let $f:(a,b)\to\mathbb{R}$ be a monotonically decreasing convex invertible function. is $f^{-1}$ convex as well? is the condition that f is monotone is essential?

I tried to take some examples and find a contradiction, but none of my examples is fit. I think that the statement is true and that the condition is essenetial (we can take $x^2$ as an example for a contradiction).

Now, after understanding that the statement is probably true, I tried to go from definitions of convexity and monotony, but got nowhere.

How may someone prove the statement?

Please help, thank you!


Yes, the statement is true. If you assume strict monotonicity the proof can be found here. The condition of monotonicity is essential as your example with $x^2$ shows.

For the non-strict case you can found a proof here in Proposition 2(2). According to the author if $f$ is two times differentiable there is an easier proof which he gives in Proposition 1. (2)

  • $\begingroup$ e^x is not monotonically decreasing... $\endgroup$ – Galc127 Feb 26 '14 at 18:12
  • $\begingroup$ @Galc127 true, sorry. I will delete it $\endgroup$ – Jimmy R. Feb 26 '14 at 18:12
  • $\begingroup$ @Galc127 The proof is for strict monotonic. That is not what you want, is it? $\endgroup$ – Jimmy R. Feb 26 '14 at 18:26
  • $\begingroup$ No, the function is defined to be monotonically decreasing, convex and invertible. $\endgroup$ – Galc127 Feb 26 '14 at 18:28
  • $\begingroup$ @Galc127 I think that is exactly what you want in Proposition 2(2). You have additionally that $f^-1$ is also decreasing $\endgroup$ – Jimmy R. Feb 26 '14 at 18:37

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