Inverse of Strictly Increasing Convex Real Function is Concave

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Theorem

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

Let $J = f \sqbrk I$.


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


Proof

Let:

$X = \map f x \in J$
$Y = \map f y \in J$.

From the definition of convex:

$\forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} \le \alpha \map f x + \beta \map f y$


Let $f$ be strictly increasing on $I$.

From Inverse of Strictly Monotone Function it follows that, $f^{-1}$ is strictly increasing on $J$.

Thus:

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

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

$\blacksquare$


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