masters phd real analysis abridged

8/6/2019 Masters PhD REAL ANALYSIS Abridged

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Masters/ Ph.D maths course: Real Analysis

Witness Simbanegavi (Ph.D)School of Economics, UCT

February 2009

This module (Real Analysis) is meant to prepare students for Ad-vanced Microeconomics. Modern microeconomics is mathematical in na-ture: it draws extensively from real analysis and calculus hence our needto study this course. The pace at which we cover the material in this moduleis rather hectic (5 days). Consequently, what the course hopes to achieveis for students to have an operational understanding of the concepts cov-ered rather than an ability to produce elaborate proofs. However, easy andstraight forward proofs are fair game. Below are the topics covered in thismodule.

Propositions and theorems may be incorrectly referenced!!

OBS! Topic 7 will not be covered in this course (real analysis) but in theMicro course

TOPICS

1: Introduction

2: Sets and Mappings

Notation Set Theory Mappings

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3: Geometry and Convexity

Basic algebraic operations Convexity Inner product, Norm, Metric Concavity and quasi-concavity of real valued functions

4: Topology

Openness, closedness of sets

Boundary points and cluster points Compactness Connectedness

5: Sequences

Denition and notation

Convergent/divergent sequences

Subsequence

6: Continuity

Characterization of continuity Weierstrass Maximum Theorem

7: Lagrange, Kuhn-Tucker and Envelope Theorems

Partial and directional derivatives Constrained optimization Envelope Theorem

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1 INTRODUCTION

Why study mathematics?

Quantitative analysis Qualitative analysis Training in logical and analytical thinking (intellectual gym) Language course (precise and powerful language)

Contemporary economic paradigm

Paradigm is microeconomic in nature Economy system of economic agents linked to each other via markets

and other economic institutions.

Examples: Problems ofwhat to produce, how to produce and for whomto produce.

Assumptions: Agents are rational and behave "as if" each of them were to max-

imize some goal function.

Goal functions preferences (utility), prots, etc.

Agents generally operate in a given environment. Environment determines the constraint set the agent faces. For

example, market prices determine the consumers budget set orrms revenues and costs.

All this can be summarized in the following mathematical optimizationprogram:

maxx2X(a)

f(x; a) ([M])

where:

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a = agents environment (beyond the control of the agent) x = choice or action available to agent. X(a) = agents opportunity set/ constraint set in environment a

set of all available courses of action.

f = agents goal function.

Solution

Unique solution: There exist x 2 X(a) such that f(x; a) f(x; a) 8x 2X(a) :

Multiple solutions: There exist at least one x 2 X(a) such thatf(x; a) f(x; a) 8x 2 X(a) :

Let X (a) = arg maxx2X(a)

f(x; a) : X (a) is the solution set (solution

correspondence). Obviously, X (a) X(a) :

Value function

Achieved value of the goal function. v (a) max

x2X(a)f(x; a) = f(x (a) ; a)

Examples

Utility maximization Prot maximization

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Given program [M], our task is to address the following:

Q1: Does a solution exist? Q2: If it exists, is it unique? Q3: How can we nd a solution/ characterize it? Q4: How does solution depend on environment, a? Q5: How does the achieved value depend on a?

The tools that we are going to learn in this course should enable us toanswer these questions!

What if more than one agent (i.e., if we are studying the interactionbetween agents)?

USe equilibrium analysis.

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2 SETS AND MAPPINGS

BASIC CONCEPTS

What is a set?

A collection of objects.

Ways of describing a set

List all the "elements"/ members of the set:

For instance, A =

f0; 1; 2; 3

g Specify a common property that characterizes the objects in the set:

For instance: B = fx : x is a nonnegative integer not exceeding 3g

Elements: objects belonging to a set (x 2 X).

x =2 X =) x is not a member/ element of X:

Singleton set: a set containing exactly one element (just a point).Let A and B be some arbitrary sets. One writes A = B if they contain

exactly the same elements.

A B (A contained in B). That is, A is a subset of B: Equivalently,B A:

A B and A 6= B =) A B (proper subset).

Empty set: An empty set is a set that contains no element: denoted, ;:

Example: X = fx : x is a round squareg = ;:

OBS!!;

A for every set A: In particular,; ;

:

As we will see later, the empty set ; and the real line R have someinteresting properties. In particular, they are the only two "clopen" sets!!

Finite set: has nitely many elements.

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Let A be any set. A is nite i#A < +1 (#A denotes the number ofelements in A).

It follows therefore that ; is nite!

Examples:

Innite sets: sets with innitely many elements

SOME USEFUL SETS

R =

fx : x is a real number

g Can you list the real numbers for me???

R+ = R++ =

Observe that we distinguish between "non-negative" and "positive"real numbers!!

Z = fx : x is an integerg N = fx : x is a positive integerg Q = set of rationals = fx 2 R : x = m=n for some m 2 Z and n 2 Ng :

Which of the above sets are nite (innite)?

Exercise 1 How are the setsR;Z;N andQ related?

Exercise 2 Consider the set D = [0; 1]. Is this set nite or innite?

Intervals

Consider the sets A and B given by:

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A = fx 2 R : a x bg ; and B = fx 2 R : a < x < bg where aand b are real numbers such that a < b: The sets A and B arecalled a (bounded) intervals.

Why bounded?

The set D = fx 2 R : a x < +1g = [a; +1) is unbounded. Open, half open, closed intervals: Examples

X = fx 2 R : a < x < bg = (a; b) :X = fx 2 R : a x bg = [a; b] :X =

fx

2R : a < x

b

g= (a; b]

X = fx 2 R : a x < bg = [a; b)

We will discuss in detail the idea of closedness and openness ofsets in Lecture 3.

OPERATIONS ON SETS

For any sets A and B;

A

\B =

fx : x

2A and x

2B

g A listing of only those elements that appear in both sets. A [ B = fx : x 2 A or x 2 B or bothg

A listing of all the elements of the sets A and B.

Two sets A and B are said to be disjoint if their intersection is empty,i.e., they have no element in common.

Generalizations

Let I be an arbitrary nonempty set to be called an index set.For each index i 2 I; let Ai be a set. Then\i2IAi = fx : x 2 Ai8i 2 Ig[i2IAi = fy : y 2 Ai for some i 2 Ig :

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Examples

\n2N[n 1; n] = 0[n2N[n 1; n] = R+[n2N[ 1n+1 ; nn+1 ] = (0; 1)

Why (0; 1) and not [0; 1]??

Compliment of a set

Given any set A X; the set B = fx 2 X : x =2 Ag is called thecompliment of A (relative to X).

B = A

c

B =eAB = XA:

Proposition 3 Let I be an arbitrary set and suppose Ai; for each i 2 I is aset. Then e[i2I Ai = \i2IeAi ande\i2I Ai = [i2IeAi:Proof. In class.

CARTESIAN PRODUCTS

Read in notes!

Binary relations

A binary relation P on a nonempty set X is a subset of its Cartesianproduct with itself, X2 = X X:

Example: the ordering

of real numbers.

Let P = f(x; y) 2 R2 : x yg :x y i(x; y) 2 P:

Observe that the relation P is complete and transitive!

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Completeness: For any x; y 2 R; either (x; y) 2 P or (y; x) 2 Por both. That is, either x y or y x:Transitivity: If (x; y) 2 P and (y; z) 2 P; then (x; z) 2 P:Reexivity: (x; x) 2 P:

Consider the binary relation > on R: Is this relation complete?,transitive, reexive?

SUPREMUM PROPERTY OF R

Denition 1: Suppose X R: A number b 2 R is called an upper bound for X ifx b8x 2 X:

A number a 2 R is called a lower bound for X if a x8x 2 X:A set which has both an upper bound and a lower bound is calledbounded.

Denition 2: Suppose X R:

A number b 2 R is called a least upper bound for X; or a supre-mum of X; written b = sup X; if b is an upper bound for X

and no lower upper bound exists (i.e., if x b8x 2 X; andx b8x 2 X =) b b).A number a 2 R is called a greatest lower bound for X; or aninmum of X; written a = infX; if a is a lower bound for Xand no greater lower bound exists (i.e., if a x8x 2 X; anda x8x 2 X =) a a).

A set X R can have at most one supremum and one inmum!! Ifaand b are the inmum and supremum ofX R; then X [a; b] :

Can I say then that X = [a; b]?

Axiom 4 (The Supremum Property): There exists a supremum for everynonempty set X R which has an upper bound.

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Is a(b) 2 X?

Not necessarily!!

Examples: Find the upper bounds and supremae and inmae ofthe following sets.

X = [0; 1]

X = R

X = (0; 1)

X = R+

If b = sup X

2X; then b = max X:

Likewise, if a = infX 2 X; then a = min X:

Nested intervals property: Any nested collection of closed and boundedintervals has at least one element in common. More formally,

Proposition 5 Suppose that, for each i 2 N; Xi is a nonempty, closed andbounded interval such that Xi+1 Xi8i: Then the intersection of these setsis nonempty: \i2NXi 6= ;:

Illustration!

MAPPINGS

What is a function?

A rule that assigns to each element x of one set, X; a unique elementy = f(x) of a set, Y:

f : X ! Y Here, X is the domain of f (D(f) = X) and Y is codomain of f:

The subset of points y 2 Y which are images of points in X iscalled the range off (R(f)). Thus, R(f) = f(X) = fy 2 Y : y = f(x) for some x

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Examples:

Consider the function f : R! R dened by f(x) = 2x

f(x) = x2

What is the domain of f; the codomain of f; the range of f and thegraph of f??

For any set A X; f(A) = fy 2 Y : y = f(x) for some x 2 A \ Xg

Why x 2 A \ X? f(A) = direct image of set A under f:

Inverse image (pre-image) of set B in Y :

f1 (B) = fx 2 X : f(x) 2 Bg

Surjection: f is surjective (onto) if the range of f is all of Y:

Examples:

f(x) = x; X = Y = R?

f(x) = x2; X = Y = R?

f(x) = x2; x 0 and Y = R+?

Injection: f is injective (one-to-one) if it sends distinct elements of Xto distinct elements of Y: That is, x 6= x0 =) f(x) 6= f(x0):

Examples:

f(x) = a; X = Y = R?

f(x) = x; X = Y = R?

f(x) = x2; X = Y = R?

f(x) = x2; x 0 and Y = R+?

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A function f is called bijective if it is both injective and surjective!

OBS!! f bijective implies f has an inverse. That is, there exists g : Y !X s.t. y = f(x) ix = g(y) = f1(y):

One can easily send back mail to the original sender!

OBS! f1(B) is always well dened even if f does not have an inverse.

Example: f : R! R dened by f(x) = x2:

f doesnt have an inverse. Why?

See example 5 in book!Level sets/ contour sets for a real valued function

For any 2 R; the set A = fx 2 X : f(x) = g is the isoquant off:

The upper level set associated with any level 2 R is the subset Xf ()where the value of f is at least :

Xf () = fx 2 X : f(x) g Examples: utility function/ production function.

Increasing and decreasing functions

A function f is (strictly) increasing if x < x0 =) f(x) < f(x0) A function f is (stricly) decreasing if x < x0 =) f(x) > f(x0) A function f is non-increasing if x < x0 =) f(x) f(x0) A function f is non-decreasing if x < x0 =) f(x) f(x0)

Illustrations!!

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3 GEOMETRY AND CONVEXITY

VECTOR SPACE

Denition 6 A vector space is a set X (whose elements are called vectors)equipped with two binary operations, called vector addition and scalarmultiplication.

A vector space is closed under addition and scalar multiplication.

Properties(i) x + y = y + x(ii) (x + y) + z = x + (y + z)(iii) + x = x; where = (0; 0;:::; 0) is the origin(iv) (x) + x = 0(v) 1x = x(vi) a (bx) = (ab) x(vii) a (x + y) = ax + ay

The operations vector addition and scalar multiplication are denedcomponent wise.

Let x = (x1;::;xn) and y = (y1; : : ;yn) be vectors and

2R a real

number. Thenx + y = (x1 + y1;:::;xn + yn)

andx = (x1;:::;xn)

What is x y?

Denition 7 A subspace ofRn is a set X Rn such that x + y 2 X and 2 R:

OBS! Straight line through the origin in Rn is a subspace while a straightline that does not go through the origin is not a subspace.

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Example 8 Example

INNER PRODUCT, NORM AND METRIC

Inner product multiplication between vectors from the same space.

Denition 9 For any vectors x; y 2 Rn;x:y =nPi=1

xiyi:

What if the vectors are not from the same space. That is, let x 2 Rn andy 2 Rk; k 6= n: What then?

More formally, if X is a vector space, then an inner product (or dotproduct) is a function on X X to R; denoted s (x; y) = x:y; with thefollowing properties:

s (x; x) = 0 ix = s (x; y) = s (y; x) 8x; y 2 X s (x; y + z) = s (x; y) + s (x; z) 8x;y;z 2 X s (x;y) = s (x; y) If x; y 2 Rn are two non-zero vectors, then x:y = 0 i x and y are

orthogonal (perpendicular).

The Norm

Denition 10 If X is a vector space, then a norm on X is a function onX to R; denoted q : x 2 X ! kxk

The Euclidean norm of any vector x 2 Rn; denoted kxk ; is a nonnegativereal number. In particular,

kxk = px:x = " nXi=1

x2i#1

2

:

OBS! The length of the vector (norm) is measured from the origin.

Properties

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q(x) = 0 ix = q(x) = jj q(x) q(x + y) q(x) + q(y) (Triangle inequality).

Proposition 11 (Cauchy-Schwartz inequality) (a) x:y kxk kyk 8x; y 2 Rnand (b) For x; y 6= ;x:y = kxk kyk i y = x for some 2 R++

Proof of triangle inequality.

The Metric

Closely related to the concept of norm. Measures the distance between vectors

Formally, a metric is a function d : RnRn ! R+ satisfying the followingproperties:

d (x; y) = 0 i x = y d (x; y) = d (y; x) d (x; z) d (x; y) + d (y; z)

Balls

Denition 12 Given any vector x

2Rn and scalar

2R++; theopen ball

with center x and radius is the subset:

B (x) = fy 2 Rn : d (x; y) < g = fx 2 Rn : ky xk < g :

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Closed ball similarly dened.

Boundedness

Denition 13 A set X R is bounded if there exist ; 2 R s.t. X [; ] :

Euclidean spaces

Denition 14 A set X Rn is bounded if there exist i; i 2 R s.t. X ni=1 [i; i] :

A set X Rn is bounded i9 such that kxk 8x 2 X:

In other words, no point of X is at a distance greater than from theorigin.

CONVEXITY

Denition 15 A set X Rn is convex if, for all vectors x and y in X andall scalars

2[0; 1] ; also the vector z = x + (1

) y

2X:

z is called a convex combination of x and y: Rn and ; are convex sets!

OBS! There is a dierence between a linear combination and a convexcombination.

For any vectors (points) x and y in X, the relation: z = x+y ; ; 2R is called a linear combination of x and y.

The only requirement of and is that they are scalars.

Convexity is preserved under the operation of intersection.

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Proposition 16 Let be any collection of convex sets X Rn: Then,X = \X2X is a convex set.

Convex hull of a set X; written co(X); is the smallest convex set con-taining X:

Illustration!

It follows that X = co(X) iX is convex!

Hyperplanes

Denition 17 A set X Rn

is a hyperplane if there exists some vectorp 2 Rn; p 6= ; and scalar 2 R such that

X =

(x 2 Rn :

nXi=1

pixi =

):

Vector p is the normal vector to the hyperplane.

ExamplesBudget sets, costs, etc

Theorem 18 Any two disjoint convex sets inRn can be separated by somehyperplane inRn

CONCAVITY AND CONVEXITY OF REAL VALUEDFUNCTIONS

What do you understand by real-valued function?

Denition 19 SupposeX Rn is convex. A functionf : X ! R is concaveif

f(x + (1 ) x0) f(x) + (1 ) f(x0) for all vectorsx; x0 2 X and all scalar 2 [0; 1] :

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OBS! f is convex iff is concave.

Denition 20 For any domainX Rn; subgraph(f) = f(x; y) 2 XR : y f(x)g : These are the points lying on and below the graph (curve).

Concavity of a function can be characterized in terms of the convexity ofthe subgrapgh of f: Equally, a function f is convex if and only if its epigraph(dened as: f(x; y) 2 X R : y f(x)g) is convex.

Strict concavity/ convexity straight forward

Illustrations!!

QUASI-CONCAVITY

Upper level (contour) set for a function f : X ! R is:Xf () = fx 2 X : f(x) g

Denition 21 Suppose f : X ! R where X R is convex. Then fis quasi-concave if Xf () is convex for all 2 R:

Proposition 22 f concave implies f quasi-concave.

Proof. See hand outs.

Proposition 23 If f : X (convex) ! R is quasi-concave and F is increas-ing, then F(f(x)) is quasi-concave.

First, understand the denition of an increasing function!Second, understand the denition of quasi-concavity!

Proof. Suppose f is quasi-concave and F is increasing. Then F(f(x))

F (f(x0)) =) f(x) f(x0) (since F is increasing). By denition, f quasi-concave =) f(x + (1 ) x0) f(x0) : So, F (f(x + (1 ) x0)) F (f(x0)) (F is increasing). Hence F is quasi-concave.

OBS! D (f) = X while D (F) = R (f) : Thus F is a composite function!

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4 TOPOLOGY

OPEN SETS AND CLOSED SETS

Denition 24 A set X Rn is open if there for each x 2 X exists somepositive scalar such that B (x) X: A set Y Rn is closed if its comple-ment is X =eY is open.

Idea is simple: X Rn is open if it is the case that for any ele-ment x 2 X; all the "neighbours" of x are also in X: That is, all the"neighbours" at distance > 0 (no matter how small is) are in X:

Rn and ; are both open and closed!!

Is the set X = fx 2 R : 0 x 1g open or closed? (Use the denition)Properties of open sets

The whole space Rn and the empty set are both open. Arbitrary unions of open sets are open. The intersection of nitely many open sets is open.

Properties of closed sets

A set is closed if and only if its complement is open. The whole space Rn and the empty set are both closed. Arbitrary intersections of closed sets are closed. The union of nitely many closed sets is closed.

Example 25 Let Xi = 1i+1 ; 1i+1 ; i 2 N: Then, \i2NXi = f0g ; a closedset!!

Example 26 LetXi = i

i+1 ;i

i+1

; i 2 N: Then, [i2NXi = (1; 1) ; an open

set!!

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BOUNDARY AND CLUSTER POINTS

Denition 29 A point x 2 Rn is a boundary point of a set X Rn if everyneighbourhood A of x contains some point from X and some point from itscomplement,eX:

No matter how small you make the neighbourhood, at least one pointin the neighbourhood is from X and at least one is not from X:

OBS! x need not belong to X! Same test for open and closed sets.

@X = set of all the boundary points of set X = boundary of X:

Formally:

@X = fx 2 Rn : A \ X 6= ; and A \eX 6= ; for each nbd A of xgExamples: X = [a; b); A = f0; 1g

For any set X Rn; a point x 2 X is either an interior point or aboundary point but not both.

Formally:

int(X) \ @X = ; and X int(X) [ @XExercise 30 What is int(X) [ @X equal to??

Denition 31 A point x 2 Rn is a cluster point of X Rn if every neigh-bourhood A of x contains some point y 2 X such that y 6= x:

OBS! x need not belong to X!

It follows that, for any set X

Rn; every interior point ofX is a cluster

point!

Let X+ denote the set of cluster points of a given set X Rn: Then,for any set X Rn;

int(X) X+:

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Why containment and not equality?

Examples: Find the boundary points, interior points and cluster pointsof the following sets.X = (0; 1)X = f0; 1; 2gX = R+

Proposition 32 Let X Rn: Boundary points which are not in X arecluster points: @X \eX X+: Cluster points which are not interior areboundary points: X+ \

eint(X) @X: Moreover, X[ @X = X = X[ X+:

Corollary 33 X Rn is closed () @X X () X+ X:

Theorem 34 (Balzano-Weierstrass) Every innite and bounded set inRn has at least one cluster point.

CONNECTEDNESS

A set X Rn is connected if it cannot be contained in (covered by)two disjoint open sets, each of which contains some elements from X:

Denition 35 Suppose X Rn: A separation of X is a pair of disjointopen sets A and B such that A \ X 6= ;; B \ X 6= ; and X A [ B:Denition 36 A set X is connected if there exists no separation of X:

Examples:X = fx 2 R : x 6= 3gThe set N R is not connected.

Theorem 37 The set X = Rn is connected.

Proof. Sketch proof in class.

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5 COMPACTNESS AND SEQUENCES

5.1 COMPACTNESS

Crucial property for establishing existence of solutions to agents deci-sion problems.

Crucial for establishing existence of equilibria

Companion to continuity concept.

A covering of any given set X is a collection of open sets such that theseopen sets together cover X:

Denition 38 Let X be any set inRn: A collection of open sets B Rnis called a covering of X ifX is contained in the union of all B 2 : A subset0 is called a subcovering from if it is a covering of X:

Finite covering: A covering consisting of nitely many open sets

Denition 39 A set X

Rn is compact if every covering of X contains a

nite subcovering.

Example 37X =

x1; x2;:::;xk

Rn; k 2 N:Innite collection = fIn R : (1 n; n + 1) ; n 2 Ng is a covering ofX

(in fact R)Can you see that??

contains a subcovering of X; given by In for n > max jxij

One of the intervals the largest one.

OBS! contains no nite subcovering ofR!

Example 38Let X = R2 and let = fB (; n) : n 2 Ng : Then is a covering of X:

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However, has no nite subcovering. However, the covering = [ fR2g does contain nite subcovering,

for instance, 0 = fR2g ; 0 = fB (; n) ;R2g

Denition 40 A set X Rn is compact if every covering of X containsa nite subcovering.

To establish compactness one needs to show that every covering of X

contains a nite subcovering (not easy).

To establish noncompactness, it suces to nd one covering that doesnot contain a nite subcovering.

Revisit example 38

Is X compact??

Example 39Let X = R and let I = fIk R : Ik = (k; k) ; k 2 Ng :

Is I a covering of X? Is X compact?

Example 40Let X = (0; 1) and let I =

Ik R : Ik =

1

k+3 ;k

k+3

; k 2 N

Is I a covering of X?

Does I contain a nite subcover? Is X compact?

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Proposition: Every nite set is compact

OBS! ; is compact while X = Rn is not!OBS! The union of nitely many compact sets is compact.

Proposition 41 Every closed subset of a compact set is compact.

Illustrate!

Heine-Borel Theorem

Denition of compactness dicult to apply in practice. Good news: In Euclidean space, compactness is equivalent with being

both bounded and closed.

Theorem 42 (Heine-Borel) A set X Rn is compact i it is closed andbounded.

Reconsider example 37X =

x1; x2;:::;xk

Rn; k 2 N:

What is int(X)?What is @X?Is X bounded?

5.2 SEQUENCES

Denition 43 A sequence from any set X is a function f : N! X

The function assigns an element xt = f(t)

2X to each positive integer

t:OBS! X here is the range; that is, D (f) = N and R (f) = X:

Notationsfxtg1t=1

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fxtgt2Nfxtg = fxt : t 2 Ng X is the functions range.

Constant sequence: xt = x18t 2 N

ExampleLet f : N! [0; 1] be dened by f(t) = 1

t8t:

Give the terms of this sequence Is there any peculiar features?

Limit point of a sequence

Denition 44 Let X Rn and let fxtg be a sequence from X: A pointx 2 Rn is a limit point of fxtg if there for every nbd A of x exists someT 2 N such that xt 2 A8t T:

Equivalently;

Denition 45 A sequence fxtg inRn converges to a point x if for each" > 0 there exists a natural number T(") such that xt 2 B (x

; ") 8t T orequivalently, d (xt; x

) ! 0 as t ! 1:

Idea: The terms of the sequence eventually cluster (concentrate aroundx)

A sequence which has a limit point is called convergent, otherwise it isdivergent.

Claim 46 A sequence can have at most one limit point (prove it!!)

Subsequence

An innite sampling from a given sequence.

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OBS! Not every sampling yields a subsequence!!

Let fxtg be any sequence in Rn: Ift1 < t2 < t3 < :::: is a strictly increasingsequence of natural numbers, then

fykg1k=1 = fxtkgkis called a subsequence.

Since the sequence tk is strictly increasing, tk k8k:

OBS! The terms of the subsequence are all present in the original se-quence.

Let fxtg = x1; x2; x3;x4;:::: be the original sequence.

Sampling at even times yields the (sub)sequence: fykg = x2; x4;x6;:::

ExampleConsider the sequence: xt =

1t8t 2 N:

Give the terms of the sequence.

Can you nd subsequences of xt? Consider the sequence: xtk = 14 ; 13 ; 12 ; 1; 15 ; 16 ;:::

Is this a subsequence of xt? Why or why not?

OBS! Any sequence is a subset of itself!

Examplef(t) = xt = (1)t 8t 2 N:

Is this sequence convergent? If so, what is the limit point? Is this sequence constant?

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How would you sample to get a convergent subsequence?

Proposition 47 A sequence from a set X Rn converges to x 2 Rn i allits subsequences converge to x 2 Rn:

Intuition:Suppose xt ! x: Then for any nbd A of x; there exists T such that

f(t) 2 A8t T (clustering of the terms of the sequence). Since the termsof the subsequence are all present in the original sequence and since tk k8

k; xtk must lie in A8t

T: (just pull out some terms from xt and leave

everything the same).

Balzano-Weierstrass

Convergence CriteriaA sequence fxtg (of real numbers) is said to(a) Increasing (or non-decreasing) if xt xt+1; for t = 1; 2;:::(b) Strictly increasing if xt < xt+1; for t = 1; 2;:::

(c) decreasing (or non-increasing) if xt xt+1; for t = 1; 2;:::(b) Strictly decreasing if xt > xt+1; for t = 1; 2;:::

Monotone convergence criteria

Gives necessary and sucient condition for convergence of sequencesfrom R:

Cauchy Convergence criterion

Gives necessary and sucient condition for convergence of sequencesfrom any Euclidean space.

A sequence is bounded from above if there exists b 2 R : xt b8t:

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i.e., range of sequence, when viewed as a function, is bounded fromabove.

Theorem 48 (Monotone convergence) An increasing sequencefxtg fromX = R is convergent i it is bounded from above.

Denition 49 A sequence fxtg is a Cauchy sequence if, for every " > 0;there exists a T(") 2 N such that kxt xsk < "8t;s > T(") :

Points suciently "far out" in the sequence are "arbitrarily close toeach other".

Theorem 50 (Cauchy Convergence) A sequence from a Euclidean spaceis convergent i it is a Cauchy sequence.

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6 CONTINUITY

In this chapter, we reach our TARGETED PEAK in the mountain rangeof REAL ANALYSIS.

FunctionsSuppose X Rn and Y Rm: A function f : X ! Y is said to be

continuous at x if its values at points x0 near to x are near to its value at x(= f(x)).

Idea: a "small movement" in the domain does not cause a "big jump" inthe range. That is, nearby points in the domain of the function should map

into nearby points in the range.

Denition 51 f is continuous at x if there for every nbd B of y = f(x)exists a nbd A of x such that f(x0) 2 B8x0 2 A \ X

Denition 52 f is continuous at x if for every " > 0; 9 > 0 such thatd (f(x0) ; f(x)) < "8x 2 X with d (x0; x) < ; or equivalently, such thatf(B (x) \ X) B" (f(x)) :

Continuity and Sequences

Denition 53 A sequencefxtg converges to x 2 Rn if for every" > 0 9T(")such that d (xt; x) < " whenever t > T(") :

Proposition 54 Suppose f : X ! Y andx 2 X: f is continuous at x if andonly if f(xt) ! f(x) for every sequence fxtg of points in X that convergesto x:

Proof. Suppose f is continuous at x and let fxtg be a sequence in X thatconverges to x: Let " > 0 be given. Then there exists a > 0 such thatd (f(x0) ; f(x)) < " whenever x0

2B (x)

\X: Because xt

!x; there exists

a number T such that d (xt; x) < for all t > T: But then xt 2 B (x) \ Xand so d (f(xt) ; f(x)) < " for all t > T: In other words, f(xt) converges tof(x) :

A characterization of Continuity

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Proposition 55 Suppose f : Rn ! Rm: Then, the following are equivalent;(a) f is continuous(b) The pre-image of every open set is open(c) The pre-image of every closed set is closed.

Bad news!OBS! Openness/ closedness are not generally preserved under continuous

mappings

Example 56 f : R! R dened by f(x) = 18x:

Good news!!The properties of compactness and connectedness are always preserved

under continuous mappings.

This is so great news that it warrants being stated as a theorem.

Theorem 57 Suppose f : Rn ! Rm is continuous. If C Rn is compact,then so is f(C) Rm: If D Rn is connected, then so is f(D) Rm:

Weierstrass Maximum Theorem

Let m = 1; so that f is a real valued function. If X Rn is non-emptyand compact, then so is V = f(X) : In particular, since V is bounded, it hasan inmum v0 and a supremum v1. Moreover, because V is closed, v0; v1 2 V:

Theorem 58 (Weierstrass Maximum Theorem) If X Rn is non-empty and compact and f : X ! R is continuous, thenf(x) = supx2Xf(x)

for at least one x 2 X:

Theorem 59 Weierstrass Maximum Theorem (general). Let f : X !R be continuous, X is a non-empty and compact subset of Rn. Then thereexists vectors x, bx 2 X such that:

f(bx) f(x) f(x) 8x 2 X:32


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Proof. In class.

Let us revisit our microeconomic maximization program:

maxx2X(a)

f(x; a) (1)

What do we need on order for our program to have a solution?(i) X(a) to be non-empty and compact(ii) f(x; a) is continuous in x

Under these conditions, X (a) is of necessity compact.

it is the pre-image of a closed set, namely the singleton set that containsf(x) It is a subset of a compact set, X(a) :

To summarize:

Proposition 60 If the opportunity set X(a) in [M] is non-empty and com-pact and the maximand f(x; a) is continuous in x, then the solution setX (a) = arg max

x2X(a)f(x; a) is non-empty and compact.

Exercise 61 Consider a continuous function f over the set X = (1; 2) :Does a maximum exist for f? What then can you conclude about WeierstrassMaximum Theorem?

Lets revisit Chapter 3Observe that ifX(a) Rn is convex and f(x; a) is quasi-concave (w.r.t.

x), then X (a) = arg maxx2X(a)

f(x; a) is convex (though possibly empty).

This follows from the fact that X (a) is either empty (and hence con-vex) or it is the intersection of two convex sets, the upper contour setand the opportunity set.

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If X(a) Rn is convex and f(x; a) is strictly quasi-concave (w.r.t. x),then X

(a) has at most one element (OBS! it can still be empty!)

uniqueness of solution.

THE PEAKCombining existence and uniqueness results

Theorem 62 If X(a) Rn is convex, non-empty and compact and f(x; a)is strictly quasi-concave and continuous inx; then the setX (a) = arg max

x2X(a)f(x; a)

is a non-empty singleton set. That is, a unique solution exists for program

[M]:

A recapFrom Lecture 1, we had the following:Given program [M], our task is to address the following:

Q1: Does a solution exist? Q2: If it exists, is it unique?

Q3: How can we nd a solution/ characterize it?

Q4: How does solution depend on environment, a? Q5: How does the achieved value depend on a?

We have managed to answer Q1 and Q2. In eect we have done thedicult bit. We next try to address Q3-Q5.

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7 SOME USEFUL (MATHEMATICAL) RESULTS

7.1 Lagrange theorem; Kuhn-Tucker theorem, Enve-lope theorem

7.1.1 Partial derivatives and directional derivatives

Let y = f(x1;:::;xn) : The partial derivative of f w.r.t. xi is dened as:

@f (x1;:::;xn)

@xi lim

h!0

f(x1;:::;xi + h;:::;xn) f(x1;:::;xn)h

:

Exactly the same way we dene the derivative of a function of one vari-able!

The partial derivative,@f (x1;:::;xn)

@xi; measures the rate of change of

f(x1;:::;xn) in the direction of the xiaxis (here we are holding all the otherxj ; j 6= i constant).

Example 63 f : R2 ! R dened by f(x) = px1x2 for all x 2 R2+ andf(x) = 0 otherwise. f01 (x1x2) exists at any point x

0

2R2++: Note that rst

(second) partial derivative does not exist along the x2 axis (x1 axis),while the rst partial derivative is zero on the x1 axis and the second iszero on the x2 axis (the function is zero along the axes).

Any dierentiable function f : Rn ! R has n partial derivatives. Theconverse however is not true. A function can have n partial derivatives at x0

without being dierentiable at that point. Why and How?

Reason: Partial derivatives are directional derivatives along the axes.Thus, in order for partial derivatives to exist, it suces that the function has

a well dened slope in the direction of each coordinate axis, while it need notbe well dened in other directions.

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Example 64 f : R2 ! R dened by f(x) = 1 for all x such that x1x2 = 0and f(x) = 0 for all other x: Note that f takes the value 1 along the axesand otherwise its vale is zero. In particular, the function is not continuous atx = (0; 0) ; and thus is not dierentiable at that point. However, its partialsat that point are well dened (= 0).

Proposition 65 If all partial derivatives f0i (x) =@f (x)

@xi; for i = 1;:::;n

exist for all x in some neighbourhood of x0; and if all partial derivatives arecontinuous in x at x = x0; then f is dierentiable at x0:

What about the rate of change of f in other directions?

Fix x0 = (x01;:::;x0n) and consider the rate of change off as we move away

from x0 in the direction ofz = (z1;:::;zn) : Dene a directional function g by:

g (t) = f

x0 + tz

; t 2 R:

What happens to the function g as t increases? Use the chain rule to get:

g0 (t) =n

Xi=1@f (x0 + tz)

@xi:@(x0 + tz)

@t=

n

Xi=1@f (x0 + tz)

@xizi

Evaluating at t = 0 yields

g0 (0) =nXi=1

@f (x0)

@xizi (2)

This is the directional derivative of f at x0 in the direction z:

The vector of partial derivatives

@f (x0)

@x1;:::;

@f (x0)

@xn

is called the gra-

dient off at x

0

and is denoted by rf(x0

) : This vector (rf(x0

)) points inthe direction of steepest ascent off at x0: Thus, from (2) above we can write

g0 (0) = rfx0 z:36


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Observe that@f (x1;:::;xn)

@xi

is a function ofn variables. One can easily de-

rive the second order partials of f (provided@f (x1;:::;xn)

@xiis dierentiable).

Properties of the gradient:

If rf(x) 6= (why rf(x) 6= ), then:1. rf(x) is orthogonal to the level surface f(x1;:::;xn) = c (level curve

in R2).2. rf(x) points in the direction of maximal increase of f(x1;:::;xn) :

7.1.2 Constrained Optimization (One constraint): Lagrange Method

Consider the problem:maxx2X

f(x)

where X R and f : Rn ! R is dierentiable. Suppose that (the constraintset) X = fx 2 Rn : g (x) 0g for a dierentiable function g : Rn ! R suchthat g (x) > 0 for x 2 int(X) and g (x) = 0 for all x 2 @X: Let X denotethe (possibly empty) solution set.

Interior Solutions (unconstrained optimization) The constraint not

binding (i.e., constraint is slack).

Denition 66 The vector x 2 X is a local maximizer of f() if there isan open neighbourhood of x; A X such that f(x) f(x) 8x 2 A: Iff(x) f(x) 8x 2 X (i.e., neighbourhood equal to the set X), then we saythat x is a global maximizer off() :[Local and global minimizer are denedanalogously]

Proposition 67 Suppose that f is dierentiable. If x 2 X is an interiorpoint of X; then rf(x

) = :

Put dierently, x must be a critical point of f: If not, then we must

have that@f (x1;:::;x

n)

@xi6= 0 for some i: That is, a small movement in the

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direction of increasing xi or the opposite direction raises the function value.

But then x

=2 X

:

Note that the F.O.Cs in the proposition above consists of n equations inn unknowns, x1;:::;x

n:

OBS! The rst order conditions are necessary but not sucient for aninterior point to be a solution. Why not sucient?

Example 68 Let f : R ! R dened by f(x) = x2 (x 1) ; and let X =[0; 2] : Find the solution to the maximization program?

x2 (x 1)

0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

-2

-1

0

1

2

3

4

5

6

x

y

Boundary Solutions More relevant case for microeconomics (agent faces

constraints). Recall that (the constraint set) X = fx 2 Rn : g (x) 0g for adierentiable function g : Rn ! R such that g (x) > 0 for x 2 int(X) andg (x) = 0 for all x 2 @X:

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Note that the gradient rg (x) at any point on the boundary of X isorthogonal to the boundary (the zero level curve of g) and points inwardsfrom the boundary (remember g (x) > 0 for x 2 int(X)).

It follows that if a solution x is a boundary of X and f is dierentiableat x; then rf(x) is either the zero vector (constraint not binding andtherefore x is a critical point of f) or rf(x) points in opposite directionto the gradient rg (x)

In other words, rf(x) should be some non-positive scalar of the gradientrg (x) at x:1

Theorem 69 (Lagranges Theorem) Suppose f : Rn ! R and g : Rn !R are continuously dierentiable, and X = fx 2 R

n

: g (x) 0g : If x

2X; g (x) = 0 andrg (x) 6= ; then9 2 R+ such thatrf(x)+rg (x) =:

It is customary to write the rst order conditions as:

rf(x) + rg (x) = (3)g (x) = 0 (4)

consisting of n + 1 equations in n + 1 unknowns, x1;:::;xn; :

In fact, the rst order conditions can be written as:

@L (x; )@xi

= 0; i = 1;:::;n

@L (x; )@

= 0:

L : Rn+1 ! R is dened by

L (x; ) = f(x) + g (x)

is the Lagrangian function.

1 Caveat: This point depends on the denition of the set X: If the set X is dened sothat g (x) 0; then rf(x) should be some non-negative scalar of the gradient rg (x)at x:

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A necessary condition for a boundary point x 2 Rn to be a solution isthat (x; ) 2 R

n+1

is a critical point of the Lagrangian for some 2 R+:OBS! The n equations @L=@xi = 0; i = 1;:::n together are equivalent to

rf(x) + rg (x) = while @L=@ = 0 is equivalent to g (x) = 0:The scalar 2 R+ is called the Lagrange multiplier.Whats the condition rg (x) 6= about?This condition is known as the constraint qualication at x: In order that

exist, it is necessary that the gradient rg (x) at x is non-zero. Note thatwhen the constraint is binding (g (x) = 0), x is not necessarily a criticalpoint off: Hence, gradient

rf(x) need not be zero at x (i.e.,

rf(x)

6= ).

But then, ifrg (x) = ; there is no scalar such that (3) holds!Let XCQ denote the set of all choice variables satisfying the constraint

qualication. Then

XCQ = fx 2 X : rg (x) 6= g :

Recipe for solving Lagrangian problems

Consider the problem:

max(min)f(x1; x2) s.t. g (x1; x2) = c

Step 1. Write down the Lagrangian function

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

Step 2. Take partials of L (x1; x2; ) w.r.t. x1; x2 and and equate themto zero.

Step 3. Solve the three rst order conditions for x1; x2 and :

Example 70 A rm uses inputs K and L of capital and labor, respec-tively, to produce a single output Q according to the production functionQ = f(K; L) = K

1

2 L1

4 : The prices of capital and labour are r and w re-spectively.

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Find the cost minimizing inputs of K and L and also the minimum cost

C, as functions of r, w and Q.This generalizes straight-forwardly to the case of several variables and

several constraints (see for instance, MWG. Section MK!)

7.1.3 Nonlinear Programing

Optimization subject to inequality constraints.

Consider again the problem:

maxx2X

f(x)

where X R and f : Rn ! R is dierentiable. Suppose that (the constraintset) X = fx 2 Rn : g (x) 0g for a dierentiable function g : Rn ! R suchthat g (x) > 0 for x 2 int(X) and g (x) = 0 for all x 2 @X: Let X denotethe (possibly empty) solution set.

For any x 2 X; letI(x) = fi 2 I : gi (x) = 0g

and let

X

CQ

= fx 2 X : frgi (x) : i 2 I(x)g is linearly independentgObserve that we worry only about those constraints that bind (Why??).

The requirement that the subset of gradients frgi (x) : i 2 I(x)g be lin-early independent is known as the constraint qualication at x:

If I(x) = ;, that is, all constraints are met with strict inequality, thenthe constraint qualication is automatically met (no gradients to be tested!).

Theorem 71 (Kuhn-Tucker). Ifx 2 X; then there exists scalars; 1;:::;m;not all zero such that

rf(x) +mXi=1

irgi (x) =

igi (x) = 08i 2 I

. If x 2 X \ XCQ ; then can be taken to be 1 and all i 0:

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Recipe for solving non-linear problems Consider the problem:

max(min)f(x1; x2) s.t. g (x1; x2) 0 (5)Step 1. Write down the Lagrangian function

L (x1; x2; ) = f(x1; x2) + (g (x1; x2)) (6)Step 2. Take partials of L (x1; x2; ) w.r.t. x1; x2 and equate them to

zero.

L1 = f01 (x1; x2) + g01 (x1; x2) = 0 (7)

L2 = f

0

2 (x1; x2) + g0

2 (x1; x2) = 0 (8)

Step 3. Introduce complimentary slackness condition

0 (= 0 if g (x1; x2) > 0) ; or g (x1; x2) = 0 (9)Step 4: Require (x1; x2) to satisfy the constraint.

g (x1; x2) 0: (10)Steps (7)-(10) are generally called the Kuhn-Tucker conditions. These

are necessary conditions for the solution of problem (5) (they are generally

not sucient).

Under what conditions will the rst order necessary conditions also besucient for an optimum to problem (5)?

What we pick out from the recipe is a saddle point of the Lagrangian(x1; x

2; ). L (x1; x2; ) is maximized in x1 and x2 while it is minimized

in :

Recipe for solving non-linear problems with non-negativity con-straints Consider the problem:

max f(x1;:::;xn) s.t.

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The Kuhn-Tucker necessary conditions are:

@f (x)

@xi+

mXj=1

j@g (x)

@xi 0 (= 0 if xi > 0) ; i = 1;:::;n

j 0 (= 0 if gj (x) > 0) ;j = 1;:::;mor, using vector notation:

rf(x) +mX

j=1

jrg (x) 0 and x"rf(x) +

mXj=1

jrg (x)#

= 0

j 0 and jgj (x) = 0 8j 2 I(x) :

The second expression in each line is the complimentary slackness condi-tion.

Exercise 72 A single output rm produces outputqusing nonnegative amountsof inputs z1 and z2 according to the production function f(z1; z2) = z

21 + z

22 :

a. Set up the rms cost minimization problem. [2]

b. Dene conditional factor demands. [2]

c. Using appropriate techniques, derive the rms conditional factor de-mand. What is the rms minimized cost, c (w; q)? [13]

d. What is the economic interpretation of in this case? (Support youranswer algebraically!!) [3]

Exercise 73 maxx1;x2

f(x1; x2) = x21 + x

22 + x2 1 s.t. g (x1; x2) 0; where

g (x1; x2) = x21 x22 + 1:

Exercise 74 maxx1;x2

u (x) = x1x2 subject to

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7.1.4 Envelope Theorem

Consider the maximization program:

v (a) maxx2X(a)

f(x; a)

where X(a) Rn; a 2 Rk and f : Rn Rk ! R:

Suppose you have computed the maximum value function, call it v (a) andit turns out that v (a) is a continuous function ofa; where a = (a1;:::;ak) is theenvironment.2 You may then want to know what happens to the maximumvalue, v (a) as the environment changes. Do we need to redo the optimization

altogether to gure out the eect of a change in the environment?

If the solution set X (a) is a singleton set and the unique solution isa dierentiable function of the environment, then the Envelope Theoremgives the answer!! Consider the case k = 1;i.e., the environment varies inonly one dimension. The treatment obviously diers depending on whetherthe solution is interior or a boundary solution.

Interior solution As we saw above, ifx 2 X (a0) is an interior pointof X(a0) ; then, assuming dierentiability, x must be a critical point of f

(here a0

is the starting values of a; and thus is xed): That is, f0

i (x

; a0

) =@f (x; a0)

@xi= 08i: Suppose x (a) is unique and dierentiable in a: Let

@f (x; a0)

@a= f0a (x

(a0) ; a0) :

By denition:v

a0

= f

x

a0

; a0

:

Question: What happens to v (a) if a moves slightly away from a0?

2 These properties of the value function are guaranteed by Berges Maximum Theorem.

We dont need to worry about it though!

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Use the chain rule of dierentiation:

v0

a0

= df(x (a) ; a)da

ja=a0

= f0a

x

a0

; a0

+nXi=1

f0i

x

a0

; a0 dxi (a0)

da

= f0a

x

a0

; a0

;

where we used the fact x (a0) is a critical point of f:

Observe that a small change in the environment, a aects the valuefunction, v (a) in two ways; a direct eect via fs dependence on a and an

indirect eect via the induced change in the vector x (a) :

What the envelope theorem tells us is that this latter eect is zero, sincethe original decision x (a0) was optimal. In other words, in the neighbour-hood of a0; x is still optimal. Thus, small shifts in the solution point haveno rst order eects on the value function.

Boundary solutions Suppose x (a) is a boundary point, for all a inthe neighbourhood A of a0: By Lagrange Theorem,

f0i x; a0+ g0i x; a0 = 0; i = 1;::;n (12)By hypothesis, x (a) is a boundary point of X(a) for every a 2 A; so,

in particular, g (x (a) ; a) 0. Since its an identity, we can dierentiate theL.H.S. and the R.H.S. and equate the derivatives. Doing so we get:

dg (x (a) ; a)

daja=a0 = g0a

x

a0

; a0

+nXi=1

g0i

x

a0

; a0 dxi (a0)

da= 0

This gives

g0ax a0 ; a0 = nX

i=1

g0ix a0 ; a0 dxi (a0)

da(13)

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Now consider the value function: v (a0) = f(x (a0) ; a0) : Dierentiating

with respect to a gives:

v0

a0

=df(x (a) ; a)

daja=a0

= f0a

x

a0

; a0

+nXi=1

f0i

x

a0

; a0 dxi (a0)

da

= f0a

x

a0

; a0 nX

i=1

g0i

x

a0

; a0 dxi (a0)

da

= f0a

x

a0

; a0

+ g0a

x

a0

; a0

= @L@a a=a0 :where the rst equality follows from the denition of v and the third equalityfollows from (12) while the penultimate equality follows from (13).

Obs! In the special case when f0a (x (a0) ; a0) = 0 for all x and a (the

environment does not directly aect the evaluation function) and the con-straint function g is of the form g (x; a) = a h (x) ; for some dierentiablefunction h : Rn ! R; then X = fx 2 Rn : a h (x)g ; and thus

v0 (a) = :

The Lagrangian multiplier then is the shadow price of the constrainth (x) a; in the sense that a small increase in the constraining quantity a0results in an increase in the maximum value of f by the amount 4v = 4a:In this sense, is the price, expressed in units of the evaluation functionf that the economic agent would be willing to pay per unit for a marginalrelaxation of the constraint.

Theorem 75 (Envelope Theorem) Suppose x0 is the unique solution whenthe environment is a = a0: Under the dierentiability assumptions above:

@v (a0

)@a

= @f (x0

; a0

)@a

if x0 2 intXa0@v (a0)

@a=

@f (x0; a0)

@a+ 0

@g (x0; a0)

@aif x0 2 @Xa0

masters phd real analysis abridged

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