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Foundations of Mathematical Economics 1st edition Michael Carter - Solutions
Let y be a boundary point of a nonempty, convex set S in a Euclidean space X. There exists a supporting hyperplane at y; that is, there exists a continuous linear functional f ∈ X* such that f (y) ≤ f (x) for every x ∈ S
Generalize exercise 3.182 to dispense with the assumption that S is closed. That is, let S be a nonempty, convex set in a Euclidean space X and y ∉ S. There exists a continuous linear functional f ∈ X* such that f (y) ≤ f (x). for every x ∈ S
Let S be an open convex subset of a linear space X and f ∈ X' a nonzero linear functional on X. Then f (S) is an open interval in ℜ.
Prove theorem 3.2 (assuming that X is Euclidean). [Hint: Apply exercise 3.184 to separate 0 from S = B + (- A) Then use exercise 3.185.]
Let Hf (c) be a bounding hyperplane of a cone C in a normed linear space X, that is, f (x) ≤ c for every x ∈ C. Then f (x) ≥ 0 for every x ∈ C
A linear function f: X → Y is nonsingular if and only if kernel f = {0} and f(X) = Y.
Let Hf (c) be a bounding hyperplane of a subspace Z of a normed linear space X, that is, f (x) ≤ c for every x ∈ Z. Then Z is contained in the kernel of f, that is, f (x) = 0 for every x ∈ Z
Prove corollary 3.2.3. Separation theorems are so pervasive in mathematical economics that it is necessary to have a range of variations in the armory. In the following sections we develop some refinements of the basic separating hyperplane theorem that are useful in applications.
Let A and B be nonempty, convex subsets in a normed linear space X with int A and int AB . Then A and B can be properly separated.Frequently stronger forms of separation are required. Two sets A and B are strictly separated by a hyperplane Hf. if A
Prove proposition 3.14 for a finite-dimensional space X. [Hint: Apply exercise 3.182 to the set B - A. Compactness of A is necessary to ensure that B - A is closed.] The following exercise shows that compactness of A is essential in proposition 3.14.
1. Let A and B be nonempty, disjoint, convex subsets in a normed linear space X. A and B can be strongly separated if and only if there exists a convex neighborhood of U of 0 such that (A + U) ⋂B = ∅ 2. Prove proposition 3.14.
Let A and B be convex subsets in a finite-dimensional normed linear space X. A and B can be strongly separated if and only if P(A, B) = inf{||x - y|| : x ∈ A, y ∈ B} > 0 Combining proposition 3.14 with corollary 3.2.3 gives the following important result, a geometric form of the Hahn-Banach
Let M be a nonempty, closed, subspace of a linear space X and y ∉ M. Then there exists a continuous linear functional f ∈ X* such that f (y) > 0 and f (x) = 0 for every x ∈ As an application of the previous result, we use it in the following exercise to provide an alternative derivation of
Let g1; g2; . . . ; gm be linear functional on a linear space X, and letS ={x X : gj(x) = 0, j = 1,2,..., m}=Suppose that f ^ 0 is another linear functional such that such that f (x)= 0 for every x e S. Show that Figure 3.20 The Fredholm alternative via separation 1. The set Z . {f} x,
Show that the set L(X, Y) of all linear functions X → Y is a linear space.
The inverse of a (nonsingular) linear function is linear.
Show the converse; that is, if f (x) =then f .(x). 0 for every x S where S . {x X : gj(x) = 0 = j . 1; 2 . . .m}.
Let g1, g, . . . , gm be linear functionals on a linear space X. For fixed numbers cj, the systems of equationsgj(x) = cj, j = 1, 2, . . . ,mis consistent if and only iffor every set of numbers λ1, λ2, . . . , λm.
Let K be a closed convex cone in a finite-dimensional linear space X and M a subspace with K ∩ M = {0}. Then there exists a linear functional f ∈ X* such that f(x) > 0 for every x ∈ K\{0} and f(x) = 0 for every x ∈ M
Suppose that f0 is a linear functional defined on a subspace Z of X such thatf0(x) ¤ g(x) for every x Zwhere g X* is convex. Show that1. The setsand are convex subsets of the linear space Y = X Ã (figure 3.21) 2. int A
Let f0 be a bounded linear functional on a subspace Z of a normed linear space X. Then f0 can be extended to a linear functional on the whole space X without increasing its norm, that is,
Let x0 be an element of a normed linear space X. There exists a linear functional f ∈ X* such that ||f|| = 1 and f(x0) = ||x0||.
Let x1, x2 be distinct points in a normed linear space X. There exists a continuous linear functional f X* that evaluates them differently, that is, such that
Let S be a nonempty compact convex subset of a normed linear space X. 1. Let
Let S be a compact convex set. Every supporting hyperplane to S contains an extreme point of S.
Every compact, convex set is the closed convex hull of its extreme points.
If f, g are nonsingular linear functions, then so is their composition g ∘ f with (g ∘ f)-1 = f-1 ∘ g-1 The following converse of 3.14 is the linear version of the important implicit function theorem (theorem 4.5).
Let {S1, S2, . . . , Sn} be a collection of nonempty compact subsets of an m-dimensional linear space, and let x convWe consider the Cartesian product of the convex hulls of Si, namely Every point in P is an n-tuple (x1, x2, . . . , xn) where each xi belongs to the corresponding conv
Let S1 = {0, l}, S2 = {2, 3}, and x = 2.5. Illustrate the sets conv Si, P, and P(x).
Extend exercise 3.210 to allow for noncompact Si, thus proving the Shapley-Folkman theorem.Exercise 3.210Let {S1, S2, . . . , Sn} be a collection of nonempty compact subsets of an m-dimensional linear space, and let x convWe consider the Cartesian product of the convex hulls of Si,
Prove proposition 3.17.Proposition 3.17 (Minkowski's theorem)A closed, convex set in a normed linear space is the intersection of the closed halfspaces that contain it.Minkowski's theorem is illustrated in figure 3.22.
Show that 1. V*(y) is a closed convex set containing V(y) 2. If V(y) is convex and monotonic, then V*(y) = V(y)
Assume that the cost function for a convex, monotonic technology V(y) is linear in y, that is, c(w, y) = (w)y. Then the technology exhibits constant returns to scale.
Every polyhedral set is closed and convex.
Let A be an m × n matrix. The set of solutions to the system of linear inequalities Ax ≤ c is a polyhedron in ℜn.
Show that a polytope can be defined alternatively as • The convex hull of a finite set of points • A nonempty compact polyhedral set That is, show the equivalence of these two definitions.
If f: X → Y and h: X → Z are linear functions with kernel f ⊆ kernel h, then there exists a linear function g: f X → Z such that h = g ∘ f.
Let S be a nonempty set in a normed linear space X. Show 1. S* is a closed convex cone in X* 2. S** is a closed convex cone in X 3. S ⊆ S**
Let S1, S2 be nonempty sets in a normed linear space.Whatever the nature of the set S, S* and S** are always cones (in X* and X respectively) and S S**. Under what conditions is S identical to its bipolar S** (rather than a proper subset)? Clearly, a necessary condition is that S is a
Let S be a nonempty convex cone in a normed linear space X. Then S = S** if and only if S is closed.
Use proposition 3.14 directly to prove the Farkas lemma when X is a Hilbert space.
Let A be an m × n matrix. A necessary and sufficient condition for the system of linear equations AT y = c to have a nonnegative solution y ∈ ℜm+ is that cTx ≤ 0 for every x ∈ ℜn satisfying Ax ≤ 0.
Let S be convex set in ℜn which contains no interior points of the nonnegative orthant ℜn+. Then there exists a hyperplane with nonnegative normal p ≩ 0 (that is p ≠ 0, pi ≥ 0) such that pTx ≤ 0 for every x ∈ S.
Suppose that the production possibility set Y is convex. Then every efficient production plan y ∈ Y is profit maximizing for some nonzero price system p ≥ 0.
Let S be convex set in ℜn that contains no interior points of the non positive orthant ℜn-. Then there exists a hyperplane with nonnegative normal p ≩ 0 such that pTx ≥ 0 for every x ∈ S
Suppose that x* = (x*1, x*2,..., x*n) is a Pareto efficient allocation in an exchange economy with l commodities and n consumers (example 1.117). Assume that¢ Individual preferences are convex, continuous and strongly monotonic.¢ x*Show that 1. The set is the set of all
Suppose that x* = (x*1, x*2,..., x*n) is a Pareto-efficient allocation in an exchange economy (example 1.117) in which each of the n consumers has an endowment wi l+ of the l commodities. Assume that¢ Individual preferences ¿i are convex, continuous
Suppose that f: X → Y is a linear function and B ⊆ X is a basis for X. Then f (B) spans f (X).
Let K be a closed convex cone that intersects the nonnegative orthant at 0, that is, K ∩ ℜn+ = {0}. Then there exists a hyperplane with positive normal p > 0 (i.e., pi > 0 for every i) such that pTx ≤ 0 for every x ∈ K.
There is no arbitrage if and only if there exist state prices.
Let A be an m × n matrix and c be a nonzero vector in ℜn. Then exactly one of the following systems has a nonnegative solution: Either I Ax ≤ 0 and cx > 0 for some x ∈ Rn+ or II ATy ≥ c for some y ∈ Rm+. Still other variants can be derived by creative reformulations of the alternative
Let A be an m × n matrix and c be a nonzero vector in ℜn. Then exactly one of the following systems has a solution: Either I Ax = 0 and cTx > 0 for some x ∈ ℜn+ or II ATy ≥ c for some y ∈ ℜm.
Let g1, g2, ..., gm be linear functionals on n. For fixed numbers c1, c2,..., cm, the system of inequalitiesgj(x) ¥ cj, j = 1, 2,..., m (31)is consistent for some x n if and only ifimplies that for every set of nonnegative numbers λ1,
Let A be an m × n matrix and c be a nonzero vector in ℜn. Then exactly one of the following systems has a solution: Either I Ax = 0 and cx > 0 or II ATy = c
Derive the Fredholm alternative (exercise 3.237) directly from exercise 3.199.
Let A be an m × n matrix. Then exactly one of the following systems as a solution: Either I Ax > 0 for some x ∈ ℜn or II ATy = 0, y ≩ 0 for some y ∈ ℜm
Suppose that f: X → Y is a linear function. If X is finite-dimensional, then rank f + nullity f = dim X
Derive Gordan's theorem from the Farkas alternative. [Apply the Farkas alternative to the matrix B = (A, 1), the matrix B augmented with a column of ones.]
Let S be a subspace in ℜn and S⊥ its orthogonal complement. Either I S contains a positive vector y > 0 or II S⊥ contains a nonnegative vector y ≩ 0
Let A be an m × n matrix. Then exactly one of the following systems has a solution: Either I Ax ≩ 0 for some x ∈ ℜn or II ATy = 0, y > 0 for some y ∈ ℜm++
Deduce Stiemke's theorem directly from exercise 3.230.
Let A be an m × n matrix. Then exactly one of the following systems has a nonnegative solution: Either I Ax > 0, x > 0 for some x ∈ ℜn or II ATy ≤ 0, y ≩ 0 for some y ∈ ℜm.
Let A be an m × n matrix. Then exactly one of the following systems has a nonnegative solution: Either I ATx > 0, x ≩ 0 for some x ∈ ℜm or II Ay ≤ 0, y ≩ 0 for some y ∈ ℜn.
Assume the system Ax ≩ 0 has a solution, while Ax > 0 has no solution. Then A can be decomposed into two consistent subsystems Bx > 0 and Cx = 0 such that Cx ≩ 0 has no solution.
Let A be an m × n matrix. The dual pair Ax ≥ 0 and ATy = 0; y ≥ 0
Let A be an m × n matrix. The dual pair Ax ≥ 0; x ≥ 0 and ATy ≤ 0; y ≥ 0
Show that Gordan's and Stiemke's theorems are special cases of Tucker's theorem.
Derive the Farkas lemma from Tucker's theorem.
Let A, B, and C be matrices of order m1 × n, m2 × n and m3 × n respectively with A nonvacuous. Then either
Let S be a nonempty convex set in n with 0 S.1. For every point a S, define the polar set S*a = {x n : ||x|| = 1; xTa ¥ 0}. S*a is a nonempty closed set.2. For any finite set of points {a1, a2, ..., am} in S, let A be the m
In any finite two person zero-sum game, at least one of the players has a mixed strategy that guarantees she cannot lose. That is, either v1(p) ≥ 0 or v2(q) ≤ 0.
Prove proposition 3.20 by showing 1. For every c ∈ ℜ, either v1 ≥ c or v2 ≤ c 2. v1 = v2
Let v denote the value of a two person zero-sum game. Show that 1. Player 1 has an optimal strategy p* that achieves the value v. Similarly player 2 has an optimal strategy q*. 2. The strategy pair (p*, q*) constitutes a Nash equilibrium of the game.
In the context of the previous example, prove that v1(p) Previous exampleConsider a zero-sum game in which player 1 has two strategies {s1, s2}, while player 2 has 5 strategies {t1, t2, t3, t4, t5}. The payoffs to player 1 are
Let A be a m à n matrix which represents (exercise 3.253) the payoff function of a two-person zero-sum game in which player 1 has m pure strategies and player 2 has n strategies. Let Z be the convex hull of the columns of A, that is, Z = {z = Aq : q În-1}.
Prove the minimax theorem by extending the previous exercise to an arbitrary two-person zero-sum game with v2 = c ≠ 0. Previous exercise Let A be a m × n matrix which represents (exercise 3.253) the payoff function of a two-person zero-sum game in which player 1 has m pure strategies and player
Assuming that X = ℜn and Y = ℜm, prove proposition 3.1 for the standard basis (example 1.79).
Show that the set of optimal strategies for each player is convex.
Let f be a bilinear functional on the product of two simplices, that is f: Îm à În R. Then
A classic example in coalitional game theory is the three-person majority game, in which the allocation of $1 among three persons is decided by majority vote. Let N = {1, 2, 3}. The characteristic function is w({i}) = 0, i = 1, 2, 3 w({i, j}) = 1, i, j ∈ N, i ≠ j w(N) = 1 1. Show that the
Show that cohesivity is necessary, but not sufficient, for the existence of a core.A partition {S1, S2,..., SK} is a family of coalitions in which each player I belongs to one and only one of the coalitions. A balanced family of coalitions B is a set of coalitions {S1, S2,..., Sk} together with a
List three other balanced families of coalitions for the three-player game.
Find a nontrivial (i.e., not a partition) balanced family of coalitions for the four-player game with N = {1, 2, 3, 4}.
A collection B of coalitions is balanced if and only if there exists positive weights {λS : S B} such thatwhere eS is the characteristic vector of the coalition S (example 3.19) and gS: X represents the share of coalition S at the allocation x
Why can we normalize so that μ = 1 in the previous proof?
The set of balanced games forms a convex cone in the set of all TP-coalitional games GN.
To construct a suitable space for the application of a separation argument, consider the set of pointswhere eS is characteristic vector of the coalition S (example 3.19) and w(S) is its worth. Let A be the conic hull of A0, that is, Let B be the interval Clearly, A and B are convex and nonempty. We
A TP-coalitional game (N, w) is called 0-1 normalized ifw({i}) = 0 for every i N and w(N) = 1Show that1. To every essential game (N, w) there is a corresponding 0-1 normalized game (N, w0)2. Core (N, w) = α core (N, w0) + w, where w = (w1, w2,..., wn), wi = w({i}),3.
Let (N, w) be a 0-1 normalized TP-coalitional game, and let A be the set of all nontrivial coalitions A = {S N : w(S) > 0}. Consider the following two-player zero-sum game. Player 1 chooses a player i N and player 2 chooses a coalition S A. The payoff (from 2
Let (N, w) be a 0-1 normalized TP-coalitional game, and let G be the corresponding two-person zero-sum game described in the previous exercise with value δ. Show that δ ¥ 1 if (N, w) is balanced.Previous exerciseLet (N, w) be a 0-1 normalized TP-coalitional game,
Core(N, w) (N, w) balanced.Combining the three previous exercises establishes the cycle of equivalences(N, w) balanced δ ¥ 1 (core(N, w) ) (N, w) balanced and establishes
Describe the matrix representing the Shapley value. Specify the matrix for three-player games (|N|) = 3). If f: X → Y is a nonsingular linear function with matrix representation A, then the representation of the inverse function f -1 with respect to the same bases is called the matrix inverse of
A linear function f: X → Y is continuous if and only if it is continuous at 0. A linear function f: X → Y is bounded if there exists a constant M such that || f(x) || ≤ M ||x|| for every x ∊ X Note that boundedness does not imply that the range f (X) is bounded but rather that f (S) is
Show that f in example 3.1 is linear. The function f : ℜ2 → ℜ2 defined by F( x1, x2) = (x1 cos θ - x2 sin θ ,x1 sin θ + x2 cos θ), 0 ≤ 0 2π rotates any vector in the plane counter clock wise through the angle θ (figure 2.3). It is easily verified that f is linear. Linearity implies
A linear function f: X → Y is continuous if and only if it is bounded. Fortunately every linear function on a finite-dimensional space is bounded and therefore continuous.
A linear function f: X Y is bounded if X has finite dimension.Rewriting (4), we have that a linear function f is bounded if there exists a constant MFor every x X The smallest constant M satisfying (5) is called the norm of f. It is given by Clearly,
If f is a bounded linear function, an equivalent definition of the least upper bound is
The space BL(X, Y) of all bounded linear functions from X to Y is a normed linear space, with normIt is a Banach space (complete normed linear space) if Y is complete. The following proposition is an important result regarding bounded linear functions.
Prove proposition 3.2 assuming that X is finite-dimensional as follows: Let B be the unit ball in X, that is, B = {x: ||x|| < 1} The boundary of B is the unit sphere S = {x: ||x|| < 1}. Show that 1. f (S) is a compact subset of Y which does not contain 0Y. 2. There exists an open ball T ⊆ (f
Let X be a finite-dimensional normed linear space, and {x1; x2,..., xn} any basis for X. The function f: n X defined byis a linear homeomorphism (remark 2.12). That is, ¢ f is linear ¢ f is one-to-one and onto ¢ f and f-1 are continuous
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