TITLE: convexity of inverse function
QUESTION [11 upvotes]: I have a question on the reverse of a convex function.
Let $f(x)$ be a convex function. 
Is the reverse function, say $g(x)=f(x)^{-1}$, is necessarily a concave function ? Considering that such function $g$ does exist ($f$ is invertible): 
-Is necessarily a concave function? 
-What about the opposite $-f(x)$, is concave?
Any hints?

REPLY [8 votes]: I just stumbled upon this question, wondering the same thing as the OP. YourAdHere's answer was what I was looking for, but I also needed a formal proof, so I thought I might try and add that to complement his/her answer.
Consider a continuous, convex, bijective function $f: A \rightarrow B$ and its inverse $f^{-1}: B \rightarrow A$. By definition of convexity we have for $\lambda \in [0, 1]$ and $x, y \in A$:
\begin{align}
  f(\lambda x + (1-\lambda)y) \leq \lambda f(x) + (1-\lambda) f(y).
\end{align}
For $u, v \in B$ define
\begin{align}%\label{}
  a :=& f^{-1}(\lambda u + (1-\lambda) v) 
  \\
  b :=& \lambda f^{-1}(u) + (1-\lambda)f^{-1}(v) \,.
\end{align}
If $a \leq b$ then $f^{-1}$ is convex, else it is concave. We have 
\begin{align}%\label{}
  f(a) =& f(f^{-1}(\lambda u + (1-\lambda) v) ) 
  \\ =&
  \lambda u + (1-\lambda) v = \lambda f(f^{-1}(u)) + (1-\lambda)f(f^{-1}(v))
  \\ \geq&
  f(\lambda f^{-1}(u) + (1-\lambda) f^{-1}(v)) = f(b) \,,
\end{align}
where the inequality is due to the convexity of $f$. So if $f' \geq 0$ we have
$f(a) \geq f(b) \Leftrightarrow a \geq b$ and hence $f^{-1}$ is concave. If
conversely $f' \leq 0$, then $f(a) \geq f(b) \Leftrightarrow a \leq b$ and 
$f^{-1}$ is convex.