Is it possible to have an inferior good under the assumption of a utility function which is strictly monotonically increasing and strictly quasiconvex?

I do not think so, but I could not find a formal proof.

*My underlying idea is that whit an inferior good, there exist a point where the utility consuming less of one good is better and then its derivative negative at that point, moreove, I think that this behaviour would break the concavity shape.

  • $\begingroup$ I was reading those examples and they do not fulfill the convexity assumption. $\endgroup$ Dec 12, 2016 at 21:53
  • $\begingroup$ Perhaps you have not spent enough time on the first example $$ U(x,y) = \alpha_1 \ln(x-\gamma_x)- \alpha_2 \ln(\gamma_y - y) $$ which does fulfill all your conditions for $0 < \alpha_1 < \alpha_2$. $\endgroup$
    – Giskard
    Dec 12, 2016 at 21:59
  • $\begingroup$ But the second derivative of U(x,y) over y is: alpha2/(gamma-y)^2 which is positive in its domain. $\endgroup$ Dec 12, 2016 at 22:42
  • $\begingroup$ That is not the definition of convexity for multivariate functions. Please read the paper linked in the answer, it actually explains why convexity holds. $\endgroup$
    – Giskard
    Dec 13, 2016 at 6:28

1 Answer 1


I could not add images in the comments, therefore I used this method.

I think that the convexity definition does not hold for the function, since there are pairs of points that when they are conected, produce a segment which are out of the set. Because of it, the set is not convex.

Here there is an example of such situation with this function, which does not show a convex set: enter image description here


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