Let $c > 1$ be a real number. A function $f: [0 ,1 ] \to R$ is called c-friendly if $f(0) = 0, f(1) = 1$ and $|f(x) -f(y)| \le c|x - y|$ for all the numbers $x ,y \in [0,1]$. Find the maximum of the expression $|f(x) - f(y)|$ for all c-friendly functions $f$ and for all the numbers $x,y \in [0,1]$.
Problem
Source: Rioplatense Olympiad 2016 level 3 P4
Tags: function, algebra, maximum, Functional inequality
gejay
03.07.2019 11:35
Do you have the solution to the problem?
abc1234561
03.07.2019 13:14
parmenides51 wrote: Let $c > 1$ be a real number. A function $f: [0 ,1 ] \to R$ is called c-friendly if $f(0) = 0, f(1) = 1$ and $|f(x) -f(y)| \le c|x - y|$ for all the numbers $x ,y \in [0,1]$. Find the maximum of the expression $|f(x) - f(y)|$ for all c-friendly functions $f$ and for all the numbers $x,y \in [0,1]$. Is it possible to use Lagrange theorem?
deadflowers
03.07.2019 13:19
Try it, it may be possible
Gryphos
04.07.2019 19:12
We show that the maximum is $\frac{c+1}{2}$.
Let $0 \leq x < y \leq 1$, then by assumption $|f(x)-f(y)| \leq c(y-x)$. On the other hand,
$$|f(x)-f(y)| \leq |f(0)-f(x)|+|f(1)-f(0)|+|f(y)-f(1)| \leq cx+1+c(1-y)=c+1+c(x-y).$$Adding the two inequalities yields $2|f(x)-f(y)| \leq c+1$.
The maximum is attained for the piecewise linear function $f$ given by
$$f(x)=\begin{cases}
-cx & x \leq \frac{c-1}{4c} \\
cx-\frac{c-1}{2} & \frac{c-1}{4c} \leq x \leq 1- \frac{c-1}{4c} \\
-cx+c+1 & x \geq 1- \frac{c-1}{4c}
\end{cases} $$for the points $x= \frac{c-1}{4c}$, $y=1- \frac{c-1}{4c}$.