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?

  • $\begingroup$ 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 '19 at 13:49
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    $\begingroup$ Your question seems fairly straightforward. Have you tried directly applying the definitions of convex preferences and quasi-concave utility functions? $\endgroup$ – Giskard Nov 3 '19 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 '19 at 15:48
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    $\begingroup$ 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 '19 at 17:10
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    $\begingroup$ 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 '19 at 17:11

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".

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    $\begingroup$ 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 '19 at 13:22

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