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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!

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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)

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  • $\begingroup$ 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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