Consider an integer $n \geq 1$, $a_1,a_2, \ldots , a_n$ real numbers in $[-1,1]$ satisfying \begin{align*}a_1+a_2+\ldots +a_n=0 \end{align*}and a function $f: [-1,1] \mapsto \mathbb{R}$ such \begin{align*} \mid f(x)-f(y) \mid \le \mid x-y \mid \end{align*}for every $x,y \in [-1,1]$. Prove \begin{align*} \left| f(x) - \frac{f(a_1) +f(a_2) + \ldots + f(a_n)}{n} \right| \le 1 \end{align*}for every $x$ $\in [-1,1]$. For a given sequence $a_1,a_2, \ldots ,a_n$, Find $f$ and $x$ so hat the equality holds.
Problem
Source: Balkan MO ShortList 2008 A5
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Tintarn
07.04.2020 18:25
W.l.o.g. $a_1 \ge a_2 \ge \dots \ge a_n$. Let $k$ be the index with $a_k \ge x \ge a_{k+1}$ (with the understanding that $k=0$ and $k=n$ is possible). W.l.o.g. $k \le \frac{n}{2}$ (otherwise replace $a_k$ by $-a_k$ and $x$ by $-x$). Then \[\left\vert nf(x)-f(a_1)-f(a_2)-\dots-f(a_n)\right\vert \le \vert f(x)-f(a_1)\vert+\dots+\vert f(x)-f(a_n)\vert \le (a_1-x)+\dots+(a_k-x)+(x-a_{k+1})+\dots+(x-a_n)=2(a_1+\dots+a_k)+(n-2k)x \le n\]by the triangle inequality. For the equality case, there are two cases: 1) $n=2k$. Then we need $a_1=\dots=a_k=1$ and hence $a_{k+1}=\dots=a_n=-1$. Then it's easy to see that equality can only hold if $f(x)=x+c$ for some $c$ or $f(x)=-x+c$ for some $c$, and then equality already holds for all $x$. 2) $n \ne 2k$. Then we need $a_1=\dots=a_k=1$ as well as $x=1$. We then need $f(y)=y+c$ or $f(y)=-y+c$ for some $c$ and all $y=a_{k+1},\dots,a_n$. Of course the same equality case exists when $x=-1$ and some $n-k$ of the $a_i$ are equal to $-1$.