Inverse of Strictly Decreasing Convex Real Function is Convex

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

Let $J = f \left[{I}\right]$.

If $f$ be strictly decreasing on $I$, then $f^{-1}$ is convex on $J$.

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

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

Let $f$ be strictly decreasing on $I$

Then from Inverse of Strictly Monotone Function it follows that $f^{-1}$ is strictly decreasing on $J$.

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 convex on $J$.

Also see

 * Inverse of Strictly Increasing Convex Real Function is Concave


 * Inverse of Strictly Increasing Concave Real Function is Convex
 * Inverse of Strictly Decreasing Concave Real Function is Concave