Inverse of Strictly Increasing Convex Real Function is Concave

Theorem
Let $f$ be a real function which is convex on the open interval $I$.

Let $J = f \left({I}\right)$.

Then:
 * If $f$ be strictly increasing on $I$, then $f^{-1}$ is concave on $J$.
 * If $f$ be strictly decreasing on $I$, then $f^{-1}$ is convex on $J$.

Corollary
Let $f$ be a real function which is concave on the open interval $I$.

Let $J = f \left({I}\right)$.

Then:
 * If $f$ be strictly increasing on $I$, then $f^{-1}$ is convex on $J$.
 * If $f$ be strictly decreasing on $I$, then $f^{-1}$ is concave on $J$.

Proof
Let $X = f \left({x}\right) \in J, Y = f \left({y}\right) \in J$.

From the definition of convex:
 * $\forall \alpha, \beta \in \R: \alpha > 0, \beta > 0, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$


 * Suppose $f$ is strictly increasing on $I$, $f^{-1}$ is strictly increasing on $J$ from Inverse of Strictly Monotone Function.

Thus:
 * $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \le f^{-1} \left({\alpha X + \beta Y}\right)$

Hence $f^{-1}$ is concave on $J$.


 * Suppose $f$ is strictly decreasing on $I$, $f^{-1}$ is strictly decreasing on $J$ from Inverse of Strictly Monotone Function.

Thus:
 * $\alpha f^{-1} \left({X}\right) + \beta f^{-1} \left({Y}\right) = \alpha x + \beta y \ge f^{-1} \left({\alpha X + \beta Y}\right)$

Hence $f^{-1}$ is convex on $J$.

Proof of Corollary
The nature of the inverses of strictly monotone concave functions follow directly from the above.