<p>This makes use of <a href="https://en.wikipedia.org/wiki/Extreme_value_theorem" rel="nofollow">Weierstrass</a> theorem: </p>

<p>"Any continuous function mapping from a non-empty compact set to a a subset of $\mathbb{R}$ attains a maximum and minimum. </p>

<p>So, for example: </p>

<p>We know by <a href="https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem" rel="nofollow">Hein-Borel</a> that a $B(p,w)$ (the budget set) is compact. Since it is compact, we know by Weierstrass theorem that there exists a solution to the utility maximization problem on B(p,w). </p>

<p>In general, optimization on a compact set will have a solution if you have a continuous objective function.</p>

CC BY-SA 3.0
2016-06-29T14:10:35.100Z
123, when you say by Hein-Borel, the budget set is compact, what is the line of reasoning? It probably takes up the space for the entire thing to be spewed out, but a brief outline of proof would be greatly appreciated!!
Frank Swanton
CC BY-SA 3.0
2016-06-30T20:05:31.080Z
When dealing with the topology of metric spaces, Heine-Borel theorem tells us that a subset S of some nth-dimensional Euclidean space is compact iff it is closed and bounded. So, because the budget set is closed and bounded, we know it is compact. Then, any continuous function over this budget set (i.e., any continuous utility function) must attain a maximum or minimum over this range.
123
CC BY-SA 3.0
2016-07-01T13:27:55.257Z
123, thanks for the comment.  I would have to let this sit in and go back to the analysis book to complete absorb what you said. Thank you for the comment!
Frank Swanton
CC BY-SA 3.0
2016-07-01T14:50:54.560Z
No worries. Also - I should correct my comment to say "a maximum and a minimum". Sorry - typo.
123
CC BY-SA 3.0
2015-10-06T17:09:31.353Z
2015-10-06T17:32:06.377Z
2015-10-06T17:32:06.377Z
123
<p>It is most likely that the statement is in the form of 'A utility maximization problem on a compact set has a solution when the utility function is continuous.' In the one dimensional case, this is a direct corollary of the extreme value theorem. A reference to this theorem can be found at <a href="https://en.wikipedia.org/wiki/Extreme_value_theorem">https://en.wikipedia.org/wiki/Extreme_value_theorem</a> </p>

<p>However, discontinuity of the utility function may break this result. Consider the following function.
$$
u(x) =  \begin{cases} x \mbox{ if $x\in[0,\frac{1}{2})$,} \\ \frac{1}{4} \mbox{ if $x \in [\frac{1}{2},1].$   } \end{cases} 
$$
It should be immediate to recognize that this function does not have attain a maximum for any $x\in [0,1]$ even though we have a compact set.</p>

<p>A book that I had used before is <a href="http://rads.stackoverflow.com/amzn/click/0521497701">here</a> which deals with optimization problems of the sort of an economist would encounter in a rigorous way. Also, <a href="http://rads.stackoverflow.com/amzn/click/0393957330">this book</a> provides a more intuitive way of establishing the main results, which would be a better start.</p>

CC BY-SA 3.0
2015-10-06T05:56:42.913Z
Am I reading this correctly? What you are saying is that as $x$ approached $\frac{1}{2}$, the utility increases. As soon as it reaches this point, utility drops to $\frac{1}{2}$ If the first equation were $x\in[0,\frac{1}{2}]$ we would have a maximum?
Jamzy
CC BY-SA 3.0
2015-10-06T06:05:47.057Z
Yes that is correct. For reference, an example where u(x)=x whenever $x\in [0,1)$ and $u(1)=0$ also does not have a maximum. When the answer to the question 'is it true that for every $x$, there exists $x'$ such that $u(x)<u(x')$?' is affirmative, then we would not have a maximum.
ramazan
CC BY-SA 3.0
2015-10-06T05:31:40.377Z
2015-10-06T05:31:40.377Z
ramazan
<p>I'm a first year grad student and we're learning about the utility maximization problem and the producer's problem now. We have been assuming that a solution to these problems exists. But every now and then there's an exercise problem asking us to show that a solution exists. I have encountered once that "A maximization problem on a compact set has a solution". I wonder if there's any good reference on this? We're using MGW's Microeconomic Theory as our textbook; I thought I could find something in the appendix but I failed.  </p>

CC BY-SA 3.0
2015-10-06T05:00:09.620Z
If a set is bounded, there has to be a solution. I like to think of maximization problems like topography. You are looking for the highest point. If you have an area, there has to be a highest point within that area somewhere.
Jamzy
CC BY-SA 3.0
2015-10-06T06:25:13.523Z
MWG's mathematical appendice M.J and M.K contain an overview of the main theorems you'll need for maximization. As for the statement you quoted, that's just theorem M.F.2 (ii). So MWG does have a good enough reference. Another, perhaps a bit more accessible, source of reference would be Jehle and Reny (2011).
Herr K.
CC BY-SA 3.0
2015-10-06T12:18:06.353Z
@Jamzy Yes you're right about it!
Kenneth Chen
<microeconomics>
Existence of solution to maximization problem
Kenneth Chen
# Existence of solution to maximization problem

(Asked by: username_0 on 2015-10-06)

I'm a first year grad student and we're learning about the utility maximization problem and the producer's problem now. We have been assuming that a solution to these problems exists. But every now and then there's an exercise problem asking us to show that a solution exists. I have encountered once that "A maximization problem on a compact set has a solution". I wonder if there's any good reference on this? We're using MGW's Microeconomic Theory as our textbook; I thought I could find something in the appendix but I failed.

**Tags:** microeconomics
> **Comments:**
> **username_1**: If a set is bounded, there has to be a solution. I like to think of maximization problems like topography. You are looking for the highest point. If you have an area, there has to be a highest point within that area somewhere.  
> **username_2**: MWG's mathematical appendice M.J and M.K contain an overview of the main theorems you'll need for maximization. As for the statement you quoted, that's just theorem M.F.2 (ii). So MWG does have a good enough reference. Another, perhaps a bit more accessible, source of reference would be Jehle and Reny (2011).  
> **username_0**: @Jamzy Yes you're right about it!  

---

## 2 answers

### Answer by username_5 on 2015-10-06. Score: 5 | Accepted answer

It is most likely that the statement is in the form of 'A utility maximization problem on a compact set has a solution when the utility function is continuous.' In the one dimensional case, this is a direct corollary of the extreme value theorem. A reference to this theorem can be found at <https://en.wikipedia.org/wiki/Extreme_value_theorem>

However, discontinuity of the utility function may break this result. Consider the following function.
$$
u(x) = \begin{cases} x \mbox{ if $x\in[0,\frac{1}{2})$,} \\ \frac{1}{4} \mbox{ if $x \in [\frac{1}{2},1].$ } \end{cases}
$$
It should be immediate to recognize that this function does not have attain a maximum for any $x\in [0,1]$ even though we have a compact set.

A book that I had used before is [here](http://rads.stackoverflow.com/amzn/click/0521497701) which deals with optimization problems of the sort of an economist would encounter in a rigorous way. Also, [this book](http://rads.stackoverflow.com/amzn/click/0393957330) provides a more intuitive way of establishing the main results, which would be a better start.

> **Comments:**
> **username_1**: Am I reading this correctly? What you are saying is that as $x$ approached $\frac{1}{2}$, the utility increases. As soon as it reaches this point, utility drops to $\frac{1}{2}$ If the first equation were $x\in[0,\frac{1}{2}]$ we would have a maximum?  
> **username_5**: Yes that is correct. For reference, an example where u(x)=x whenever $x\in [0,1)$ and $u(1)=0$ also does not have a maximum. When the answer to the question 'is it true that for every $x$, there exists $x'$ such that $u(x)<u(x')$?' is affirmative, then we would not have a maximum.  

---

### Answer by username_3 on 2015-10-06. Score: 4

This makes use of [Weierstrass](https://en.wikipedia.org/wiki/Extreme_value_theorem) theorem:

"Any continuous function mapping from a non-empty compact set to a a subset of $\mathbb{R}$ attains a maximum and minimum.

So, for example:

We know by [Hein-Borel](https://en.wikipedia.org/wiki/Heine%E2%80%93Borel_theorem) that a $B(p,w)$ (the budget set) is compact. Since it is compact, we know by Weierstrass theorem that there exists a solution to the utility maximization problem on B(p,w).

In general, optimization on a compact set will have a solution if you have a continuous objective function.

> **Comments:**
> **username_4**: 123, when you say by Hein-Borel, the budget set is compact, what is the line of reasoning? It probably takes up the space for the entire thing to be spewed out, but a brief outline of proof would be greatly appreciated!!  
> **username_3**: When dealing with the topology of metric spaces, Heine-Borel theorem tells us that a subset S of some nth-dimensional Euclidean space is compact iff it is closed and bounded. So, because the budget set is closed and bounded, we know it is compact. Then, any continuous function over this budget set (i.e., any continuous utility function) must attain a maximum or minimum over this range.  
> **username_4**: 123, thanks for the comment.  I would have to let this sit in and go back to the analysis book to complete absorb what you said. Thank you for the comment!  
> **username_3**: No worries. Also - I should correct my comment to say "a maximum and a minimum". Sorry - typo.  

---

