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 >Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, b, c \in I} with >Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a <; b <; c} we have that >Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f : I \to \mathbb{R}} is convex if and only if whenever we take three points Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, b, c \in I} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a <; b <; c} we have that the slope of the line connecting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a, f(a))} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (b, f(b))} is less than or equal to the sope of the line connecting Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (b, f(b))} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (c, f(c))} . In other words, the slope of the line segments connecting consecutive pairs of points on the graph of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f} is increasing.

We can combine theorems 1 and 2 to get a nice chain of inequalities. That is, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f : I \to \mathbb{R}} is convex if and only if for all Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a, b, c \in I} with >Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a < b < c} we have that:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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}}

Licensing

Content obtained and/or adapted from:

• Maxima and minima, Wikipedia under a CC BY-SA license
• Convex and Concave Functions, Mathonline under a CC BY-SA license