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 $J$.
$\blacksquare$
Also see
- Inverse of Strictly Increasing Concave Real Function is Convex
- Inverse of Strictly Increasing Strictly Concave Real Function is Strictly Convex
- Inverse of Strictly Decreasing Convex Real Function is Convex
- Inverse of Strictly Decreasing Strictly Convex Real Function is Strictly Convex
- Inverse of Strictly Decreasing Concave Real Function is Concave
- Inverse of Strictly Decreasing Strictly Concave Real Function is Strictly Concave
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 12.21 \ (3)$