Definition:Convex Function

Definition

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

Then $f$ is convex on $I$ iff:

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

Equivalently:

$\forall x, y \in I: \forall t \in \left(0 . . 1\right) : f \left({tx + (1-t)y}\right) \le t f\left({x}\right) + \left(1-t\right) f\left({y}\right)$

The function $f$ is strictly convex on $I$ if, in the above inequalities, equality cannot hold unless $x = y$.

The geometric interpretation is that any point on the chord drawn on the graph of any convex function always lies on or above the graph.

Alternative Definition

A real function $f$ defined on a real interval $I$ 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:

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

The function $f$ is strictly convex on $I$ if, in the above inequalities, equality cannot hold.

Hence a geometrical interpretation:

• In the left hand image above, the slope of $P_1 P_2$ is less than that of $P_2 P_3$.
• In the right hand image above, the slope of $P_1 P_2$ is less than that of $P_1 P_3$.

Equivalence of Definitions

These two definitions can be seen to be equivalent from Equivalence of Convex and Concave Definitions.

Also see

• Compare concave function. It is immediately obvious from the definition that $f$ is convex on $I$ iff $-f$ is concave on $I$.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 12.13$