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, $$\forall \alpha, \beta \in \mathbb{R}: \alpha > 0, \beta > 0, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$$.


 * 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.