Relative Extrema and Convex Functions

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:

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

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

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

• Solving for $t$ gives us:

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

• Therefore:

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

• Similarly, we compute $1-t$ to be:

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

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

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

• And hence:

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

• Therefore:

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

• $\Leftarrow$ 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:

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

Licensing

Content obtained and/or adapted from:

• [1] under a CC BY-SA license
• [2] under a CC BY-SA license