Real Function is Concave iff its Negative is Convex

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Theorem

Let $f$ be a real function.

Let $I \subseteq \R$ be an interval of $\R$.


Then $f$ is concave on $I$ if and only if $-f$ is convex on $\R$.


Proof

Necessary Condition

Let $f$ be concave on $I$.

Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$.

Then:

\(\ds \map f {\alpha x + \beta y}\) \(\ge\) \(\ds \alpha \map f x + \beta \map f y\) Definition of Concave Real Function
\(\ds \leadsto \ \ \) \(\ds -\map f {\alpha x + \beta y}\) \(\le\) \(\ds -\paren {\alpha \map f x + \beta \map f y}\)
\(\ds \) \(=\) \(\ds \alpha \paren {-\map f x} + \beta \paren {-\map f y}\)

and so $-f$ is convex by definition.

$\Box$


Sufficient Condition

Let $-f$ be convex on $I$.

Let $\alpha, \beta \in \R_{>0}$ such that $\alpha + \beta = 1$.

Then:

\(\ds -\map f {\alpha x + \beta y}\) \(\le\) \(\ds \alpha \paren {-\map f x} + \beta \paren {-\map f y}\) Definition of Convex Real Function
\(\ds \leadsto \ \ \) \(\ds \map f {\alpha x + \beta y}\) \(\ge\) \(\ds - \paren {\alpha \paren {-\map f x} + \beta \paren {-\map f y} }\)
\(\ds \) \(=\) \(\ds \alpha \map f x + \beta \map f y\)

and so $f$ is concave by definition.

$\blacksquare$


Sources

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