Mathematics II, course

Mathematics II, course 2013-2014 Juan Pablo Rincón Zapatero October 24, 2013 Summary: The course has four parts that we describe below. (I) Topology in Rn is a brief review of the main concepts and properties of the Euclidean space, including open and closed sets, compact sets, sequences, limits, continuity and differentiability of functions, and separation of convex sets; (II) Metric Spaces generalize topological concepts of the Euclidean space to arbitrary sets endowed with a metric, with the objective of understanding Theorem of Banach for contraction mappings; (III) Parametric optimization problems studies parametric continuity and parametric monotonicity of solutions of optimization problems. (IV) Fixed point theorems analyzes the theorems of Brower and Kakutani for functions and correspondences, respectively and the Theorem of Banach for contraction mappings in complete metric spaces. In parts (II), (III) and (IV), several applications to economic problems are given, as the existence of Nash equilibria in noncooperative games, consumer theory, or the existence of equilibria in a pure exchange economy. Bibliography: Apostol, T.M. Mathematical Analysis, Second Edition, Addison-Wesley, 1974. Berge, C. Espaces Topologiques. Fonctions Multivoques, Deuxime dition, Dunod, 1966. Rudin, W. Principles of Mathematical Analysis, Third Edition, McGraw-Hill, 1976.. Real and Complex Analysis, Third Edition, McGraw-Hill, 1987. Ok, F.A. Real Analysis with Economic Applications, Princeton University Press, 2007. Sundaram, R.K. A First Course in Optimization Theory, Cambridge University Press, 2005. Sydsaeter, Hammond, Seierstad, Strom. Further Mathematics for Economic Analysis, Second Edition, Prentice Hall, 2008. 1

Contents 1 Topology in R m 3 1.1 The Euclidean norm................................. 3 1.2 Sequences and limits................................. 4 1.3 Open, closed and bounded sets........................... 4 1.4 Completeness of R m................................. 6 1.5 Continuous functions................................. 7 1.6 Convex and concave functions............................ 8 1.7 Separation Theorems................................. 10 1.8 Problems........................................ 11 1.9 Homework....................................... 17 2 Parametric Optimization Problems 19 2.1 Continuity of correspondences............................ 19 2.2 The Theorem of the Maximum........................... 23 2.3 Problems........................................ 26 3 Fixed Point Theorems in R m 33 3.1 Theorems....................................... 33 3.2 Application to Game Theory............................. 33 3.3 Application to strictly positive matrices....................... 35 3.4 Problems........................................ 36 4 Metric spaces 39 4.1 Complete metric spaces................................ 42 4.2 Fixed points. Contractions and Banach s Theorem................. 44 4.3 Application to the existence and uniqueness of solutions of ordinary differential equations....................................... 47 4.4 Problems........................................ 50 4.5 Homework....................................... 56 2

1 Topology in R m 1.1 The Euclidean norm We start with some concepts that are probably very familiar to you. If x, y R m, the inner product of the two vectors is m x y = x i y i. The main properties of the inner product are: i) x x 0 and x x = 0 if and only if x = 0; ii) x y = y x; iii) (λx) y = λ(x y), λ R; iv) (x + y) z = x y + y z. We define the Euclidean norm of x by i=1 x = (x x) 1/2. Lemma 1.1 (Cauchy Schwartz inequality). If x, y R m, then x y x y. Proof. Let a = x 2, b = y 2, and c = x y. If b = 0, there is nothing to prove, so we suppose that b > 0. Then, for any λ R 0 n (x i λy i ) 2 = a 2λc + λ 2 b. i=1 Letting λ = c/b, we obtain 0 a c 2 /b and so c 2 ab. Lemma 1.2. The Euclidean norm satisfies 1. x 0; 2. If x = 0, then x = 0; 3. If λ R, then λx = λ x ; 4. (The triangle inequality) x + y x + y. Proof. The triangle inequality is a consequence of the Cauchy Schwartz inequality x + y 2 = (x + y) (x + y) = x 2 + 2x y + y 2 x 2 + 2 x y + y 2 = ( x + y ) 2. If we set x = a b and y = b c, then the triangle inequality becomes a c a b + b c that means that the length of one side of a triangle is less than or equal to the sum of the lengths of the other two sides. 3

1.2 Sequences and limits Definition 1.3. A sequence x 1, x 2,... tends in the Euclidean norm of R m to a limit x as n tends to infinity x n x as n, if, given any ɛ > 0, there exists an n 0 (ɛ) such that x n x < ɛ for all n n 0 (ɛ). Lemma 1.4. We work in R m with the Euclidean norm. 1. The limit is unique; 2. If x n x and n(1) < n(2) < < n(j) <, then x n(j) x as j. 3. If x n x and y n y, then x n + y n x + y; 4. If x n x and λ n, λ R with λ n λ, then λ n x n λx. Next theorem establishes that any bounded sequence has a convergent subsequence. Theorem 1.5 (Bolzano Weierstrass). If x n R m and there exists a K such that x n K for all n, then we can find n(1) < n(2) < < n(j) < and x R m such that x n(j) x as j. 1.3 Open, closed and bounded sets Definition 1.6. Let x R m and r > 0. The open ball of center x and radius r is the set B(x, r) = {y R m : x y < r}. The closed ball of center x and radius r is the set B(x, r) = {y R m : x y r}. Definition 1.7. A set U R m is open if, whenever x U, there exists an r > 0 such that B(x, r) U. Thus every point of an open set is surrounded by a ball consisting only of points of the set. Definition 1.8. The set N is a neighborhood of the point x if we can find an r > 0 such that B(x, r) N. Now we identify sets that are closed under the operation of taking limits. Definition 1.9. A set F R m is closed if whenever x n F for each n and x n x as n, then x F. Lemma 1.10. A subset U of R m is open if and only if its complement R m \U is closed. 4

Proof. Necessity: Suppose that U is open and let x n R m \U with x n x as n. Arguing by contradiction, if x / R m \U, then x U; since U is open, B(x, ɛ) U for some ɛ > 0. As x n x, there exists N such that x N x < ɛ, and so x N B(x, ɛ) U, contradicting the statement that x n R m \U, arriving to a contradiction. Thus, x R m \U and R m \U is closed. Sufficiency: Suppose that R m \U is closed. Arguing by contradiction, if U is not open, then there must exist an a U such that, given any ɛ > 0, there exists y / U with y a < ɛ. In particular, we can find x n R m \U such that x n a < 1/n. This means that x n a as n, and so, since R m \U is closed, a R m \U, hence a U R m \U, which is absurd. Lemma 1.11. Let x R m and r > 0. 1. The open ball B(x, r) is open. 2. The closed ball B(x, r) is closed. Lemma 1.12. Consider the collection τ of open sets in R m. 1. τ, R m τ; 2. If U α τ for all α A, then α A U α τ; 3. If U 1, U 2,..., U n τ, then n j=1 U j τ. Lemma 1.13. Consider the collection F of closed sets in R m. 1. F, R m F; 2. If F α F for all α A, then α A F α F; 3. If F 1, F 2,..., F n F, then n j=1 F j F. Example 1.14. The intersection of open sets need not be open. In R we have ( j 1, 1) = [0, 1). j=1 To show this, let x belongs to the intersection set at the left of the inequality. Then, j 1 < x < 1 for every j, so 0 x < 1. Conversely, if 0 x < 1, then j 1 < x < 1 for any j and then x j=1 ( j 1, 1). Definition 1.15. An open cover of a set E in R m is a collection {U α } of open subsets of R m such that E α U α. Definition 1.16. A subset K of R m is said to be compact if every open cover of K contains a finite subcover. This means that if {U α } is an open cover of K, then there are finitely many indices α 1,..., α n such that K U α1 U αn. Theorem 1.17 (The Heine Borel Theorem ). A subset K in R m is compact if and only if it is closed and bounded 1. 1 I quote here the following poem of Conway: 5

Theorem 1.18. 1. If K is a compact subset in R m, then every sequence in K has a subsequence converging to a point of K; 2. If K is a subset of R m such that every sequence in K has a subsequence converging to an element of K, then K is a compact set. Proof. 1. If x n K, then it is bounded and so, by the Theorem of Bolzano Weierstrass, has a convergent subsequence x n(j) x K since K is closed. 2. We argue by contradiction, supposing first that K is not closed. Then, we can find x n K and x / K such that x n x as n. Since any subsequence of a convergent sequence converges to the same limit, no subsequence of the x n can converge to a point of K: Second, if K is not bounded we can find x n K such that x n > n. If x is any point in R m, then the inequality x n x x n x > n x shows that no subsequence of the x n can converge. 1.4 Completeness of R m One of the most useful criteria of convergence of sequences is Cauchy s Principle of Convergence, which allows us to prove the convergence of a sequence in R m without any knowledge about the limit. Definition 1.19. A sequence {x n } defined in R m is a Cauchy sequence if for every ε > 0 there exists an integer n 0 (ɛ) such that x p x q < ε for all p, q n 0 (ɛ). The followings lemma holds in any metric space, not only in R m with the Euclidean metric. Lemma 1.20. Any convergent sequence in R m is a Cauchy sequence. Proof. Suppose that x n x as n. Let ɛ > 0. We can find n 0 (ɛ) such that x n x < ɛ 2 n n 0 (ɛ). Then x p x q x p x + x q x < ɛ 2 + ɛ 2 = ɛ, p, q n 0(ɛ). The following result, however, is not true in arbitrary metric spaces. If K is closed and bounded, say Heine Borel, And also Euclidean, then we can tell That, if it we smother With a large open cover, There s a finite refinement as well! 6

Lemma 1.21. Any Cauchy sequence in R m converges. Proof. It is easy to see that a Cauchy sequence is bounded. By definition, given any ɛ > 0, there exists n 0 (ɛ) such that x p x q < ε for all p, q n 0 (ɛ). In particular, if n n 0 (1), we have x n x n x n0 (1) + x n0 (1) < 1 + x n0 (1). Hence x n max 1 r n0 (1) x r + 1. Then, by the Bolzano Weierstrass theorem, there is a subsequence x n(j) x as j. Thus, given any ɛ > 0, one can find J(ɛ) such that x n(j) x < ɛ for all j J(ɛ). Take now n n 0 (ɛ/2), j J(ɛ/2) and n(j) > n 0 (ɛ/2). Then x n x x n x n(j) + x n(j) x < ɛ/2 + ɛ/2 = ɛ. Theorem 1.22 (Cauchy s Principle of Convergence). A sequence in R m converges if and only if it is a Cauchy sequence. 1.5 Continuous functions Definition 1.23. Let D R m. We say that a function f : D R p is continuous at x D if, given ɛ > 0, there exists δ(ɛ, x) > 0 such that, whenever y D and x y < δ(ɛ, x), we have f(x) f(y) < ɛ. If f is continuous at every x D, we say that f is a continuous function on D. Lemma 1.24. Let D R m and f : D R p. Let x n, x D such that x n x as n. Then f is continuous at x if and only if f(x n ) f(x) as n. Proof. If f is continuous at x, for every ɛ > 0 there exists δ(ɛ, x) > 0 such that for any y satisfying x y < δ(ɛ, x), we have f(x) f(y) < ɛ. But if x n converges to x as n, there exists an integer n 0 (δ(ɛ, x)) such that n n 0 implies x x n < δ(ɛ, x) and hence that f(x) f(x n ) < ɛ. Hence the sequence f(x n ) converges to f(x) as n. To prove the sufficient condition, suppose that it is satisfied but that f is not continuous at x. Then there exists ɛ > 0 such that for any δ > 0 we can find y satisfying x y < δ and f(x) f(y) ɛ. Let δ n 0 as n and x n with x x n < δ and f(x) f(x n ) ɛ. The sequence x n converges to x as n, hence the sequence f(x n ) converges to f(x) as n. But this is in contradiction with the inequality f(x) f(x n ) ɛ for all n. The assumption that f is not continuous at x is thus false. Another way of looking at continuity, which is crucial to define continuity in topological spaces without a metric, is given in the following result. Lemma 1.25. The function f : R m R p is continuous if and only if f 1 (U) is open whenever U is an open set in R p. Theorem 1.26 (Weierstrass Theorem). Let K be a compact subset of R m and f : K R p a continuous function. Then f(k) is closed and bounded. 7

Proof. By Theorem 1.18 we need only prove that any sequence in f(k) has a subsequence converging to a limit in f(k). Let y n f(k). By definition, x n K with f(x n ) = y n. Since K is closed and bounded, there is n(j) and x K such that x n(j) x, thus y n(j) = f(x n(j) ) f(x) f(k) since f is continuous. Theorem 1.27 (Extreme Value Theorem). Let K be a compact subset of R m and f : K R a continuous function. Then there exists x 1, x 2 K such that for all x K. f(x 2 ) f(x) f(x 1 ) Proof. Since f(k) is bounded, it has a supremum, M = sup f(k). Since f(k) is closed, M f(k), that is to say, M = f(x 1 ) for some x 1 K. We obtain x 2 similarly. That is, a real valued continuous function on a closed bounded set attains global minimum and maximum on that set. 1.6 Convex and concave functions Definition 1.28. A subset C R m is convex if λx + (1 λ)y C for all x, y C and each 0 λ 1. A function f : C R is said to be convex provided for all x, y C and each 0 λ 1. The function f is concave if f is convex. f(λx + (1 λ)y) λf(x) + (1 λ)f(y) Example 1.29 (Young s inequality). For any 1 < p, q <, 1 p + 1 q = 1 ab ap p + bq q (a, b > 0). The function x e x is convex, thus using the definition of convexity we have ab = e ln a+ln b = e 1 p ln ap + 1 q ln bq 1 p eln ap + 1 q eln bq = ap p + bq q. Theorem 1.30. Suppose f : R m R is convex. Then for each x R m there exists p R m such that the inequality f(y) f(x) + p (y x) (1) holds for all y R m. Definition 1.31. The vector p in (1) above is called a subgradient of f at x. The set of all subgradients at x is called the subdifferential of f at x and this set is denoted f(x). Example 1.32. { 1}, if x < 0; ( x ) = [ 1, 1], if x = 0; {1}, if x > 0. 8

Remark 1.33. (i) The mapping y f(x) + p (y x) determines the supporting hyperplane of f at x. Inequality (1) says the graph of f lies above each supporting hyperplane. If f is differentiable at x, then p = f(x). (ii) If f is of class C 2, then f is convex (concave) if and only if the Hessian matrix Hf(x) is positive (negative) semidefinite or positive (negative) definite. This means m f xi x j (x)h i h j 0( 0) x, h R m. i,j=1 Example 1.34. The function f(k, L) = K α L 1 α with 0 < α < 1 is concave in R + R +. It is continuous in R + R + and of class C 2 in R ++ R ++. The derivatives are: f K = αk α 1 L 1 α, f L = (1 α)k α L α f KK = α(α 1)K α 2 L 1 α, f KL = f LK = α(1 α)k α 1 L α, f LL = α(1 α)k α L α 1. The Hessian matrix is ( ) L Hf(K, L) = α(1 α)k α 2 L α 1 2 KL KL K 2 The first principal minor have negative sign, and the second is the determinant, which is 0, thus the quadratic form is negative semidefinite for every (K, L) R ++ R ++, thus f is concave in this region. Since it is continuous in R + R +, we can conclude that f is also concave here. Theorem 1.35 (Jensen s Inequality I). Let f : C R m R be convex. Then ( n ) n f λ j x j λ j f(x j ) for all x 1,..., x n C, for all λ 1,..., λ n 0 and n j=1 λ j = 1. j=1 Example 1.36. Using Jensen s inequality one can show that for any strictly positive real numbers x 1,..., x n the geometric mean is below the arithmetic mean: j=1 (x 1 x n ) 1/n x 1 + + x n. n For, let f(x) = ln x, which is convex in R ++ and let λ j = 1/n. Then ( ) x1 + + x n ln 1 n n (ln x 1 + + ln x n ) = ln (x 1 x n ) 1/n and taking exponentials we are done. Remark 1.37. Suppose X is a random variable with a discrete distribution with finitely many pairwise distinct x 1,..., x n. The expectation of X is EX = n x j Prob{X = x j }. j=1 Jensen s inequality implies that for any convex function f : R R Ef(X) f(ex). 9

There exists useful generalizations. Here we present the following: Let λ be the Lebesgue measure on R m and let u : U R be a measurable function such that the Lebesgue integral U u(x) dx is finite, where U is an open and bounded subset of Rm. Theorem 1.38 (Jensen s Inequality II). Let f : R R be a convex function. Then ( ) 1 f u(x) dx 1 f(u(x)) dx. λ(u) U λ(u) U Proof. Since f is convex, for each x R there exists p R such that f(y) f(x) + p(y x) for all y R. Let x = 1 u(x) dx, y = u(x). λ(u) U Then ( ) ( 1 f(u(x)) f u(x) dx + p u(x) 1 ) u(x) dx. λ(u) U λ(u) U Integrating with respect to x over U we get the desired inequality, since U ( u(x) 1 λ(u) U ) u(x) dx dx = U u(x) dx λ(u) λ(u) U u(x) dx = 0 1.7 Separation Theorems Separation theorems refer to the possibility of separating disjoint sets in R m by means of a linear variety. In general, this is not possible for arbitrary sets, unless they are convex. Separation theorems useful in establishing optimality conditions without differentiability requirements in convex programming. They are also fundamental in welfare economics. Given a nonzero vector u in R m and a scalar c, the set H = {x R m : u x = c} is a hyperplane in R m, with u as a normal vector. The associated closed half-spaces are H + = {x : u x c}, H = {x : u x c}. Given two subsets A and B of R m, we say that H separates A and B if for instance A H + and B H. In other words, A and B can be separated by a hyperplane if there exists u nonzero and a scalar c such that u x c u y for all x A and all y B. If both inequalities are strict, then the hyperplane strictly separates A and B. Theorem 1.39. Let A be a closed, convex set of R m and let y be a point in R m such that y / A. Then there exists a nonzero vector u R m and a scalar α such that A and {y} are strictly separated by the hyperplane H = {x R m : u x = c}. 10

Given a set A in R m, we say that a point x is in the boundary of A if for any ɛ > 0, A B(x, ɛ) and A c B(x, ɛ). We will write in this case x A and say that x is a boundary point of A. Note that boundary points are never interior points. A hyperplane H is said to be a support (or supporting hyperplane) for a convex set A at a point y A if H separates A and {y}. The existence of a supporting hyperplane to a convex set is assured in the following result. Theorem 1.40. Let A be a convex set in R m and suppose that y is not an interior point of A. Then there exists a nonzero vector u R m and a scalar α such that A and {y} are separated by the hyperplane H = {x R m : u x = c}. Theorem 1.41. Let A and B be two disjoint nonempty convex sets in R m. Then there exists a nonzero vector u R m and a scalar α such that A and B are separated by the hyperplane H = {x R m : u x = c}. Consider the following generalization of the above theorem. Let us denote by Å the set on interior points of the subset A. Theorem 1.42. Let A and B two nonempty convex sets in R m such that Å is nonempty and Å and B are disjoints. Then there exists a nonzero vector u R m and a scalar α such that A and B are separated by the hyperplane H = {x R m : u x = c}. 1.8 Problems We work in R m with the Euclidean norm. 1. (a) Show that the limit is unique, that is, if x n x and x n y, then x = y (b) If x n x, then x n x 0 and x n x. However, x n x does not imply x n x. (c) Suppose x n R m, x R m, λ n R and λ R. If x n x and λ n λ, then λ n x n λx. (d) If x n x and y n y, show that x n y n x y. Solution: (a) Since x n x and x n y, for any ɛ > 0 there exists n 0 (ɛ) such that for any n n 0 (ɛ) x n x < ɛ/2 and x n y < ɛ/2. By way of contradiction, if x y, then taking ɛ = x y > 0 and using the triangle inequality we obtain x y x n x + x n y < ɛ/2 + ɛ/2 = x y, a contradiction. Thus, the limit is unique. (b) The definition establishes that convergence of the sequence of vectors x n to x is equivalent to convergence of the sequence of real numbers x n x to 0. To show 11

the second part, observe that for arbitrary x, y, x x y + y and y x y + x. Hence we have the two inequalities x y x y and y x x y, which means x y x y. Since x n converges to x, for any ɛ there exists n 0 such that for any n n 0 x n x < ɛ, thus xn x xn x < ɛ, hence x n x. (This implies that, as function, the norm : R m R is continuous). However, if we consider the sequence (1, 1, 1, 1,...), x n 1 but x n does not converge. (c) Observe that by triangle inequality and homogeneity of the norm λ n x n λx λ n x n λ n x + λ n x λx = λ n x n x + λ n λ x. Since the norm is continuous, taking limits we get λ n x n λx 0, hence λ n x n λx. (d) Observe that by triangle inequality and the Cauchy Schwartz inequality x n y n x y x n (y n y) + y (x n x) x n y n y + y x n x. Since the norm is continuous, taking limits we get x n y n x y 0, hence x n y n x y. This means that the inner product mapping is continuous. 2. Show that the open ball is open and that the closed ball is closed. 3. (a) Show that the arbitrary union of open sets is open. (b) Show that the finite intersection of open sets is open. (c) Show that arbitrary intersection of closed sets is closed. (d) Show that the finite union of closed sets is closed. 4. Let U 1, U 2,... be open sets in R such that U 1 U 2.... Show, by means of examples that j=1 U j may be (a) open but not closed, (b) closed but not open, (c) open and closed or (d) neither open nor closed. Solution: For (a) let U j = (0, 1), for (b) let U j = ( 1/j, 1/j), for (c) let U j = (0, 1/j) and for (d) let U j = ( 1/j, 1). Then j=1 U j is (a) (0, 1), (b) {0}, (c) and (d) [0, 1), respectively. 12

5. If A is open in R m and B is open in R p, show that A B is open in R m R p. 6. Show that a function f : R m R p is continuous if and only if f 1 (U) is open whenever U is an open set in R p. Solution: Recall that f 1 (U) = {x R m : f(x) U}. Necessity: Suppose f is continuous and U is open in R p. Let x f 1 (U), so f(x) U and hence there exists ɛ > 0 such that B(f(x), ɛ) U. Since f is continuous at x, there exists δ > 0 such that for any y B(x, δ) Thus, we have shown which means thus f 1 (U) is open. f(x) f(y) < ɛ or, in other way, f(y) B(f(x), ɛ). f(b(x, δ)) B(f(x), ɛ) U, B(x, δ) f 1 (U), Sufficiency: Suppose that f 1 (U) is open if U is open in R p. Let x R m and ɛ > 0. Since B(f(x), ɛ) is open, f 1 (B(f(x), ɛ)) is open and x f 1 (B(f(x), ɛ)), so there exists δ > 0 such that B(x, δ) f 1 (B(f(x), ɛ)), or which means hence f is continuous. f(b(x, δ)) B(f(x), ɛ), x y < δ f(x) f(y) < ɛ, 7. (a) Let A = (0, 1). Show that the map f : A R given by f(x) = 1/x is continuous but f(a) is unbounded (the continuous image of a bounded set need not be bounded). (b) Let A = [1, ) and let f : A R be given by f(x) = 1/x. Show that A is closed and f is continuous but f(a) is not closed (the continuous image of a closed set need not be closed). (c) Show that the projection functions π 1, π 2 : R 2 R defined by π 1 (x, y) = x and π 2 (x, y) = y are continuous. Show that the set A = {(x, 1/x) : x > 0} is closed in R 2 but that π 1 (A) is not. (d) Show, however, that if F is a closed set in R 2 with π 2 (F ) bounded, then π 1 (F ) must be closed. Solution: 13

(a) f(a) = (0, ). (b) f(a) = (0, 1]. (c) If (x n, y n ) (x, y) 0, then x n x and y n y, thus both projection functions are continuous. Now, if (x n, y n ) A and (x n, y n ) (x, y) 0, then y n = 1/x n 1/x and y n y, hence y = 1/x and (x, y) A, thus A is closed. However, π 1 (A) = (0, ) is not closed since 1/n π 1 (A) and 1/n 0 / π 1 (A) as n. (d) If x n π 1 (F ) and x n x we can find y n π 2 (F ) such that (x n, y n ) F. Since π 2 (F ) is bounded, by the Bolzano Weierstrass Theorem there exists a convergent subsequence y n(j) of y n converging to some y as j. Since x n(j) x and F is closed, we have (x, y) F, so that x π 1 (F ). 8. Let A R m. A function f : A R is said to be upper semicontinuous on A if, given any x A and any ɛ > 0, we can find δ(ɛ, x) > 0 such that f(y) f(x) < ɛ for all y A with y x < δ(ɛ, x). (a) Give an example of a function f : R R which is upper semicontinuous but not continuous. (b) If A R m and f : A R is such that both f and f are upper semicontinuous on A, show that f is continuous on A. (c) For a function f : A R, the upper sections are defined as the subsets of R m given by S λ (f) = {x A : f(x) λ}, where λ R. Suppose that the set A is closed. Show that f is upper semicontinuous if and only if S λ is closed for any λ. (d) If K R m is compact and f : K R is upper semicontinuous, show that f has a global maximum on K. (e) Is it necessarily true that an upper semicontinuous function is bounded below? Is it necessarily true that, if f is bounded below, it attains a global minimum? (f) A function f such that f is upper semicontinuous is called lower semicontinuous. Show that a lower semicontinuous functions attains global minimum on compact sets. Solution: (a) The function f(x) = 0 for x 0 and f(0) = 1 is upper semicontinuous but not continuous. (b) If f is upper semicontinuous on A, then given any x A and any ɛ > 0, we can find δ(ɛ, x) > 0 such that f(y) f(x) < ɛ for all y A with y x < δ(ɛ, x). Thus, since f is also upper semicontinuous (f)(y) ( f)(x) < ɛ for all y A with y x < δ(ɛ, x). 14

Thus we have the two inequalities that means hence f is continuous on A. f(y) f(x) < ɛ and (f(y) f(x)) < ɛ, f(y) f(x) < ɛ for all y A with y x < δ(ɛ, x), (c) Necessity: Note that S λ (f) = A {x R m : f(x) λ}. If we prove that F = {x R m : f(x) λ} is closed, we are done since we have an intersection of two closed sets. Let x / F. Then ɛ = λ f(x) > 0 and since f is upper semicontinuous, there exists δ > 0 such that y x < δ implies f(y) < f(x) + ɛ = λ, thus y / F. We have shown that B(x, δ) R m \F, hence R m \F is open and F is closed. Sufficiency: Let x A and let ɛ > 0. Define λ = f(x) + ɛ. Obviously, x R m \S λ (f), which is open by assumption. Thus, there exists δ > 0 such that B(x, δ) R m \S λ (f), that is, for y x < δ we have f(y) < λ = f(x) + ɛ, which proves that f is upper semicontinuous. (d) Let us show that f is bounded above. Reasoning by contradiction, if f is not bounded above, then we can find x n K such that f(x n ) n. Since K is closed and bounded, we can extract a convergent subsequence x n(j) x K as j. Let ɛ = 1 and choose j 0 such that for any j j 0 Since f is upper semicontinuous, x n(j) x < δ(1, x). f(x n(j) ) < f(x) + 1, thus we get the contradiction n(j) f(x) + 1 for any j, which is absurd, since n(j) as j. Thus, f is bounded above. Let M be the supremum of the set f(k), that is, M = sup f(x), x K and let us show that there is some x 1 K such that it is attained, M = f(x 1 ). Consider a sequence x n K approaching the supremum, f(x n ) M as n. Since K is closed and bounded, there is a subsequence x n(j) converging to some x 1 K. Since f is upper semicontinuous, by a reasoning similar as above, for ɛ = 1/j it is possible to select j 0 such that for any j j 0 one has Taking limits as j we have Hence, M = f(x 1 ). f(x n(j) ) < f(x 1 ) + 1/j. M f(x 1 ). (e) The function f : [0, 1] R given by f(x) = 1/x for x 0, f(0) = 0 is upper semicontinuous but not bounded below. 15

(f) The function f : [0, 1] R given by f(x) = x for x 0, f(0) = 1 is upper semicontinuous and bounded below, but it does not attain a global minimum. (g) Since f is upper semicontinuous, it attains a global maximum on any compact set K R m, hence there exists x 1 K such that ( f)(x) ( f)(x 1 ) for all x K, thus f(x) f(x 1 ) for all x K, which means that x 1 is a global minimum of f. 9. Let A R m be closed and unbounded and let f : A R be an upper semicontinuous function such for any sequence x n A with x n as n, it is possible to extract a subsequence x n(j) such that f(x n(j) ) as j. Show that f attains a global maximum on A. Solution: Let us show that the upper sections S λ (f) are bounded. If not, then there is some sequence x n S λ (f) such that x n. It is possible then to extract a subsequence such that f(x n(j) ) j. Then j λ for any j, which is absurd. Hence, the upper sections are bounded. On the other hand, we know from a previous problem that S λ (f) is closed since f is upper semicontinuous, thus for any λ the set S λ (f) is closed and bounded. Fix any λ 0 for which S λ 0 (f). Let x 0 S λ 0 (f) a global maximum of f in the set S λ 0 (f). We have sup{f(x) : x A} = sup{f(x) : x S λ 0 (f)} = f(x 0 ), hence f attains its global maximum on A. 10. Let A R m be closed and let b / A. Show that there is y A such that b y b x x A, that is, there is some point y A which distance to b is minimum among all points of A. Solution: Consider the function g(x) = b x. It is continuous since the norm is continuous. If A is bounded, then g attains a global minimum y A by Weierstrass Theorem. If A is unbounded, then there is some sequence x n A such that x n, hence g(x n ) = x n b x n b as n. According to the previous problem, there is y A minimizing g on A and we are done. 11. (a) Give an example of a continuous and bounded function that does not attain maximum and minimum on a bounded set. (b) Give an example of a continuous and bounded function that does not attain maximum and minimum on a bounded set. 16

(c) Give an example of a function which is neither upper semicontinuous nor lower semicontinuous in the interval [0, 1]. Solution: (a) f(x) = x and A = (0, 1). (b) f(x) = arctan x and A = R. (c) f(x) = x if x 0, 1, and f(0) = f(1) = 1/2. 1.9 Homework We work in the Euclidean space R m. 1. Show that the open ball is open and that the closed ball is closed. 2. (a) Show that the arbitrary union of open sets is open. (b) Show that the finite intersection of open sets is open. (c) Show that arbitrary intersection of closed sets is closed. (d) Show that the finite union of closed sets is closed. 3. If A is open in R m and B is open in R p, show that A B is open in R m R p. 4. Show that if F K with F closed and K compact, then F is compact. 5. Which of the following statements are true and which are false? Give proofs or counterexamples. (a) If A is an open (closed) subset of R m+1, then is open (closed) in R m. {x R m : (0, x) A} (b) If A is an open (closed) subset of R m, then is open (closed) in R m+1. {(0, x) R m+1 : x A} 6. Let A R m and let f : A R p be a function. We call G = {(x, f(x)) : x A} the graph of f. 17

(a) Show that, if G is bounded, then A is. (b) Show that it is possible to have G closed and f continuous but A not closed. (c) Show that, if A is closed and f is continuous, then G is closed. (d) Show that G and A may be closed but f discontinuous. 7. Give an example of a subset S of R and a function f : S R continuous and strictly monotonous increasing in S such that its inverse, f 1, is not continuous on f(s). (Hint: look for a suitable S) 18

2 Parametric Optimization Problems 2.1 Continuity of correspondences Definition 2.1. A correspondence Γ : X R m R p is a mapping Γ : X R m 2 Rp, where 2 Rp is the collection of all subsets of R p, including the empty set. Definition 2.2. We say that Γ is upper hemicontinuous (uhc) at x 0 X iff for every open set U R p such that Γ(x 0 ) U, there is an open set N R m such that x 0 N and for every x X N, Γ(x) U. Remark 2.3 (Not uhc). If Γ is not uhc at x 0, then there is an open set U Γ(x 0 ), a sequence x n X B(x 0, 1 n ) and a sequence y n Γ(x n ) for which y n / U for all n 1. Definition 2.4. We say that Γ is lower hemicontinuous (lhc) at x 0 X iff for every open set U R p such that Γ(x 0 ) U, there is an open set N R m such that x 0 N and for every x X N, Γ(x) U. Remark 2.5 (Not lhc). If Γ is not lhc at x 0, then there is y 0 Γ(x 0 ), an open set U y 0 and a sequence x n X B(x 0, 1 n ) such that for any element y Γ(x n), y / U for all n 1. Proposition 2.6. If Γ is uhc or lhc at x 0 X and single valued in B(x 0, r) X for some r > 0, then it is a continuous mapping at x 0. Proof. We can consider that a Γ is a mapping in B(x 0, r) X. Let U be an open set such that y 0 = Γ(x 0 ) U. Then, if Γ is uhc at x 0, then there is an open set N with x 0 N and Γ(x) U for any x X N. Take N B(x 0, r), so that Γ is singled valued. Then, Γ(x) U for any x X N, so that X N Γ 1 (U) and since X N is open in X, we have shown that Γ is continuous at x 0. The argument for the lhc case is the same, as Γ(x) U and Γ(x) U are identical conditions for a mapping. Reciprocally, if f : X R p is a continuous mapping, then Γ defines as Γ(x) = {f(x)} is both usc and lhc. Definition 2.7. We say that the correspondence Γ is continuous at x 0 X iff it is borh uhc and lhc at x 0. Definition 2.8. We say that the correspondence Γ is uhc, lhc or continuous at X iff it is uhc, lhc or continuous at every x X, respectively. Exercise 2.9. Study if Γ : R + R + is uhc/lhc in the following cases. 1. Γ(x) = [0, x]. 2. Γ(x) = [0, x). 3. Γ(x) = (0, x]. 4. Γ(x) = (0, x) if x > 0, Γ(0) = {0}. 19

1. Γ is uhc: Let U be open with [0, x 0 ] U. Then there is δ > 0 such that ( δ, x 0 + δ) U. Let N = (x 0 δ/2, x 0 + δ/2). Since f(x) = x is increasing, Γ(x) [0, x 0 + δ/2) for any x R + N, thus Γ(R + N) [0, x 0 + δ/2) ( δ, x 0 + δ) U, thus Γ is uhc. Γ is lhc: Let U be open with [0, x 0 ] U. Then there is some point y 0 (0, x 0 ) U. Let δ = x 0 y 0 > 0 and consider N = (x 0 δ/2, x 0 + δ/2). Let us check that for any x N one has y 0 Γ(x) = [0, x], so that Γ(x) U. The smallest of these intervals contain [0, x 0 δ/2) and x 0 δ/2 = x 0 /2 + y 0 /2 > y 0, thus y 0 Γ(x) U for all x N and Γ is lhc. 2. Γ is not uhc: it suffice to show an U open with [0, x 0 ) U for some x 0 and Γ(R + N) U for any open set N containing x 0. Let U = ( 1, x 0 ), so that Γ(x 0 ) U and U is open. For any N neighborhood of x 0, there is some δ > 0 such that (x 0 δ, x 0 + δ) N. But x + δ/2 (x 0 δ, x 0 + δ) and Γ(x 0 + δ/2) = [0, x 0 + δ/2) U. Γ is lhc: the same proof as above. 3. Γ is uhc: for any x 0 > 0, and any U (0, x 0 ], there is some δ > 0 such that (0, x 0 ] (0, x 0 + δ) U. Take N = (x 0 δ/2, x 0 + δ/2). Then for any x R + N one has Γ(x) (0, x 0 + δ/2) (0, x 0 + δ) U, thus Γ is uhc. Γ is lhc: the same proof as above. 4. Γ is not uhc at x 0 > 0: the same proof as above. Γ is lhc at x 0 > 0: the same proof as above. Γ is both uhc and lhc at 0. Part 1 of the above exercise is a particular case of the following result. Exercise 2.10. Let Γ(x) = [g(x), f(x)], where functions g, f : R R are continuous and g(x) f(x) for all x. Show that Γ is a continuous set valued map. Proof. Γ is uhc: Let U be open with [g(x 0 ), f(x 0 )] U. There exists ɛ > 0 such that [g(x 0 ) ɛ, f(x 0 ) + ɛ] U. Given that g (f) is continuous, there exists δ > 0 (δ > 0) such that if x (x 0 δ, x 0 + δ ), (x (x 0 δ, x 0 + δ )), then g(x) > g(x 0 ) ɛ (f(x) < f(x 0 ) + ɛ). Taking δ = min{δ, δ } we have g(x 0 ) ɛ < g(x) f(x) < f(x 0 ) + ɛ, thus [g(x), f(x)] U. Γ is lhc: Let U be open with [g(x 0 ), f(x 0 )] U. There is some y 0 U (g(x 0 ), f(x 0 )) because U is open. Let ɛ > 0 such that ɛ y 0 g(x 0 ) and ɛ f(x 0 ) y 0. Then, since both g and f are continuous, there exists δ > 0 such that for any x (x 0 δ, x 0 + δ) g(x) < g(x 0 ) + ɛ g(x 0 ) + y 0 g(x 0 ) = y 0, f(x) > f(x 0 ) ɛ f(x 0 ) f(x 0 ) + y 0 = y 0, thus y 0 (g(x), f(x)), and [g(x), f(x)] U, hence Γ is lhc. Exercise 2.11. Show that Γ(x) = [ x, x ] with x R is continuous. Proof. The function x x is continuous and apply the result above. Definition 2.12. The domain of Γ is the set domγ = {x X : Γ(x) }. The image of the set A domγ by the correspondence Γ is the set Γ(A) = x A Γ(x). 20

The graph of Γ is the set Ω = {(x, y) : x domγ, y Γ(x)} X Γ(X). Definition 2.13. We say that Γ has open, closed, bounded, compact or convex values iff for every x X the sets Γ(x) are open, closed, bounded, compact or convex, respectively. Definition 2.14. We say that Γ has the closed graph property at x 0 X iff for any sequence (x n ) in X converging to x 0 and for any sequence (y n ), y n Γ(x n ), converging to some y 0, one has y 0 Γ(x 0 ). Definition 2.15. We say that Γ has the closed graph property iff it has the closed graph property at every x X. This means that Ω is relatively closed in X R p, so that Ω is closed in R m R p if X is closed. Proposition 2.16. If Γ has the closed graph property at x 0 and it is locally bounded at x 0 X (there exists an open set N with x 0 N and Γ(X N) is a bounded set), then Γ is uhc at x 0. Proof. Suppose N as in the theorem, and let B(x 0, r) N. We argue by contradiction, supposing that Γ is not uhc at x 0. Then, according to Remark 2.3 there is an open set U Γ(x 0 ), a sequence x n X B(x 0, 1 ) and a sequence y n n Γ(x n ) for which y n / U for all n 1. However, y n Γ(x n ) Γ(B(x 0, 1 )) Γ(B(x n 0, 1)) for all n 1. Since Γ is locally bounded, Γ(B(x 0, 1)) is a bounded set. By the Bolzano Weierstrass Theorem, there is a subsequence y n(j) that converges to some y 0. But Γ has the closed graph property at x 0, thus given that x n x 0, y 0 Γ(x 0 ). On the other hand, y n(j) R p \U that is closed, hence y 0 R p \U or y 0 / U and hence y 0 / Γ(x 0 ), a contradiction. Corollary 2.17 (Compact graph uhc test). If the graph of Γ is compact, then Γ is uhc. Exercise 2.18. Study whether the correspondence { [0, 1/x], x > 0; Γ(x) = {0}, x = 0. is uhc/lhc or has the closed graph property. Is it closed, bounded, compact, convex valued? Proof. For points x > 0, Γ is cns by Example 2.10. At x = 0 is not uhc, but it is lhc. Since X = R + is closed in R, to have the closed graph property is equivalent to the graph of Γ to be closed in R 2. Obviously, the graph is not closed (the vertical segment y = 0 is no in the graph), hence Γ has no the closed graph property. It is closed, bounded, compact, convex valued. Exercise 2.19. Study whether the correspondence { [0, 1/x], x > 0; Γ(x) = R +, x = 0. is uhc/lhc or has the closed graph property. Is it closed, bounded, compact, convex valued? Is the graph of Γ convex? 21

Proof. For points x > 0, Γ is cns by Example 2.10. At x = 0 is both uhc, and lhc. It is uhc since for any U open with Γ(0) U we must have Γ(x) U for any x, since Γ(x) is compact valued. It is lhc at x = 0 since for any open set U with Γ(0) U, let y 0 + ɛ Γ(0) U. Let N = (0, δ) with 0 < δ < 1/(y 0 + ɛ). Then, for 0 < x < δ, 1/x > 1/δ > y 0 + ɛ, thus y 0 Γ(x) for any x R + N. It has the closed graph property, since the graph of Γ is closed in R 2. It is closed and convex valued, and bounded valued for x > 0. The graph of Ω is not convex, since the points (1, 1), (2, 1 2 ) are in Ω, but 1 2 (1, 1) + 1 2 (2, 1 2 ) = ( 3 2, 3 4 ) / Ω, since 3 4 (3/2) 1 = 2 3. Proposition 2.20. If f : X Y R m R p R is continuous and Y is closed and bounded, then the correspondence Γ defined as is uhc in its domain. Γ(x) = {y Y : f(x, y) 0} Proof. Notice that the set F = {(x, y) X Y : f(x, y) 0} is closed since f is continuous and F = f 1 ((, 0]) is the inverse image of the closed set 2 (, 0], Hence, the graph of Γ Ω = {(x, y) : x X, f(x, y) 0} = F (X Y ) is relatively closed in X Y and given that Y is closed, F is relatively closed in X R p. Thus Γ has the closed graph property. Since Γ(x) Y for all x, it is also locally bounded, thus it is uhc by Proposition 2.16. Example 2.21. Let X = [0, 1] and Y = [0, 2] and let f : X Y [0, 2] be the continuous function defined as { x + y 1, if 0 y 1; f(x, y) = xy, if 1 y 2. The associated correspondence Γ in Proposition 2.20 is uhc according to the above result. Its graph is 2 Y 1 Ω 0 1 X This example shows that Γ is not in general lhc. 2 Recall that f is continuous if and only if inverse images of closed (open) sets are closed (open). 22

Proposition 2.22. If f : X R m R is continuous and Y is compact, then the correspondence Γ defined as Γ(x) = {y Y : y f(x)} is continuous in its domain. 2.2 The Theorem of the Maximum Given the function f : X Y R m R p R and the correspondence Γ : X Y consider the problem v(x) = sup f(x, y). y Γ(x) Function v is the value function and the set valued map is the optimal correspondence. Γ (x) = {y Y : v(x) = f(x, y)} Proposition 2.23. Suppose that Γ is uhc with compact values and that K X is compact. Then, Γ(K) = x K Γ(x) is compact. Proof. Γ(K) is bounded: If not, there is x n K and y n Γ(x n ) such that y n n for all n. Since K is compact, there is a subsequence x n(j) x 0 K. Let U open and bounded such that Γ(x 0 ) U (this is possible since Γ(x 0 ) is bounded). Since Γ is uhc, Γ(x n(j) ) U for any j, thus for some M, n(j) y n(j) M for any j, which is absurd. Thus, Γ(K) is bounded. Γ(K) is closed: Let y n Γ(K) such that y n y 0. There are x n K such that y n Γ(x n ) and a subsequence x n(j) x 0 K. Let U open and bounded such that Γ(x 0 ) U (this is possible since Γ(x 0 ) is bounded). Since Γ is uhc, Γ(x n(j) ) U for any j, thus y n(j) is a bounded sequence. By the Bolzano Weierstrass Theorem, y n(j) y 0. Suppose y 0 / Γ(x 0 ). Then, since Γ(x 0 ) is compact, there is some open set W such that y 0 W and U W =. Hence y n(j) R p \W that is closed, thus the limit y 0 / W, a contradiction. Hence y 0 Γ(x 0 ) and since the limit is unique, y 0 Γ(x 0 ), thus y 0 Γ(K) and Γ(K) is therefore closed. Theorem 2.24 (Theorem of Berge or Theorem of the Maximum). Assume that v(x) < + for every x X. 1. If f is lsc and Γ is lhc, then v is lsc. 2. If f is usc and Γ is uhc with compact values, then v is usc. 3. If f is continuous and Γ is continuous with compact values, then v is continuous and Γ is uhc with compact values. Proof. 1. Let x 0 X and ɛ > 0. By definition of v(x 0 ), there exists y 0 Γ(x 0 ) such that v(x 0 ) f(x 0, y 0 ) + ɛ/2. Since f is lsc, there is δ > 0 such that f(x 0, y 0 ) < f(x, y) + ɛ 2 (2) 23

for any (x, y) (x 0, y 0 ) < δ. Let U = B(y 0, r). Since Γ is lhc, there is B(x 0, r ) such that Γ(x) U for any x X B(x 0, r ). Take r + r < δ. Then there is y Γ(x) such that (2) holds, so that f(x 0, y 0 ) < f(x, y) + ɛ 2 v(x) + ɛ 2, by the definition of v. Hence for any x x 0 < r and v is lsc. v(x 0 ) f(x 0, y 0 ) + ɛ/2 < v(x) + ɛ 2. Let us show that the upper sections S λ (v) are closed. Let x n S λ (v), x n x 0. Since f is usc and Γ(x n ) is compact for each n, there exists y n Γ(x n ) such that f(x n, y n ) = v(x n ) λ n. This means that (x n, y n ) S λ (f). Since x n x, the set K = (x n ) {x 0 } is compact, thus by Proposition 2.23 Γ(K) is also compact. Hence there is a subsequence y n(j) y 0 Γ(K). But y 0 Γ(x 0 ), because otherwise it is possible to find an open set W such that y 0 W and y n / W by the uhc of Γ, reaching a contradiction with the fact that the set R p \W is closed (see the proof of Proposition 2.23 above). Since S λ (f) is closed, (x n(j), y n(j) ) (x 0, y 0 ) S λ (f), thus f(x 0, y 0 ) λ. Since y 0 Γ(x 0 ) we have that is, x 0 S λ (v) and v is usc. v(x 0 ) f(x 0, y 0 ) λ, 3. By parts (a) and (b) v is continuous. On the other hand Γ (x) = {y : v(x) f(x, y) = 0} Γ(x) is the intersection of a closed set (because both v and f are continuous) and a compact set (because Γ has compact values), thus Γ (x) is compact. Let us show that Γ has the closed graph property. Let (x n ) x 0, (y n ) y 0 and y n Γ(x n ). Then y 0 Γ(x 0 ): otherwise, there is some open set W such that y 0 W and y n / W since Γ is uhc and Γ(x 0 ) is compact (this argumentation has been used several times above). Let us show that in fact y 0 Γ (x 0 ). Let z 0 Γ(x 0 ) arbitrary. Let the open set U j = B(z 0, 1/j). Since x n x 0 and Γ is lhc, for each j there exists n(j) such that x n(j) x 0 and z n(j) Γ(x n(j) ) with z n(j) U j, thus z n(j) z 0. Then and taking limits, since f is continuous f(x n(j), z n(j) ) f(x n(j), y n(j) ) f(x 0, z 0 ) f(x 0, y 0 ), thus y 0 Γ (x 0 ) and Γ has the closed graph property. Since Γ is also locally bounded, then it is uhc by Proposition 2.16. 24

Theorem 2.25. Suppose that X R m is convex, f : X R p R is continuous and concave and that Γ : X R p is a continuous set valued mapping with convex graph Ω and with compact values. Let v be the value function v(x) = max y Γ(x) f(x, y). Then we have 1. The value function v is concave and the optimal correspondence in convex valued. 2. If f(x, ) is strictly concave in y for every x X such that (x, y) Ω, then the optimal correspondence is single valued and it is continuous as a function. Proof. We now that v is well defined and continuous and that the optimal correspondence Γ exists and is uhc. 1. Let x 1, x 2 X and λ [0, 1]. We want to show v(λx 1 + (1 λ)x 2 ) λv(x 1 ) + (1 λ)v(x 2 ). Let y i Γ (x i ), i = 1, 2, and let y λ = λy 1 + (1 λ)y 2. Since Ω is convex we have (for x λ = λx 1 + (1 λ)x 2 ) that is, y λ Γ(x λ ), hence (x λ, y λ ) = λ(x 1, y 1 ) + (1 λ)(x 2, y 2 ) Ω, v(x λ ) f(x λ, y λ ) f concave λf(x 1, y 1 ) + (1 λ)f(x 2, y 2 ) = λv(x 1 ) + (1 λ)v(x 2 ), hence v is concave. moreover, it is well known that the maximum set of a concave function is convex, thus given that for any x, f(x, y) is concave in y, we have Γ (x) is convex. 2. it is well known that the maximum set of a strictly concave function is empty or single valued. Since Γ (x) is non empty, it is single valued and since it is uhc, it defines a continuous function. Example 2.26. Let the consumers problem max x 1 + x 2 s.t.: 0 2x 1 + 3x 2 R. What can you say about this problem without solving it explicitly? Proof. The utility function u(x 1, x 2 ) = x 1 + x 2 is continuous as it is the budget correspondence Γ : R + R 2 + defined by Γ(R) = {(x 1, x 2 ) R 2 + : 0 2x 1 + 3x 2 R}. Thus, according to the Theorem of Berge, the value function or indirect utility function depends continuously on R. Since R 1 < R 2 implies Γ(R 1 ) Γ(R 2 ), v(r 1 ) v(r 2 ), and since u is strictly increasing in both components, v is strictly increasing. Moreover, u is strictly concave in R 2 + and Ω is convex (why?), thus v is concave and the demand correspondences, x 1(R), x 2(R), are singletons and they are continuous functions of R. 25

2.3 Problems 1. For the following correspondences R R, study whether they are continuous and with compact or convex values. Study whether the graph is closed or convex. (a) Γ 1 (x) = [0, 1) if x = 0; Γ 1 (x) = {0} otherwise. (b) Γ 2 (x) = [0, 1] if x = 0; Γ 1 (x) = {0} otherwise. (c) Γ 3 (x) = {0} if x = 0; Γ 1 (x) = [0, 1] otherwise. Solution: (a) It is continuous at any x 0; at 0 it is uhc but not lhc. It has convex values but not compact values, since Γ(0) is not compact. The graph is not closed, not convex. (b) It is continuous at any x 0; at 0 it is uhc but not lhc. It has convex and compact values. The graph is closed, not convex. (c) It is continuous at any x 0; at 0 it is lhc but not uhc. It has convex and compact values. The graph is not closed (the sequence (1/n, 1) is in the graph and converges to (0, 1) as n, which is not in the graph), not convex. 2. Let Γ : [0, 1] [0, 1] be defined by Γ(x) = { x } [, if x 2 [ x 2, x ], if x 0, 1 ] ; 2 ( ] 1 2, 1. Determine if Γ is uhc and/or lhc. Does it have compact and/or convex sections? Does it have a convex graph? Solution: It is uhc and lhc for x 1. At 1 is lhc but not uhc. It has compact and convex 2 2 values (or sections), but does not have a convex graph. Ω 0 1 1 2 26

3. Let Γ 1, Γ 2 : R R be two lhc (uhc) correspondences. Show that the correspondence Γ defined by Γ(x) = Γ 1 (x) Γ 2 (x) is also lhc (uhc). Solution: We only prove the lhc case. Let x 0 R and let U be an open set such that Γ(x 0 ) U. Then Γ 1 (x 0 ) U or Γ 2 (x 0 ) U. Suppose, without loss of generality, that Γ 1 (x 0 ) U. Since Γ 1 is lhc, there is an open set N 1 with x 0 N 1 and such that for any x N 1, Γ 1 (x) U, thus (Γ 1 (x) Γ 2 (x)) U, showing that Γ is lhc. 4. Show that if Γ 1, Γ 2 : X R p are continuous and Γ 1 (x) Γ 2 (x) for every x X, then the correspondence given by Γ(x) = Γ 1 (x) Γ 2 (x) is also continuous. Solution: Let x 0 X and let U be open with U Γ(x 0 ). Let U i open such that Γ i (x 0 ) U i, i = 1, 2 and U 1 U 2 U. There are open sets N 1 x 0 and N 2 x 0 such that Γ 1 (X N 1 ) U 1 and Γ 2 (X N 2 ) U 2. Hence, for N = N 1 N 2 which is open, we have Γ(X N) U 1 U 2 U, hence Γ is uhc. Now, let U Γ(x 0 ). Let U i be open such that Γ i (x 0 ) U i, i = 1, 2 and U 1 U 2 U. The there are open sets N i x 0 such that Γ i (X N i ) U i i = 1, 2. Hence, for N = N 1 N 2 which is open, we have Γ(X N) U, hence Γ is lhc. 5. (Sequential characterization of lower hemi continuity). Let Γ : X R m R p be a correspondence and let x 0 X. Prove: Γ is lhc at x 0 if and only if [ ] y 0 Γ(x 0 ) x n X with x n x 0, n 0 N and y n Γ(x n ) for n n 0 with y n y 0. Solution: ] Let y 0 Γ(x 0 ) and let x n X with x n x 0. Let B j = B(y 0, 1/j) for j 1. Since Γ is lhc at x 0, ɛn j open, x 0 N j, such that x N j X, B j Γ(x). Since x n x 0, j n j such that x n N j for n j. One can take n 1 < n 2 < < n j <. By the lhc of Γ at x 0 y j n B j Γ(x n ) for n = n j, n j + 1,.... Let now y n = y j n if n j n < n j+1. Then y n Γ(x n ) and y n y 0, since n implies j. ] By contradiction, if Γ is not lhc at x 0, then y 0 Γ(x 0 ) and an open set U with U Γ(x 0 ) such that n x n B(x 0, 1/n) and U Γ(x n ) =. Let y 0 Γ(x 0 ). Since x n x 0, there exists n 0 and y n Γ(x n ) for n n 0 such that y n y 0, so there exists 27

n 1 such that y n U for any n n 1. Then, for n max{n 0, n 1 } we get y n U Γ(x n ) with x n B(x 0, 1/n), that contradicts U Γ(x n ) =. 6. (Sequential characterization of upper hemi continuity). Let Γ : X R m R p be a correspondence with graph Ω and let x 0 X. Prove: Γ is uhc at x 0 if [ ] (x n, y n ) Ω with x n x 0, n(j), y n(j) y 0 Γ(x 0 ). The converse is also true if Γ is compact valued. Solution: We argue by contradiction supposing that Γ is not uhc at x 0. Then U open Γ(x 0 ) U, x n B(x 0, 1/n) and y n Γ(x n ) such that y n / U. Since x n x 0, by [**] there exists a subsequence (y n(j) ) of (y n ) converging to y 0 Γ(x 0 ). But y n(j) R m \U which is closed, thus y 0 / U, contradicting Γ(x 0 ) U. To show the converse, let x 0 X and a sequence x n X with x n x. The set K = (x n ) {x 0 } is compact, thus the image of K under Γ, Γ(K), is also compact since Γ is uhc and compact valued (Proposition 2.18). Let y n Γ(x n ), so that (y n ) Γ(K) and hence (y n(j) ) y 0 Γ(K) as j. In fact y 0 Γ(x 0 ): if not, it is possible to find open sets U and W with U W = and Γ(x 0 ) U, y 0 W. Since Γ is uhc at x 0, N open, x 0 N Γ(x) U x N X. Since x n x 0, n 0 such that x n N X n n 0, thus Γ(x n ) U n n 0 and hence y n(j) U, y n(j) / W for j large enough. Since the set R p \W is closed, y 0 R p \W and thus y 0 / W, a contradiction. 7. Let Γ : X R m R p be a correspondence with an open graph Ω. Show that Γ is lhc. Solution: Let x 0 X and let x n X with x n x. Let y 0 Γ(x 0 ). Since Ω is open, there exists n 0 N such that for all n n 0 the ball B((x 0, y 0 ), 1/n) Ω. Since x n x, there exists n 0 N and y n Γ(x n ) such that (x n, y n ) B((x 0, y 0 ), 1/n) for all n n 0. Then y n y 0 and thus Γ is lhc at x 0 in virtue of the sequential characterization of lower hemi continuity of correspondences. 8. (Budget correspondence) Consider m commodities x R m ++ with prices p R m ++. A consumer receives income I > 0. Let for (p, I) the set of affordable commodity vectors { } m B(p, I) = x R m ++ : p x p i x i I. The correspondence B : R m+1 ++ R m ++ is the budget correspondence. Show that B is continuous. Solution: 28 i=1