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So, the problem is actually from a microeconomics class. The problem is this:

If preferences are represented by a utility function $u(x,y)=xy$, show that these preferences are convex.

Now in case you don't know, in economics, "convex preferences" means preferences such that the set of preferences that are at least as preferred to some bundle is convex. So basically what this means is I need to show this:

Let $0\le t\le 1$

if $u(x_1,y_1)=u(x_2,y_2)$, then $u(tx_1+(1-t)x_2,ty_1+(1-t)y_2)\ge u(x_1,y_1)$.

so $x_1y_1\le (tx_1+(1-t)x_2)(ty_1+(1-t)y_2)$.

Now, I have tried expanding this out and factoring all kinds of different ways, and I feel like I'm not getting anywhere. Am I going about this incorrectly by trying to expand this? Is there some simpler way? If anyone could give me some kind of hint that would be amazing.

Thanks!

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  • $\begingroup$ Are you sure that the inequality is in that direction? A convex function is usually defined as $$f(tx+(1-t)y)\le tf(x)+(1-t)f(y)$$ for $0\le t\le1$. $\endgroup$
    – robjohn ♦
    Oct 12, 2011 at 6:20
  • $\begingroup$ @robjohn, Colin asks that the sets $\{u\geqslant h\}$ are convex for every $h$ (that is, yes, that the function $u$ is concave). $\endgroup$
    – Did
    Oct 12, 2011 at 6:27
  • $\begingroup$ @Didier: Thanks, that makes sense. $\endgroup$
    – robjohn ♦
    Oct 12, 2011 at 6:50
  • $\begingroup$ The implication "it has sets $\Rightarrow$ [elementary-set-theory] or [set-theory] fits as tags" is incorrect. I removed the unneeded tag. :-) $\endgroup$
    – Asaf Karagila ♦
    Oct 12, 2011 at 7:13

1 Answer 1

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Since $x_1y_1=x_2y_2$, what you want to prove is equivalent to $$tx_1y_1+(1-t)x_2y_2\leqslant(tx_1+(1-t)x_2)(ty_1+(1-t)y_2). $$ Expanding the RHS, one sees that the RHS minus the LHS is $$t(1-t)(x_1y_2+x_2y_1-x_1y_1-x_2y_2)=t(1-t)(x_1-x_2)(y_2-y_1). $$ Using $x_1y_1=x_2y_2$ once again, you know that $x_1>x_2$ implies $y_1<y_2$ and that $x_1<x_2$ implies $y_1>y_2$ hence the last product is always nonnegative. Done.

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  • $\begingroup$ Holy moly. I thought about trying to use the fact that $$x_1 y_1 = x_1 y_1$$ but it never occurred to me to do it the way that you did. How did you even think of that? Anyway, thank you very much for your help, I've got it. $\endgroup$
    – crf
    Oct 12, 2011 at 7:32
  • $\begingroup$ Not enough time to elaborate just now but the answer is: symmetry. $\endgroup$
    – Did
    Oct 12, 2011 at 7:36

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