# Equivalence of Convex and Concave Definitions

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## Theorem

Let $f$ be a real function which is defined on a real interval $I$.

### Convex

These two definitions of convex function are equivalent:

• $f$ is convex on $I$ iff:
$\forall \alpha, \beta \in \R: \alpha > 0, \beta > 0, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \le \alpha f \left({x}\right) + \beta f \left({y}\right)$

wherever $x, y \in I$.

• $f$ is convex on $I$ iff:
$\displaystyle \forall x_1, x_2, x_3 \in I: x_1 < x_2< x_3: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

or:

$\displaystyle \forall x_1, x_2, x_3 \in I: x_1 < x_2< x_3: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$

### Concave

These two definitions of concave function are equivalent:

• $f$ is concave on $I$ iff:
$\forall \alpha, \beta \in \R: \alpha > 0, \beta > 0, \alpha + \beta = 1: f \left({\alpha x + \beta y}\right) \ge \alpha f \left({x}\right) + \beta f \left({y}\right)$

wherever $x, y \in I$.

• $f$ is concave on $I$ iff:
$\displaystyle \forall x_1, x_2, x_3 \in I: x_1 < x_2< x_3: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

or:

$\displaystyle \forall x_1, x_2, x_3 \in I: x_1 < x_2< x_3: \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \ge \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$

## Proof

Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$ in the first definition.

As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \left({1 - \alpha}\right) x_3$.

Thus:

$\displaystyle \alpha = \frac {x_3 - x_2} {x_3 - x_1}, \beta = \frac {x_2 - x_1} {x_3 - x_1}$

So doing the substitution for $\alpha$ and $\beta$:

• Convex:
$\left({x_3 - x_1}\right) f \left({x_2}\right) \le \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)$
• Concave:
$\left({x_3 - x_1}\right) f \left({x_2}\right) \ge \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)$

### Proof for Convex

Now take the inequality:

$\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_1}\right)} {x_3 - x_1}$

This is equivalent to:

$\left({x_3 - x_1}\right) f \left({x_2}\right) - \left({x_3 - x_1}\right) f \left({x_1}\right) \le \left({x_2 - x_1}\right) f \left({x_3}\right) - \left({x_2 - x_1}\right) f \left({x_1}\right)$

... that is:

$\left({x_3 - x_1}\right) f \left({x_2}\right) \le \left({x_3 - x_1 - x_2 + x_1}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)$

This is the same thing as:

$\left({x_3 - x_1}\right) f \left({x_2}\right) \le \left({x_3 - x_2}\right) f \left({x_1}\right) + \left({x_2 - x_1}\right) f \left({x_3}\right)$

A similar argument establishes the inequality:

$\displaystyle \frac {f \left({x_2}\right) - f \left({x_1}\right)} {x_2 - x_1} \le \frac {f \left({x_3}\right) - f \left({x_2}\right)} {x_3 - x_2}$

$\blacksquare$

### Proof for Concave

Exactly the same argument can be used as for convex, by reversing the inequalities appropriately.

$\blacksquare$