## Problem 3

Let $\mathrm{S}$ be the set of functions $\mathrm{f}$ defined on reals in the closed interval $[0,1]$ with non-negative real values such that $\mathrm{f}(1)=1$ and $\mathrm{f}(\mathrm{x})+\mathrm{f}(\mathrm{y}) \leq \mathrm{f}(\mathrm{x}+\mathrm{y})$ for all $\mathrm{x}, \mathrm{y}$ such that $\mathrm{x}+\mathrm{y} \leq 1$. What is the smallest $\mathrm{k}$ such that $\mathrm{f}(\mathrm{x}) \leq$ $\mathrm{kx}$ for all $\mathrm{f}$ in $\mathrm{S}$ and all $\mathrm{x}$ ?


## Solution

Answer: $\mathrm{k}=2$.

Consider the function $f(x)=0$ for $0 \leq x \leq 1 / 2,1$ for $1 / 21 / 2$, then $2 x>1$, so $f(x)<2 x$. If $x$ $\leq 1 / 2$, choose $\mathrm{n} \geq 1$, so that $1 / 2^{n+1}<\mathrm{x} \leq 1 / 2^{n}$. Then $2^{\mathrm{n}} \mathrm{f}(\mathrm{x}) \leq \mathrm{f}\left(2^{n^{n}} \mathrm{x}\right) \leq 1$, so $\mathrm{f}(\mathrm{x}) \leq 1 / 2^{\mathrm{n}} \leq 2 \mathrm{x}$.

