If $f$ satisfies Lipschitz condition then f is bounded variation in $\mathbb{R}$?

Question

If $f$ satisfies Lipschitz condition then $f$ is bounded variation in $\mathbb{R}$? A function $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfies a Lipschitz condition at if $\exists M>0$ such that $\forall x,y \in \mathbb{R}$, it holds that $|f(x)−f(y)|≤M|x−y|$

My answer is no, I try to find a example.I know that Dirichlet function is not of bounded variation on any interval in $\mathbb{R}$, but $f(x)=1$ if $x \in\mathbb{Q}$, $f(x)=0$ if $f \in \mathbb{R- Q}$ satifies Lipschitz condition?

Do you have another example?. Thanks for your help.

Answer 1

Take $$f(x)=x \;\; for \;\; x\in \Bbb R$$

then

$$|f(x)-f(y)|=|x-y|\le \color{red}{1}|x-y|$$

$ f $ is Lipschitz and unbounded.