Prove $|f(x) -f(y)| \le 1 \ \forall x,y$ 
Let $f : [-1, 1] \rightarrow \mathbb{R}$ such that:
  
  
*
  
*$|f(x) - f(y)| \le |x-y| \ \forall x,y$ 
  
*$f(1)=f(-1)$
Prove $|f(x) -f(y)| \le 1  \ \forall x,y$

Making $y=0$ I get $|f(x) -f(0)| \le |x| \le 1$ but I cannot extend it to the required conclusion.
 A: Hint : 
Let $x$, $y$ such that $|f(x)-f(y)|>1$. From 1. it comes $|x-y|\geq|f(x)-f(y)|>1$, so without loss of generality we can suppose that $$x<0<y \tag{3}$$ 
In the other hand, 
\begin{align*}
1<|f(x)-f(y)|=|f(x)-f(y)-f(1)+f(1)|&\leq |f(x)-f(1)|+|-f(y)+f(1)|\\
&=|f(x)-f(-1)|+|-f(y)+f(1)| \tag{from 2.}\\
&\leq |-1-x|+|1-y|
\end{align*}
Now, what do you think about the distances between $-1$ and $x$, $1$ and $y$, since $-1\leq x<0<y\leq 1$ and $|x-y|>1$ ? 
A: Hint.
Assume that $|f(x)-f(y)|>1$ for some $x<y$ and show that $$|x-y|>1\Longrightarrow |x-(-1)|+|y-1|<1$$
Then use the fact that
$$|f(x)-f(y)|= |f(x)-f(-1)+f(1)-f(y)|$$
implies $$|f(x)-f(y)|\leq |f(x)-f(-1)|+|f(1)-f(y)|$$
A: Try filling in the details of the following reasoning.
First note that for all $f : [-1, 1] \to \mathbb{R}$ satisfying (1.) and (2.), any vertical translation of $f$'s graph, that is, any $\varphi : [-1, 1] \to \mathbb{R} : x \mapsto f(x) + c$ for some $c \in \mathbb{R}$, also satisfies (1.) and (2.). So we always can choose the convenient translation $\varphi(x) = f(x) - f(1)$. Thus $\varphi(-1) = \varphi(1) = 0$. Therefore without loss of generality we consider that $f$ satisfies (2.) with $f(-1) = f(1) = 0$.
Then from (1.) we obtain
$$ |f(x) - f(\pm1)| \le |x \pm 1| $$
And since $x \in [-1, 1]$
$$ |f(x)| \le 1 - |x|$$
Then
$$ \text{graph}(f) = \{(x, f(x)) : x \in [-1,1]\} \subseteq \{ (x,y) \in \mathbb{R}^2 : |x| + |y| \le 1 \} $$
Now let $m = \min\{f(x) : x \in [-1,1]\}$ and show that $\max\{f(x) : x \in [-1,1]\} \le 1 + m$. Therefore $f([-1,1]) \subseteq [m, m + 1]$, whence follows the claim.
See if you can interpret the following figure according to this reasoning.

