# Convex Preference but Convex Utility

Can preference be convex when utility is not a concave function (e.g. $$U=x_1^2 + x_2^2$$)?

It's well known that a convex preference implies quasiconcave utility functions. Since quasiconcavity need not imply concavity, it's easy to find examples of a non-concave utility function representing a convex preference.

For example: $$u(x,y)=(x+y)^3$$. The preference this function represents is convex (though not strictly so), as can be seen from its linear indifference curves. The function is quasiconcave, as evidenced by the convex upper contour sets. Lastly, the function is not concave, as betrayed by the exponent.

• Thanks! A follow up question: Let's say the utility function is not even quasiconcave, is using Lagrangian to maximize the function possible?
– megg
Aug 30 '20 at 9:44
• @megg: What exactly do you mean by not quasiconcave? Strictly convex? Or neither quasiconcave nor quasiconvex? At any rate, it may be worth asking a separate question. Aug 30 '20 at 17:14
• The result is stronger, isn't it? preferences are convex iff utility is quasiconcave. So I guess the question can basically be reduced to: does there exist a function which is quasiconcave and convex? Counting out linear functions, I guess the answer is no. However, there can be funky functions that are quasi concave but NOT concave.
– user28372
Aug 31 '20 at 4:03
• @megg, you can always use Lagrangian for con-convex preferences. In that case, you just need the KKT conditions to check out tfor corner solutions. Having said that, its sometimes much easier to find out the optimum intuitively rather than set up the KKT conditions (e.g. $u=x^2+y^2$ is so easy to solve using diagrams.
– user28372
Aug 31 '20 at 4:08
• @Tomcat: Yes, you're right: a utility function is quasiconcave iff the preference it represents is convex. But regarding your reinterpretation of OP's question: $(x+y)^3$ is both convex and quasiconcave on $\mathbb R^2_+$. Likewise in $\mathbb R$, any convex function that is also monotonic, e.g. $\mathrm e^x$, is also quasiconcave (as well as quasiconvex). Aug 31 '20 at 13:51