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Homeproof of criterion for convexity

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# proof of criterion for convexity

###### Theorem 1.

Suppose $f\colon(a,b)\to\mathbb{R}$ is continuous and that, for all $x,y\in(a,b)$,

$f\left({x+y\over 2}\right)\leq{f(x)+f(y)\over 2}.$ |

Then $f$ is convex.

###### Proof.

We begin by showing that, for any natural numbers $n$ and $m\leq 2^{n}$,

$f\left({mx+(2^{n}-m)y\over 2^{n}}\right)\leq{mf(x)+(2^{n}-m)f(y)\over 2^{n}}$ |

by induction. When $n=1$, there are three possibilities: $m=1$, $m=0$, and $m=2$. The first possibility is a hypothesis of the theorem being proven and the other two possibilities are trivial.

Assume that

$f\left({m^{{\prime}}x+(2^{n}-m^{{\prime}})y\over 2^{n}}\right)\leq{m^{{\prime}% }f(x)+(2^{n}-m^{{\prime}})f(y)\over 2^{n}}$ |

for some $n$ and all $m^{{\prime}}\leq 2^{n}$. Let $m$ be a number less than or equal to $2^{{n+1}}$. Then either $m\leq 2^{n}$ or $2^{{n+1}}-m\leq 2^{n}$. In the former case we have

$\displaystyle f\left({1\over 2}{mx+(2^{n}-m)y\over 2^{n}}+{y\over 2}\right)$ | $\displaystyle\leq{1\over 2}\left(f\left({mx+(2^{n}-m)y\over 2^{n}}\right)+f(y)\right)$ | ||

$\displaystyle\leq{1\over 2}{mf(x)+(2^{n}-m)f(y)\over 2^{n}}+{f(y)\over 2}$ | |||

$\displaystyle={mf(x)+(2^{{n+1}}-m)f(y)\over 2^{{n+1}}}.$ |

In the other case, we can reverse the roles of $x$ and $y$.

Now, every real number $s$ has a binary expansion; in other words, there exists a sequence $\{m_{n}\}$ of integers such that $\lim_{{n\to\infty}}m_{n}/2^{n}=s$. If $0\leq s\leq 1$, then we also have $0\leq m_{n}\leq 2^{n}$ so, by what we proved above,

$f\left({m_{n}x+(2^{n}-m_{n})y\over 2^{n}}\right)\leq{m_{n}f(x)+(2^{n}-m_{n})f(% y)\over 2^{n}}.$ |

Since $f$ is assumed to be continuous, we may take the limit of both sides and conclude

$f(sx+(1-s)y)\leq sf(x)+(1-s)f(y),$ |

## Mathematics Subject Classification

52A41*no label found*26A51

*no label found*26B25

*no label found*

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