Determine the smallest number P

Question

I have here a hard problem, which I couldn't solve.

Denote $M$ the set of all functions $f:[0,1]\to\mathbb{R}$ with the following properties:

1. $f(x)\ge0, \forall x$ in $[0,1]$,
2. $f(1)=1$,
3. $f(x+y)\ge f(x)+f(y), \forall x,y$ and $x+y$ in $[0,1]$.

The problem is to determine the smallest real value $C$ such that $f(x)\le Cx,$ for all functions $f\in M$ and $x\in[0,1]$.

Answer 1

We have $f(x)\le 1$ for all $x$, for $f(x)+0\le f(x)+f(1-x)\le f(1)$.

By induction we see that $0\le x\le 2^{-n}$ implies $f(x)\le 2^{-n}$. In fact we have just shown the base case $n=0$. If $0\le x\le2^{-(n+1)}$ then $0\le x+x\le 2^{-n}$, hence using the induction hyxpothesis $2f(x)\le f(2x)\le2^{-n}$ as was to be shown.

Therefore $f(x)\le 2x$ fore all $x$: If $x=0$ this follows from $f(0)\le f(1+0)-f(1)=0$. For $x>0$ let $n\in \mathbb N_0$ be minimal with $2^{-n-1}<x$, i.e. $n=\lfloor -\log_2x\rfloor$. Then $2^{-n-1}<x\le2^{-n}$, hence $f(x)\le 2^{-n}<2x$.

The inequality is sharp as can be seen from $$f(x)=\begin{cases}0&\text{if }0\le x\le \frac12\\ 1&\text{if }\frac12<x\le 1 \end{cases} $$

Answer 2

By your conditions: $$ f(1) \ge f(1 - x) + f(x) $$ and thus $f(x) \le f(1) = 1$ for all $x$