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.
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1$\begingroup$ 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. $\endgroup$– JamzyOct 6, 2015 at 5:00
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1$\begingroup$ 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). $\endgroup$– Herr K.Oct 6, 2015 at 6:25
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$\begingroup$ @Jamzy Yes you're right about it! $\endgroup$– Kenneth ChenOct 6, 2015 at 12:18
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2 Answers
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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 which deals with optimization problems of the sort of an economist would encounter in a rigorous way. Also, this book provides a more intuitive way of establishing the main results, which would be a better start.
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$\begingroup$ 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? $\endgroup$– JamzyOct 6, 2015 at 5:56
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$\begingroup$ 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. $\endgroup$– ramazanOct 6, 2015 at 6:05
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This makes use of Weierstrass 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 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.
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$\begingroup$ 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!! $\endgroup$ Jun 29, 2016 at 14:10
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$\begingroup$ 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. $\endgroup$– 123Jun 30, 2016 at 20:05
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$\begingroup$ 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! $\endgroup$ Jul 1, 2016 at 13:27
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$\begingroup$ No worries. Also - I should correct my comment to say "a maximum and a minimum". Sorry - typo. $\endgroup$– 123Jul 1, 2016 at 14:50