User:TakuyaMurata/Sandbox

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

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

(a) $\phi$ is convex.
(b) The difference quotient
${\phi (x+h)-f(x) \over h}$ increases as $h$ does.
(c) If $d\mu \geq 0$ 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 $I$ .
$\phi \left(\int x(t)d\mu (t)\right)\leq \int \phi \circ x(t)d\mu (t)$ (Jensen's inequality)
(d) If the interval $[a,b]\subset I$ , then
$\phi ((1-\lambda )a+\lambda b)\leq (1-\lambda )\phi (a)+\lambda \phi (b)$ for any $\lambda \in [0,1]$ .
(e) $\phi$ 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 ${\mbox{b}}[a,b]=\{a,b\}$ and $(f-A)(a)=0=(f-A)(b)$ ,

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

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

$(b-x)f(x)+(x-a)f(x)$	$\leq (b-x)f(a)+(x-a)f(b)$
${f(a)-f(x) \over a-x}$	$\leq {f(b)-f(x) \over b-x}$

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

\lim _{h>0,h\to 0}f(x+h)=f(x)+\lim _{h>0,h\to 0}h{f(x+h)-f(x) \over h}

.

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

${f(a)-f(x) \over a-x}$	$\leq {f(b)-f(x) \over b-x}$
$f(x)-f(a)$	$\leq f(b)-f(x)$
$f\left({a+b \over 2}\right)$	$\leq {f(a)+f(b) \over 2}$

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

f\left({1 \over 2^{n}}\sum _{1}^{2^{n}}x_{j}\right)\leq {1 \over 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

$f\left({1 \over 2^{n}}\sum _{1}^{2^{n}}x_{j}\right)$	$=f\left({1 \over 2}\left({1 \over 2^{n-1}}\sum _{1}^{2^{n-1}}x_{j}+{1 \over 2^{n-1}}\sum _{1}^{2^{n-1}}x_{2^{n-1}+j}\right)\right)$
	$\leq {1 \over 2}f\left({1 \over 2^{n-1}}\sum _{1}^{2^{n-1}}x_{j}\right)+{1 \over 2}f\left({1 \over 2^{n-1}}\sum _{1}^{2^{n-1}}x_{2^{n-1}+j}\right)$
	$\leq {1 \over 2^{n}}\sum _{1}^{2^{n-1}}f(x_{j})+{1 \over 2^{n}}\sum _{1}^{2^{n-1}}f(x_{2^{n-1}+j})$
	$={1 \over 2^{n}}\sum _{1}^{2^{n}}f(x_{j})$

Let $x_{1},x_{2}\in [a,b]$ and $\lambda \in [0,1]$ . There exists a sequence of rationals number such that:

\lim _{j\to \infty }{p_{j} \over 2q_{j}}=\lambda

.

It then follows that:

$f(\lambda x_{1}+(1-\lambda )x_{2})$	$\leq \lim _{j\to \infty }{p_{j} \over 2q_{j}}f(x_{1})+\left(1-{p_{j} \over 2q_{j}}\right)f(x_{2})$
	$=\lambda f(x_{1})+(1-\lambda )f(x_{2})$
	$\leq \lambda y_{1}+(1-\lambda )y_{2}$ .