满足\(\small x< y\implies f(x)\le f(y),\,f(1-x)=1-f(x),\,f(5x)=2f(x)\) 的函数

elim

题：求具有下列三性质的函数\(\,f:\mathbb{R}\to\mathbb{R}\):
\((1)\quad x< y\implies f(x)\le f(y)\),
\((2)\quad f(1-x) = 1-f(x),\)
\((3)\quad f(5x)=2f(x).\)

elim

题：固定\(\,2< p\in\mathbb{N},\,\)试证\(\,\exists! f:\mathbb{R}\to\mathbb{R}\,\)满足
\((1)\quad x< y\implies f(x)\le f(y),\)
\((2)\quad f(1-x)=1-f(x),\)
\((3)\quad f(px)=2f(x).\)
证：假定所论\(f\)存在,则
\(\qquad\small f(0)\overset{(3)}{=}2f(0),\,f(1)\overset{(2)}{=}1-f(0),\,f(x)\overset{(1),\,\exists n}{\ge}f(p^{-n})\overset{(3)}{=}2^{-n}\;(x>0)\)
\(\qquad\small f(\frac{1}{p}-x)=\frac{1}{2}f(1-px)=\frac{1}{2}(1-f(px))<\frac{1}{2}=f(\frac{1}{p})=f(\frac{p-1}{p})\)
\(\qquad\small f(\frac{p-1}{p}+x)=1-f(\frac{1}{p}-x)>1-\frac{1}{2}\;(x>0)\)
\(\underset{\,}{\therefore}\quad\small\boxed{f(0)=0,\,f(1)=1\,\;f^{-1}(\{2^n\})=[p^n,p^n(p-1)]\;(n\in\mathbb{Z})}\)
\(\quad\)对\(\,n\in\mathbb{N}^+,\,k\in\mathbb{Z},\,\gcd(k,p)=1,\,\)定义：
\(\quad f(\frac{k}{p^n})=\begin{cases}{\small 1-}f({\small 1-}\frac{k}{p^n}),& {\small k< 0},\\0,& {\small k=0},\\ \frac{1}{2}f(\frac{k}{p^{n-1}}),& {\small 1\le k< p^{n-1}},\\ \frac{1}{2},&{\small p^{n-1}\le k\le (p-1)p^{n-1}},\\1-\frac{1}{2}f(\frac{p^n-k}{p^{n-1}}),& {\small p^{n-1}< k< (p-1)p^{n-1}},\\ 1,& {\small k=p^n},\\ 2f(\frac{k}{p^{n+1}}),& {\small k>p^n}. \end{cases}\)
\(\quad\)这是一个递归性定义,把负自变量的函数值用正自变量的
\(\quad\)函数值,单位区间外的函数值用单位区间内的函数值表出,
\(\quad r={\frac{k}{\large p^{\,n}}}{\small\in(0,1)-}[\frac{1}{p},\frac{p-1}{p}]\,\)时\(f(r)\)由\(\,f(r')\,(r'=\frac{k'}{p^{\,m-1}}\in(0,1))\)
\(\quad\)表出,最后\( f(r)=\frac{1}{2}\,(r=\frac{k}{p^n}\in[\frac{1}{p},\frac{p-1}{p}])\).
\(\qquad\)下面证明\(\,f\,\)在\(\small\,D_p=\{{\large\frac{k}{p^n}}:\, n,|k|\in\mathbb{N}\}\cap[0,1]\)上满足\((1\sim 3)\).
\((\dagger)\quad f(\frac{k}{p^n})-f(\frac{k-1}{p^n})\in\{0,\frac{1}{2^n}\}\;(k=\overline{1,p^m})\,\)对\(\,n=1\,\)成立.
\(\quad\)设\((\dagger)\)对某\(\small\,n\ge 1\,\)成立,则\(\,\small f(\frac{k}{p^{n+1}})-f(\frac{k-1}{p^{n+1}})=\frac{1}{2}(f(\frac{k}{p^n})-f(\frac{k-1}{p^n})),\)
\(\quad\small=f(\frac{p^{n+1}-k+1}{p^{n+1}})-f(\frac{p^{n+1}-k}{p^{n+1}})\in\{0,\frac{1}{2^{n+1}}\}\).同理\((2\sim 3)\)在\(\small D_p\)上成立.
\((\ddagger)\quad f(\{\frac{k}{p^n}\mid k=\overline{1,p^n}\})=\{\frac{k}{2^n}\mid k=\overline{1,2^n}\}\;\;\small(n=1,2,\ldots)\)
\(\qquad\)这是因为\(\small\,f(0)=0,f(1)=1\)以及\(\small\,(\dagger).\)
\(\quad\)最后定义\(\,f(x)=\sup\{f(r)\mid x\ge r{\small\in D_p}\}\;\small(x\in\mathbb{R}).\quad\square\)

elim

楼上是目前能够想到的较简明的构造. 我本想通过函数图像\(\,\{(x,f(x)\mid x\in\mathbb{R}\}\)来给出\(\,f\,\)的构造.直观上看,所论图像是平面上一大堆水平线段,其构造可说是一种分形几何(fractal,递归二维水平加密线段序列)目前尚未找到好的表述和解读方式.希望大家参与.