Math Methods for Economicsand Microeconomic Theory

Cesar E. TamayoDepartment of Economics, Rutgers University

ctamayo@econ.rutgers.edu

Class notes: fall 2010

Contents

I Math Methods 4

1 Topology and analysis 51.1 Metric spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 Series, sequences and subsequences . . . . . . . . . . . . . . . . . . . . . . 61.3 Open and closed sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Completeness, boundedness and compactness . . . . . . . . . . . . . . . . . 91.5 Continuity of functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.6 Continuity of correspondences (set-valued mappings) . . . . . . . . . . . . 141.7 Maximum theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161.8 Fixed point theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2 Convex optimization 212.1 Convex sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212.2 Separating hyperplanes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222.3 Convexity of functions and subgradients . . . . . . . . . . . . . . . . . . . 242.4 Support function theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292.5 Kuhn-Tucker theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

2.5.1 Introduction to nonlinear programming . . . . . . . . . . . . . . . . 322.5.2 The Kuhn-Tucker conditions . . . . . . . . . . . . . . . . . . . . . . 322.5.3 Constraint qualications . . . . . . . . . . . . . . . . . . . . . . . . 352.5.4 Non-negativity and equality constraints . . . . . . . . . . . . . . . . 362.5.5 Su ciency conditions . . . . . . . . . . . . . . . . . . . . . . . . . . 38

II Microecomic Theory 40

3 Producer and consumer theory 413.1 Producer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

3.1.1 Production sets and technologies . . . . . . . . . . . . . . . . . . . 413.1.2 Cost minimization . . . . . . . . . . . . . . . . . . . . . . . . . . . 433.1.3 Prot maximization . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.2 Consumer theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473.2.1 Utility maximization . . . . . . . . . . . . . . . . . . . . . . . . . . 47

1

3.2.2 Expenditure minimization . . . . . . . . . . . . . . . . . . . . . . . 49

4 Game Theory and General Equilibrium 514.1 Game theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

4.1.1 Zero-sum games . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 514.1.2 Non-zero-sum games and Nash Equilibrium . . . . . . . . . . . . . 514.1.3 The generalized game . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4.2 General Equilibrium theory . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2.1 Prelminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.2.2 Existence of equilibrium in pure exchange economies . . . . . . . . 564.2.3 Existence of equilibrium in production economies . . . . . . . . . . 584.2.4 Welfare theorems (pure exchange economies) . . . . . . . . . . . . . 614.2.5 Relaxing the Walrasian assumptions . . . . . . . . . . . . . . . . . 66

5 Decision making under uncertainty 705.1 Expected utility hypothesis . . . . . . . . . . . . . . . . . . . . . . . . . . 705.2 Risk aversion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71

5.2.1 Application: portfolio choice . . . . . . . . . . . . . . . . . . . . . . 725.3 Comparative risk aversion . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5.3.1 Application: portfolio choice . . . . . . . . . . . . . . . . . . . . . . 765.3.2 Application: insurance . . . . . . . . . . . . . . . . . . . . . . . . . 76

5.4 First order stochastic dominance (FOSD) . . . . . . . . . . . . . . . . . . . 795.4.1 FOSD and precautionary savings . . . . . . . . . . . . . . . . . . . 815.4.2 FOSD and portfolio choice . . . . . . . . . . . . . . . . . . . . . . . 83

5.5 Likelihood ratio stochastic dominance . . . . . . . . . . . . . . . . . . . . . 845.6 Concave and second order stochastic dominance (SOSD) . . . . . . . . . . 85

5.6.1 Concave order SD and prot maximization . . . . . . . . . . . . . . 885.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

A Review of functions, dierentiation and integration 90

B Review of vectors and matrix algebra 93

2

Summary

These notes cover the rst semester mathematics and microeconomics material of thePhD program at Rutgers University. The range of mathematical tools presented belowis wide, including topology, real analysis, convex optimization, xed point theory andstochastic dominance, but attention is limited to their use in economic theory. The notesare almost entirely based on lectures by Professor Richard P. McLean, but all errorsand omissions are my sole responsability. Ocasionally, I also refer to examples anddenitions found in Carters (2001) "Foundations of Mathematical Economics", Mas-Colellet. als (1995) Microeconomic Theory and Varians (1992) Microeconomic Analysis.

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Part I

Math Methods

4

Chapter 1

Topology and analysis

1.1 Metric spaces

Denition 1 Let X be a non-? set. A metric on X is a function d : X X! R satisfyingthe following 8 x; y; z 2 X :

1:d (x; y) 02:d (x; y) = d (y; x)3:d (x; y) = 0, x = y4:d (x; y) d (x; z) + d (z; y)

For the case when X = Rn we have:

d1 (x;y) =nXi=1

jxi yij

d2 (x;y) =

vuut nXi=1

(xi yi)2

...

d1 (x;y) = maxi=1:::n

fjxi yijgDenition 2 A metric space is a pair (X; d) where X is a non-? set and d is a metric onX:

Denition 3 Let (X; d) and (Y; ) be metric spaces. The Box metric on X Y is ((x^; y^); (x; y)) =maxfd(x; x^); (y; y^)gDenition 4 The point-to-set distance d(x;A) = miny2A d(x; y)

Notation 1 The symbol B"(x) denotes the open ball of radius " centered at x:

Denition 5 Given A;B compact (see denitions 25-26 below), the Hausdor distanceis: Hd(A;B) = maxfmaxx2A d(x;B);maxx2B d(x;A)g: Note that each max function is welldened since A;B are compact (see Theorem 25 below)

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1.2 Series, sequences and subsequences

Denition 6 A sequence in X is a function f : N! X (a 1-to-1 mapping from N to X).

Denition 7 A sequence is called convergent with limit x if, for every " > 0; 9 N s.t.whenever n > N; xn 2 B"(x).

Lemma 1 A convergent sequence fxng in a metric space (X; d) has exactly one limit.

Proof. Suppose that xn ! x1 and xn ! x2. Then we must show that d (x1; x2) = 0.To see this, note that d (x1; x2) = 0 , d (x1; x2) < " 8 " > 0. Choose " > 0. Sincelimn!1 fxng = x1 we know that 9 n^1 such that n > n^1 ) xn 2 B "

2(x1). Likewise,

since limn!1 fxng = x2 we know that 9 n^2 such that n > n^2 ) xn 2 B "2(x2). So

choose m = max fn^1; n^2g and note that xm 2 B "2(x1) and xm 2 B "

2(x2); to complete the

argument, use the triangle inequality to conclude:

d (x1; x2) d (x1; xm) + d (x1; xm) 0 and n^ such that n^ > 1=".Now, n > n^ implies:

d (0; xn) = j0 xnj = 1n 0; n > n^) d (xn; 1) < " or d (xn;1) < ":

Denition 8 fxng1n=1 is a Cauchy Sequence if for any " > 0; 9 N s:t: 8 i; j N; d(xi; xj) 0 and choose K such that K > 1

". Then

k > K ) xk 2 B 1k(x). Also, k > K ) 1

K> 1

kso that:

B 1k(x) B 1

K(x) B"(x)

) fxkg is convergent with limit x

Lemma 3 (sandwich I) If yn ! !, zn ! ! and yn xn zn; 8 n then xn ! !.

Denition 10 If xk; x 2 Rn we say that xk ! x() xk is pairwise convergent to x inR1 that is, if. x1k ! x1; x2k ! x2:::

1.3 Open and closed sets

Denition 11 Let (X; d) be a metric space (m:s:) and let S X. If, for x 2 X; 9 " >0 s:t: B"(x) S, then x is an interior point of S. The set of all interior points of S isintS.

Denition 12 x is NOT an interior point of S if B"(x) \ X n S 6= 0;8 " > 0.

This esures that an (n 1)-dimensional subset of Rn cannot be an open set since an-dimesional ball around any of its points would contain points outside the subset.

Denition 13 Let (X; d) be a metric space (m:s:):S X.is an open set if intS = S:

Denition 14 Suppose that (X; d) is a metric space, x 2 X and " > 0. Then an open ballcentered at x with radius " is the set:

B" (x) = fy 2 X j d (x; y) < "g

7

Example 5 Let X = R1 and d1 (x; y) above, the open ball aroun x is:

B" (x) =y 2 R1 j d1 (x; y) < "

=

y 2 R1 j x " < y < x+ "

Example 6 Let X = R2 and d2 (x; y) ; the open ball aroun x is:

B" (x) =y 2 R2 j d2 (x; y) < "

=

y 2 R2 j

q(x1 y1)2 + (x2 y2)2 < "

Denition 15 Two metrics d; d0 are equivalent if an open set in (X; d) is open in (X; d0),that is, if for x 2 X, B"d(x) B"d0 (x) and vice versa.Example 7 If S = ? then S is open. If S = X then S is open in (X; d):

Example 8 Let X = R1 and d1 (x; y). Let S = (0; 1). We can claim that S is an open set.To see this, choose any x 2 S and " = min f1 x; xg : Then: B"(x) (0; 1). To see whychoose any y 2 B"(x) and note:

y 2 B"(x)) jx yj < ") " < y x < " (denition of abs value)) x " < y < x+ ") x x < x " < y < x+ 1 x (by def of ")) 0 < y < 1

) y 2 B"(x)) y 2 S = (0; 1)) B"(x) S = (0; 1)) S is open

Example 9 Let X = R1 and d1 (x; y). Let S = Q =rational numbers.Then intS = ?:

Remark 3 Between any two real numbers, there are at least one rational and one irrationalnumber.

Proposition 1 Every open ball is an open set.

Proof. Let (X; d) be a metric space. Choose x 2 X and pick r > 0. Next, choose y 2 Br(x).Now let " = r d (x; y) and note that " > 0 since r d (x; y) > 0: We must show that anypoint in B"(y) is also a point in Br(x). To do this, choose z 2 B"(y). Then:

d (z; x) d (z; y) + d (y; x)< "+ d (y; x)

= [r d (x; y)] + d (y; x)= r

) d (z; x) < r) z 2 B"(y)) z 2 Br(x)) B"(y) Br(x)

8

Denition 16 x is a closure point of S , B"(x) \ S 6= ? 8 " > 0: In turn, x is NOT aclosure point of S , B"(x) Sc 8 " > 0.

Note that it is always true that: S clS, since x 2 S ) x 2 B"(x) \ S, thus,B"(x) \ S 6= ;, hence, every x 2 S is a closure point of S.

Denition 17 S is closed , clS = S, which by the remark above implies that S is closed, clS S.

Example 10 S = X is closed in (X; d). S = ? is closed.

Example 11 Let X = R1 and S = f0; 1g. Then nothing outside [0; 1] is a closure point.To se why, note that there will always 9 " > 0 su ciently small such that B"(x)\S = ? forx 2 (1; 0)[(1;1). Likewise, nothing in (0; 1) is a closure point of S. Thus, clS = f0; 1g.) S is closed.

Denition 18 (sequential characterization of closure point) x is a closure point of S if9 fxkg s:t: xk 2 S 8 k and fxkg ! x.

Denition 19 (sequential characterization of closed set) S in Rm is a closed set if when-ever fxng is a convergent sequence completely contained in S, it follows that limfxng 2 S.

Denition 20 Discrete metric: TBC

Remark 4 The intersection of a nite collection of open sets is an open set.

Remark 5 The union of any collection of open sets is an open set.

Remark 6 The intersection of any collection of closed sets is a closed set.

Remark 7 The union of a nite collection of closed sets is a closed set.

Denition 21 x is a boundary point of S (denoted x 2 @S) if 8 " > 0; B"(x)\S 6= ? andB"(x) \ Sc 6= ? (i.e., if x 2 clS but x =2 intS).

1.4 Completeness, boundedness and compactness

Denition 22 A m.s.(X; d) is complete if any cauchy sequence in (X; d) is convergent withlimit 2 X.

9

Example 12 Let X = [0; 1) and d =Euclidean. Claim: "(X; d) is not complete". To seewhy consider the sequence xk = kk+1 and note that fxkg is a Cauchy sequence. This canbe shown by choosing " > 0 and noting that 9 N such that k > N ) d (xk; 1) < "=2 andm > N ) d (xm; 1) < "=2; thus, by the triangle inequality:

d (xk; xm) d (xm; 1) + d (xk; 1)< "=2 + "=2

= "

) xk is a Cauchy sequence

Furthermore, xk 2 X 8 k. However, limk!1 xk = 1 =2 X:

Denition 23 A m.s. (X; d) is totally bounded if 9 nite set fx1:::xmg X s:t: X B"(x1) [ ::: [B"(xm)

Denition 24 A set S is bounded if 9 B s:t: 8 x 2 S; kxk B.

Lemma 4 If set S is bounded above, it has a smaller upper bound, or supremum: orK = supS.

Denition 25 A m.s. (X; d) is compact () it is complete and totally bounded

Denition 26 Let (X; d) be a m.s. and S X; then S is compact in (X; d)() (S; d) isa compact m.s.

Denition 27 A m.s. (S; d) is compact () every sequence fxkg contains a subsequencefxkmg1m=1 whose limit 2 (S; d) (i.e.every sequence contains a convergent subsequence in(S; d))

Denition 28 Let (X; d) be a metric space. We say that S X is compact set if thefolowing holds: whenever C is a collection of open sets in (X; d) whose union contains S,9 a nite subcollection from C whose union contains S:The collection C is called an opencover. Conversely, S is not compact if 9 an open cover (or collection), C from which wecannot extract a subcollection whose union contains S:

Remark 8 C is not a subset of X but a collection of subsets of X: That is, C is not a unionof sets.

Example 13 Let X = [0; 1), d =Euclidean metric and S = [0; 1). Claim: S is not com-pact. To see why we must exhibit an open cover from where no nite subcovers containingS can be extracted. Consider:

C =0;

n

n+ 1

j n 1

10

and note that each0; n

n+1

is open and:

1[n=1

0;

n

n+ 1

= [0; 1)

i.e., every point in [0; 1) belongs to at least one of the sets0; n

n+1

. Thus, C is an open

cover for S. However, we cannot extract a nite subcollection of C whose union containsS:

Remark 9 Every closed subset of a compact set is compact.

Remark 10 A nite subset of a metric space is compact.

Remark 11 The real numbers (Rn) with the Euclidean metric is a complete metric space.

Theorem 14 (Heine-Borel) Let X = Rn, d = Euclidean metric. We say that S Rn iscompact , S is closed and bounded.

Proof. ()) Suppose that S is compact. If S = ? then S is closed and bounded. If S 6= ?then S compact) (S; d) is compact. That it, (S; d) is complete and totally bounded. Since(S; d) is complete, every Cauchy sequence is convergent in (S; d) ; since every convergentsequence is a Cauchy sequence, every convergent sequences has its limit in (S; d), thus, Sis closed. Since (S; d) is totally bounded it follows that S is bounded.(() Suppose that S is closed and bounded. If S = ? then trivially S is compact. IfS 6= ? then by the remark above, we know that X = Rn ) (X; d) is complete and S Xplus (X; d) complete, implies that (S; d) is complete. Also, S bounded implies that (S; d)is totally bounded. Thus, (S; d) is a compact metric space. Hence, S is compact in (X; d).

Corollary 1 If S is a compact subset of the metric space (X; d), then S is closed andbounded in (X; d) :

Proposition 2 A metric space (S; d) is compact () every sequence fxkg contains asubsequence fxkmg1m=1 whose limit 2 (S; d) (i.e.fxkmg1m=1 is convergent in (S; d))

Proposition 3 Let qk = pkkpkk : Then kqkk = 1 and the set fqkj kqkk = 1g is compact, i.e.9 qkm and q s.t. kqk = 1 and qkm ! q

1.5 Continuity of functions

Notation 15 Denoting the function f : X! Y we call X the domain and Y the co-domain.The co-domain is a subset of the range.

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Notation 16 Let d be a metric dened on X and a metric on Y then a function isdenoted f : (X; d)! (Y; ) :

Denition 29 A function f : (X; d) ! (Y; ) is continuous at x if, for every " > 0, 9 > 0 such that whenever x 2 B(x)) f(x) 2 B"(f(x)).

Denition 30 A function f : (X; d) ! (Y; ) is continuous if it is continuous at everyx 2 X.

Proposition 4 (sequential characterization of continuity) A function f : (X; d) ! (Y; )is continuous at x ,, whenever xk ! x it follows that f (xk)! f (x)

Proof. ()) Suppose that f is continuous at x 2 X. Choose fxng such that xn ! x:We must show that f (xn) is convergent in (Y; ) with limit f (x) : Equivalently, we mustshow that 8 " > 0 9 > 0 such that k > k^ ) f (xn) 2 B" (f (x)). To see this choose" > 0. Since f is continuous, at x it follows that 9 > 0 such that x 2 B (x) )f (x) 2 B" (f (x)) : Next, since xk ! x we know that 9 k^ such that k > k^ ) xn 2 B (x).) k > k^ ) xn 2 B (x)) f (xn) 2 B" (f (x)) :(() (contrapositive) Now suppose that f is not continous at x. Since f is not continous9 some " > 0 such that 8 > 0 we can nd x 2 B (x) with the property that f (xn) =2B" (f (x)). Therefore, 9 " > 0 such that 8 n we can nd xn 2 B1=n (x) but f (xn) =2B" (f (x)). Thus from lemma 34, we conclude that xn ! x but f (xn) =2 B" (f (x)). Thusf (xn) is not convergent with limit f (x) :

Denition 31 f is uniformly continuous if for every "; 9 a that "works" for any arbi-trary point x (i.e., does not depend upon x as in mere continuity)

We note that a continuous function on a compact domain is uniformly continuous.Also, a linear combination of continuous functions is continuous. Finally, the compositionof continuous functions is continuous.

Proposition 5 Every uniformly continuous function is continuous.

Proof. Suppose that f : (X; d) ! (Y; ) is uniformly continuous. Choose " > 0: Then9 some > 0 such that (f (x) ; f (y)) < " whenever x 2 X, y 2 Y and d (x; y) < : Inparticular, (f (x) ; f (y)) < " whenever d (x; y) < so that f is continuous at x. Since f iscontinuous at some arbitrary x it is continuous at every point and therefore is continuous.

Exercise 17 (A.11, McLean) Again let f : (X; d) ! (Y; ) and suppose that X =Y =(0;1) and d = = Eclidean. Show that the function f (x) = 1=x is continuousbut not uniformly continuous.

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Proof. To see that f is continuous, choose " > 0 and x > 0. Next, choose:

0 < < min

x2"

2;x

2

so that:

jx yj < ) jx yj < x2) x y < x

2) x2< y

consequently:

jx yj < )1x 1y

= jx yjxy xy < 2x2 < "However, f is not uniformly continuous. To see why, choose " = 1

2: For each integer m > 0

choose xm = 1m and ym =12m: Then:

jxm ymj = 12m

but: 1m 12m = m

If > 0, then choose m so that 12m

< . Then jxm ymj < but 1m 1

2m

= m > 12:

Denition 32 f : (X; d) ! (Y; ) is uniformly continuous if, whenever xk is a Cauchysequence in (X; d) ; it follows that f(xk) is Cauchy sequence in (Y; ) :

Example 18 (2.82, Carter) Let f : [0; 1)! R be dened by:

f (x) =x

1 xthen f is continuous but not uniformly continuous. To see why, note that the sequencexk = 1 1k is a Cauchy sequence in X but f (xk) is not a Cauchy sequence in Y:

Example 19 If xk ! x then jxkj ! jxj i.e., the absolute value function is continuous.

Lemma 5 (sandiwch II) If g(x)! l; h(x)! l as x ! x and g(x) f(x) h(x), thenf(x)! l as x! x.

Denition 33 The function f : Rn ! Rm dened as:

f (x) :

0BBB@f1 (x)f2 (x)...

fm (x)

1CCCAis continuous if and only if fi : Rn ! R1 is continuous for each i = 1; :::m

13

Remark 12 The function f(x) = 1=x is unbounded from above but is bounded from below.

Denition 34 The mapping f : (X; d) ! (Y; ) is upper semi continuous if the setfx 2 X j f (x) g is closed in (X; d) :

Denition 35 The mapping f : (X; d) ! (Y; ) is lower semi continuous if the setfx 2 X j f (x) g is closed in (X; d) :

Example 20 Consider the function:

f (x) =

1 if x 01 if x < 0

we can claim that f is USC but not LSC. To se why, note that if > 1 the set fx 2 X j f (x) g =? which is closed. Next, if 1 < 1 the set fx 2 X j f (x) g = fx 2 X j x 0gwhich is also closed (the set [0;1) is closed). Finally, if < 1 the set fx 2 X j f (x) g =R, also a closed set. Thus, for any value of , the set fx 2 X j f (x) g is closed. Onthe other hand, if e.g., 0 < < 1 then the set fx 2 X j f (x) g = (1; 0) which is nota closed set.

Remark 13 The distribution function of a random variable is USC.

Remark 14 A function is continuous if and only if it is USC and LSC.

Proposition 6 Suppose that f : (X; d)! (Y; ) is continuous. If S is compact in (X; d),then f (S) is compact in (Y; ) :

Proof. Choose some sequence fykg such that yk 2 f (S) 8 k: This in turn implies that foreach k, 9 xk 2 S such that yk = f (xk). Since fxkg is a sequence in S and S is compact,9 a subsequence fxkmg and x such that xkm ! x. Moreover f continuous implies thatf (xkm) ! f (x). But f (xkm) = ykm so that ykm ! f (x) :Thus, fykmg is a subsequence offykg convergent with limit in f (S). Hence, f (S) is compact.

1.6 Continuity of correspondences (set-valued map-pings)

Denition 36 Suppose that X 6= ?. A relation in X is a "rule" that associates with eachx 2 X a non-? subset of X, denoted (x) :

Denition 37 (preference relation) Let : X X be a relation.Then: i) is reexiveif x 2 (x) ; ii) is transitive if y 2 (x) and z 2 (y)) z 2 (x) and iii) is completeif x 6= y ) x 2 (y) and/or y 2 (x) : We say that is a preference relation if it satsesi) iii):

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Denition 38 Let : X X be a relation. A function u : X! R is a representation for if u (x) u (y), x 2 (y) :

Denition 39 A correspondence or set-valued mapping, : (X; d) (Y; ); is a rule thatassigns to every element x 2 X, a non-? subset (x) of Y.

Proposition 7 Let (X; d) is a m.s. and if: i) : X X is reexive, transitive andcomplete, ii) (x) is closed in (X; d), then (x) has an USC representation.

Denition 40 : (X; d) (Y; ) is upper hemi continuous (UHC) at x if for every openset U (Y; ) s.t. (x) U 9 > 0 s.t. whenever x 2 B(x)) (x) U

Denition 41 is lower hemi continuous (LHC) at x if for every open set U (Y; )s.t. (x) \ U 6= ?, 9 > 0 s.t. whenever x 2 B(x)) (x) \ U 6= ?:

Denition 42 The graph of : (X; d) (Y; ) is dened as: Gr() = f(x; y) 2 X Y j y 2 (x)g

Denition 43 : X Y is closed graph (w.r.t. box metric) at x if i) xk ! x ii) yk ! yand iii) yk 2 (xk) 8 k ) y 2 (x):

Denition 44 Alternatively is closed graph at x if Gr() is a closed set, i.e., whenever(xk; yk)! (x; y) and (xk; yk) 2 Gr() 8 k then it follows that (x; y) 2 Gr().

Lemma 6 is continuous , it is UHC and LHC.

Remark 15 A constant correspondence (x) = K 8 k is trivially continuous (i.e. UHCand LHC).

Lemma 7 If is closed graph ) is closed-valued

Example 21 (2.88, Carter) The correspondence : R+ R dened by:

(x) =

1x

if x > 0

f0g if x = 0

has closed graph but is not UHC. To see why, note that Gr() = f(x; 1=x) 2 R2jx > 0g [(0; 0) which is a closed set in R2; however, for every sequence fxkg such that x ! 0; thesequence yk 2 (xk) does not converge.

Example 22 The constant correspondence (x) = (0; 1) is UHC but not closed-valuedand therefore does not have closed graph (closed graph)closed valued, so qclosed valued)qclosed graph).

Proposition 8 If 1:(X; d) (Y; ) is closed graph and 2:(X; d) (Y; ) is UHC andcompact-valued, then 1 \ 2 is compact-valued.

15

Proposition 9 The product of UHC correspondences is UHC. Thus, the product of UHCand compact correspondences is UHC and compact.

Proposition 10 If : (X; d) (Y; ) is closed graph, then is UHC whenever Y is acompact set.

Example 23 (2.89, Carter) Let P denote the domain of the budget correspondence, thatis, the set of all prices and incomes pairs for which some consumption is feasible:

P =

((p;m) 2 RnRjmin

x2X

nXi=1

pixi m)

where X is the consumption set, the graph of the budget correspondence X (p;m) is givenby :

Gr (X (p;m)) =

((p;m;x) 2 P X j

nXi=1

pixi m)

which is a closed set in P X. To see why, let fxkg be a sequence of consumption bundlesX (p;m). Since X (p;m) is bounded, xk ! x for some x 2 X: Thus,

p1x1k + :::+ pnxnk m

so that:p1x1 + :::+ pnxn m

so that (p;m;xk) ! (p;m; x) and (p;m; x) 2 Gr (X (p;m)). Therefore, Gr (X (p;m))is closed. Consequently, if the consumption set X is compact, the budget correspondenceX (p;m) is UHC.

Theorem 24 Suppose that :(X; d) (Y; ) is UHC and compact valued. If K is compactin (X; d) then (K) is compact in (Y; ).

Proof. Choose yk 2 (K). We must show that 9 a subsequence ykm ! y and y 2 (K).To see this, choose xk 2 K such that yk 2 (xk) : Now, K compact implies that evenif xk is not convergent, it must contain a subsequence fxkmg with xkm ! x and x 2 K:Summarizing: 1) xkm ! x, 2) yk 2 (xk) so ykm 2 (xkm). Now, UHC and compactvalued implies that 9 ykmt1t=0 and y such that ykmt ! y. Finally, x 2 K so that (x) (K). Therefore, y 2 (K), so we have constructed a subsequence of yk which isconvergent and whose limit belongs to (K) :

1.7 Maximum theorems

First we introduce the main theorem that relates topological assumptions of a maximizationproblem with the existence of optima.

16

Theorem 25 (Weierstrass) Suppose that (X; d) is a metric space and f : X! R is acontinuous function. If S X is non-? and compact, then f attains a maximum and aminimum in S:

Proof. First, note that f (S) is non-? and compact in R. Hence, by the Heine-Boreltheorem f (S) is bounded and in particular, bounded from above. Next, we notice thatany set in R has a least upper bound or sup. Let sup f (S) = . Now, let Ak = f (S) \y 2 R j y 1

k

. Note Ak 6= ? 8 k since there must be something between 1k and

which belongs to f (S) or else would not be the least upper bound. So, choose fykgsuch that yk 2 f (S) \

y 2 R j y 1

k

which implies that yk 1k . Thus,

yk ! :. But f (S) is compact and therefore closed so it must be that 2 f (S) : Now if 2 f (S), then 9 x 2 S such that f (x) = . Therefore, f (x) = f (x) for every x 2 S.Thus x is a maximizer for f in S. An identical argument can be applied to show that fattains a minimum in S:

Next, we study the properties of the maximizer correspondence and the optimal valuefunction of maximization prtoblems. Consider the general constrained maximization prob-lem:

max f(x; y)

s.t. y 2 (x)

Theorem 26 (The continuous maximum) If f : X Y! R is continuous (w.r.t. boxmetric) and : X Y is non-?, UHC, LHC, compact valued, then: 1) : X Ywhere (x) = argmaxy2(x) f(x; y) is non-?; UHC and compact valued (hence UseqC)correspondence and 2) V : X! R where V (x) = maxff(x; y)j y 2 (x)g is a continuousfunction of x.

Proof. First note that f continuous w.r.t. box metric) y 7! f(x; y) is continuous. Since(x) is compact valued, the Weierstrass theorem implies that (x) 6= ?. Next, to se that closed-valued...

Theorem 27 (The Concave Maximum) If f : XY! R is continuous and quasicon-cave and : Y X is convex valued, then: 1) : X Y where (y) = argmaxx2(y) f(x; y)is convex-valued and 2) if f is strictly concave, then V : X! R where V (y) = maxff(x; y)jx 2 (y)g is concave in x.

Proof. To see that is convex-valued, choose x1; x2 2 (y). Naturally, these two maxi-mizers yield the same maximum value. So let = f(x1; y) = f(x2; y). Choose t 2 [0; 1] andnote that f quasiconcave implies that the set tx1 + (1 t)x2 2 fx 2 (y)jf(x1; y) g.Thus, f (tx1 + (1 t)x2) . But (y) is a convex set and x1; x2 2 (y) (y). Hence,tx1+(1 t)x2 2 (y) which implies that f (tx1 + (1 t)x2) . Therefore, we concludethat f (tx1 + (1 t)x2) = and tx1 + (1 t)x2 2 (y):

17

1.8 Fixed point theory

We survey two main classes of xed point theorems (FPT). The rst class of FPT allowsfor exible domain sets, but requires heavy structure on the objectve function. The secondclass of FPT requires a highly structured domain sets but requires only mild assumptionsabout the objective function.

Denition 45 f : (X; d) ! (X; d) is a contraction mapping if 9 2 [0; 1[ s.t.d(f(x); f(y)) d(x; y) 8 x; y 2 X:

Remark 16 A contraction is always a continuous function.

Theorem 28 (Banach Fixed Point) Let f : (X; d)! (X; d) be a contraction mappingand (X; d) be a complete metric space, then 9 a unique xed point f (x) = x and if xk is asequence satisfying xk+1 = f(xk) for each k 0, then, xk is convergent with limit x:

Corollary 2 Every contraction mapping f : (X; d) ! (X; d) on a complete m.s. has aunique xed point.

The Banach FPT guarantees the existence of a stationary point x under the appropriatecontracion and completeness assumptions. However, stability around such stationary pointis another matter. A xed point is globally stable if, starting from any point in the domainof f , we can generate a sequence that converges to x. On the other hand, a xed pointis called locally stable, if, starting from a neighborhood of x (say B" (x)) we can generatea sequence converging towards x: The following proposition outlines the main conditionsunder which local stability can be established.

Proposition 11 Suppose that f : Rn ! Rn is a contraction dened in a complete metricspace (i.e., it has a xed point x). Then if Jf (x) ; the Jacobian or matrix of partial deriva-tives at x has all its eigenvalues lying inside the unit circle the stationary point x is locally(asymptotically) stable.

Remark 17 In one dimension this condition trivially reduces to jf 0 (x)j < 1:

Theorem 29 (Browers Fixed Point) Let f : C ! C be a continuous function. IfC Rn is non-?, compact and convex, then f has a xed point x = f(x).

Remark 18 Note this is a su ciency theorem, so one can make examples that violatethese assumptions and have a xed point.

Exercise 30 (Excess demand theorem 2.6.1 Carter) Dene the relative price setn1 =fp 2 Rnjpi 0 and

Pni=1 pi = 1g and dene z : n1 ! Rn to be a continuous function

satisfying p z (p) = 0 8 p 2n1. Then there exists p 2 n1 such that z (p) 0.

18

Proof. First dene the following "adjustment function" gi : n1 ! :

gi (p) =pi +max f0; zi (p)g

1 +Pn

j=1max f0; zj (p)gand the corresponding g : n1 ! n1 as:

g (p) =g1 (p) g2 (p) ::: gn (p)

0then p 7!gi (p) is continuous for each p and therefore p 7!g (p) is a continuous function.Since n1 is convex, compact non-? and g () is continuous, we can apply Browers theo-rem and justify the existence of p such that g (p) = p. That is, for each i :

pi =pi +max f0; zi (p)g

1 +Pn

j=1max f0; zj (p)gor:

pi

nXj=1

max f0; zj (p)g = max f0; zi (p)g

next, multiply both sides by zi (p) :

zi (p) pi

nXj=1

max f0; zj (p)g = zi (p)max f0; zi (p)g

and summ over i :nXi=1

zi (p) pi

nXj=1

max f0; zj (p)g =nXi=1

zi (p)max f0; zi (p)g

now, since by assumptionPn

i=1 zi (p) pi = 0, the LHS is zero so that:

nXi=1

zi (p)max f0; zi (p)g = 0

now, every term of this sum is nonegative since it is either 0 or [zi (p)]2. Thus, for a sum of

nonnegative numbers to be equal to zero, it must be the case that all the terms of the sumare zero so that max f0; zi (p)g = 0 and zi (p) 0 for each i implying that z (p) 0:Theorem 31 (Kakutanis Fixed Point) Let : C C be UHC, non-?; convex andcompact-valued. If CRn is non-?, compact and convex, then has a xed point x 2(x):

Exercise 32 (A.52, McLean) If f : C ! Rn is continuous, the Variational InequalityProblem for

f;Rn+

is the following: nd x 2Rn+ such that f (x) (x x) 0 8 x 2Rn+.

Suppose that C Rn is compact, convex, non-?. Show that the VIP for f;Rn+ has asolution.

19

Proof. Dene the constant correspondence (C) = C. And note that is compact,non-?, UHC and LHC (see remark (15) and theorem (24)). Next, dene the problemminz2(C) z f (x) and note that z 7! z f (x) is continuous and quasiconcave for eachx. Therefore, the maximum theorems imply that (x) = minz2(C) z f (x) is non-?(Weierstrass), compact, UHC (Continuous Max) and convex (Concave Max): Next notethat (x) C for each x: That is, : C C satises all the assumptions required forKakutanis theorem to apply and we conclude that has a xed point; i.e., 9 x 2 (x) =minz2C z f (x) so that f (x) x f (x) x 8 x 2Rn+ and we conclude that x solves the VIPforf;Rn+

:

Theorem (28) establishes that, under the required topological conditions, the existenceof a unique xed point is ensured. However, theorems (29) and (31) are merely existencetheorems. Uniqueness of xed points, equilibria and solutions to system of equations require"monotonicity" conditions that generalize to multidimensional problems the notion of astrictly increasing function of one variable. For denition and examples of monotonicitysee section (2.3).

20

Chapter 2

Convex optimization

2.1 Convex sets

Denition 46 S R is a convex set if x+(1)y 2 S whenever x; y 2 S and 2 [0; 1].

Remark 19 Note that the convex combination x+ (1 )y is a particular linear combi-nation.

Denition 47 The convex hull of S, denoted convS is the intersection of all convex su-persets of S :

convS =

(y 2 Rn j y =

mXi=1

ixi

)forxi 2 S

and i > 0;mXi=1

i = 1

Remark 20 The intersection of any collection of convex sets is convex so that, trivially,convS is a convex set.

Remark 21 S is convex () S = convS

Exercise 33 (1.232, Carter) The consumers budget set (correspondence) is convex.

Remark 22 The closure of a convex set is convex.

Lemma 8 The Minkowskis sum of two convex sets A B = fa+ b j a 2 A and b 2 Bgis a convex set

Lemma 9 Also, if A;B are convex, then the cartesian product AB is a convex set.

21

2.2 Separating hyperplanes

Remark 23 Recall that the equation of the tangent of a curve at x is y = f(x)+f 0(x)(xx)Denition 48 If p is a non-0 vector, the vector p x = 0 is orthogonal to p and p isnormal to p x = 0Remark 24 p "points" in the direction where p x > 0Theorem 34 (Separating Hyperplane) (aka Minkowskis theorem): Let S be non-?,closed and convex, and let x =2 S, then 9 p 6= 0 and x0 2 S s.t.: p x > p x0 p x; 8x 2 SProof. (step 1) Dene f (x) = (x x)(x x) and claim 9 x02 S such that f (x) f (x0)8 x 2S: To see why this is true note that f as dened above is continuous. Morover, chooser > 0 and the closed ball Br (x) so that Br (x)\ S is a compact set (recall S is closed andBr (x) is compact). Thus, Weierstrass theorem ensures the existence of the minimizer x0on Br (x)\S. This in turn implies that x0 is actually a minimizer for f on S (if x^ =2 Br (x)the function f (x^) is even greater than f (x0)):(step 2) Next, let p = x x0 and note that p 6= 0 since by assumption x =2S while x02S.Therefore p p > 0 implies:

p p = (x x0) (x x0)= p (x x0)= p x p x0> 0

) p x > p x0

(step 3) Finally we show that p x p x0 8 x 2S: To do so, choose x 2S and t 2 (0; 1).Let xt = tx+(1 t)x0 and note that so that xt2S since S is convex by assumption. Now:xt x xt x = t2 (x x0) (x x0) + 2t (x x0) (x0 x) + (x0 x) (x0 x)

or:xt x xt x (x0 x) (x0 x) = t2 (x x0) (x x0) + 2t (x x0) (x0 x)

now, since (xt x) (xt x) (x0 x) (x0 x) 0 it follows that:t2 (x x0) (x x0) + 2t (x x0) (x0 x) 0t (x x0) (x x0) + 2 (x x0) (x0 x) 0

and since limt!0 t (x x0) (x x0)+2 (x x0) (x0 x) = 2 (x x0) (p) we concludethat:

(x x0) (p) 0) p x p x0

22

Corollary 3 Under the same assumptions it is also true that 9 q 6= 0 and x0 2 S s.t.:q x > q x0 q x; 8 x 2 S:Denition 49 If A is a m n matrix, posA =

ny 2 Rm j y =Pnj=1 ajxj and xj 0o :

Remark 25 Note that aj are the columns of A so posA is the cone formed by the column-vectors of A

Theorem 35 (Farkas lemma) Let A be n m matrix; then b 2 posA , y 2 Rm andATy 0 imply bTy 0Proof. ()) Suppose b 2 posA then 9 x 2 Rm+ such that b =

Pnj=1 ajxj = Ax. Now, if

ATy 0 then 0 xT ATy = (Ax)T y = bTy:(() (contrapositive and separating hyperplane theorem) Suppose that b =2 posA. SinceposA is closed convex and non-? there exists p 6= 0 and y0 2 posA such that p b < p y0 p y 8 y 2 posA. Now since posA is a cone, it follows that 0 2 posA and 2y0 2 posA. Thusp y0 p 0 = 0 and p y0 p2y0 so that p y0 0 and we conclude that p y0 = 0which in turn implies that p b < 0:Therefore, we conclude that ATy 0 so ATp 0 butpTb < 0. Summarizing, b =2 posA ) y 2 Rm and ATy 0 but bTy < 0:Proposition 12 (supporting hyperplane) Let S be non-? and convex, and supposethat xk =2 clS 8 k with xk ! x. Then 9 p 6=0 such that p x p x 8 x 2 S.Proof. Recall that S convex ) clS convex. Thus all the conditions required to applythe separating hyperplane theorem are satised. However, note that each xk requires adierent qk. So we conclude that for each xk, 9 qk 6=0 such that qkxk qkx 8 x 2 clSand since S cls it follows that qkxk qkx 8 x 2 S. Now, since qk 6=0 then kqkk 6= 0 sothat:

qkkqkk xk

qkkqkk x 8x 2 S

and since

qkkqkk

= 1 it follows that:

qkkqkk 2 A = fy 2R

nj kyk = 1g

note that A is the pre-image of f (y) = kyk = 1. Now this function is trivially continuousso that the set A is closed (apply sequential characterization of closedness and continuity).Moreover, A is bounded so the Heine-Borel theorem ensures that A is compact. Thuse, wecan extract a convergent subsequence:

qkmkqkmk such that

qkmkqkmk ! p and p 2A

Next, note that:

qkmkqkmk

xk qkmkqkmk x 8x 2 S ) qkmkqkmk

x qkmkqkmk x 8x 2 S ) p x p x

23

Corollary 4 Let S be non-?; closed and convex, and suppose that x 2 @S then 9 p 6=0such that p x p x:

2.3 Convexity of functions and subgradients

Denition 50 f is a convex function if: f(x)+(1)f(y) f(x+(1)y) wheneverx; y 2 S and 2 [0; 1].( for concavity)Denition 51 Let f : S ! R and S Rn. The epigraph of f is epi (f) = f(x; y) 2 Rn+1jx 2S and y f (x)gDenition 52 Let f : S ! R and S Rn. The hypograph of f is hyp (f) = f(x; y) 2 Rn+1jx 2S and y f (x)gLemma 10 f is convex , epi (f) is a convex set.Lemma 11 f is concave, hyp (f) is a convex set.Remark 26 Note that Grf epif and that (x; f(x)) 2 @epifProposition 13 If S Rn is closed and convex, and is f : S ! R is continuous andconvex on S then epif is closed and convex.

Proof. That epif is convex follows from the assumption that f is convex. Next, to see thatepif is closed note that epif Rn+1 so choose (xk; yk) 2epif such that (xk; yk)! (x; y) :If (xk; yk) 2epif 8 k then yk f (xk) which, given that f is continuous, implies thaty f (x). Summarizing, (xk; yk) 2epif; (xk; yk) ! (x; y) and (x; y) 2epif so weconclude that epif is a closed set.

Denition 53 Let f : S ! R , then p is a subgradient for f at x 2 S if f(x) f(x) + p (x x) 8 x 2 S:Denition 54 The subdierential @f(x) is the set of all the subgradients of f at x

Proposition 14 Let f : S ! R with S a convex subset of Rn. If @f(x) 6= ? for eachx 2S then f is convex.Proof. Choose x1;x2 2 S and 2 [0; 1]. Let xt = tx1+(1 t)x2 and note that @f(xt) 6=?. Next choose p 2 @f(xt) and note:

f(x1) f(xt) px1 xt

f(x2) f(xt) p

x2 xt

nally multimply the rst of these weak inequalities by t and the second one by 1 t; addthem together and obtain:

tf(x1) + (1 t) f(x2) f (tx1 + (1 t)x2)so we conclude that f is convex.

24

Proposition 15 Let g : Rn ! R and g(x) g(x) whenever x 2 B"(x) for some " > 0; ifg is dierentiable. at x, then rg(x) = 0.Corollary 5 Now let g(x) = q x if q x q x 8 x 2 B"(x) then q = 0 = rg(x):Proposition 16 Let S be non-?; and convex and let f : S ! R be convex and continuousat x, if x 2 intS then @f(x) 6= ? (i.e., there exists at least one subgradient).Proof. Let A =epif . By proposition (13) we know that A is convex. Next we can claimthat (x;f(x)) 2 @A: To see why, notice that (x;f(x)) 2 A since Gr(f) epif = A. Thus,B" (x;f(x)) \ A 6= ?. However, (x;f(x) 1=k) =2 A so that B" (x;f(x)) \ Ac 6= ?. Thusx;f(x) 2 A but x;f(x) 2 intA so x;f(x) is a boundary point and the claim is true. Now,using corollary (4) it follows that (x;f(x)) 2 @A) 9 (q; r) 6= 0 such that:

q x+rf(x) q x+ryfor each (x;y) 2 A; but since Gr(f) epif = A it is also true that:

q x+rf(x) q x+rf(x) (2.1)for each x 2S. These two facts can only mean that r 0. In fact we can deduce thatr < 0 since r = 0 implies q x q x, which using corollary (5) implies that q =0 whichcontradicts the hypothesis that (q; r) 6= 0. Now, since r < 0 we can divide (2.1) by r toobtain:

f(x) f(x) +1rq

| {z } subgradient

(x x)

thus,1

rq 2 @f(x) and therefore we conclude that @f(x) 6= ?:

Lemma 12 (from subgradient to gradient) Suppose that S Rn is convex and x 2intS. If @f(x) 6= ? and if f : S ! R is dierentiable at x, then @f(x) = frf (x)gProof. Since @f(x) 6= ? choose q 2@f(x). Then:

f(x) f(x) + q (x x) 8 x 2SNow, since x 2 intS then 9" > 0 such that B"(x) S. hence:

f(x) f(x) + q (x x) 8 x 2B"(x)q xf(x)| {z }

g(x)

q xf(x)| {z }g(x)

8 x 2B"(x)

so that g (x) g (x) 8 x 2B"(x). Finally, note that g is dierentiable at x since f isdierentiable at x so that applying corollary (5) we conclude that rg (x) = 0 or:

0 = rg (x) = qrf (x)) q =rf (x)

25

Denition 55 (strong monotonicity) A function f : Rn ! Rn+ is said to be (strongly)monotone if [f (x) f (y)] (x y) 0 (> for strong)

Example 36 If f : Rn ! Rn+ is strongly monotone, the system f (x) = 0 has at most onesolution.

Proof. (by contradiction) Suppose that x2 and x2 solve the system f (x) = 0 and x2 6=x2. Then strong monotonicity and x2 6= x2 imply:

[f (x1) f (x2)] (x1x2) > 0

but by hypothesis f (x1) = 0 =f (x2) since x2 and x2 solve the system. Therefore:

0 (x1x2) > 0) 0 > 0

a contradiction.

Lemma 13 (homothetic function) Let u : Rm ! R be a strictly monotone increasingfunction. Then u is homothetic if and only if, 8 x;x0 2 Rm and 8 t > 0:

u (x) u (x0) () u (tx) u (tx0)

Lemma 14 (homothetic function II) : Suppose f : Rm ! R is homothetic and C1.Then for any x 2 R and > 0, there is a k > 0 such that rf(x)=rf(x) = k

Proposition 17 (monotonicity of subgradients) Let f : S ! R: If p12@f(x1) andp22@f(x2) then (p1 p2) (x1 x2) 0:

Proof. p12@f(x1)) f(x2) f(x1) + p1 (x2 x1) while p22@f(x2)) f(x1) f(x2) +p2 (x1 x2). Adding these two inequalities together we obtain (p1 p2) (x1 x2) 0:

Corollary 6 (monotonicity of gradients) Let f : S ! R and S Rn open, convexand non-?: If f is dierentyiable and convex, then (rf (x1)rf (x2)) (x1 x2) 0 foreach x1;x2 2 S:

Remark 27 Note that if n = 1 then (rf (x1)rf (x2))(x1 x2) 0 becomes (f 0 (x1) f 0 (x2))(x1 x2) 0 so that x1 x2 ) f 0 (x1) f 0 (x2), i.e., the derivative is nondecreasing.

Denition 56 (single crossing) A function f : X Y ! R is said to satisfy singlecrossing if s < t and x < y imply:

f (y; s) f (x; s) < f (y; t) f (x; t)

Remark 28 Supermodularity implies single crossing.

26

Proposition 18 (A.45, McLean) If f is dierentiable, the supermodularity property isequivalent to @

2f@x@t

> 0:

Proof. Suppose that f : [a; b] [0; 1] ! R is continuously dierentiable. Choose x < yand s < t: Then for each z 2 [a; b], @2f

@x@t> 0 implies:

0 f(x1) +rf(x1) (x2x1)f is convex , f(x2) f(x1) +rf(x1) (x2x1)

Proof. ()) Suppose f is convex. Choose x1;x2 2 S and 2 [0; 1]. then:f(x2 + (1 )x1) f(x2) + (1 ) f (x1)

f [ (x2 x1)] + x1 f (x1) [f(x2) f (x1)]f [ (x2 x1)] + x1 f (x1)

f(x2) f (x1)

if > 0 this is a Newton quotient dening the directional derivative of f in the directionof (x2 x1) so that:

rf(x1) (x2x1) f(x2) f (x1)(() Now choose x2;x1 2 S and suppose that f(x2) f(x1)+rf(x1)(x2x1) 8x2;x1 2 S.Pick 2 [0; 1], denote x = x1 + (1 )x2 and note that S convex) x 2 S. Thus:

f(x2) f(x) +rf(x) (x2x)= f(x) +rf(x) ((1 ) (x2x1))

f(x1) f(x) +rf(x) (x1x)= f(x) +rf(x) ( (x1x2))

27

multiply the rst inequality by , the second by (1 ) and add them together to get:f(x2 + (1 )x1) f(x2) + (1 ) f (x1)

so that f is convex.

Proposition 20 If S Rn open and convex, and if f : S ! Rn is dierentiable, then fis convex if and only if its gradient is monotone i.e.:

f is strictly convex , [rf(x2)rf(x1)] (x2 x1) > 0f is convex , [rf(x2)rf(x1)] (x2 x1) 0

Remark 29 In general, the product of two convex functions is not convex. However, iff; g are both convex, monotone and positive, f g is convex.Denition 60 Let A be n n. We say that A is PSD if xTAx 0 for all x 6= 0Remark 30 In R1 we say that f() is non-decreasing , f 0() 0: In Rn we replacenon-decreasing by monotonicity, f 0() by the Jacobian, Jf (); and 0 with positive-semi-deniteness (PSD).

Proposition 21 g : Rn S ! Rn is monotone ,its Jacobian (Jg) is PSD.Proof. (() Assume that Jg is PSD. Choose x2;x1 2 S and for each t 2 [0; 1] dene thefunction (t) = g ((1 t)x1 + tx2) (x2 x1). Since (t) is a function of t and a scalarfor each t, we can invoke the Mean Value Theorem (MVT) to deduce that 9 t 2 (0; 1)such that:

(1) (0)1 0 =

0 (t)

writing out this expression for 0 (t) we get:

[g (x2) g (x1)] (x2 x1) = (x2 x1)T Jg ((1 t)x1 + tx2) (x2 x1)| {z }0(t)

and since Jg is PSD we conlcude that:

[g (x2) g (x1)] (x2 x1) 0()) Now suppose that g is monotone. Choose x 2 S. Next, choose such that x +y 2 B" (x). Not that if 0 < < then x+y 2S. Now dene (t) = g ((1 t) (x+y) + tx)(x+y x). Again using the MVT we obtain:

[g (x+y) g (x)] (x+y x1)| {z }0 since g is assumed to be monotone

= (y)T Jg (z) (y)

where z = (1 t) (x+y) + tx. We want to show that (y)T Jg (x) (y) 0 or equiv-alently yTJg (x)y 0. So let ! 0 so that z ! x.and we arrive at the conclusion thatyTJg (x)y 0 so that Jg is PSD.

28

Proposition 22 f is convex , the Jacobian (Jf ) of the gradient is PSD.

Corollary 7 Equivalently, f is convex,the Hessian (Hf ) of the original function is PSD.

Corollary 8 Equivalently, if C Rn is convex and Q : C ! Rn with Q (x) = xTAx+ bxthen Q (x) is convex whenever A is PSD (NSD for concavity)

2.4 Support function theory

Many optimization problems can be cast in the form:

maxq xsubject to x 2 S

which is the cannonical form of a type of convex optimization problem; the support functionproblem.

Denition 61 K Rn is a pointed cone if x 2 K whenever x 2 K and 0

Remark 31 If K is a cone, 0 2 K.

Denition 62 Remark 32 0 is a cone and Rn is a cone.

Exercise 37 (A.52, McLean) If f : Rn ! Rn is continuous, the Variational InequalityProblem (see example ()) for

f;Rn+

is the following: nd x 2Rn+ such that f (x)(x x)

0 8 x 2Rn+. Show that x solves the VIP forf;Rn+

, x 2Rn+, f (x)2Rn+ and f (x) x = 0.Proof. ()) Suppose that x solves the VIP for f;Rn+. Then f (x) (x x) 0 8 x 2Rn+.Since 0 2 Rn+ it follows that f (x) (0 x) 0 ) f (x) x 0. But Rn+ is a cone sox 2Rn+ ) 2x 2Rn+. Therefore, f (x) (2x x) 0) f (x) x 0 from which we concludethat f (x) x =0. Finally, to see that f (x)2Rn+ note that x solves the VIP for

f;Rn+

implies f (x) x f(x) x = 0 so that f (x) x 0 and since x 2Rn+ we conclude thatf (x)2Rn+:(() Suppose that x 2Rn+, f (x)2Rn+ and f (x) x = 0. Then f (x) (x x) = f (x) xf(x) x =f (x) x 0 since f (x)2Rn+ and x 2Rn+. Moreover, since x 2Rn+ we concludethat x solves the VIP for

f;Rn+

:

Denition 63 (barrier cone) Let S Rn be non-?. The barrier cone of S, b (S) isdened as follows: p 2 b(S) if and only if the problem maxx p x subject to x 2 S has asolution.

Remark 33 b (S) Rn and 0 2 b (S) since maxq x = 0 x makes any x a maximizer.

Example 38 Suppose that S = fx 2R2j kxk < 1g, then b (S) = f0g.

29

Example 39 On the other hand, if S = fx 2R2j kxk 1g, then b (S) = R2.Example 40 Suppose that S = fx 2R2j x1 0 ;x2 0g (i.e., the third quadrant). To ndb (S) in this case, rst note that if p has a negative coordinate, @ a maximizer. Thus, pmust have both non-negative coordinates, i.e., b (S) = fp 2R2j p1 0 ;p2 0g. Naturally,if pi > 0 and pj = 0 the problem would have innitely many maximizers. On the otherhand, if p1 > 0 and p2 > 0 the problem has a unique maximizer. The support function isfound by plugging any of the maximizers in the objective function.

Lemma 15 b (S) is a cone in Rn

Proof. Suppose that p 2b (S) and choose 0. Then we know that p 2b (S), 9 x suchthat p x p x 8 x 2S. Therefore, 0) (p x) (p x) or (p) x (p) x sothat q =p 2b (S) so we conclude that b (S) is a cone.Denition 64 (support function) If p 2b (S) then the function s (p) = max [p x j x 2S]is well dened. The function s : b (S)! R is called the support function of S:Remark 34 The cost and prot functions are support functions.

Remark 35 s is homogeneous of degree one, i.e., if p 2b (S) then s (tp) = ts (p) :Remark 36 s is a convex function, whenever b (S) is a convex set.

Theorem 41 (support function) Let S Rn convex and p 2 b(S) then x 2 argmaxx2S p x , x 2 @s(p):Proof. ()) Suppose that x 2 arg maxx2S p x. Then s(p) = p x. Next choose q 2 b(S)and notice that

s(q) q x

= q x+=0z }| {

(s(p) p x)= s(p) + x (q p)) x 2@s(p)

(() Suppose that x 2@s(p). This implies that p 2 b(S): Now, by the denition ofsubgradient:

s(p) s(p) + x (p p) 8 p 2 b(S)choosing p = 0 and p =2p we note that 2p 2 b(S) since p 2 b(S) and b(S) is a cone.Thus:

s(0) = 0

s(p) + x (0 p)) x p s(p)

30

while:

s(2p) = 2s(p) (by homogeneity)

s(p) + x (2p p)= s(p) + x p) x p s(p)

therefore, x p = s(p) and hence, x p x p 8 x 2 S. To complete the argument wemust show that x 2S. We do this by contradiction and using the separating hyperplanetheorem. Suppose that x =2S then 9 w 6= 0 such that:

w x > w x0 w x;8x 2S

which in turn implies that x0 2 arg maxx2S w x. But w x > w x0 so we would havethat:

s(w) < w x

= w x+=0z }| {

(s(p) p x)= s(p) + x (w p)

a contradiction since x 2@s(p):

Corollary 9 Let S Rn convex and suppose b (S) is a convex set. If p 2intb (S) and sis dierentiable at p then argmaxx2S x p = frs(p)g = x is the unique solution to themaximization problem.

Example 42 Let S =x 2R2+j x1x2 1

then b (S) = fp 2R2jp1 < 0,p2 < 0g [ f0g and

intb (S) = fp 2R2jp1 < 0, p2 < 0g. Also, s(p1; p2) = 2pp1; p2. Notice that s(p1; p2) isdierentiable at intb (S) so that:

x1 =@s(p1; p2)

@p1=

rp2p1

x2 =@s(p1; p2)

@p2=

rp1p2

and (x1; x2) =q

p2p1;q

p1p2

is the solution to the max problem.

So far we have dened all these objects in terms of a maximization problem. There isa natural equivalence with the minimization problem:

Denition 65 (supergradient) Let f : S ! R , then p is a supergradient for f at x 2 Sif f(x) f(x) + p (x x) 8 x 2 S:

31

Denition 66 (barrier cone min) Let S Rn be non-?. The barrier cone of S, ~b (S)is dened as follows: p 2 ~b(S) if and only if the problem minx2S p x subject to x 2 S hasa solution.

Denition 67 (support function min) If p 2~b (S) then the function ~s (p) = min [p x j x 2S]is well dened. The function ~s : ~b (S)! R is called the support function of S:

Remark 37 ~s is also homogeneous of degree one, is a concave function, whenever ~b (S)is a convex set.

2.5 Kuhn-Tucker theory

2.5.1 Introduction to nonlinear programming

Suppose that we face the NLP problem with equality constraint:

min f (x1; x2)

s:t: g (x1;x2) = 0

with f; g dierentiable. To nd optima, we set-up the Lagrangian:

L = f (x1; x2) g (x1;x2)

obtain FOC:

rf (x1; x2) ="

@f(x1;x2)@x1

@f(x1;x2)@x2

#=

"@f(x1;x2)@x1

@f(x1;x2)@x2

#= rg (x1x2)

which, says that, at the optimum, the gradient of the objective is colinear with the gradientof the restriction. After eliminating this method yields the familiar result:

@f (x1; x2) =@x1@f (x1; x2) =@x2

=@g (x1; x2) =@x1@g (x1; x2) =@x2

2.5.2 The Kuhn-Tucker conditions

Next introduce inequality constraints. Suppose that f : Rn ! R; g : Rn ! R with f; gdierentiable. The cannonical NLP problem is:

min f(x)

s:t: g1(x) 0... (2.2)

gm(x) 0

32

A Kuhn-Tucker pair for this problem is a pairx;

2 Rn Rn satisfying:1) gi(x) 0 8 i = 1; :::;m (2.3)2) i 0 8 i = 1; :::;m (2.4)3) rf(x) =

Xi

irgi(x) (2.5)4) i gi(x) = 0 8 i = 1; :::;m (2.6)(complementary slackness)

Example 43 (linear programming) Let the objective function be f (x1; x2) = c1x1 +c2x2 and the constraints be g1 (x1; x2) = a11x1+a12x2b1 and g2 (x1; x2) = a21x1+a22x2b2.Then the LP problem is:

min c xs:t: aT1 x b1aT2 x b2

a KT pair for this problem would bex;

2 R2 R2 satisfying:c =1a1 + 2a2; 2R2+aT2 xb2 0; 1

aT2 xb2

= 0

aT2 xb2 0; 2aT2 xb2

= 0

Theorem 44 (Kuhn-Tucker for LP) Let c 2Rn; A be mn and b 2Rm: Then x solvesthe LP problem:

min c xs:t: Ax b 0

if and only if, 9 2Rm+ such that:c =AT (2.7)

Ax b 0 (2.8) [Ax b] = 0 (2.9)

Proof. (() Suppose that x; 2 Rn Rm satises the KT conditions. We must showthat any other x such that Ax b 0) c x c x: Now, from (2.7):

c =AT ) cTx = AT T xso that:

cTx =AT

Tx

=AT

Tx+ bT bT

= (Ax b)T + bT bT = AT

= cTx

33

()) Now suppose that x solves the problem. To begin, dene the set of active constraints:I (x) =

ijaTi x b = 0

. If I (x) = ? then there are no bonding constraints so to sat-

isfy rcTx = 0) c =AT we simply need = 0:Now, if I (x) 6= ?, suppose that y 2Rnsatises aTi y 0. Then we can claim that cTy 0; to see this, choose t > 0 such thataTi x+ta

Ti y >bi if i =2 I (x) and if i 2 I (x) it follows that:

aTi (x+ty) = aTi x+ta

Ti y

= bi + taTi y

biSummarizing, A (x+ty) b ) cT (x+ty) cTx) cTy 0. Now, applying Farkaslemma (proposition (35)) we conclude that for each i 2 I (x) ;9 i 0 such that c 2posAthat is:

c =Xi2I(x)

aii

or c =AT : To complete the argument let i = 0 8 i =2 I (x) :Theorem 45 (duality of linear programming) Let A be m n; b 2Rm and c 2Rn:The problem:

min c x (2.10)s:t: Ax b 0

has a solution if and only if the problem:

maxb y (2.11)s:t: ATx = c

yi 0has a solution.

Proof. It su ces to show that (x; y) is a KT pair for the problem in (2.10) if and only if(y; x) is a reduced KT pair for the problem in (2.11). Suppose that (x; y) is a KT pair forthe problem in (2.10). Then:

c =ATy, Ax b 0y 0 and y [Ax b] = 0

therefore, since max f (x) is equivalent to minf (x) ; it follows that (y; x) is a reducedKT pair for the problem:

minb y (2.12)s:t: ATx = c

yi 0

34

where the complementary slackness condition for the problem in (2.10) becomes the colin-earity of gradients condition for the problem in (2.12), the colinearity of gradients conditionfor the problem in (2.10) guarantees that the equality restriction in (2.12) is satised andthe condition that the multiplier is nonnegative in (2.10) guarantees that the nonnegativityrestriction in (2.12) are satised.

2.5.3 Constraint qualications

Denition 68 (general constraint qualication) Suppose that x solves the problem

min f(x)

s:t: g1(x) 0:::gm(x) 0Then x satises the general constraint qualication (GCQ) if I (x) 6= ? ) x solves theproblem:

minrf(x) xs:t: rgi(x) (x x) 0 8i 2 I (x)

that is, if x solves the local-linearized problem.

Theorem 46 (Kuhn, Tucker; 1951) Supose that x solves the problem

min f(x)

s:t: g1(x) 0:::gm(x) 0

If x satises the GCQ then 9 2Rm such that x; is a KT pair; that is, x; satisfythe KT conditions (2.3)-(2.6).

Example 47 (GCQ is not satised) Let the problem be:

minx1

s:t: x2 0; x31 x2 0the unique solution to this problem is x =(0; 0) and:

rf (x) =10

, rg1 (x) =

01

, rg2 (x) =

3x211=

01

Then it is not possible to express rf (x) as a linear combination of rg1 (x) and rg2 (x).The problem here is that x =(0; 0) does not solve the local-linearized problem, i.e., x =(0; 0)does not satisfy the GCQ.

Condition 48 (Cottle CQ) A NLP satises the Cottle CQ if I(x) 6= ? 9 z 2 Rn s.t.rgi(x) z > 0 8 i 2 I(x)

35

Condition 49 (linear independence CQ) A NLP satises linear independence if I(x) 6=? then rgi(x) and rgj(x) are linearly independent whenever i; j 2 I(x):

Condition 50 (Slater CQ) A NLP satises the Slater CQ if each gi() is concave and 9x0 2 Rn s.t. gi(x0) > 0 8 i (all the constraint sets are convex)

Remark 38 In the last example, none of these three constraint qualications is satised.

2.5.4 Non-negativity and equality constraints

Non-negativity constraints

If we modify the NLP problem to include non-negativity constraints we look at:

min f(x)

s:t: g1(x) 0:::gm(x) 0 (2.13)xi 0 8 i

In order to ignore the multipliers for these restrictions, use the (reduced) K-T system:

rf(x)Pmi=1 irgi(x) 0 (1)i 0 (2) gi(x) 0 (3)

xj 0 (4) igi(x) = 0 (5)h@fxJ(x)Pmi=1 i @gxJ (x)i xj = 0 (6)

Example 51 Let the problem be:

minx1 + x2

s:t: x21 + 4x22 4

xi 0 8 i

the KT conditions are:

0, xi 0 8 ix21 + 4x

22 4 0; (x21 + 4x22 4) = 0

1 2x1x1 = 0;

1 8x2

x2 = 0

11

2x18x2

0

36

There are three cases to consider:Case 1: x2 = 0, x1 > 0. In this case, we replace the last condition with x2 = 0 and solve:

1 2x1 = 0) x1 = 12

1

2

2 4 = 0

so that: =

1

4and x1 = 2

Case 2: x1 = 0, x2 > 0. Now we replace the second-to-last condition with x1 = 0 and solve:

1 8x2 = 0) x2 = 18

4

1

8

2 4 = 0

so that: =

1

8and x2 = 1

Case 3: Finally we look for an interior solution (i.e. x1 > 0, x2 > 0) so we solve:

1 2x1 = 0) x1 = 12

1 8x2 = 0) x2 = 18

1

2

2+ 4

1

8

2= 4

so that: =

p5

8, x2 =

4p5, x2 =

1p5

However, only the KT pair of the second case is a solution to the problem. To see why,replace this candidates for solutions in the objective function and note that:

1 < 2 1. Next, since Y satises NIRS then y 2 Y and k 2 [0; 1] ) ky 2 Y .Summarizing, y 2 Y ) ky 2 Y 8 k 0 so that Y is a cone. To see that Y is convex choosey; y0 2 Y . Since Y satises NIRS it follows that k 2 [0; 1] ) ky 2 Y and (1 k) y0 2 Y ;and by additivity ky+ (1 k) y0 2 Y so Y is convex.(() Now suppose that Y is a convex cone:Since Y is a cone, ky 2 Y 8 k 0 so to see thatY exhibits NIRS simply choose k 2 [0; 1] : Next, since Y is convex it follows that 8 y; y0 2 Yand k 2 [0; 1], ky+ (1 k) y0 2 Y . But Y is a cone so ky 2 Y and (1 k) y0 2 Y . Thereforewe see that whenever ky 2 Y and (1 k) y0 2 Y it follows that ky+ (1 k) y0 2 Y so Ysatises additivity.

Example 56 (Carter, 1.91) Suppose that an economy contains m producers and n com-modities. The technology of each producer is summarized by its production possibility setY j Rn. Aggregate production (set) y is the sum of the net outputs of each of the producersyj, that is:

y = y1 + :::+ ym

and the set of feasible aggregate production plans or the aggregate (economy-wide) produc-tion possibility set is:

Y = Y1 + :::+ Ym RnA su cient condition for Y to be convex is that each Yj be convex.as is recorded in lemma(8)

Remark 42 Note that although production posibility sets Yj;Y are xed and given bytechnology, actual production yj and y depend on the price vector p. Thus, the sets yj andy can in fact be thought of as correspondences yj(p) and y(p) where:

y(p) =X

yj(p) =ny 2Rnjy =

Xyj for some yj 2 yj (p)

o3.1.2 Cost minimization

Support function approach

Using the input requirement set notation, and allowing for multi-product rms (so thaty 2Rm); the cannonical cost minimization problem can be cast as:

minw xs:t:x 2V (y)

wherew 2Rn is a vector of input prices. Note that ifw 2eb(V (y)) then argminx2V (y) w x 6= ?:In fact the (reverse) support function @~V (y)(w) is the optimal (cost) value function.

43

Proposition 24 If V (y) is closed-convex and non-? and w 2 eb(V (y)) \ Rn+ = K convexset, then:

c(;y):K ! R is concave and homogeneous of degree 1x 2 arg min

x2V (y)w x , x 2 @c(w;y) = @~V (y)(w)

Moreover, if w 2 intK and c(w;y) is dierentiable:arg min

x2V (y)w x = frc(w; y)g = x| {z }

(Shephard0s Lemma)

Example 57 (perfect substitutes) Let V (y) =x 2 R2+jx1 + x2 y

: This produc-

tion technology results in either unique conrner solutions if the input price vector w 2 R2+has coordenates wi > wj; or in innitely many solutions if wi = wj: In fact, c(w; y) =min fw1; w2g y. Also, note that whenever c(w; y) is dierentiable, say when w1 > w2 Shep-hards lemma holds since c(w; y) = w2y and therefore:

rc(w; y) =0y

so that theres a unique supergradient equal to the gradient. Intuitively, since input 1 isrelatively more expensive and inputs are perfect substitutes, the rm produces y using onlyx2:

Kuhn-Tucker approach

A similar method to solve cost minimization problems is to replace x 2 V (y) with g (x) ywhere g () is a production function. This way we face an optimization problem underinequality and nonegativity constraints and a natural approach to deal with it is the KTtheory developed in section 2.5.

Example 58 Suppose that the rm has production function V (y) =x 2 R2+jx1 y and x11 x22 y

where i > 0 and 1 + 2 = 1. Then the problem can be cast as:

minw1x1 + w2x2

s:t: x1 y and x11 x22 yxi 0 8 i

the KT conditions for this problem are:

w1 1 21x1112 x22 0; w1 22x112 x212 0w1 1 21x1112 x22

x1 = 0;

w1 22x112 x212

x2 = 0

xi 0 , i 0 i = 1; 2; x11 x22 y, x1 y2 (x

11 x

22 y) = 0; 1 (x1 y) = 0

44

rst note that if there is to be any production in this problem, xi > 0 for both i = 1; 2. Werst look for a solution with i > 0 for i = 1; 2. Thus, we solve the system:

w1 1 21x1112 x22 = 0; w1 22x112 x212 = 0x11 x

22 = y x1 = y

which yields the solution:

1 = w1 12w2; 2 = w22 ; x1 = y = x2this will be a solution to the problem i w1

1 w2

2. On the other hand, if w1

1< w2

2then 1

becomes infeasible so we look for a solution with 1 = 0 and rewrite the system as:

w1 21x1112 x22 = 0; w1 22x112 x212 = 0x11 x

22 = y

solving this new system yields:

2 =

w22

2 w11

1, x1 = y

w11w22

2and x2 = y

w22w11

1now with 1 = 0, the condition 1 (x1 y) = 0 is trivially satised. However, we still needto check that condition x1 y is also satised. Since we are looking a the special case whenw11< w2

2it follows that w11

w22> 1 so that:

x1 y = yw11w22

2 y =

w11w22

2 1y 0

and we have a solution to the problem even when w11< w2

2:

Example 59 (no interior solution) Suppose that V (y) =x 2 R2+j1x1 + 2x2 y

.

Then the problem can be cast as:

minw1x1 + w2x2

s:t:1x1 + 2x2 yxi 0 8 i

Now, since this is a linear programming problem, we can use theorem (44) to conclude thata KT pair for this LP will be a solution to the problem. Thus we look for

x; 2 R2 R

such that:w1 1 0 w2 2 0

w1 1x1 = 0;

w2 2

x2 = 0

xi 0 8 i; 0 (1x1 + 2x2 y) = 0 1x1 + 2x2 y 0

this system has no solution with both x1; x1 > 0. The intuition is simple; if the priceof input i relative to its productivity is smaller than the price of input j relative to itsproductivity, the rms optimal plan is to use only input i. That is, if w1

1< w2

2then solving

the system above would yield x2 < 0 so we replace conditionw2 2

x2 = 0 with x2 = 0:

The solution to the problem would then be: x2 = 0, x1 =y1and = w1

1:

45

3.1.3 Prot maximization

Support function approach

Let y 2 Rm , x 2 Rn and let p 2 Rm+ and w 2 Rn+ then the prot maximization problemis:

maxfp y w xgs:t: x 2 V (y) and y 2 Rn+

To make this a traditional support function problem, dene T = f(z;y)jz 2 V (y)g (thatis, z = x) and note that (x; y) solves the problem above , (x; y) solves the problem:

maxfp y +w zgs:t: (z;y) 2 T

Remark 43 The proles (w;p) for which the max problem has a solution is the barriercone b(T ):Furthermore, if (w; p) 2 b(T ) then (w;p) = T (w;p):Proposition 25 Suppose T; b(T ) are convex. LetK = b(T )\[Rn+Rm+ ] so that (!;p) 2 K:Then:

(): K ! R is convex and homogeneous of degree 1

(x; y) 2 arg maxx2V (y)

fp y w xg , (x; y) 2 arg max(z;y)2T

fp y +w zg, x 2 @(w;p) = @T (w;p)

moreover, if is dierentiable:

arg max(z;y)2T

fpy+wzg = frw(w;p);rp(w;p)g = (x; y)| {z }(Hotelling0s Lemma)

Example 60 Let V (y) = fx 2 R+jpx yg. Note that if w = 0, theres no solution to

this problem. On the other hand, if p; w 2 R++ the problem is:max p

px wx

s:t: x 0with solution:

x = p2w

2and y =

p

2wso that:

(w;p) = pp

2w w

p2w

2=

p2

4w

46

Note that (w;p) is convex as it should. To see why we can use corollary (7):

r(w;p) = p

2w

2p2w

) H =

p2

2w3 p2w2 p

2w212w

and since H.is PSD we conclude that (w;p) is convex. Finally note that (Hotelingslemma): "

@(w;p)@w

@(w;p)@p

#=

p2w

2p2w

=

xy

Example 61 V (y) = fx 2 R+jx yg. In this setup, constant returns to scale complicatesmatters. Thus, if w > p the unique solution to the prot max problem is (0; 0). If p > wthe problem has no solution and if p = w there are innitely many solutions all of whichlead to (w;p) = 0:

Remark 44 Every function which is convex and homogeneous of degree one is a protfunction for some Y . Thus, we can recover the production technology set of a rm from itsprot function.

3.2 Consumer theory

3.2.1 Utility maximization

The cannonical utility maximization problem:

maxu (x) (3.1)

s:t: p x yxi 0

is not a linear problem and therefore we cannot apply the methods of support functiontheory. However, we can still use the Kuhn-Tucker approach for solving NLP optimizationproblems. Moreover, if the objective function u is concave, a straightforward extension oftheorem (52) implies that a KT pair would be a solution to the optimization problem. Witha solution at hand, we can compute the optimal value function which is usually referred asthe indirect utility function. In this case, the (reduced) KT conditions for optima become:

ru (x) + p 0; y p x 0 [y p x] =0; xi 0

xj

h@u@xj(x) pi

i= 0; 0

Remark 45 Notice that maxu (x) is equivalent to minu (x) so that the colinearity ofgradients condition is:

ru (x) p 0 ru (x) + p 0ru (x) + p 0

47

As in the cases for cost minimization and prot maximization, the optimal value func-tion for the utility maximization problem:

v (p; y) = maxpxy

u (x)

which is called the indirect utility function, also has a nice interpretation. It gives themaximum level of utility attainable for any given level of income at the prevailing prices.The IUF also has some interesting properties described in what follows.

Proposition 26 The indirect utility function v : Rn+R+ ! R is quasiconvex. That is:

S =(p; y) 2 Rn+1+ jv (p; y)

is convex 8

Proof. Choose (p^; y^) and (p; y) both 2 S. Then choose t 2 [0; 1] let (pt; yt) = t (p^; y^) +(1 t) (p; y). Now (p^; y^) 2 S ) v (p^; y^) and (p; y) 2 S ) v (p; y) :Next, choose:

xt 2 arg maxptxyt

u (x)

Now we can claim that xt is feasible for at least one of the (p^; y^) or (p; y) problems. Ifthis were not the case p^ xt > y^ and p xt > y, which, if multiplied by t and (1 t),respectively, and added together imply:

tp^ xt+ (1 t) p xt > ty^ + (1 t) y

pt xt > yt

a contradiction since by hypothesis xt solves the (pt; yt) problem. Thus, xt is feasible forat least one of the (p^; y^) or (p; y) problems:Therefore:

uxt max fv (p; y) ; v (p^; y^)g

vpt; yt

and we conclude that pt; yt 2 S so that S is a convex set and v () is quasiconvex.The following theorem helps us recover the solution to a utility maximization problem

from the indirect utility function:

Example 62 Suppose that a consumer lives in an ncommodity world and has preferencesu (x) = x1 + ::: + x

n with 0 < < 1; then the utility maximization problem for each level

of income y is:

maxx1 + :::+ xn

s:t:

nXi=1

pixi y

48

This problem is simplied by the fact that u () is strictly increasing and strictly concaveon each of its arguments. Strict concavity will guarantee that at the optimum, each xi > 0;strict increasingness will guarantee that the constraint will bind so that > 0. Hence weonly need to look for an interior solution to the KT system:

x1i pi = 0 and y nXi=1

pixi = 0

which yield solution:

=

yPn

i=1 p

1i

!1and xi =

24 p 11iPni=1 p

1i

35 ythis yields an indirect utility function:

v (p; y) = ynXi=1

24 p 11iPni=1 p

1i

35Theorem 63 (Roys identity) Suppose that x solves the problem in (3.1). If xi > 0 8i,x;is a (reduced) KT pair, > 0 and x is regular, then:

xi =@v (p; y) =@pi@v (p; y) =@y

3.2.2 Expenditure minimization

Suppose that agents have a certain welfare target and want to nd the minimum amountof spending required to attain such level of utility. Then they face the problem:

minp xs:t: u (x)

xi 0

Since the objective function is linear, this is a support function problem and we can usethe methods from section 2.4 to nd a solution. The problem is identical to that of costminimization and, in this case yields an optimal value function e (p; ). As in any supportfunction problem, e (p; ) is homogeneous of degree one and concave. Moreover, from theoptimal value function we can recover the solution to the cost minimization by:

@e (p; )

@pi= xi (p; )

49

Example 64 Suppose that Suppose that u : Rn+ ! R is continuous and monotonic in thefollowing sense: u(x) > u(y) whenever x > y where x > y means that xi > yi for eachi = 1; :::; n. If x solves the utility maximization problem:

maxu (x) s:t: p x y, xi 0

show that x solves the expenditure maximization problem:

minp x s:t: u (x) v (p; y) , xi 0

where v (p; y) = u (x).

Proof. Suppose that x solves the utility maximization problem but does not solve theexpenditure maximization problem. Since x solves the utility maximization problem thenp x y and u (x) u (x) 8 x such that p x y. Thus x is trivially feasible for theexpenditure min problem. But the hypothesis is that x does not solve the this problemso there exists x0 such that u (x0) u (x) and p x0 < p x: This in turn implies thatp x0 < y or that x0 is feasible for the utility maximization problem. To complete theargument, choose " > 0 small enough so that: p x0 < p (x0 + ") < y, then x0 + " > x0which, by monotonicity of u () ; implies that u (x0 + ") > u (x0) u (x), contradicting thehypothesis that x solves the utility maximization problem.

Exercise 65 Consider a consumer with utility function u : Rn+ ! R. Furthermore, sup-pose u = f h where h : Rn+ ! R+ is homogeneous of degree one and f : Rn+ ! R+ isa strictly increasing function satisfying f(0) = 0. Suppose that and ; 0 positive numbersand that the expenditure minimization problems associated with utility levels and and0 are well defned for all nonnegative price vectors. Let e (p; ) denote the value of theexpenditure function for output price vector p and utility level : If u(x) ; x 0 andp x = e (p; ); show that there exists a positive number such that p (x)= e (p; 0).

Proof. First note that x solves the expenditure min problem for the utility level . Nex,note also that by the assumptions of the problem f1 is well dened. So, let:

=f1 (0)f1 ()

50

Chapter 4

Game Theory and GeneralEquilibrium

4.1 Game theory

4.1.1 Zero-sum games

Denition 76 The standard n1 simplex is the set = fx 2Rnjxi 0 andPn

i=1 xi = 1gRemark 46 The standard n 1 simplex is a closed, non-? and convex subset of Rn:Denition 77 A mixed strategy is a realization x of the random variable X that assignsprobabilities to each of the pure strategies (or actions) available to players. If M is theaction set for a player, these randomizations comprise the mixed extension of the actionset (M).

Theorem 66 (Von-Neumans minimax) In a game with two players 1 and 2, eachwith strategy set M and N, with the simplices (M) Rm and (N) Rn as the mixedextensions of M and N, respectively, and A as the matrix of payos:

maxx2(M)

miny2N

xTAy

= min

y2(N)

maxx2M

xTAy

Theorem 67 (Kakutanis minimax) Suppose that C Rn and D Rn are non-?; convex, compact. Suppose that f : C D ! R is continuous and x 7! f (x;y)is quasiconcave while y 7! f (x;y) is quasiconvex. Then 9 (x; y) 2 C D such thatf (x; y) f (x; y) f (x;y). That is, (x; y) is a saddle point.

4.1.2 Non-zero-sum games and Nash Equilibrium

Preliminaries

The following game theoretic results are obtained in the context of a game with:

51

A set of players N = f1; 2; :::ng A set of actions for each player: Ai Payo functions fi : A1 A2 ::: An ! R

Denition 78 Allowing for mized strategies, a strategy prole x^ is a vector of random-izations, one for each player, containing x^1 2 (A1) ; :::x^n 2 (An) where, as before,(Ai) =

nxi2Rmjxji 0 and

Pmj=1 x

ji = 1

o:

Denition 79 The strategy prole x^ = (x^1; x^2:::x^n) 2 A1 A2 ::: An is a Nashequilibrium if, for each i 2 N one has that: x^i 2 arg max

yi2Aif (yi; x^i)

Remark 47 Recall that the continuity properties of correspondences are inherited by their(Minkowskis) sum and (cartesian) product, i.e., if 1 : X Y and 2 : X Z are bothupper hemicontinuous (UHC), then = 1 (x) + 2 (x) and = 1 (x) 2 (x) are UHCtoo.

Remark 48 Recall that non-emptiness, compactness and convexity properties of sets arealso inherited by their (Minkowskis) sum and (cartesian) product.

Theorem 68 (Nash, 1950) Suppose that:

1) : Ai Rm is compact, convex, non-?2) : fi : A1 A2 ::: An ! R is continuous3) : xi 7! fi (xi;xi) is quasiconcave

then the game has a Nash Equilibrium.

Note that in order to nd Nash equilibria, we need to be able to obtain the expectedpayo of each player for each strategy prole. Suppose that g () describes the payofunction. Then we need to compute:

E [f (xi;xi)] =m1Xi1=1

:::

mnXin=1

gi (i1; :::in) xi11 ... x

inn

:

Theorem (68) is the most widely used to justify the existence of equilibrium in strategicform games. However, this result does not address the issue of uniqueness of equilibria.There are mainly two avenues towards ensuring uniqueness, both of which rely on xedpoint arguments; the rst one uses monotonicity of the objective function (as was hintedin section 1.8) while the second uses a contraction mapping feature.

52

Exercise 69 (uniqueness via monotonicity) Let (u1; u2; A1; A2) comprise a 2-personstrategic form game. Suppose that Ai = R and that ui : R R! R is twice dierentiableand concave in xi for each xi. If the following two conditions hold:

(1) :@u1@x1

(x1; x2) = 0 =@u2@x2

(x1; x2)

(2) :

@2u1@x21

(z) @2u1

@x1@x2(z)

@2u1@x1@x2

(z) @2u1@x22

(z)

!is negative denite 8z 2R2

show that (x1; x2) is the unique Nash equilibrium of the game.

Proof. First note that xi 7! ui (x1; x2) concave along with the two conditions (1) (2)imply that (x1; x2) is a Nash equilibrium for the game. Next, to see that it is unique, letz =(x1; x2) and dene fi (z) = @ui (z) =@xi: Thus, condition (2) implies (by proposition(21)) that f is strongly monotone. Therefore, by example (36) we know that the systemf (z) = 0 has a unique solution so that f (z) = 0 also has a unique solution and weconclude that z =(x1; x2) is the unique Nash equilibrium of the game.

Exercise 70 (uniqueness via contraction) Consider the setup of theorem (68) and mod-ify only the following: for each i 2 N , xi 7! fi (xi;xi) is strictly quasiconcave and'i : A1 ::: An ! Ai is the best response mapping. Note that in this case strict quasi-concavity implies that 'i is single-valued, i.e. it is a best response function. Show that if' is a contraction, the game has a unique Nash equilibrium.

Proof. First note that the conditions for the existence of an equilibrium are satised.Next, suppose that x and x0 are two Nash equilibria and x 6= x0. Since ' is a contractionwe know that 9 < 1 such that d (' (x) ; ' (x0)) d (x;x0). However, if x and x0 are bothNE, then x =' (x) and x0 = ' (x0) implying that d (x;x0) d (x;x0) which is possibleonly if x = x0.

Next we provide Debreus (1952) version of the main theorem for the existence of Nashequilibria in so-called "generalized" games. This is the key step towards the original proofof the existence of equilibrium in pure exchange economies by Arrow and Debreu presentedin section 4.2.2.

4.1.3 The generalized game

Consider the case of a generalized game, also known as an "abstract economy". In thissetup, feasible actions are not independent as in the simple nperson game, but thereexists a transition correspondence i : Ai Ai which maps action by players i intofeasible actions by player i; that is, it yields the set of actions that player i can take, givenwhat every one else is doing. Under the appropriate topological assumptions regarding ithe existence of equilibrium can be ensured as the following theorem states.

53

Theorem 71 (Debreu, 1952) Suppose that:

1) : Ai Rmi is compact, convex, non-?2) : fi : A1 A2 ::: An ! R is continuous3) : xi 7! fi (xi;xi) is quasiconcave4) : i : Ai Ai is UHC, LHC, compact,convex, non-?-valued

then a Nash equilibrium for this generalized game (abstract economy) exists, i.e., 9 a prolex^ = (x^1; x^2:::x^n) 2 A1 A2 ::: An such that, for each i 2 N :

x^i 2 arg maxyi2i(x^i)

f (yi; x^i)

Proof. (step 1) First note that Ai Rmi compact, convex, non-? ) A1 ::: An Rm1 ::: Rmn compact, convex, non-?.(step 2) Next, for each (x1; :::xn) 2 A1 ::: An, let:

i (xi) = arg maxyi2i(xi)

fi (yi;xi)

now dene: (x) = 1 (x2; :::xn) ::: n (x1; :::xn1)

and note: (x) A1 ::: An

(step3) Note that since fi (xi;xi) is continuous and i () is non-?, UHC, LHC,compact-valued, by the Continuous Maximum theorem (Theorem 26) we have that i (xi)is non-?;compact-valued and UHC for each i. Furthermore, since xi 7! fi (xi;xi) isquasiconcave and i () is convex-valued, by the Concave Maximum theorem (Theorem27) i (xi) is convex-valued for each i. Therefore, (x) is also convex-valued.(step 4) Since the set A1:::An is a non-?, compact, convex set and : A1:::An

A1 ::: An is a non-?, compact, convex-valued, UHC correspondence, by Kakutanistheorem, has a xed point, i.e. 9 x^ =(x^1; :::; x^n) A1 ::: An such that:

x^ 2 (x^)x^i 2 i (x^i) 8 i 2 Nx^i 2 arg max

yi2i(xi)fi (yi;xi) 8 i 2 N

so the "generalized" game has a Nash equilibrium.

4.2 General Equilibrium theory

4.2.1 Prelminaries

Denition 80 (pure exchange economy) A pure exchange economy is dened by a setof consumers N = f1; 2; :::ng, a set of endowments for each consumer !i 2 RL+ and payosfor each player/consumer ui : RL+ ! R:

54

Denition 81 (improvable by coallition) Suppose that (x1; :::;xn) is a feasible alloca-tion of the pure exchange economy. Let S N . We say that S can improve upon x if, foreach i 2 S; 9 x0i 2 RL+ such that: X

i2Sx0i

Xi2S!i

ui (x0i) > ui (xi)

Denition 82 (core allocation) A feasible allocation is a core allocation if it cannot beimproved by any coallition.

Remark 49 When agents are given the opportunity to trade their endowments, the out-come of the exchange will lie inside the core of the economy.

Denition 83 (Walrasian equilibrium) In a pure exchange economy a (n+ 1)tuple:(x^1; :::; x^n;p) is a Walrasian equilibrium (W-Eq) if p 2 RL+n f0g and:

1) : x^i 2 arg maxpxip!i

ui (xi) (optimal choices)

2) :nXi=1

x^i nXi=1

!i (no excess demand)

Remark 50 Note that each !i is a L 1 vector. Thus, the complete set of endowmentsfor the economy ! is a L N matrix whose typical element !ki is the amount of good kthat player i is endowed with. The same note applies for the set of choices xi

Remark 51 Since !i and xi are L 1 vectors, the second condition aboves implies:0B@Pn

i=1 x^1i

...Pni=1 x^

Li

1CA 0B@Pn

i=1 !1i

...Pni=1 !

Li

1CAthat is, the sum of choices of a particular commodity accross players must be less than orequal to the economys total endowment of such commodity.

Example 72 (W-Eq does not exist) Consider an economy with two agents. The econ-omy is characterized by the following endowments and preferences:

!1 = (a1; b1) = (1; 1) u1 (x1; y1) = x21 + y

21

!2 = (a2; b2) = (1; 1) u2 (x2; y2) = x2y2

A W-Eq for this economy does not exist. To see why, let p =(p; q) be the price vector andnote that the we can use the KT conditions to solve the problems:

maxx21 + y21 maxx2y2

s:t: px1 + qy1 p !1 s:t: px2 + qy2 p !255

now an interior solution to the KT system would be:

2x1 p1 = 0 x2 q2 = 02y1 q1 = 0 y2 p2 = 0

px1 + qy1 = p+ q px2 + qy2 = p+ q

whose solution is:1 = 2

p+qp2+q2

2 =p+q2pq

x1 = p

p+qp2+q2

x2 =

p+q2p

y1 = q

p+qp2+q2

y2 =

p+q2q

note that if p > q then p

p+qp2+q2

+ p+q

2p> 2 so that there is excess demand for good x.

Likewise, if p < q then q

p+qp2+q2

+ p+q

2q> 2 so that there is excess demand for good y.

Finally, if p = q there are two solutions to agent 1s problem in either of which there wouldbe excess demand for one good. Thus, a W-Eq for this economy does not exist.

So what is wrong with the last example? The main issue is that the utility functionof agent one (x1; y1) 7! u1 (x1; y1) is convex. In the following section we explore what isrequired from preferences in order for a Walrasian equilibrium to exist.

4.2.2 Existence of equilibrium in pure exchange economies

In this section we present a version of the main existence theorem for the case of pureexchange economies. This is nested in the more general "main theorem" found in Arrowand Debreu (1954).

Theorem 73 (Arrow and Debreu, 1954) Suppose a pure exchange economy with !i 2RL+n f0g and ui : RL+ ! R where ui () is continuous and strictly quasiconcave on xi. Thena Walrasian equilibrium exists for this economy.

Proof. (step 1: set up a generalized game) First, dene the L-simplex of relative prices:

=

p 2 RL+ j

LPk=1

pk = 1

and:

K =z 2 RL+ j z

nPi=1

!i + (1)

where (1) is a L 1 vector of ones. Next, for each p 2 dene the zero-player with setof feasible actions:

=0 (x1; :::;xn)

56

and payo function:

u0 (p;x) =nPi=1

p (xi !i)

and note that now we have (n+ 1) players: n consumers and the price-setter (auctioner).Next, for a typical consumer i dene the budget set correspondence:

i (p) =xi 2 RL+ j p xi p !i

and the feasible action correspondence:

i (p) = i (p) \ K

and note that the feasible action correspondence for player i is parametrized only by pwhich is chosen by player zero.(step 2: show the generalized game has a NE) Next, notice that 0 (x) = is obviously

non-? compact, convex, UHC, LHC. Note also that K compact and i (p) closed) i (p)\K =i (p) compact. Finally, one can easily show that i () is UHC, LHC. On the otherhand u0 () is continuous and quasiconcave in p since its linear and recall that ui () iscontinuous and quasiconcave on xi 8 i by assumption.Summarizing, since all the objective functions are continuous and quasiconcave on the

decision variable, all the feasible action correspondences are UHC, LHC, convex, non-?, compact-valued, the resulting optimal-value (best-response) correspondences is non-?;compact, convex-valued, UHC and dened on non-?, compact, convex sets. Therefore, byKakutanis theorem, the game has a NE, i.e., 9 an (n+ 1)-tuple (x^1; :::x^n; p^) such that:

p^ 2 arg maxp20(x^)

u0 (p; x^) =Pn

i=1 p (xi !i)x^i 2 arg max

x2i(p^)uip^; x^i;xi

8 i 2 N(step 3: show the NE for the generalized game is a W-Eq) Now the claim is that

(x^1; :::x^n; p^) is a Walrasian equilibrium to the pure exchange economy. To see why, rstnote that:

x^i 2 i (p^) :) x^i 2 i (p^)) p^ (xi !i) 0)Pni=1 p (xi !i) 0

and therefore:

maxp2

Pni=1 p (xi !i) 0

maxp2

PLk=1 pk

Pni=1

xki !ki

0Pni=1 (xi !i) 0

57

and therefore excess demand for all commodities in the economy is less than or equal tozero. Next, to show that all consumers are optimizing, one must show that in this particularcase:

x^i 2 arg maxxi2i(p^)

uip^; x^i;xi

) x^i 2 arg maxxi2i(p^)

uip^; x^i;xi

to see this, suppose 9 x0i 2 i (p^) such that ui

p^; x^i;x

0i

> ui

p^; x^i; x^i

: Then 9 such

that whenever 2 (0; ):

x0i + (1 ) x^i ui

p^; x^i; x^i

) ui p^; x^i; x0i + (1 ) x^i > ui (x^i)which contradicts the result of step 2 that:

x^i 2 arg maxx2i(p^)

uip^; x^i;xi

8 i 2 NSummarizing, x^i 2 argmaxx2i(p^) ui

p^; x^i;xi

, therefore all the consumers are making

optimal choices andPn

i=1 (xi !i) 0 so there is no excess demand. Hence, the pureexchange economy has a Walrasian equilibrium.

4.2.3 Existence of equilibrium in production economies

In this section we present a reduced-form version of Debreus (1959) celebrated theoremfor the existence of equilibrium in private ownership production economies. It is a reducedversion because we assume the existence of a "well behavied" net supply correspondence(see below).

Denition 84 A private ownership production economy is dened by a set of consumersN = f1; 2; :::ng, a set of endowments for each consumer !i 2 RL+; a set of rms M =f1; 2; :::mg owned by consumers with ij as the share of rm j owned by consumer i andPn

i=1 ij = 1 and payos for each player/consumer ui : RL+ ! R:

58

Denition 85 A Walrasian equilibrium in a private ownership production economy is a(n+m+ 1)tuple: (x^1; :::; x^n; y^1; :::; y^m;p) such that p 2 RL+n f0g and for all i 2 N andall j 2M :

1) : x^i 2 arg maxx2i(p)

ui (xi) (consumers maximize)

2) : y^j 2 arg maxyj2Yj

p yj (rms maximize)

2) :nXi=1

x^i nXi=1

!i +

mXi=1

y^j (no excess demand)

where i (p) =nxi 2 RL+ j p xi p !i +

Pmj=1 ij

p y^j

oand Y = Y1+ :::+Ym is the

aggregate production possibility set.

Remark 52 Note that the vector p contains the prices of all commodities in the economy,both inputs and nal goods.

Remark 53 Finally, a pure exchange economy is a special case of a production economywhere yj = f0g 8 j

Example 74 (decreasing returns) Suppose that there are two commodities and twoconsumers with problems:

!1 = (6; 0)