0
$\begingroup$

It is well-known that convex indifference curve (i.e. the function is convex)/ preference would imply quasi-concave utility function. But does quasi-concave utility function imply convex indifference curve?

It seems that this answer give a brief sketch of the proof, but how can I show it in a more formal way?

$\endgroup$
  • $\begingroup$ Seems like that answer had little effect, because I tried to point out in it that the indifference curve is usually not "convex". $\endgroup$ – Giskard Nov 3 at 13:49
  • 1
    $\begingroup$ Your question seems fairly straightforward. Have you tried directly applying the definitions of convex preferences and quasi-concave utility functions? $\endgroup$ – Giskard Nov 3 at 13:50
  • $\begingroup$ You said that "What you probably mean is that the IC curve implicitly defines a convex function f where f(x)=y." That is what I meant. I am not sure what do you mean by " IC curve is not convex in the usual meaning of the word convex when applied to sets". $\endgroup$ – Aqqqq Nov 3 at 15:48
  • 1
    $\begingroup$ Would you mind typing in your quasi-concave utility function to convex preference proof (edit your question, do not add more comments), so we can see why it is not reversible? $\endgroup$ – Giskard Nov 3 at 17:10
  • 1
    $\begingroup$ On convexity: Most curves are not convex sets. If you pick two points from the curve you can usually find a convex combination of them that is not on the curve. $\endgroup$ – Giskard Nov 3 at 17:11
5
$\begingroup$

Does quasi-concave utility function imply convex indifference curve?

No that is not true. Consider $u(x, y) = -x^2 - y^2$ defined on $\mathbb{R}^2_+$. Since $u$ is concave it is quasiconcave. Observing the graph of the indifference curves, we see that ICs of $u$ are not "convex".

enter image description here

$\endgroup$
  • 2
    $\begingroup$ The answer linked in the question does say that monotonicity is required. You are right that this information is not contained in the body of this question. $\endgroup$ – Giskard Nov 4 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.