Optimisation theory
M. Baltovic
MT3042, 2790042
2011
Undergraduate study in
Economics, Management,
Finance and the Social Sciences
This is an extract from a subject guide for an undergraduate course offered as part of the
University of London International Programmes in Economics, Management, Finance and
the Social Sciences. Materials for these programmes are developed by academics at the
London School of Economics and Political Science (LSE).
For more information, see: www.londoninternational.ac.uk
This guide was prepared for the University of London International Programmes by:
Mark Baltovic, Department of Mathematics, London School of Economics and
Political Science.
This is one of a series of subject guides published by the University. We regret that due to
pressure of work the author is unable to enter into any correspondence relating to, or arising
from, the guide. If you have any comments on this subject guide, favourable or unfavourable,
please use the form at the back of this guide.
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Reprinted with minor revisions 2011
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Contents
Contents
1 Introduction
1
1.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.3
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
1.4
How to use this subject guide . . . . . . . . . . . . . . . . . . . . . . . .
2
1.5
Structure of the subject guide . . . . . . . . . . . . . . . . . . . . . . . .
3
1.6
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.7
Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.8
Online study resources . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
1.8.1
The VLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.8.2
Making use of the Online Library . . . . . . . . . . . . . . . . . .
5
Examination advice . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
1.10 Syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.11 Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
1.12 List of abbreviations used in this subject guide . . . . . . . . . . . . . . .
7
1.9
2 Mathematical preliminaries
9
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Further reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
2.2
Vector calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
2.2.1
Inner products . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
2.3
Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
11
2.4
Basic set theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2.4.1
Subsets of R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
2.4.2
Bounded subsets of Rn . . . . . . . . . . . . . . . . . . . . . . . .
14
2.4.3
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
2.5.1
15
2.5
Open and closed sets . . . . . . . . . . . . . . . . . . . . . . . . .
i
Contents
2.5.2
Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
17
Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.6.1
Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
18
2.6.2
Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
2.6
3 Weierstrass’ Theorem
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
21
3.2
The Theorem of Weierstrass . . . . . . . . . . . . . . . . . . . . . . . . .
22
3.3
Proving the theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
23
3.3.1
Continuity and compactness . . . . . . . . . . . . . . . . . . . . .
23
3.3.2
Compactness and boundedness . . . . . . . . . . . . . . . . . . .
26
Worked examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
27
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
30
3.4
4 Unconstrained optimisation
39
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
39
4.1
The basic optimisation problem in Rn . . . . . . . . . . . . . . . . . . . .
39
4.2
First-order conditions for optima . . . . . . . . . . . . . . . . . . . . . .
41
4.3
Definite and semidefinite matrices . . . . . . . . . . . . . . . . . . . . . .
42
4.4
Second-order conditions for optima . . . . . . . . . . . . . . . . . . . . .
43
4.5
Worked example
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
43
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
45
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
46
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
47
5 Optimisation under equality constraints
ii
21
53
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
5.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
53
5.2
Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . .
53
5.2.1
55
The standard form . . . . . . . . . . . . . . . . . . . . . . . . . .
Contents
5.3
The Theorem of Lagrange . . . . . . . . . . . . . . . . . . . . . . . . . .
56
5.3.1
Necessary versus sufficient conditions . . . . . . . . . . . . . . . .
57
5.3.2
Local versus global optimal points . . . . . . . . . . . . . . . . . .
58
5.4
Proof of Lagrange’s Theorem . . . . . . . . . . . . . . . . . . . . . . . .
58
5.5
The constraint qualification . . . . . . . . . . . . . . . . . . . . . . . . .
61
5.6
The Lagrangean multipliers . . . . . . . . . . . . . . . . . . . . . . . . .
63
5.7
The cookbook approach . . . . . . . . . . . . . . . . . . . . . . . . . . .
64
5.7.1
The Lagrangean . . . . . . . . . . . . . . . . . . . . . . . . . . . .
65
Worked examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
66
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
71
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
5.8
6 Optimisation under inequality constraints
83
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
6.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
83
6.2
Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . .
84
6.2.1
The standard form . . . . . . . . . . . . . . . . . . . . . . . . . .
84
6.2.2
Effective constraints . . . . . . . . . . . . . . . . . . . . . . . . .
84
The Theorem of Kuhn and Tucker . . . . . . . . . . . . . . . . . . . . . .
85
6.3.1
Necessary versus sufficient conditions . . . . . . . . . . . . . . . .
86
6.4
Proof of the Kuhn-Tucker Theorem . . . . . . . . . . . . . . . . . . . . .
86
6.5
The mixed constraint problem . . . . . . . . . . . . . . . . . . . . . . . .
88
6.6
The constraint qualification . . . . . . . . . . . . . . . . . . . . . . . . .
88
6.7
The Kuhn-Tucker multipliers . . . . . . . . . . . . . . . . . . . . . . . . .
90
6.8
The cookbook approach . . . . . . . . . . . . . . . . . . . . . . . . . . .
91
6.8.1
The Lagrangean . . . . . . . . . . . . . . . . . . . . . . . . . . . .
92
Worked examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
93
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
98
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
100
6.3
6.9
7 Finite Horizon Dynamic Programming
127
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
127
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
127
iii
Contents
7.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
127
7.2
Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . .
131
7.3
Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
132
7.4
Value functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
134
7.5
The method of backwards induction . . . . . . . . . . . . . . . . . . . . .
135
7.6
The Bellman equations . . . . . . . . . . . . . . . . . . . . . . . . . . . .
137
7.6.1
Suitable conditions for the Bellman equations . . . . . . . . . . .
140
7.7
A cookbook approach . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
141
7.8
Worked examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
142
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
148
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
149
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
152
8 Infinite Horizon Dynamic Programming
169
Essential reading . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
Aims and objectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
8.1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
169
8.2
Statement of the problem . . . . . . . . . . . . . . . . . . . . . . . . . .
170
8.3
Strategies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
171
8.4
Existence of an optimal strategy . . . . . . . . . . . . . . . . . . . . . . .
173
8.5
Worked example
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
175
Learning outcomes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
178
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
179
Solutions to exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
182
A Examination advice
191
General remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
191
Showing your working . . . . . . . . . . . . . . . . . . . . . . . . . . . .
191
Covering the syllabus . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
191
Expectations of the examination paper . . . . . . . . . . . . . . . . . . .
191
Examination technique . . . . . . . . . . . . . . . . . . . . . . . . . . . .
192
B Sample examination paper 1
193
C A guide to answering Sample examination paper 1
199
D Sample examination paper 2
215
iv
1
Chapter 1
Introduction
1.1
Introduction
Optimisation theory is a Level 3 course (also known as a ‘300 course’) offered on the
Economics, Management, Finance and the Social Sciences (EMFSS) suite of
programmes. It is based on a similar course at LSE where it is offered as a second or
third year course in the Department of Mathematics.
Optimisation problems occur a lot in finance, economics, and operations research, as
well as a lot of applied mathematics courses. Because of this, students feel quite familiar
with the nature of the problems, which is sometimes an obstacle. Many students tend to
see optimisation problems as exercises in calculus and algebra, whereas a great deal of
the work involved in the problems seen here is the application of a theory, and not just
a method.
One of the obstacles of this course then is that students may have extensively used the
methods presented here without questioning their validity, accepting the fact that they
will always yield a solution to a given optimisation question. Here, we will not only ask
ourselves when this could fail, but explore the reasons why as well. Instances of failure
prompt us to ask why the theory has failed us, and what can be done to obtain a
solution, and you should expect to be given optimisation problems which offer no
solution.
So this course is not just about learning and applying a method, it is also about
understanding why and when the theory works and in what situations it will fail. You
will also get to see and practise how we can begin to translate the real world into
variables and quantifiable relationships, and hence use mathematics to model, and then
solve, real problems.
I hope that you enjoy studying this course.
1.2
Aims and objectives
This half course is designed to:
enable you to obtain a rigorous mathematical background to optimisation
techniques used in areas such as economics and finance
enable you to understand the connections between the several aspects of continuous
optimisation, and about the suitability and limitations of optimisation methods for
different purposes.
1
1
1. Introduction
1.3
Learning outcomes
At the end of this half course and having completed the Essential reading and activities
you should:
have knowledge and understanding of important definitions, concepts and results,
and how to apply these in different situations
have knowledge of basic techniques and methodologies in the topics covered
have basic understanding of the theoretical aspects of the concepts and
methodologies covered
be able to understand new situations and definitions, including combinations with
elements from different areas covered in the course, investigate their properties, and
relate them to existing knowledge
be able to think critically and with sufficient mathematical rigour
be able to express arguments clearly and precisely.
1.4
How to use this subject guide
This subject guide is not supposed to be a textbook to the course. It is here to help you
interpret the syllabus and give practical applications of the theory. You should read each
chapter to find out what you are expected to know for each area of the syllabus, and
use the questions provided to test your understanding. If you experience any difficulties
with the text, each chapter recommends relevant readings from the set textbook.
Unlike many of the courses available on the International Programmes there is only one
set textbook which you must read for this course; much of the information you need to
learn and understand is contained in examples and activities within the subject guide
itself.
It is important that you appreciate that different topics are not self-contained. There is
a degree of overlap between them as each chapter introduces a new topic in the light of
topics you will have seen in earlier chapters.
In terms of studying this course, the chapters of this subject guide are designed as
self-contained topics of study, but for examination purposes you need to have an
understanding of the subject as a whole, so it is important that you get a good
understanding (including being able to identify your weaknesses) of a topic before
progressing from one chapter to the next. In particular, Chapters 2, 3 and 4 provide
background material for the entire course, while Chapters 5 and 6 and Chapters 7 and 8
are two pairs of subjects which clearly exhibit overlap in nature and content. Thus you
are encouraged to see the common points between equality-constrained and
inequality-constrained optimisation problems, as well as those between finite and
infinite horizon dynamical programming problems.
Every chapter which introduces new material has exercises, with solutions, for you to
test your understanding. A good way to proceed is to read through an exercise and see
if you can answer it without help. If not, try to first use the solutions as suggestions for
your own answer. If this is not possible, then work through the solutions as you would
the text, making sure you question why each step is done. Remember, a solution you do
2
1.5. Structure of the subject guide
not understand is of no use to you. While this course attempts to systematise the
theory into a readily applicable recipe or ‘cookbook’ procedure, you must still
understand the theory to successfully practise the methods here.
At the end of each chapter you will find a checklist of your learning outcomes which is a
list of the main points that you should understand, once you have covered the material
in the chapter and the associated readings. If you still have problems with these
concepts, then you should refer back to Sundaram or other texts for further guidance.
1.5
Structure of the subject guide
Chapter 2 surveys the mathematical preliminaries that students should be familiar
with.
Chapter 3 introduces the Theorem of Weierstrass, which is a key first step in
solving the optimisation problems of the subsequent chapters.
Chapter 4 considers unconstrained optimisation of differentiable functions and
introduces the theorems for first-order and second-order conditions for local optima.
Chapter 5 introduces constrained optimisation, beginning with equality-constrained
optimisation problems and the Theorem of Lagrange, which we then use to develop
a cookbook approach to the solution of such problems.
Chapter 6 continues with the material seen in Chapter 5 and introduces
inequality-constrained optimisation problems and the Theorem of Kuhn and
Tucker. This is then used to develop a cookbook approach to the solution of such
problems.
In Chapter 7 we introduce time into our optimisation problem. Contrasted with the
single-period, single-decision problems of the previous chapters, our problems now
require a sequence of decisions to be made, with earlier decisions affecting what is
optimal at later stages. Necessary properties for a solution are introduced in the
Bellman equations. With these, we develop a program for solving these problems.
Chapter 8 extends the ideas of Chapter 7 a little further and considers what
happens when the the time span becomes infinite. Many of the ideas in this part of
the course are quite technical, and we prefer to give a brief overview of the ideas in
favour of a more practical approach to solving such problems.
1.6
R
Essential reading
You should purchase:
Sundaram, R.K. A First Course in Optimisation Theory. (Cambridge: Cambridge
University Press, 1996) [ISBN 9780521497701].
Each chapter of the subject guide commences by identifying the appropriate chapters
from this textbook.
It should be noted that this subject guide builds on previous knowledge and
understanding you will have gained in studying for the prerequisite courses if you are
studying this course as part of a BSc degree.
Detailed reading references in this subject guide refer to the editions of the set
3
1
1
1. Introduction
textbooks listed above. New editions of one or more of these textbooks may have been
published by the time you study this course. You can use a more recent edition of any
of the books; use the detailed chapter and section headings and the index to identify
relevant readings. Also check the VLE regularly for updated guidance on readings.
1.7
Further reading
Please note that as long as you read the Essential reading you are then free to read
around the subject area in any text, paper or online resource. To help you read
extensively, you have free access to the virtual learning environment (VLE) and
University of London Online Library (see below).
R
For the pure mathematics background, a good book to refer to is:
Rudin, W. Principles of Mathematical Analysis (International Series in Pure &
Applied Mathematics). (New York; London: McGraw-Hill Publishing Co., 1976)
[ISBN 0070856133].
However, I recommend that you find a textbook that you are comfortable with, as most
will provide the necessary topics needed for this chapter. Different authors have
different writing styles, and you should take some time in finding one whose voice suits
your needs. Of the authors listed in Sundaram’s text,ones which have been useful to my
students are :
R
R
R
R
Bartle, R.G. and D.R. Sherbert Introduction to Real Analysis. (New York: Wiley,
1999) [ISBN 0471321486].
Hewitt, E. and K. Stromberg Real and Abstract Analysis. (New York:
Springer-Verlag, 1965) [ISBN 0387901388].
Royden, H.L. Real Analysis. (New York: Macmillan, 1988) [ISBN 0024041513].
I would also recommend:
Bryant, V. Yet Another Introduction to Analysis. (Cambridge: Cambridge
University Press, 1990) [ISBN 052138835X].
A useful test is to try and identify a particular concept you had difficulty in
understanding and using this as a litmus test for various texts. Finally, do not expect
the answers to all your questions to be found in one textbook. In any course you are
advised to read extensively.
Having said that, there are a number of different textbooks on optimisation theory and,
more generally, on mathematical economics (which tends to cover equality-constrained
and inequality-constrained optimisation problems, and sometimes a little dynamic
programming). You should be aware that some texts will develop procedures which can
be different from the ones here, though they share the same name.
1.8
Online study resources
In addition to the subject guide and the Essential reading, it is crucial that you take
advantage of the study resources that are available online for this course, including the
4
1.8. Online study resources
VLE and the Online Library.
You can access the VLE, the Online Library and your University of London email
account via the Student Portal at:
http://my.londoninternational.ac.uk
You should have received your login details for the Student Portal with your official
offer, which was emailed to the address that you gave on your application form. You
have probably already logged in to the Student Portal in order to register! As soon as
you registered, you will automatically have been granted access to the VLE, Online
Library and your fully functional University of London email account.
If you forget your login details at any point, please email [email protected]
quoting your student number.
1.8.1
The VLE
The VLE, which complements this subject guide, has been designed to enhance your
learning experience, providing additional support and a sense of community. It forms an
important part of your study experience with the University of London and you should
access it regularly.
The VLE provides a range of resources for EMFSS courses:
Self-testing activities: Doing these allows you to test your own understanding of
subject material.
Electronic study materials: The printed materials that you receive from the
University of London are available to download, including updated reading lists
and references.
Past examination papers and Examiners’ commentaries: These provide advice on
how each examination question might best be answered.
A student discussion forum: This is an open space for you to discuss interests and
experiences, seek support from your peers, work collaboratively to solve problems
and discuss subject material.
Videos: There are recorded academic introductions to the subject, interviews and
debates and, for some courses, audio-visual tutorials and conclusions.
Recorded lectures: For some courses, where appropriate, the sessions from previous
years’ Study Weekends have been recorded and made available.
Study skills: Expert advice on preparing for examinations and developing your
digital literacy skills.
Feedback forms.
Some of these resources are available for certain courses only, but we are expanding our
provision all the time and you should check the VLE regularly for updates.
1.8.2
Making use of the Online Library
The Online Library contains a huge array of journal articles and other resources to help
you read widely and extensively.
To access the majority of resources via the Online Library you will either need to use
5
1
1
1. Introduction
your University of London Student Portal login details, or you will be required to
register and use an Athens login:
http://tinyurl.com/ollathens
The easiest way to locate relevant content and journal articles in the Online Library is
to use the Summon search engine.
If you are having trouble finding an article listed in a reading list, try removing any
punctuation from the title, such as single quotation marks, question marks and colons.
For further advice, please see the online help pages:
http://www.external.shl.lon.ac.uk/summon/about.php
1.9
Examination advice
Important: the information and advice given here are based on the examination
structure used at the time this guide was written. Please note that subject guides may
be used for several years. Because of this we strongly advise you to always check both
the current Regulations for relevant information about the examination, and the VLE
where you should be advised of any forthcoming changes. You should also carefully
check the rubric/instructions on the paper you actually sit and follow those instructions.
The examination paper for this course is two hours in duration and you are expected to
answer four questions, from a choice of five. The Examiner attempts to ensure that all
of the topics covered in the syllabus and subject guide are examined. Some questions
could cover more than one topic from the syllabus since the different topics are not
self-contained. A Sample examination paper appears as an appendix to this subject
guide, along with a sample Examiners’ commentary.
The Examiners’ commentaries contain valuable information about how to approach the
examination and so you are strongly advised to read them carefully. Past examination
papers and the associated reports are valuable resources when preparing for the
examination.
In general, it is advisable to read through all the questions in the paper first before
attempting to answer any questions. You should also allow yourself equal time for all
questions which you attempt since they will have equal marks attached.
You are advised to attempt all the exercises at the end of the chapters in order to gain
sufficient mastery of the topics for the examination. In particular, the exercises provided
in Chapters 5–8 are indicative of the level (if not the style) of the examination questions
for this course. A fully worked out optimisation question as presented in one of these
chapters is indicative of the content of an examination question in these topics.
However, please note that the solutions given in these chapters give far more
explanation than is required in an examination answer. For guidance on the length and
content expected in an answer to an examination question, please refer to Appendix B.
Remember, it is important to check the VLE for:
up-to-date information on examination and assessment arrangements for this course
where available, past examination papers and Examiners’ commentaries for the
course which give advice on how each question might best be answered.
6
1.10. Syllabus
1.10
Syllabus
Prerequisite: if taken as part of a BSc degree, then you must have passed 116
Abstract mathematics before this half course may be attempted. You are also
strongly encouraged to take 41 Advanced mathematical analysis.
This course aims to bring together several parts of the wide area of mathematical
optimisation, as encountered in many applied fields. The course concentrates on
continuous optimisation, and in this sense extends the theory studied in standard
calculus courses. The emphasis here will be on the mathematical ideas and theory used
in continuous optimisation. This course covers the following topics:
Introduction and review of relevant parts from real analysis, with emphasis on
higher dimensions.
Weierstrass’ Theorem on continuous functions on compact set.
Review with added rigour of unconstrained optimisation of differentiable functions.
Lagrange’s Theorem on equality-constrained optimisation.
The Kuhn-Tucker Theorem on inequality-constrained optimisation.
Finite and infinite horizon dynamic programming.
1.11
Acknowledgments
I would like to acknowledge the work of Prof. Jan van den Heuvel, Prof. Graham
Brightwell and Prof. Steve Alpern who lectured this course at the LSE before I taught
it. Between them they laid much of the foundation of this course, and hence this subject
guide would not exist without their efforts.
1.12
List of abbreviations used in this subject guide
ECOP
Equality-constrained optimisation problem
FHDP
Finite Horizon (Markovian) Dynamical Programming Problem
FOC
First-order conditions
ICOP
Inequality-constrained optimisation problem
MCOP
Mixed-constraint optimisation problem
SDP
Stationary Discounted Programming Problem
SOC
Second-order conditions
7
1
1
1. Introduction
8
Chapter 2
Mathematical preliminaries
2
Essential reading
R
Sundaram, L.R. A First Course in Optimisation Theory. (Cambridge University
Press, 1996)Chapter 3.
Further reading
R
Anthony, M. Subject guide 41 Advanced mathematical analysis.
Aims and objectives
In this chapter we will review the basic concepts that are used in the theorems (and
proofs) of the later chapters. A lot of these will have been seen in 116 Abstract
mathematics, while much of the rest builds upon these ideas. Students who are also
studying 41 Advanced mathematical analysis will recognise all the topics here.
All of the material here is covered more extensively in Chapter 3 of Sundaram, and
students who would like some insight into the definitions and results we simply state
here should look there first. More detailed discussions can be found in the references
given in the previous chapter.
2.1
Introduction
You are expected to have seen 116 Abstract mathematics, and may also be studying
41 Advanced mathematical analysis. What follows is a brief review of the subject
matter we assume you are familiar with.
Here we will introduce the basic definitions and results that we will use without
question in this subject guide. They are therefore examinable only to the extent they
are used in the statement of results and their proofs. Learning activities have been
provided to highlight some aspects of these concepts which are interesting or which have
proven tricky for students. While no answers have been provided, these questions are
tackled in almost any undergraduate textbook on real analysis.
9
2. Mathematical preliminaries
2.2
2
Vector calculus
A word on notation. We will not distinguish between scalars and vectors in this
course. Thus x can be used to refer to either a real number or to a point in Rn and the
you are expected to understand from the context which it is. Because we will need to
talk about sequences of vectors, we will modify our vector notation and use
superscripts to indicate vector indices. Thus a point y ∈ Rn is a n-tuple of
co-ordinates y i , i = 1, . . . , n, which is written as:
y = y1, . . . , yn .
Notice that we write vectors in rows but they are to be considered as column vectors. If
we wish to refer to y as a row vector, we will denote it y 0 .
2.2.1
Inner products
The inner, or dot, product of two vectors x, y ∈ Rn will be denoted x · y instead
of the
√
more cumbersome hx, yi notation. The norm of a vector x, is then kxk = x · x. We can
define the distance d(x, y) between two vectors x, y ∈ Rn as the norm of their difference:
d(x, y) = kx − yk. We now have enough to state a very important result for vector
analysis:
Theorem 2.1 (The Triangle Inequality) For any vectors x, y, z ∈ Rn we have
d(x, z) ≤ d(x, y) + d(y, z).
Equivalently, for any vectors u, v ∈ Rn , we have
ku + vk ≤ kuk + kvk .
Activity 2.1 It should not be beyond you to prove the Triangle Inequality above,
and the equivalence of the two forms. Once you have done this, use the result to
prove the following inequality: for any x, y, z ∈ Rn , we have
|d(x, y) − d(y, z)| ≤ d(x, z).
Equivalently: for any x, y ∈ Rn we have
|kxk − kyk| ≤ kx − yk .
Finally, a useful result you should have seen in either an analysis course or a linear
algebra course:
Theorem 2.2 (The Cauchy-Schwartz Inequality) For any x, y ∈ Rn we have
|x · y| ≤ kxk kyk .
10
2.3. Analysis
2.3
Analysis
Because we will often be working in Rn , the notation we will use throughout this course
is slightly different from that of 41 Advanced mathematical analysis. Subscripts
will once again be used to denote indices in a sequence; however, vector coordinate
indices will now be indicated by superscripts. Furthermore, vectors will be denoted by
the traditional parentheses, while sequences are denoted by braces.
A sequence of real numbers is denoted {xk }, where xk ∈ R, ∀k ∈ Z
We can also denote a sequence of n-dimensional vectors with {xk }; however, we
now have xk ∈ Rn , i.e. xk = (x1k , . . . , xnk ), ∀k ∈ Z.
Definition 2.1 (Bounded sequences) A sequence {xk } is
1. bounded above if and only if there exists an M such that xk ≤ M for all k ∈ N.
2. bounded below if and only if there exists a L such that xk ≥ L for all k ∈ N.
If {xk } is bounded both above and below, then it is said to be bounded.
Definition 2.2 (Convergence of a sequence) A sequence {xk } of points xk in Rn
converges to a point x (as k tends to ∞) if and only if: for any ε > 0 there exists an
integer K such that kxk − xk < ε for all k > K.
In this case, we write xk −→ x, where the point x is said to be the limit of the sequence
{xk }.
We can approach this definition another way by first defining a null sequence:
Definition 2.3 (Null sequences) A sequence of real numbers {xk } is a null
sequence if and only if: for any ε > 0 there exists an integer K such that |xk | < ε for
all k > K.
That is to say a null sequence is a sequence of real numbers which converges to 0 in the
sense of Definition 2.2. With this definition, Activity 2.2 should be a trivial task to
perform.
Activity 2.2 Prove that a sequence of points {xk } in Rn converges to a point x, i.e.
xk −→ x if and only if the sequence {xk − x} is a null sequence, i.e. xk − x −→ 0.
In the real number line, we have the particular example of a non-convergent sequence: a
sequence which diverges to ±∞.
Definition 2.4 (Divergence to infinity) A sequence of real numbers {xk } diverges
to +∞ if and only if: for all M ∈ R there exists an integer K such that |xk | > M for all
k > K.
However, there is no concept of this for general Rn spaces.
We list some of the basic results of sequences you should already know; their proof is
left as an activity.
11
2
2. Mathematical preliminaries
Theorem 2.3 A sequence in Rn has at most one limit.
2
Theorem 2.4 A sequence {xk } in Rn converges to x if and only if xik −→ xi , for all
i = 1, . . . , n.
Activity 2.3 Prove Theorems 2.3 and 2.4.
One last property of sequences which should be familiar to you is:
Theorem 2.5 Suppose a sequence {xk } of real numbers is monotone
non-decreasing (so that x1 ≤ x2 ≤ . . .) and there exists some number M ∈ R such
that xk ≤ M for all k ∈ N. Then the sequence has a limit x, where x ≤ M .
That is to say, an increasing sequence that is bounded above has a limit. Similarly, a
decreasing sequence that is bounded below has a limit.
2.4
2.4.1
Basic set theory
Subsets of R
Definition 2.5 (Upper and lower bounds) Suppose we have a non-empty set A ⊆ R.
An upper bound of A is a real number u such that u ≥ a for all a ∈ A.
A lower bound of A is a real number l such that l ≤ a for all a ∈ A.
If a set has an upper bound, we say it is bounded above, and similarly for bounded
below. If a set is both bounded above and below, then we simply say that the set is
bounded.
Activity 2.4 Prove that a non-empty set A ⊆ R is bounded if and only if there
exists an M > 0 such that |a| < M for all a ∈ A.
Definition 2.6 (Supremum and infimum) Suppose we have a non-empty set A ⊆ R.
If A has at least one upper bound, then there exists a least upper bound, which is
called the supremum of A, and is denoted sup(A).
If A has at least one lower bound, then there exists a greatest lower bound, which
is called the infimum of A, and is denoted inf(A).
Definition 2.7 (Maxima and minima) Suppose we have a non-empty set A ⊆ R.
If A has an element which is greater than or equal to all elements a ∈ A, then this
element is called the maximum of A, and is denoted max A. Thus max A ≥ a for
all a ∈ A.
12
2.4. Basic set theory
If A has an element which is less than or equal to all elements a ∈ A, then this
element is called the minimum of A, and is denoted min A. Thus min A ≤ a for all
a ∈ A.
The definitions of supremum and infimum are not easy to use when answering
questions, so it is helpful to identify certain properties of them.
Property 2.1 If A is non-empty subset of R that is bounded above, then the
supremum of A is the unique point σ such that:
1. for all x > σ, x ∈
/ A;
2. for all x < σ there exists an τ ∈ A such that x < τ < σ.
A further property, which is often taken as an axiom for the real numbers, is
Property 2.2 If non-empty subset of A ⊆ R has an upper (lower) bound, then its
supremum (infimum) is well defined.
If we follow the convention that:
if a set A admits no upper bounds, then we set sup A = +∞, and
if a set A admits no lower bounds, then we set inf A = −∞,
then we obtain the result that any non-empty subset of R has an infimum and a
supremum.
From Property 2.2 we can deduce that every Cauchy sequence of real numbers has a
limit. It is more useful to state this property in the following form:
Property 2.3 Let A ⊆ R be a non-empty set with an upper bound. Then there exists
a sequence {ak } in A such that ak −→ sup(A).
Finally, we note that while the supremum (infimum) of a bounded subset of R always
exists, the maximum (minimum) does not have to, though when it does, they are equal.
Property 2.4 Let A ⊆ R be a non-empty set.
If sup(A) ∈ A, then max(A) exists and max(A) = sup(A).
If max(A) exists, then max(A) = sup(A).
A similar property holds for infima and minima.
Activity 2.5 Prove Properties 2.1-2.4 above.
Activity 2.6 Determine which of the following sets are bounded, giving brief
reasons for your answer. Furthermore, indicate where possible the: infimum,
supremum, minimum and maximum of each set.
1. A = {a ∈ Z | a < 4}.
13
2
2. Mathematical preliminaries
2. B = {b ∈ N | b < 4}.
3. C = {c ∈ R | c2 < 4}.
2
4. D = {d ∈ R \ Q | d ∈ [0, 1]}.
5. E = {e ∈ R | d2 − 2d − 2 ≤ 0}.
Activity 2.7 Suppose S, A ⊆ R are non-empty sets such that S ⊆ A. Show that
sup A ≥ sup S and inf A ≤ inf S.
2.4.2
Bounded subsets of Rn
Since we do not have an ordering on Rn if n > 1, we must be cautious in the way we
define boundedness in this setting. In fact, we will generalise the result you should have
proved in §2.4.
Definition 2.8 (Bounded in Rn ) A set S ⊆ Rn is bounded if and only if there exists
an M > 0 such that kxk < M for all x ∈ S.
Anticipating Definition 2.12, we see that Definition 2.8 is equivalent to saying that the
set S is a subset of some open ball B(0, M ).
2.4.3
Functions
Before we move on to a topological discussion of functions and their properties, we first
discuss some set-theoretic aspects of them. In particular, when given a function defined
by its actions on points (in Rn ), what do we mean by the image of a set of points under
this function? Or the preimage of a set of points?
Definition 2.9 (The image of a set) Suppose we are given f : X −→ Y and A ⊆ X,
then the image of the set A under f is:
f (A) = { y ∈ Y | y = f (x), x ∈ A} .
Definition 2.10 (The preimage of set) Suppose we are given f : X −→ Y and
B ⊆ Y , then the preimage of the set B under f is:
f −1 (B) = {x ∈ X | f (x) ∈ B} .
Definition 2.10 should not be confused with the much more restrictive notion of the
inverse of the function f . In particular, note that the preimage is always defined for any
set and any function, although not all functions may allow inverses on their domain.
Activity 2.8 Consider the functions
f : [0, 1] −→ R,
14
f (x) = x2 ,
2.5. Topology
and
g : [−1, 1] −→ R,
g(x) = x2 .
First, recall that a function is more than just a mapping of an input value to an
output value: a domain for the mapping must also be specified. Thus, while f and g
may share the same functional form, they are essentially different functions.
Determine the following sets where possible. If this is not possible, give a brief
indication why.
i) The preimage of [0, 1] under f .
ii) The preimage g −1 ([0, 1]).
iii) The inverse of 1 under f .
iv) The inverse of 1 under g.
v) f −1 ({1}).
vi) The preimage of {1} under g.
As Activity 2.8 indicates, preimages can behave in apparently strange ways if you are
not careful. Activity 2.9 explores some of the properties of preimages, and is
recommended if you have not worked with the definition before.
Activity 2.9 Consider the function f : Rn −→ Rm . Show that for any subsets
A1 , A2 ⊆ Rn and B1 , B2 ⊆ Rm we have:
i) f −1 (B1 ∪ B2 ) = f −1 (B1 ) ∪ f −1 (B2 );
ii) f −1 (B1 ∩ B2 ) = f −1 (B1 ) ∩ f −1 (B2 );
iii) f −1 (Rm /B1 ) = Rn /f −1 (B1 );
iv) A1 ⊆ f −1 (f (A1 ));
v) f (f −1 (B1 )) ⊆ B1 .
Finally, we can use the image set of a function to define boundedness of a function
Definition 2.11 (Bounded function) A function f : X −→ Rm is bounded (above
and below) if and only if f (X) is a bounded set.
2.5
2.5.1
Topology
Open and closed sets
We define the general notion of an open set in Rn in terms of open balls, which are in
fact a special class of open sets.
15
2
2. Mathematical preliminaries
Definition 2.12 (Open balls) For any x ∈ Rn and r > 0, the open ball B(x, r) with
centre x and radius r is the set
2
B(x, r) = {x ∈ Rn | kx − yk < r} .
Definition 2.13 (Open sets) A set U ⊆ Rn is open if and only if for all x ∈ U there
exists an r > 0 such that the open ball B(x, r) ⊂ U .
Rather than appeal to the definition throughout the course, we will usually rely on the
observation you will see in the following activity.
Activity 2.10 Suppose a region S ⊆ Rn is such that:
S = {x ∈ Rn | f (x) < a} ,
where f : Rn −→ R is continuous, and a is an arbitrary real number.
Show that S is open.
Definition 2.14 (Closed sets) A set C ⊆ Rn is closed if and only if for all sequences
{xk } of points xk ∈ C such that xk −→ x, then x ∈ C also.
We may also define closed sets in terms of open sets using the following theorem.
Theorem 2.6 A set C ⊆ Rn is closed if and only if its complement in Rn , Rn /C, is
open.
As with open sets above, the observation in the following activity is usually sufficient
for problems in this course.
Activity 2.11 Suppose a region S ⊆ Rn is such that:
S = {x ∈ Rn | g(x) = a} ,
where g : Rn −→ R is continuous and a is an arbitrary real number.
Show that S is closed.
Show that the result still holds if the equality g(x) = a is replaced with a weak
inequality g(x) ≤ a.
The following results will prove useful for future work.
Theorem 2.7 We consider open subsets of Rn .
1. The intersection of a finite collection of open sets is itself open.
2. The union of an arbitrary collection of open sets is itself open.
Similar results hold for closed sets.
16
2.5. Topology
Theorem 2.8 (Closed set operations) We consider closed subsets of Rn .
1. The intersection of an arbitrary collection of closed sets is itself closed.
2
2. The union of a finite collection of closed sets is itself closed.
Note the skewed symmetry between these two results, which can be explained by De
Morgan’s laws of set operations.
We end this subsection with two definitions which will be important when we introduce
calculus to determine optima.
Definition 2.15 (Interior points of a set) Consider a set S ⊆ Rn . The point x ∈ S is
called an interior point of S if and only if there exists some ε > 0 such that
B(x, ε) ⊂ S. That is, if there exists an open ball centred on x which is contained
entirely within S.
Definition 2.16 (Interior of a set) Consider a set S ⊆ Rn . The collection of all
interior points of S is called the interior of S and is denoted int S.
2.5.2
Compactness
Although it requires a more abstract context beyond the scope of this course for this to
be obvious, compactness of a set is essentially a statement about finiteness. Its most
important role in this course is for the Weierstrass Theorem which is an existence result
for solutions to our optimisation problems. While it should be clear to you that a
function over a finite set of points always has a maximum and a minimum, for more
general settings this may not be the case. This is where compactness and continuity
come into play and we will see this in Chapter 3.
Definition 2.17 (Compactness) A set X ⊆ Rn is compact if and only if, for any
sequence {xk } of points in X, there exists a convergent subsequence {xkn } ⊆ {xk }
whose limit lies in X.
Try to see what this definition states and doesn’t state. From the point of view of
proving a given set X is compact solely using Definition 2.17, it is not enough that you
can find a convergent sequence or, indeed, a sequence which contains a convergent
subsequence: compactness means that you can find a convergent subsequence of any
sequence given to you in X. It is therefore quite a strong requirement of a set so we
should not be surprised that most sets fail this condition. It is by virtue of this stringent
definition that compact sets prove to be so useful in the existence result of Weierstrass.
Before we proceed, there is one very useful characterisation of compact sets in Rn .
Theorem 2.9 (Compactness in Rn ) A set X ⊆ Rn is compact if and only if it is
closed and bounded.
Activity 2.12 Prove Theorem 2.9. Remember, this is an ‘if and only if’ proof, so
you must prove two things.
1. If X ⊆ Rn is compact then it is closed and bounded.
17
2. Mathematical preliminaries
2. If X ⊆ Rn is closed and bounded then it is compact.
2
You will need to be thoroughly familiar with Definition 2.14 and Definition 2.17 to
be able to do this activity.
2.6
Functions
2.6.1
Continuity
Not only do we need workable definitions of continuity, but we also need to have a
collection of continuous functions from which to build up new ones - that is, from which
we can easily ascertain the continuity of a given function. Thus polynomials,
trigonometric functions, exponential and logarithmic functions are continuous.
Furthermore, we have the composition rules of continuous functions to fall back on.
Definition 2.18 (Continuity at a point) A function f : X −→ Rm , X ⊆ Rn , is
continuous at a point a ∈ X if and only if, for any ε > 0 there exists a δ > 0 such
that, for all x ∈ X:
kx − ak < δ
⇒
kf (x) − f (a)k < ε.
An equivalent definition of continuity that is useful for us is:
Definition 2.19 (Continuity at a point) A function f : X −→ Rm , X ⊆ Rn , is
continuous at a point a ∈ X if and only if for all sequences {xk } in X such that
xk −→ x as k −→ ∞, we have f (xk ) −→ f (x).
Definition 2.20 (Continuity on a set) A function f : X −→ Rm , X ⊆ Rn , is a
continuous on a set A ⊆ X if and only if it is continuous at every point a ∈ A.
In particular, if f is continuous at every point of its domain, we then simply say that f
is continuous.
Continuity is the lowest level of ‘nice behaviour’ we like in functions in the sense that
they give ‘nice’ results when applied to other ‘nice’ objects. A very useful example of
this is when we combine continuous functions with convergent sequences (that is, ‘nicely
behaved’ sequences).
Theorem 2.10 Suppose f : X −→ Y is continuous (on X) and {xn } is a convergent
sequence in X with limit x ∈ X. Then the sequence {f (xn )} in Y , defined by applying
the function to each point of the sequence {xn }, is also a convergent sequence in Y with
limit f (x) ∈ Y .
This result can be usefully remembered as ‘the limit of the image is the image of the
limit’. Notationally, it proves that we can take the limiting process through a
continuous function, that is:
lim f (xn ) = f lim xn .
x−→∞
18
x−→∞
2.6. Functions
In Chapter 3 we will see another ‘nice’ result for continuous functions and compact sets
(that is, ‘nicely behaved’ sets).
2.6.2
2
Differentiability
We will avoid discussing derivatives at boundary points and restrict our definitions to
purely interior points. Hence, in this subsection, we will restrict our functions to open
subsets U of Rn .
Definition 2.21 (Differentiability at a point) A function f : U −→ Rm , where
U ⊆ Rn is open, is differentiable at a point a ∈ X if and only if there exists an
m × n matrix A such that for any ε > 0 there exists a δ > 0 such that, for all x ∈ U :
kx − ak < δ
⇒
kf (x) − (f (a)k
< ε.
kx − ak
If A exists, then it is called the derivative of f at a and denoted Df (a).
In the special case where n = m = 1, we will also use the notation f 0 (a).
Definition 2.22 (Differentiability on a set) A function f : U −→ Rm , where U ⊆ Rn
is open, is differentiable on a set A ⊂ U , A open, if and only if f is differentiable at
all points a ∈ A.
If f is differentiable at all points of its domain, U , then f is unambiguously said to be
differentiable.
Theorem 2.11 If f : Rn −→ Rm is differentiable then f is continuous.
Activity 2.13 It is important to note that the converse of Theorem 2.11 does not
hold, and continuous functions are not necessarily differentiable. Consider the
function f : R −→ R defined by f (x) = |x|. Prove using the definitions that
1. f is continuous (on R), but
2. f is not differentiable at 0.
Since the derivative of a function f : Rn −→ Rm is itself a function Df : Rn −→ Rn+m ,
we can discuss its continuity and differentiability.
Definition 2.23 (Continuously differentiable) If f : U −→ Rm is differentiable, and
Df : U −→ Rn+m is continuous, then f is said to be continuously differentiable on
U or simply C 1 .
We have necessary and sufficient conditions to determine whether or not a given
function f : Rn −→ R is a C 1 function; however, to quote this we must review one more
notion.
Definition 2.24 (Partial derivative) Suppose we are given the function f : U −→ R,
where U ⊆ Rn . Denote the i-th standard basis vector by ei , so that eji = 1 if j = i and
eji = 0 if j 6= i. The i-th partial derivative of f at a point a is then the limit, where
19
2. Mathematical preliminaries
it exists:
f (a + tej ) − f (a)
.
t−→0
t
lim
2
If this limit exists, we denote it:
∂f (a)
∂xi
or
∂
f (a).
∂xi
We then have the following:
Theorem 2.12 Given a function f : S −→ R, where S ⊆ Rn is open, then the function
f is C 1 on S if and only if all partial derivatives of f exist and are continuous for all
s ∈ S.
If f is C 1 , then we can write:
Df (s) =
∂f (s)
∂f (x)
,...,
∂x1
∂xn
,
∀s ∈ S.
However, note that this is not how we define the derivative Df (a) of a function f at a
point a.
While a function may be continuously differentiable (and hence continuous) it is not
necessarily the case that the derivative is a continuous function itself.
Activity 2.14 Let f : R −→ R be given by

x = 0,
0,
f (x) =
x2 sin 1 , x 6= 0.
x2
As might be expected from its definition, problems arise around the origin. However,
the function is substantially more well-behaved than you might at first think.
1. Prove, using the definition, that f is continuous at x = 0.
2. Prove that f is differentiable at x = 0.
3. Compare the behaviour of f 0 (0) and limx→0 f 0 (x).
Learning outcomes
This chapter is simply a revision of the key ideas that underpin the ideas in this course.
After working through this chapter you should know the definitions of:
convergent sequences
suprema and infima, maxima and minima
bounded, open, closed, compact subsets of Rn
images and preimages of functions f : Rn −→ Rm
continuous and differentiable functions.
20
Chapter 3
Weierstrass’ Theorem
3
Essential reading
R
Sundaram, L.R. A First Course in Optimisation Theory. (Cambridge University
Press, 1996)Chapter 3.
Aims and objectives
In this chapter we will:
introduce and prove the Theorem of Weierstrass
learn how to use it to prove the existence of solutions to optimisation problems
see what happens when it is not applicable.
3.1
Introduction
While the content and proof of Weierstrass’ Theorem should not be taxing, we will
consider it exclusively in this chapter since it is an extremely important result. First
and foremost, it is an existence result: where applicable, it guarantees that our efforts
will not be fruitless and that there is indeed a solution to the problem under
consideration. Its importance to our work here can be seen by the simple fact that it is
the very first step in our cookbook procedures to constrained optimisation: before you
begin the lengthy procedure of finding (and verifying) a solution, you should get into
the habit of checking that one exists in the first place.
Secondly, it is the first major result of the course. As such, let us take the time to see
the sort of thinking a mathematical result requires: not only in proving it, but also in
reading and understanding it.
After introducing the theorem in §3.2 we will go on to detail how it is read, and thus
how it must be applied. We will use the proof as an opportunity to review how a
student should go about proving a theorem in general.
The essential content of the Weierstrass theorem is simply ‘nice sets will always yield
maxima and minima for nicely behaved functions‘. In set theory, ‘nice sets’ are compact
sets, while from topology, ‘nicely-behaved functions’ are continuous functions. Although
real life applications of these ideas may not readily provide compact sets or continuous
functions on them, it is often more useful to make such approximations to use the full
machinery of mathematics.
21
3. Weierstrass’ Theorem
3.2
The Theorem of Weierstrass
Theorem 3.1 (Weierstrass’ Theorem)
3
Let D ⊆ Rn be a non-empty compact set, and f : D → R be continuous. Then f
has both a maximum and a minimum on D.
First note that mathematical statements should contain all necessary information and
no superfluous details. Thus, there are two suppositions that must hold for the
conclusion to be valid:
1. the set D must be compact; and
2. the function f must be continuous.
Let us consider what happens if we relax these requirements. From the definition of
compactness, Definition 2.17, we see that this means that D fails to be either closed or
bounded. Of course, this is a logical ‘or’, so both conditions can fail at the same time,
but it is conceptually simpler to consider these failures on their own.
Activity 3.1 (Relaxing the requirement of compactness)
1. Suppose a set X ⊆ R is unbounded. Show that there exists a continuous
function f : X −→ R which does not attain a maximum or minimum on X.
2. Suppose a set X ⊆ R is not closed. Show that there exists a continuous function
f : X −→ R which does not attain a maximum or minimum on X.
Thus, to successfully apply Weierstrass’ theorem, you must check that the domain is
both closed and bounded. Fortunately, Chapter 1 of Sundaram has a variety of results
to make this easy, so there is no excuse for you not doing so. Indeed, by determining
that a particular optimisation problem has no solution, you can then avoid the
somewhat arduous task of following our cookbook procedures.
Finally, not that this is simply a sufficient condition (the implication goes one way)
rather than a necessary one (which would require an ‘if and only if’ statement). Thus, it
should be possible, for example, to find a function which is not continuous on a compact
set but nonetheless attains its maximum or minimum. In fact, the following activity
asks you to test your understanding of what the theorem does and does not say.
Activity 3.2 (Understanding the Weierstrass theorem)
Give examples of sets and functions with the following properties:
1. an open set D1 ⊆ R and a continuous function f : D1 −→ R such that f has
both a maximum and a minimum on D1 ;
22
3.3. Proving the theorem
2. a compact set D2 ⊆ R and two functions g + h : D2 −→ R such that neither g
nor h have either a maximum or minimum on D2 but the sum g + h does;
3. a set D3 ⊆ R and a continuous function f : D3 −→ R such that f attains a
maximum but not a minimum on D3 ;
4. a set D4 ⊆ R which is not compact and a function f : D4 −→ R which is not
continuous, but which attains both a maximum and a minimum on D4 .
3.3
Proving the theorem
We will prove two results which, together, give Weierstrass’ Theorem.
3.3.1
Continuity and compactness
Lemma 3.2 If f : D −→ R is continuous and D is compact, then f (D) is also
compact.
Writing our premises in full, we have:
Premise 1: f : D −→ R.
Premise 2: f is continuous, so that if {xn } is a sequence in D such that xn −→ x
then f (xn ) −→ f (x).
Premise 3: D is compact, so that if {yn } is an arbitrary sequence in D then there
exists a subsequence {ynk } ⊆ {yn } such that
1. ynk is convergent, i.e. there exists a y such that ynk −→ y as nk −→ ∞.
2. y ∈ D.
With these premises, we must show that f (D) is a compact set. Specifically, according
to the definition, we must show that f (D) satisfies the following:
If {zn } is an arbitrary sequence in f (D) then there exists a subsequence
{znk } ⊆ {zn } such that:
i) znk is convergent, i.e. there exists a z such that znk −→ z; and
ii) z ∈ f (D).
So, before we begin, what are we not supposed to be proving here? That is, in what
way could we possibly misread the definition? A common mistake that students make is
to jump to the end of the definition, which seems the meatiest. There is more going on
here, so one naturally looks to this first. On the principle that the last thing we read is
the first thing on our minds, let us work back through the definition and uncover some
frequent errors.
1. We are not asked to simply show the existence of a subsequence {znk }.
What sequence {zn } would your sequence {znk } be a subsequence of exactly?
23
3
3. Weierstrass’ Theorem
2. We are not asked to show the existence of a convergent sequence.
3
In fact, every non-empty set admits a convergent sequence: a constant sequence. So
there is nothing special in this. Nor are we saying that we can find a sequence
which contains a convergent subsequence or, indeed, a divergent sequence which
contains a convergent subsequence. These are all rather trivial in spaces with more
than two points.
If these two misreadings sound forced to you, they are actually quite common. Please be
sure to read the entire definition and realise that mathematicians are very economical
when it comes to statements of definitions and theorems, neither missing important
details nor, conversely, adding extraneous ones.
Let us consider a more subtle misreading of the definition.
3. We are not asked to find a sequence which contains a convergent subsequence.
This is perhaps closer to the what the definition is saying than the previous two,
but it is still wrong. The error here is to confuse when we must show the existence
of at least one object exhibiting a certain property, and when we must show the
property holds for all objects. Here, it is not enough to show that we can find at
least one sequence with a convergent subsequence (just take a constant sequence).
We must prove the far stronger claim that given any sequence in, here, f (D) we
can find a convergent subsequence.
With that in mind, let us now consider what the definition is asking us to prove. First,
it is a statement which starts off with:
‘If {zn } is an arbitrary sequence in f (D), then . . . ’
Thus, whatever we must prove, it must be about a sequence {zn }. Put another way, if
in our proof we start working with this arbitrary {zn }, we cannot make any
assumptions about the nature of the sequence. Some things you should certainly
find yourself not saying without thinking very long and hard about include:
1. Suppose zn ≥ 0 for all n ∈ N.
2. Suppose zn −→ z as n −→ ∞.
3. Consider the sequence defined by yn =
√
zn for all n ∈ N.
This list is by no means exhaustive, but it does illustrate the sorts of errors that crop
up entirely too often. At any point, if you need to rely upon a supposed property of the
zn , you must ask yourselves what would happen to your proof in its absence: would
your proof still be valid?
In order to discourage this sort of behaviour, you will generally find textbook proofs fix
arbitrary quantities at the beginning of the proof. We will start off, then, by doing this.
Let {zn } in f (D) be given.
Once this is done, we are no longer able to adjust or prescribe its properties, it is
written in stone and whatever deficiencies it might have for our proof, we will have to
work around this. Given this arbitrary sequence, we see that the definition requires us
to prove two things.
24
3.3. Proving the theorem
We will prove that {zn } contains a convergent subsequence
At this point, we are not concerned with any properties of this subsequence, merely
that it exists.
What tools have we got to prove this? Looking around at the premise of Lemma 3.2, we
have little more information about f (D). In particular, we have no existence statements
about convergent subsequences in this set. However, we do have an existence statement
about convergent sequences for the compact set D, so our first task would be to relate
our sequence {zn } in f (D) to the set D. Specifically, from Premise 3.3.1, we see that we
need to relate it to, at the very least, a sequence in D.
However, by assumption, we have for all n ∈ N, zn ∈ f (D), the image set of D under f .
Thus, for all n ∈ N there must exist a yn ∈ D such that zn = f (yn ). This then defines a
sequence {yn } in D.
Since D is compact, by Premise 3.3.1, we know that there exists a convergent
subsequence {ynk } ⊆ {yn }. Of course, this isn’t enough. First, it is not a convergent
sequence of our original sequence {zn }, and it doesn’t necessarily live in f (D). We can,
though, relate it to a sequence in f (D). Since ynk ∈ D for all k ∈ N, then by the
definition of the image of a set, Definition 2.9, we see that f (ynk ) ∈ f (D) for all k ∈ N.
The sequence {f (ynk )} is a convergent sequence. Why? It is the image of a convergent
sequence under the continuous function f (Premise 3.3.1), so that, by Theorem 2.10, it
is a convergent sequence itself.
The sequence {f (ynk )} is a subsequence of {zn }. If zn = f (yn ) for all n ∈ N, then since
the ynk form a subset of the yn , their image points must form a subset of the zn .
Thus, we have shown that, given an arbitrary sequence {zn } in f (D), then it has a
convergent subsequence {znk }.
We will prove that the limit of this convergent subsequence lies in f (D)
Once again, we have little information about f (D) to conclude anything just yet. If we
consider what results we have regarding the limits of convergent sequences and when we
can expect them to exist in certain sets, we should recall that a closed set is such that
any convergent sequence has its limit in the set. Unfortunately, we do not yet know if
f (D) is closed: in essence, we are trying to prove that.
However, we do know that D is closed. So any convergent sequence in D has limit in D.
Can we relate our convergent sequence {znk } in f (D) to one in D? Well, clearly we can
since it was constructed as the image of a convergent sequence in D. That is, the
subsequence {ynk } is a convergent sequence in D, say with limit y. So y ∈ D. By
continuity of f , Theorem 2.10 tells us that:
lim znk = lim f (ynk ) = f
k−→∞
k−→∞
lim ynk = f (y) ∈ f (D).
k−→∞
And in fact, we are done. We have been given an arbitrary sequence in f (D), and shown
that we can find a convergent subsequence whose limit lies in f (D). Since we have not
made any assumptions about the behaviour or nature of the arbitrary sequence, we can
conclude that this must be true for any sequence in f (D), and therefore we have shown
that this set satisfies the requirements of Definition 2.17 for compactness.
That is, f (D) is compact.
25
3
3. Weierstrass’ Theorem
This was a very detailed look at the proof for compactness. The aim of this was not just
to prove the result, but to get you to think about what is needed when proving results
in general. Notice the heavy reliance on definitions and existing results. There is no
hope of doing anything powerful in mathematics if you do not learn to use the
definitions and build results upon earlier results.
3
Having said that, the above proof has far more exposition than is needed for a straight
proof. Hence, the following activity will test what you have understood of this
subsection so far.
Activity 3.3 Concisely state the proof of Lemma 3.2.
3.3.2
Compactness and boundedness
There is just one more result to conclude Weierstrass’ Theorem.
Lemma 3.3 If D ⊂ R is a non-empty compact set, then D has a maximum and a
minimum.
We will prove that a maximum exists. Here we will not go into as much detail as we did
for the first result, but we will still try to raise as many questions about the process.
First, we are talking about compact sets on the one hand, and maxima and minima on
the other. The former is defined in terms of sequences and convergence, while the latter
are defined in terms of bounds. We therefore need a common ground to discuss these.
Fortunately, we have precisely that in our alternative characterisation of real compact
sets, Theorem 2.9. D is a non-empty bounded subset of R. In particular, from Property
2.2 of suprema, we know that D has a supremum. This will ultimately give us our
maximum.
So far we have used the fact that D is bounded, and that it is non-empty, but we have
one remaining property we have not yet used. D is closed. Now, if this were not
important, then we would not have specified it in the statement of the lemma (no
extraneous detail, remember), so we must need it somewhere. Remember, a set being
closed is a statement about limits of convergent sequences. A set having a maximum is
a statement, essentially about suprema: specifically that the set contains its supremum.
Combining these two statements suggests that we could try and construct a sequence in
D with limit sup D.
From our properties of suprema, we see that Property 2.1 is the most promising. This
says that we can always find points in D as close to, and not equal to, sup D. So let us
use this to construct a sequence {xn } in the following way. For each n ∈ N, consider the
interval:
1
Dn = sup D − , sup D .
n
From Property 2.1, the intersection Dk ∩ D must be non-empty for all n ∈ N. If it were
not, say for n0 , then we would have found an upper bound sup D − n10 < sup D, which
contradicts the definition of sup D as the lowest upper bound.
26
3.4. Worked examples
For each n ∈ N pick a point xn ∈ Dn ∩ D. This forms a sequence {xn } in D. Moreover:
1
1
xn ∈ sup D − , sup D
⇒
|xn − sup D| < .
n
n
Thus, as n −→ ∞ we see that xn −→ sup D.
So {xn } is a convergent sequence in D with limit sup D. Since D is closed, we must have
that sup D ∈ D, but then by Property 2.4 of suprema, this implies that D has a
maximum.
The proof for a minimum is very similar to this.
Activity 3.4
3.4
Complete the proof of Lemma 3.3 by showing that D has a minimum.
Worked examples
Example 3.1
Suppose we wish to determine whether the function f (x, y) = x2 − y 2 has a
maximum and a minimum on the unit circle, i..e on the set:
C = (x, y) ∈ R2 x2 + y 2 = 1 .
To apply the Weierstrass Theorem, we require C to be a compact set and f to be a
continuous function.
Is C compact?
Rather than use Definition 2.17 to test for compactness, we opt to use the simpler
real space characterisation found in Theorem 2.9. That is, we must ask ourselves:
Is C closed?
Since membership of the set C is defined purely in terms of one equality, the
result of Activity 2.11 tells us that C is indeed closed.
Is C bounded?
We will appeal to the result of Activity 2.4 to determine boundedness of the set
C. There we see that we must find some upper bound M on the magnitude of
elements in C, that is, for any x ∈ C, we have kxk < M . However, looking at
the membership requirement to be in the set C:
(x, y) ∈ C
⇔
x2 + y 2 = 1
⇔
k(x, y)k = 1.
So taking M > 1 suffices to show that C is bounded.
Thus, the set C ⊆ R2 is closed and bounded, and hence compact.
Is C non-empty?
This is easily seen by giving a simple example of a point in C, say, (1, 0).
Is the function f : C −→ R continuous?
27
3
3. Weierstrass’ Theorem
f (x) is a second-order polynomial in two variables, and hence is continuous.
Thus, f is a continuous function acting on the compact set C and the Weierstrass
Theorem tells us that a maximum and minimum for this function must exist in C.
3
Example 3.2 Suppose the function f : R+ −→ R is a continuous function such
that:
1. f (0) = 1, and
2. limx−→∞ f (x) = 0
Show that f must have a maximum on R+ .
Note that the domain of f , R+ = [0, ∞) is not bounded and hence not compact.
Thus, we cannot immediately apply Weierstrass’ Theorem. However, rather than
dismiss this result entirely, we will show that we can decompose our domain into a
compact set and a set that is, in some sense, negligible in that it does not contain
any candidates for a maximum. We will do this by picking a suitable value of the
function f to split the set into two.
First, note that limx−→∞ f (x) = 0 has a very precise meaning. Specifically, for any
ε > 0, there exists an N > 0 such that for all x > N we have:
ε > |f (x) − 0| = |f (x)| .
Thus, past a finite point N the behaviour of f can be considered negligible. For the
remainder of the domain, R+ /(N, ∞) = [0, N ], we are left with a closed and
bounded interval of R. This is compact, while, by assumption, f is continuous.
Hence, by Weierstrass’ Theorem, f has a maximum on [0, N ].
Since we wish to find a maximum on R+ , and this is the only maximum whose
existence we can guarantee, we would like to show that the maximum on [0, N ] is
indeed a maximum on R+ . That is, even if f has a maximum on (N, ∞), it should
be less than or equal to that on [0, N ].
Since we know f (0) = 1, an easy way to ensure (N, ∞) is inconsequential to our
analysis, is to ensure that any value f takes on it is no greater than f (0). We can do
this by picking ε = f (0) > 0. From above, this defines a critical value N past which
the function f (x) is no greater than 1 in magnitude (and hence no greater than 1 in
value). Weierstrass’ Theorem asserts the existence of an x? such that f (x? ) is a
maximum of f on [0, N ] and for which we have:
f (x? ) ≥ f (0) = 1 > f (x),
∀x ∈ (N, ∞).
Thus, f has a maximum at x? not only on [0, N ] but on R+ .
Sadly, we simply do not have enough to ensure the existence of a minimum. For
example, consider f (x) = exp (−x2 ). This is strictly positive for all x ∈ R+ and has
limit 0 as x −→ ∞. However, it never reaches this limit, simply getting smaller and
smaller, so that a minimum does not exist.
28
3.4. Learning outcomes
Learning outcomes
At the end of this chapter and the relevant reading you should be able to:
state and prove the Theorem of Weierstrass
apply Weierstrass’ Theorem to examples
show the impossibility of maxima and minima in certain situations.
3
Exercises
The exercises here are intended to help consolidate the examinable material that you
have seen in this chapter. Thus, whatever you are asked to do or are tested upon in these
exercises should be expected in an exam. However, these exercises are not indicative of
the style of examination questions. Put simply, these exercises indicate what we might
ask you to do in an exam, though not necessarily how we will ask you to do so. For
that, you should refer to the sample examination questions in Appendices B and D.
Similarly, the solutions provided are generally of a much higher level of detail than is
expected of you in an examination, and you are referred to Appendix C for guidance.
Exercise 3.1
Suppose the set D ⊂ Rn contains a finite number of points. Show that every function
f : D −→ R has a maximum or minimum on D. Is Weierstrass’ Theorem applicable
here?
Exercise 3.2
In Example 3.2 what can you say about the existence of a minimum? Justify your
answer, hence:
if you believe that a minimum must exist, then give a proof
if you believe that a minimum may not exist, then give a counter-example.
Exercise 3.3
Suppose the function f : R −→ R is a continuous function such that limx−→±∞ f (x) = 0.
i) Show that f must have a minimum or a maximum on R.
ii) Give an example that shows that it is not true in general that f has both a
maximum and a minimum on R.
Exercise 3.4
Suppose a monopolist faces:
a downward sloping inverse-demand curve p(x) such that p(0) < ∞ and p(x) ≥ 0
for all x ≥ 0, and
29
3. Weierstrass’ Theorem
a cost curve c(x) such that c(x) ≥ 0 for all x ≥ 0 and c(0) = 0.
Assume that both functions p, c : R+ −→ R are continuous.
The monopolist must maximise profit:
π(x) = p(x)x − c(x),
3
subject to the non-negativity constraint x ≥ 0.
i) Prove that a solution exists to the firm’s maximisation problem if there exists an
x? > 0 such that p (x? ) = 0.
ii) Prove that a solution exists to the firm’s maximisation problem if there exists an
x? > 0 such that c(x) ≥ p(x)x for all x ≥ x0 .
iii) Consider the case where c(x) −→ ∞ as x −→ ∞ and the demand curve is
infinitely elastic, (that is, when p : R −→ {p̄}, for some p̄ ∈ R). Does a solution
exist in this case? If not, give a counter-example.
Exercise 3.5
Consider the expenditure minimisation problem of the consumer. Suppose, in an n-good
market we have a vector of prices p = (p1 , . . . , pn ). The consumer has a utility function
u : Rn+ −→ R and wishes to minimise their expenditure subject to attaining some fixed
minimum level of utility ū. Thus, we have the problem:
Minimise p · x subject to u(x) ≥ ū.
i) If u : Rn+ −→ R is continuous and pi > 0 for all i = 1, . . . , n, show that the
minimisation problem has a solution.
ii) What happens if either the function u fails to be continuous or at least one of the
prices is zero?
Exercise 3.6
Show that the function f : R2+ −→ R given by:
f (x, y) =
yx2
x4 + y 4 + 1
has a maximum on its domain
Solutions to exercises
Solution to exercise 3.1
We have to prove a statement about every possible function one can define on the set
D, so you certainly should not be starting your proof with something along the lines of
30
3.4. Solutions to exercises
‘Let f be the function . . . ’. The way in which we shall proceed is to let an arbitrary
function f : D −→ R be given, and prove the result for this f . Since we will not have
assumed anything about the behaviour of f , we can conclude that our result holds for
all real-valued functions defined on D.
So, let f : D −→ R be given, and write D in of the form {x1 , . . . , xk }. Applying f to all
the points in D yields the image f (D) = {f (x1 ), . . . , f (xk )}, i.e. another finite set. Thus
f (D) contains its maximum and minimum, and thus f attains its maximum and
minimum on D.
In fact, the result also follows from Weierstrass’ Theorem. Any finite set is
automatically compact, and we can show that every function defined on a finite set is
continuous. To do so, let a sequence {xk } in D such that xk −→ x be given. We must
show that f (xk ) −→ f (x).
However, as there are finitely many points in D, we claim that the convergent sequence
{xk } must be eventually constant. Since D is finite, then so must the set:
{kx − yk | x, y ∈ D, x 6= y } .
Thus, it must have a well-defined minimum, which we shall denote δ > 0. From the
definition of convergence for {xk }, we then have a K > 0 such that for all k > K, we
have kxk − xk < δ. This can only be the case if xk = x for all k > K, and hence we have
that the sequence {xk } is eventually constant.
The sequence {f (xk )} must also be eventually constant and hence trivially satisfy the
definition of convergence.
So f is continuous. Since f was an arbitrary function, we conclude that all functions
defined on D are continuous.
Finally, if all functions are continuous, the Weierstrass’ Theorem applies to all functions
defined on a finite space D.
Solution to exercise 3.2
Sadly, we simply do not have enough to ensure the existence of a minimum. For
example, consider f (x) = exp (−x2 ). This has initial value f (0) = 1, is strictly positive
for all x ∈ R+ and has limit 0 as x −→ ∞. However, it never reaches this limit, simply
getting smaller and smaller, so that a minimum does not exist.
Solution to exercise 3.3
i) First, there is some ambiguity in the statement as to what will occur. Your answer
should indicate under what conditions you expect a maximum will occur, and
under what conditions you expect a minimum to occur.
Secondly, Weierstrass’ Theorem is not immediately applicable since R is not
compact. However, as we did in the previous exercise, we can again attempt to
decompose the domain into a compact subset and a set which is negligible to our
analysis. Recall that to do this we picked a value of f to split the domain into two.
So, pick an arbitrary x̄ and consider f (x̄). We need to show that either a
maximum occurs or a minimum occurs. If the function happens to be f (x) = 0 for
31
3
3. Weierstrass’ Theorem
all x ∈ R, then we have the special case where the maximum and minimum values
are both the same and occur at all points of the domain.
However, if not, suppose we have f (x̄) > 0. Then either:
1. the function has a maximum but no minimum, such as in the case of
f (x) = exp (−x2 ), or
3
2. the function has both a maximum and a minimum, such as in the case of
x
f (x) =
.
1 + x2
Thus, if f (x̄) > 0 for some x̄ ∈ R, we claim that a maximum of f must exist on R.
Similarly, if f (x̄) < 0 for some x̄ ∈ R, we claim that a minimum of f must exist on
R.
We shall now consider the case of a maximum (the proof for a minimum can be
derived easily from this).
Supposing that f (x̄) > 0 for some x̄ ∈ R, we wish to split the domain R into:
R = {x ∈ R | f (x) ≥ f (x̄)} ∪ {x ∈ R | f (x) < f (x̄)}
{z
} |
{z
}
|
Y
Z
where we claim that the set Y is compact, while the set Z is infinite.
f (x̄)
Turning to the limiting behaviour of f , let ε =
> 0. (Why have we not used
2
ε = f (x̄) > 0, as we did in the previous exercise?) Since lim f (x) = 0, there
x−→−∞
exists an x− such that for all x < x− :
|f (x)| <
f (x̄)
2
⇔
−
f (x̄)
f (x̄)
< f (x) <
.
2
2
Similarly, since limx−→+∞ f (x) = 0, there exists an x+ such that for all x > x+ :
|f (x)| <
f (x̄)
2
⇔
−
f (x̄)
f (x̄)
< f (x) <
.
2
2
These two points thus define a compact interval [x− , x+ ]. Furthermore, by our
choice of ε, we must have that x̄ 6= x− , x+ is an element of [x1 , x2 ], which is
therefore non-empty. Thus, by Weierstrass’ Theorem, the continuous function f
has a maximum on [x− , x+ ], say at x? .
Since x? ∈ [x− , x+ ], we must have:
f (x? ) ≥
f (x̄)
> f (x),
2
∀x ∈
/ [x− , x+ ].
So x? is indeed a global maximum for the function f .
2
ii) We have already seen a simple example of such a function: f (x) = exp − x2 .
This has a global maximum at x = 0 with value 1, but no minimum.
32
3.4. Solutions to exercises
Solution to exercise 3.4
Before we start looking at the first part of the exercise, we can already write down a lot
of information about the profit function
π(x) = xp(x) − c(x).
3
So let us do that.
We are given that p(x) is such that:
1. p is continuous,
2. p(0) < ∞,
3. non-increasing, and
4. non-negative, i.e. p(x) ≥ 0, ∀x ∈ R+ .
That is, it starts from a finite point and decreases (in a continuous manner) towards
zero. Furthermore, we also have that c(x) is such that
1. c is continuous,
2. c(0) = 0,
3. c(x) ≥ 0, ∀x ∈ R+ .
Thus c is a continuous function that starts at the origin and always remains
non-negative.
Immediately we see that π(x) is the product and difference of continuous functions on
R+ and thus is itself a continuous function on R+ , which is an unbounded, and hence
non-compact, domain. Furthermore, we note that
π(0) = 0 · p(0) − c(0) = 0 − 0 = 0.
i) Suppose we have an x? > 0 such that p(x? ) = 0. We are asked to apply the
Weierstrass Theorem to obtain the desired result, but our observation above that
the function is defined on a non-compact domain defies us from immediately doing
so. What we would like to do is restrict our problem to a compact sub-domain
without drastically changing the nature of the problem. That is, can we omit a
large part of this unbounded domain?
This is possible if we can show that the maximum, if it is to occur at all, must
occur within in a compact region of the domain.
First, at the point x = x? , we are given that p(x? ) = 0, and we can then deduce
that:
π(x? ) = x? · p(x? ) − c(x? ) ≤ 0.
>0
=0
≥0
Moreover, since we know p is non-increasing and non-negative on its domain, it
must be the case that p(x) = 0 ∀x ≥ 0. Thus π(x) ≤ 0 for all x ≥ x? .
33
3. Weierstrass’ Theorem
From both this observation and the fact that π(0) = 0, we are thus able to restrict
our consideration of π(x) to the closed and bounded interval [0, x? ]. Thus, we
know have:
i π(x) is continuous on R+ , and hence continuous on [0, x? ] ⊆ R+
ii [0, x? ] is non-empty, since x? > 0
iii [0, x? ] is compact.
3
Weierstrass’ Theorem can then be applied to conclude that π must therefore have
a maximum on [0, x? ].
Furthermore, we have that:
max π(x) ≥ π(0) = 0 ≥ π(x),
x∈[0,x? ]
∀x ∈
/ [0, x? ].
Thus, this maximum is a global maximum.
ii) This can be considered in much the same way. Suppose there exists an x0 > 0 such
that c(x) ≥ xp(x) ∀x ≥ x0 . Then
∀x ≥ x0 ,
π(x) = x · p(x) − c(x) ≤ 0,
while
π(0) = 0 ≥ π(x),
∀x ≥ x0 .
As before, we can thus restrict our consideration of π to the sub-domain [0, x0 ].
i [0, x0 ] 6= ∅ since x0 > 0;
ii [0, x0 ] is compact;
iii π is continuous on [0, x0 ].
By Weierstrass, we can conclude that π attains a maximum on [0, x0 ]. Furthermore:
max0 π(x) ≥ π(0) = 0 ≥ π(x),
∀x ≥ x0 .
x∈[0,x ]
So this is indeed a global maximum.
√
iii) Consider the situation where c(x) = x. It should be easy to show that π(x)
cannot attain a maximum on an unbounded set.
Solution to exercise 3.5
i) To guarantee the existence of a minimum, we shall appeal to Weierstrass’
Theorem.
P
First, the function we wish to minimise is the linear function p · x = ni=1 pi xi ,
which should be obviously continuous to you. We are restricting this to the set:
X(ū) = x ∈ Rn+ u(x) ≥ ū = {x ∈ Rn | u(x) ≥ ū, xi ≥ 0, i = 1, . . . , n} .
Since this is defined in terms of weak inequalities, X(ū) is closed. However, it is in
general not bounded, since one can reasonably expect utility functions to be
34
3.4. Solutions to exercises
increasing in the xi , and we simply have a lower bound on utility here. Even if we
may not always have this, our problem has not given enough information to
restrict our utility function in this manner, so we must consider the general case.
As we have done in previous exercises, we shall try to find a compact subset of the
domain which we can guarantee contains the minimum of the problem. Again, this
is done by picking an arbitrary value of the function to be minimised to
decompose X(ū).
So, consider an arbitrary vector x
b ∈ X(ū). Thus, x
bi ≥ 0 for all i = 1, . . . , n, and
u (b
x) ≥ ū. The minimum of f on X(ū) can be no greater than f (b
x) = p · x
b. We
then decompose X(ū) into the union of:
Y = x ∈ Rn+ u(x) ≥ ū, p · x ≤ f (b
x) ,
Z = x ∈ Rn+ u(x) ≥ ū, p · x > f (b
x) .
We claim that Y is compact. Since it is defined by weak inequalities, it is closed.
To show that it is bounded, consider an arbitrary element y ∈ Y . This has
magnitude:
v
u n
uX
yi2 .
kyk = t
i=1
Since y ∈ Y , we have p · y ≤ f (b
x). Non-negativity of the yi and positivity of the pi
then imply, for i = 1, . . . , n:
pi y i ≤
n
X
pi yi ≤ f (b
x)
i=1
⇒
yi ≤
f (b
x)
.
pi
Here we must have pi 6= 0 for all i = 1, . . . , n for this bound to be valid. Thus,
v
v
u n
u n
uX f 2 (b
uX 1
x)
t
t
kyk ≤
=
f
(b
x
)
.
2
p
p2i
i
i=1
i=1
Since, for fixed x
b, the left-hand side is a positive constant, we can choose M to be
equal to this so that we have proven, that there exists an M > 0 such that
kyk ≤ M for all y ∈ Y . That is, Y is a bounded set.
Finally, x
b ∈ Y , so that Y is a non-empty, compact set, and Weierstrass’ Theorem
asserts the existence of a minimum of p · x on Y . Suppose this occurs at x? ∈ Y ,
then:
f (x? ) ≤ f (b
x) < f (x),
∀x ∈ Z.
So this indeed provides a minimum on X(ū).
ii) We have already seen what could go wrong if one of the prices pi were equal to
zero. In that case, our argument that Y is a bounded set fails. However, this is not
enough to indicate that problems do occur, and that a minimum might not exist,
so we should come up with a counter-example.
Suppose that a good had a zero price then our solitary expenditure constraint will
do nothing to restrict our consumption of it. If our utility function is increasing in
this good, then we have a costless way of increasing our utility without bound.
35
3
3. Weierstrass’ Theorem
More concretely, suppose we have the expenditure minimisation problem:
Minimise x2 subject to x1 x2 ≥ 1
so that p1 = 0 and p2 = 1 and X(1) = {(x1 , x2 )|x1 x2 ≥ 1}. For any ε > 0 set
x1 = ε−1 and x2 = ε. This satisfies the utility constraint with strict equality.
Furthermore, the infimum of f on X(1) is 0, but this is never attained. Hence, the
problem has no minimum.
3
If the function fails continuity, then you should already have seen situations where
anything can happen, that is minima may or may not occur, depending on the
particular function.
Solution to exercise 3.6
Since the denominator is always greater than or equal to one, so that the quotient is
always well-defined, and the remaining components are just polynomials, the function f
is continuous.
The domain R2+ is defined in terms of the weak inequalities x1 ≥ 0 and x2 ≥ 0 and is
thus closed. However, it is unbounded, so Weierstrass’ Theorem is not immediately
applicable.
As we have done before, we shall try to find a point around which a compact set can be
reasonably expected to contain a maximum for the function (on the entire domain).
Since the function f (x, y) ≥ 0 for all (x, y) ∈ R+ , picking any vector with zero
components will not help us a great deal. So, failing that, we consider integer
components; the simplest case being the point (1, 1) with value f (1, 1) = 13 .
We wish to decompose the domain R2+ into a compact set, to apply Weierstrass’
Theorem, and an unbounded set which is negligible for our analysis. Since we wish to
maximise the function f (x, y), we want the unbounded set to be those points (x, y) such
that f (x, y) < f (1, 1) . To do this, let us consider what an upper bound for f (x, y)
would be in terms of x and y. First, x, y ≤ max {x, y} so that yx2 ≤ (max {x, y})3 . For
the denominator, we also have:
x4 + y 4 + 1 > x4 + y 4 ≥ (max {x, y})4 .
Thus, for all (x, y) ∈ R2+ , we have:
f (x, y) ≤
1
(max {x, y})3
.
4 =
max {x, y}
(max {x, y})
So, if either x > 3 or y > 3 then we have:
f (x, y) <
1
= f (1, 1).
3
We can therefore decompose R2+ into the union of:
Y = (x, y) ∈ R2+ 0 ≤ x ≤ 4, 0 ≤ y ≤ 4 ,
Z = R2+ /Y.
Since Y is defined in terms of weak inequalities, it is closed. It is also bounded, for
example by the length 6. Finally, (1, 1) ∈ Y , so it is non-empty. So, for the continuous
36
3.4. Solutions to exercises
function f , Weierstrass’ Theorem asserts the existence of a maximum for f on Y , say at
the point x? ∈ Y . However, for any (x, y) ∈ Z, we have:
f (x? ) ≥ f (1, 1) =
1
> f (x, y).
3
So x? is a global maximum for the function f .
3
37
3. Weierstrass’ Theorem
3
38