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 $J$.

$\blacksquare$