ∗ ∥ = min {∥ b − z ∥ | z ∈ B } . ▶ Take any z ∈ B and any α ∈ (0 , 1) . ▶ Since (1 − α ) y ∗ + αz ∈ B , we have ∥ b − y ∗ ∥ 2 ≤ ∥ b − [(1 − α ) y ∗ + αz ] ∥ 2 = ∥ ( b − y ∗ ) − α ( z − y ∗ ) ∥ 2 = ∥ b − y ∗ ∥ 2 − 2 α ( b − y ∗ ) · ( z − y ∗ ) + α 2 ∥ z − y ∗ ∥ 2 , and therefore, ( b − y ∗ ) · ( z − y ∗ ) ≤ α 2 ∥ z − y ∗ ∥ 2 . ▶ Then let α → 0 . 3 / 28

Proof of Proposition 7.1 ▶ Let y ∗ ∈ B be as in Lemma 7.3. ▶ By Lemma 7.2, ( y ∗ − b ) · ( y ∗ − b ) > 0 . ▶ By Lemma 7.4, ( b − y ∗ ) · ( z − y ∗ ) ≤ 0 for all z ∈ B . ▶ Therefore, ( b − y ∗ ) · z ≤ ( b − y ∗ ) · y ∗ < ( b − y ∗ ) · b for all z ∈ B . ▶ Let p = b − y ∗ and c = ( b − y ∗ ) · y ∗ . 4 / 28

Dual Representation of a Convex Set For K ⊂ R N , K ̸ = ∅ , define the function φ K : R N → ( −∞ , ∞ ] by p · x, φ K ( p ) = sup x ∈ K which is called the support function of K . Proposition 7.5 Let K ⊂ R N , K ̸ = ∅ , be a closed convex set. Then K = { x ∈ R N | p · x ≤ φ K ( p ) for all p ∈ R N } . More generally, for any nonempty set K , Cl(Co K ) = { x ∈ R N | p · x ≤ φ K ( p ) for all p ∈ R N } . 5 / 28

Proof ▶ K ⊂ ( RHS ) : By definition. ▶ K ⊃ ( RHS ) : ∈ K . Let b / ▶ Since K is closed and convex, by the Separating Hyperplane Theorem, there exist ¯ p ̸ = 0 and c ∈ R such that p · z ≤ c < ¯ p · b for all z ∈ K, ¯ and hence p · z < ¯ p · b. φ K (¯ p ) = sup ¯ z ∈ K ▶ This means that b / ∈ ( RHS ) . 6 / 28

Supporting Hyperplane Theorem Proposition 7.6 (Supporting Hyperplane Theorem) Suppose that B ⊂ R N , B ̸ = ∅ , is convex, and that b / ∈ Int B . Then there exists p ∈ R N with p ̸ = 0 such that p · y ≤ p · b for all y ∈ B. For proof, we will use the following fact: Fact 1 For any convex set B ⊂ R N , Int B = Int(Cl B ) . The equality does not hold in general for nonconvex sets; for example, [0 , 1 / 2) ∪ (1 / 2 , 1] . 7 / 28

Proof ▶ Let b / ∈ Int B . Since B is convex, b / ∈ Int(Cl B ) by Fact 1. ▶ Therefore, there is a sequence { b m } with b m / ∈ Cl B such that b m → b . ▶ Since B is convex, Cl B is also convex (Proposition 4.12). ▶ Then by the Separating Hyperplane Theorem, for each m there exists p m ∈ R N with p m ̸ = 0 such that p m · y < p m · b m for all y ∈ B. ▶ Without loss of generality we assume that ∥ p m ∥ = 1 for all m . ▶ { p m } has a convergent subsequence { p m k } with a limit p , where p ̸ = 0 since ∥ p ∥ = 1 . ▶ Letting k → ∞ we have p · y ≤ p · b for all y ∈ B . 8 / 28

Separating Hyperplane Theorem Proposition 7.7 (Separating Hyperplane Theorem) Suppose that A, B ⊂ R N , A, B ̸ = ∅ , are convex, and that A ∩ B = ∅ . Then there exists p ∈ R N with p ̸ = 0 such that p · x ≤ p · y for all x ∈ A and y ∈ B. 9 / 28

Proof ▶ Since A and B are convex, A − B = { x − y ∈ R | x ∈ A, y ∈ B } is also convex (Proposition 4.5). ▶ Since A ∩ B = ∅ , 0 / ∈ A − B . ▶ Thus by the Supporting Hyperplane Theorem, there exists p ∈ R N with p ̸ = 0 such that p · z ≤ p · 0 for all z ∈ A − B, or p · x ≤ p · y for all x ∈ A and y ∈ B. 10 / 28

Separating Hyperplane Theorem Proposition 7.8 (Strong Separating Hyperplane Theorem) Suppose that A, B ⊂ R N , A, B ̸ = ∅ , are convex and closed, and that A ∩ B = ∅ . If A or B is bounded, then there exist p ∈ R N with p ̸ = 0 and c 1 , c 2 ∈ R such that p · x ≤ c 1 < c 2 ≤ p · y for all x ∈ A and y ∈ B. 11 / 28

Proof ▶ Since A and B are convex, A − B is also convex ▶ Since A and B are closed and A or B is bounded, A − B is closed. ( → Homework) ▶ Since A ∩ B = ∅ , 0 / ∈ A − B . ▶ Thus by the Separating Hyperplane Theorem, there exist p ∈ R N with p ̸ = 0 and c ∈ R such that p · z ≤ c < p · 0 for all z ∈ A − B, or p · ( x − y ) ≤ c < 0 for all x ∈ A and y ∈ B. ▶ Thus we have p · x − inf y ∈ B p · y ≤ c < 0 . sup x ∈ A Let c 1 = sup x ∈ A p · x and c 2 = inf y ∈ B p · y , where c 1 < c 2 . 12 / 28

Separation with Nonnegative/Positive Vectors Lemma 7.9 For A ⊂ R N , A ̸ = ∅ , suppose that A − R N ++ ⊂ A . For p ∈ R N , if there exists c ∈ R such that p · x ≤ c for all x ∈ A , then p ≥ 0 . Proof ▶ Assume that p n < 0 . ▶ Fix any x 0 ∈ A and any ε > 0 . We have x 0 − ( te n + ε 1 ) ∈ A − R N ++ ⊂ A for all t > 0 , while p · [ x 0 − ( te n + ε 1 )] = p · x 0 − tp n − εp · 1 → ∞ as t → ∞ , contradicting the assumption that p · x ≤ c for all x ∈ A . 13 / 28

Separation with Nonnegative/Positive Vectors Proposition 7.10 Suppose that B ⊂ R N , B ̸ = ∅ , is convex. If B ∩ R N ++ = ∅ , then there exists p ≥ 0 with p ̸ = 0 such that p · x ≤ 0 for all x ∈ B. 14 / 28

Proof ▶ Let A = B − R N ++ . ▶ Since B and R N ++ are convex, A is also convex. ▶ Since B ∩ R N ++ = ∅ , 0 / ∈ A . ▶ Thus by the Supporting Hyperplane Theorem, there exists p ∈ R N with p ̸ = 0 such that p · z ≤ p · 0 for all z ∈ A. ▶ Since A − R N ++ ⊂ A , we have p ≥ 0 by Lemma 7.9. ▶ We have p · x ≤ p · y for all x ∈ B and y ∈ R N ++ . Letting y → 0 , we have p · x ≤ 0 for all x ∈ B . 15 / 28

Separation with Nonnegative/Positive Vectors Proposition 7.11 Suppose that B ⊂ R N , B ̸ = ∅ , is convex and closed. If B ∩ R N + = { 0 } , then there exist p ≫ 0 and c ≥ 0 such that p · x ≤ c for all x ∈ B. 16 / 28

Proof ▶ Let ∆ = { x ∈ R N + | x 1 + · · · + x N = 1 } . ▶ B is convex and closed and ∆ is convex and compact. ▶ Since B ∩ R N + = { 0 } , B ∩ ∆ = ∅ . ▶ Thus by Proposition 7.8, there exist p ∈ R N with p ̸ = 0 and c ∈ R such that p · x ≤ c < p · y for all x ∈ B and y ∈ ∆ , where c ≥ 0 since 0 ∈ B . ▶ For each n , since e n ∈ ∆ , we have 0 ≤ c < p · e n = p n . 17 / 28

Efficient Production Let Y ⊂ R N be the production set of a firm. Definition 7.1 ▶ A production vector y ∈ Y is efficient if there is no y ′ ∈ Y such that y ′ ≥ y and y ′ ̸ = y . ▶ y ∈ Y is weakly efficient if there is no y ′ ∈ Y such that y ′ ≫ y . ▶ y : efficient ⇒ y : weakly efficient 18 / 28

Proposition 7.12 Suppose that Y is convex. Then for any weakly efficient production vector ¯ y ∈ Y , there exists p ≥ 0 with p ̸ = 0 such that p · ¯ y ≥ p · y for all y ∈ Y . Proof ▶ Let ¯ y ∈ Y be weakly efficient. ▶ Then ( Y − { ¯ y } ) ∩ R N ++ = ∅ , where Y − { ¯ y } is convex. ▶ Thus by Proposition 7.10, there exists p ≥ 0 with p ̸ = 0 such that p · z ≤ 0 for all z ∈ Y − { ¯ y } , or p · y ≤ p · ¯ y for all y ∈ Y . 19 / 28

From Profit Function to Production Set ▶ Let Y ⊂ R N , Y ̸ = ∅ , be the production set of a firm, and let φ Y : R N → ( −∞ , ∞ ] be the support function of Y : p · y. φ Y ( p ) = sup y ∈ Y ▶ Suppose that Y is convex and closed. Then, as we have seen, Y = { y ∈ R N | p · y ≤ φ Y ( p ) for all p ∈ R N } . ▶ What additional assumptions are needed to recover Y from the profit function, which is defined only for nonnegative, or positive, price vectors (where we allow the profit function to take values in ( −∞ , ∞ ] )? 20 / 28

▶ Free disposal : Y − R N + ⊂ Y . ▶ No free production : Y ∩ R N + ⊂ { 0 } . ▶ The ability to shut down : 0 ∈ Y . Proposition 7.13 1. If Y is nonempty, convex, and closed and satisfies free disposal, then Y = { y ∈ R N | p · y ≤ φ Y ( p ) for all p ∈ R N + } . 2. If Y is nonempty, convex, and closed and satisfies free disposal, no free production, and the ability to shut down, then Y = { y ∈ R N | p · y ≤ φ Y ( p ) for all p ∈ R N ++ } . 21 / 28

Proof 1 ▶ Y ⊂ ( RHS ) : Immediate. ▶ Y c ⊂ ( RHS ) c : Suppose that ¯ y / ∈ Y . ▶ Since Y is nonempty, convex, and closed, there exist ¯ p ̸ = 0 and c such that p · y ≤ c < ¯ ¯ p · ¯ y for all y ∈ Y , p · ¯ and hence φ Y (¯ p ) < ¯ y , by the Separating Hyperplane Theorem. ▶ Since Y satisfies free disposal, i.e., Y − R N + ⊂ Y (which implies Y − R N ++ ⊂ Y ), we have ¯ p ≥ 0 by Lemma 7.9. ▶ Hence, ¯ ∈ ( RHS ) . y / 22 / 28