View source for Relative Extrema and Convex Functions

==Convex and Concave Functions==
<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
:'''Definition''': A function <math>f : I \to \mathbb{R}</math> is said to be '''Convex''' if for every <math>x, y \in I</math> and for every <math>t \in [0, 1]</math> we have that 
:<math>f(tx + (1 - t)y) \leq tf(x) + (1 - t)f(y)</math>. 
:A function <math>f : I \to \mathbb{R}</math> is said to be '''Concave''' if for every <math>x, y \in I</math> and for every <math>t \in [0, 1]</math> we have that
:<math>f(tx + (1 - t)y) \geq tf(x) + (1 - t)f(y)</math>.</td>
</blockquote>

We now give equivalent definitions for convex and concave functions.
<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
:'''Theorem 1:''' Let <math>f : I \to \mathbb{R}</math>.<br />
:'''a)''' <math>f</math> is convex on <math>I</math> if and only if for all <math>a, b, c \in I</math> with <math>a < b < c</math> we have that <math>\displaystyle{\frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(a)}{c - a}}</math>.<br />
:'''b)''' <math>f</math> is concave on <math>I</math> if and only if for all <math>a, b, c \in I</math> with <math>a < b < c</math> we have that <math>\displaystyle{\frac{f(b) - f(a)}{b - a} \geq \frac{f(c) - f(a)}{c - a}}</math>.</td>
</blockquote>

We only prove (a) above. The proof of (b) is analogous.</em></p>
*'''Proof of a):''' Let ><math>a, b, c \in I</math> be such that <math>a < b < c</math>.</li>

*<math>\Rightarrow</math> Suppose that <math>f</math> is convex on <math>I</math>. Then for all <math>t \in [0, 1]</math> we have that:</li>
<div style="text-align: center;"><math>\begin{align} \quad f(tx + (1 - t)y) \leq tf(x) + (1 - t)f(y) \end{align}</math></div>
*Set <math>a = x</math>, <math>b = tx + (1-t)y</math>, and <math>c = y</math>. Combining the first and third equations with the second equation gives us:</li>
<div style="text-align: center;"><math>\begin{align} \quad b = ta + (1-t)c \end{align}</math></div>
*Solving for <math>t</math> gives us:</li>
<div style="text-align: center;"><math>\begin{align} \quad b = ta + c - tc \\ \quad b - c = t(a - c) \\ \end{align}</math></div>
*Therefore:
<div style="text-align: center;"><math>\begin{align} \quad t = \frac{b - c}{a - c} = \frac{c - b}{c - a} \end{align}</math></div>
*Similarly, we compute <math>1 - t</math> to be:</li>
<div style="text-align: center;"><math>\begin{align} \quad 1 - t = \frac{c - a}{c - a} - \frac{c - b}{c - a} = \frac{b - a}{c - a} \end{align}</math></div>
*From the convexity of <math>f</math> we have <math>f(tx + (1-t)y) \leq tf(x) + (1-t)f(y)</math>, or equivalently:</li>
<div style="text-align: center;"><math>\begin{align} \quad f(b) \leq \frac{c - b}{c - a}f(a) + \frac{b - a}{c - a}f(c) \\ \end{align}</math></div>
*And hence:</li>
<div style="text-align: center;"><math>\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}</math></div>
*Therefore:</li>
<div style="text-align: center;"><math>\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}</math></div>
*<math>\Leftarrow</math> Obtained by working backwards from above. <math>\blacksquare</math></li>

We state yet another important definition for convex and concave functions.</p>

<blockquote style="background: white; border: 1px solid black; padding: 1em;"> 
:'''Theorem 2:''' Let ><math>f : I \to \mathbb{R}</math>.<br />
:'''a)''' <math>f</math> is convex on ><math>I</math> if and only if for all ><math>a, b, c \in I</math> with ><math>a <; b <; c</math> we have that ><math>\displaystyle{\frac{f(b) - f(a)}{b - a} \leq \frac{f(c) - f(b)}{c - b}}</math>.<br />
:'''b)''' <math>f</math> is concave on ><math>I</math> if and only if for all ><math>a, b, c \in I</math> with ><math>a <; b <; c</math> we have that ><math>\displaystyle{\frac{f(b) - f(a)}{b - a} \geq \frac{f(c) - f(b)}{c - b}}</math>.</td>
</blockquote>

Theorem 2 gives us a nice characterization of convex functions. It tells us that a function <math>f : I \to \mathbb{R}</math> is convex if and only if whenever we take three points <math>a, b, c \in I</math> with <math>a <; b <; c</math> we have that the slope of the line connecting <math>(a, f(a))</math> and <math>(b, f(b))</math> is less than or equal to the sope of the line connecting <math>(b, f(b))</math> and <math>(c, f(c))</math>. In other words, the slope of the line segments connecting consecutive pairs of points on the graph of <math>f</math> is increasing.</p>
	
We can combine theorems 1 and 2 to get a nice chain of inequalities. That is, <math>f : I \to \mathbb{R}</math> is convex if and only if for all <math>a, b, c \in I</math> with ><math>a < b < c</math> we have that:</p>
<div style="text-align: center;"><math>\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}</math></div>
	
==Licensing==
Content obtained and/or adapted from:
* [https://en.wikipedia.org/wiki/Maxima_and_minima] under a CC BY-SA license
* [http://mathonline.wikidot.com/convex-and-concave-functions] under a CC BY-SA license