Convex and Concave Functions

Definition: A function $f : I \to \mathbb{R}$ is said to be Convex if for every $x, y \in I$ and for every $t \in [0, 1]$ we have that $f(tx + (1 - t)y) \leq tf(x) + (1 - t)f(y)$. A function $f : I \to \mathbb{R}$ is said to be Concave if for every $x, y \in I$ and for every $t \in [0, 1]$ we have that $f(tx + (1 - t)y) \geq tf(x) + (1 - t)f(y)$.

We now give equivalent definitions for convex and concave functions.

Theorem 1: Let $f : I \to \mathbb{R}$. a) $f$ is convex on $I$ if and only if for all $a, b, c \in I$ with $a < b < c$ we have that $\displaystyle{\frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(a)}{c - a}}$. b) $f$ is concave on $I$ if and only if for all $a, b, c \in I$ with $a < b < c$ we have that $\displaystyle{\frac{f(b) - f(a)}{b - a} \geq \frac{f(c) - f(a)}{c - a}}$.

We only prove (a) above. The proof of (b) is analogous.

• Proof of a): Let $a, b, c \in I$ be such that $a < b < c$.

• $\Rightarrow$ Suppose that $f$ is convex on $I$. Then for all $t \in [0, 1]$ we have that:

(1)

\begin{align} \quad f(tx + (1 - t)y) \leq tf(x) + (1 - t)f(y) \end{align}

• Set $a = x$, $b = tx + (1-t)y$, and $c = y$. Combining the first and third equations with the second equation gives us:

(2)

\begin{align} \quad b = ta + (1-t)c \end{align}

• Solving for $t$ gives us:

(3)

\begin{align} \quad b &= ta + c - tc \\ \quad b - c &= t(a - c) \\ \end{align}

• Therefore:

(4)

\begin{align} \quad t = \frac{b - c}{a - c} = \frac{c - b}{c - a} \end{align}

• Similarly, we compute $1 - t$ to be:

(5)

\begin{align} \quad 1 - t = \frac{c - a}{c - a} - \frac{c - b}{c - a} = \frac{b - a}{c - a} \end{align}

• From the convexity of $f$ we have $f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)$, or equivalently:

(6)

\begin{align} \quad f(b) & \leq \frac{c - b}{c - a}f(a) + \frac{b - a}{c - a}f(c) \\ \end{align}

• And hence:

(7)

\begin{align} \quad (c - a)f(b) \leq (c - b)f(a) + (b - a)f(c) = (c - a + a - b)f(a) + (b - a)f(c) = (c - a)f(a) - (b - a)f(a) + (b - a)f(c) \end{align}

• Therefore:

(8)

\begin{align} \quad (c - a)[f(b) - f(a)] \leq (b - a)[f(c) - f(a)] \quad \Leftrightarrow \quad \frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(a)}{c - a} \end{align}

• $\Leftarrow$ Obtained by working backwards from above. $\blacksquare$

We state yet another important definition for convex and concave functions.

Theorem 2: Let $f : I \to \mathbb{R}$. a) $f$ is convex on $I$ if and only if for all $a, b, c \in I$ with $a < b < c$ we have that $\displaystyle{\frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(b)}{c - b}}$. b) $f$ is concave on $I$ if and only if for all $a, b, c \in I$ with $a < b < c$ we have that $\displaystyle{\frac{f(b) - f(a)}{b - a} \geq \frac{f(c) - f(b)}{c - b}}$.

Theorem 2 gives us a nice characterization of convex functions. It tells us that a function $f : I \to \mathbb{R}$ is convex if and only if whenever we take three points $a, b, c \in I$ with $a < b < c$ we have that the slope of the line connecting $(a, f(a))$ and $(b, f(b))$ is less than or equal to the sope of the line connecting $(b, f(b))$ and $(c, f(c))$. In other words, the slope of the line segments connecting consecutive pairs of points on the graph of $f$ is increasing.

We can combine theorems 1 and 2 to get a nice chain of inequalities. That is, $f : I \to \mathbb{R}$ is convex if and only if for all $a, b, c \in I$ with $a < b < c$ we have that:

(9)

\begin{align} \frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(a)}{c - a} \leq \frac{f(c) - f(b)}{c - b} \end{align}