Prove $|f(x) -f(y)| \le 1 \ \forall x,y$

Question

Let $f : [-1, 1] \rightarrow \mathbb{R}$ such that:

1. $|f(x) - f(y)| \le |x-y| \ \forall x,y$

2. $f(1)=f(-1)$

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.

Answer 1

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 $$|f(x)-f(y)|\leq |f(x)-f(-1)|+|f(1)-f(y)|$$

Answer 2

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 $|x-y|>1$ ?

Answer 3

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.