Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex

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

Let $J = f \sqbrk I$.

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

Proof
Let:
 * $X = \map f x \in J$
 * $Y = \map f y \in J$.

From the definition of strictly convex:
 * $\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \map f x + \beta \map f y$

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:
 * $\alpha \map {f^{-1} } X + \beta \map {f^{-1} } Y = \alpha x + \beta y < \map {f^{-1} } {\alpha X + \beta Y}$

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

Also see

 * Inverse of Strictly Decreasing Convex Real Function is Convex


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


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


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