Introduction to Analysis/Sandbox

A function $\varphi$ from an interval $I$ to $ℝ$ is said to be convex if for every $K\subset \subset I$ and any affine function $f$ on $K$, $\varphi \le f$ on $\partial K$ implies that $\varphi \le f$ on $K$.

Theorem Let $\varphi$ be a real-valued function on an interval $I$. The following are equivalent:

• (a) $\varphi$ is convex.
• (b) The difference quotient

  $\frac{\varphi (x+h)-f(x)}{h}$ increases as $h$ does.

• (c) If $d\mu \ge 0$ and is supported by a compact set $K$ such that $\int d\mu =1$, then we have: for any $x:K\to I$ such that $\int x(t)d\mu (t)$ is in $I$.

  $\varphi (\int x(t)d\mu (t))\le \int \varphi \circ x(t)d\mu (t)$ (Jensen's inequality)

• (d) If the interval $[a,b]\subset I$, then

  $\varphi ((1-\lambda )a+\lambda b)\le (1-\lambda )\varphi (a)+\lambda \varphi (b)$ for any $\lambda \in [0,1]$.

• (e) $\varphi$ is integrable on $I$.

Proof: Suppose (a). For each $x\in [a,b]$, there exists some $\lambda \in [0,1]$ such that $x=\lambda a+(1-\lambda )b$. Let $A(x)=A(\lambda a+(1-\lambda )b)=\lambda f(a)+(1-\lambda )f(b)$, and then since $\text{b}[a,b]=\{a,b\}$ and $(f-A)(a)=0=(f-A)(b)$,

$|f(x)|=|f(x)-A(x)+A(x)|$	$\le \sup \{|f(a)-A(a)|,|f(b)-A(b)|\}+A(x)$
	$=\lambda f(a)+(1-\lambda )f(b)$

Thus, (a) $\Rightarrow$ (b). Now suppose (b). Since $\lambda =\frac{x-a}{b-a}$, for $\lambda \in (0,1)$, (b) says:

$(b-x)f(x)+(x-a)f(x)$	$\le (b-x)f(a)+(x-a)f(b)$
$\frac{f(a)-f(x)}{a-x}$	$\le \frac{f(b)-f(x)}{b-x}$

Since $a-x<b-x$, we conclude (b) $\Rightarrow$ (c). Suppose (c). The continuity follows since we have:

$\underset{h>0,h\to 0}{\lim }f(x+h)=f(x)+\underset{h>0,h\to 0}{\lim }h\frac{f(x+h)-f(x)}{h}$.

Also, let $x=2^{-1}(a+b)$ such that $a<b$, for $a,b\in K$. Then we have:

$\frac{f(a)-f(x)}{a-x}$	$\le \frac{f(b)-f(x)}{b-x}$
$f(x)-f(a)$	$\le f(b)-f(x)$
$f\left(\frac{a+b}{2}\right)$	$\le \frac{f(a)+f(b)}{2}$

Thus, (c) $\Rightarrow$ (d). Now suppose (d), and let $E=\{(x,y):x\in [a,b],y\ge f(x)\}$. First we want to show

$f\left(\frac{1}{2^n}\sum _1^{2^n}{x}_j\right)\le \frac{1}{2^n}\sum _1^{2^n}f(x_j)$.

If $n=0$, then the inequality holds trivially. if the inequality holds for some $n-1$, then

$f\left(\frac{1}{2^n}\sum _1^{2^n}{x}_j\right)$	$=f\left(\frac{1}{2}(\frac{1}{2^{n-1}}\sum _1^{2^{n-1}}{x}_j+\frac{1}{2^{n-1}}\sum _1^{2^{n-1}}{x}_{2^{n-1}+j})\right)$
	$\le \frac{1}{2}f\left(\frac{1}{2^{n-1}}\sum _1^{2^{n-1}}{x}_j\right)+\frac{1}{2}f\left(\frac{1}{2^{n-1}}\sum _1^{2^{n-1}}{x}_{2^{n-1}+j}\right)$
	$\le \frac{1}{2^n}\sum _1^{2^{n-1}}f(x_j)+\frac{1}{2^n}\sum _1^{2^{n-1}}f(x_{2^{n-1}+j})$
	$=\frac{1}{2^n}\sum _1^{2^n}f(x_j)$

Let $x_1,x_2\in [a,b]$ and $\lambda \in [0,1]$. There exists a sequence of rationals number such that:

$\underset{j\to \mathrm{\infty }}{\lim }\frac{p_j}{2q_j}=\lambda$.

It then follows that:

$f(\lambda x_1+(1-\lambda )x_2)$	$\le \underset{j\to \mathrm{\infty }}{\lim }\frac{p_j}{2q_j}f(x_1)+(1-\frac{p_j}{2q_j})f(x_2)$
	$=\lambda f(x_1)+(1-\lambda )f(x_2)$
	$\le \lambda y_1+(1-\lambda )y_2$.

Thus, (d) $\Rightarrow$ (e). Finally, suppose (e); that is, $E$ is convex. Also suppose $K$ is an interval for a moment. Then

$f(\lambda a+(1-\lambda )b)\le \lambda f(a)+(1-\lambda )f(b)$. $◻$