# Real Function is Concave iff its Negative is Convex

## 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.

$\blacksquare$