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Convex Sets and Convex Functions 1 Convex Sets Convex Sets and Convex Functions 1 Convex Sets, In this section, we introduce one of the most important ideas in economic modelling, in the theory of optimization and, indeed in much of modern analysis and computatyional mathematics: that of a convex set. Almost every situation we will meet will depend on this geometric idea. As an independent idea, the notion of convexity appeared at the end of the 19th century, particularly in the works of Minkowski who is supposed to have said: \Everything that is convex interests me." We discuss other ideas which stem from the basic definition, and in particular, the notion of a convex and concave functions which which are so prevalent in economic models. The geometric definition, as we will see, makes sense in any vector space. Since, for the most of our work we deal only with Rn, the definitions will be stated in that context. The interested student may, however, reformulate the definitions either, in an ab stract setting or in some concrete vector space as, for example, C([0; 1]; R)1. Intuitively if we think of R2 or R3, a convex set of vectors is a set that contains all the points of any line segment joining two points of the set (see the next figure). Here is the definition. Definition 1.1 Let u; v 2 V . Then the set of all convex combinations of u and v is the set of points fwλ 2 V : wλ = (1 − λ)u + λv; 0 ≤ λ ≤ 1g: (1.1) In, say, R2 or R3, this set is exactly the line segment joining the two points u and v. (See the examples below.) Next, is the notion of a convex set. 1This symbol stands for the vector space of continuous, real-valued functions defined on the closed interval [0; 1] 1 Figure 1: A Convex Set Figure 2: A Non-convex Set Definition 1.2 Let K ⊂ V . Then the set K is said to be convex provided that given two points u; v 2 K the set (1.1) is a subset of K. We give some simple examples: Examples 1.3 (a) An interval of [a; b] ⊂ R is a convex set. To see this, let c; d 2 [a; b] and assume, without loss of generality, that c < d. Let λ 2 (0; 1). Then, a ≤ c = (1 − λ)c + λc < (1 − λ)c + λd < (1 − λ)d + λd = d ≤ b: (b) A disk with center (0; 0) and radius c is a convex subset of R2. This may be easily checked (Exercise!) by using the usual distance formula in R2 namely kx − yk = p 2 2 (x1 − y1) + (x2 − y2) and the triangle inequality ku + vk ≤ kuk + kvk. n n (c) In R the set H := fx 2 R : a1x1 + ::: + anxn = cg is a convex set. For any n particular choice of constants ai it is a hyperplane in R . Its defining equation is a generalization of the usual equation of a plane in R3, namely the equation ax + by + cz + d = 0, and so the set H is called a hyperplane. To see that H is a convex set, let x(1); x(2) 2 H and define z 2 R3 by z := (1 − λ)x(1) + λx(2). Then 2 n n X (1) (2) X (1) (2) z = ai[(1 − λ)xi + λxi ] = (1 − λ)aixi + λaixi i=1 i=1 n X (1) (2) = (1 − λ) aixi + aixi = (1 − λ)c + λc i=1 = c: Hence z 2 H. (d) As a generalization of the preceeding example, let A be an m×n matrix, b 2 Rm, and let S := f(x 2 Rn : Ax = bg. (The set S is just the set of all solutions of the linear equation Ax = b.) Then the set S is a convex subset of Rn. Indeed, let x(1); x(2) 2 S. Then A (1 − λ)x(1) + λx(2) = (1 − λ)A x(1) + λA x(2) = (1 − λ)b + λb = b: (e) There are always two, so-called trivial examples. These are the empty set ;, and the entire space Rn. Note also that a singleton fxg is convex. In this latter case, as in the case of the empty set, the definition is satisfied vacuously. Note: In (c) above, we can take A = (a1; a2; : : : ; an) so that with this choice, the present example is certainly a generalization. One important fact about Rn is that the unit ball n B1 = fc 2 R j kxk ≤ 1g is a convex set. This follows from the triangle equality for norms: for any x; yinB1 and any λ 2 [0; 1] we have k (1 − λ) x + λ yk ≤ (1 − λ) jxk + λ kyk ≤ (1 − λ) + λ = 1 : Now the ball B1 is a closed set. It is easy to see that if we take its interior ◦ n B1 = fx 2 R j kxk < 1g ; this set is also convex. This gives us a hint regarding our next result. Proposition 1.4 If C ⊂ Rn is convex, the set cl(C), the closure of C, is also convexl 3 1 1 Proof: Supppose x; y 2 c`(C). Then there exist sequences fxngn=1 and fyngn=1 in C such that xn ! x and yn ! y as n ! 1. For some λ, 0 ≤ λ ≤ 1, define zn := (1−λ) xn +λ yn. Then, by convexity of C, zn 2 C. Moreover zn ! (1−λ) x+λ y as n ! 1. Hence this latter point lies in c`(C). 2 The simple example of th two intervals [0; 1] and [2; 3] on the real line shows that the union of two sets in not necessarily convex. On the other hand, we have the result: Proposition 1.5 The intersection of any number of convex sets is convex. Proof: Let fKαgα2A be a family of convex sets, and let K := [α2AKα. Then, for any x; y 2 K by definition of the intersection of a family of sets, x; y 2 Kα for all α 2 A and each of these sets is convex. Hence for any α 2 A; and λ 2 [0; 1]; (1 − λ)x + λy 2 Kα. Hence (1 − λ)x + λy 2 K. 2 While, by definition, a set is convex provided all convex combinations of two points in the set is again in the set, it is a simple matter to check that we can make a more general statement. This statement is the content of the following proposition. Notice the way in which the proof is constructed; it is often very useful in computations! p X Proposition 1.6 Let K be a convex set and let λ1; λ2; : : : ; λp ≥ 0 and λi = 1. If i=1 x1; x2; : : : xp 2 K then p X λi xi 2 K: i=1 Proof: We prove the result by induction. Since K is convex, the result is true, trivially, for p = 1 and by definition for p = 2. Suppose that the proposition is true for p = r (induction hypothesis!) and consider the convex combination λ1x1 + λ2x2 + ::: + λr+1. r r+1 r X X X Define Λ := λi. Then since 1 − Λ = λi − λi = λr+1, we have i=1 i=1 i=1 r ! r ! X X λi λ x + λ x = Λ x + (1 − Λ) x : i i r+1 r+1 Λ i r+1 i=1 i=1 λ λ Note that Pr i = 1 and so, by the induction hypothesis, Pr i x 2 K. i=1 Λ i=1 Λ i Since xr+1 2 K it follows that the right hand side is a convex combination of two points of K and hence lies in K 2 Relative to the vector space operations, we have the following result: 4 n Proposition 1.7 Let C;C1 and C2 be convex sets in R and let β 2 R. Then (a) The set β C := fz 2 Rn j z = β x ; x 2 Cg is convex. n (b) The set C1 + C2 := fx 2 R j z = x1 + x2 ; x1 2 C1; x2 2 C2g is convex. Proof: For part (a), let z1 and z2 both be elements or β C. Then there exists points x1; x2 2 C such that z1 = β x1 and z2 = β x2. Choose any λ 2 [0; 1] and form the convex combination z = (1 − λ) z1 + λ z2 : But then z = (1 − λ) β x1 + λ β x2 = β [(1 − λ) x1 + λ x2] : But C is convex so that (1 − λ) x1 + λ x2 2 C and hence z 2 β C . This proves part (a). Part (b) is proved by a similar argument by simply noting that (1 − λ)(x1 + x2) + λ (y1 + y2) = (1 − λ) x1 + (1 − λ) x2 + λ y1 + λ y2 : 2 For any given set which is not convex, we often want to find a set which is convex and which contains the set. Since the entire vector space V is obviously a convex set, there is always at least one such convex set containing the given one. In fact, there are infinitely many such sets. We can make a more economical choice if we start with the following fact. Intuitively, given a set C ⊂ V , the intersection of all convex sets containing C is the \smallest" subset containing C. We make this into a definition. Definition 1.8 The convex hull of a set C is the intersection of all convex sets which contain the set C.
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