We know that a function $f:\,\mathbb{R}\rightarrow\mathbb{R}$ is Lipschitz continuous if exists a real constant $C>0$ such that $$\left|f\left(x\right)-f\left(y\right)\right|\leq C\left|x-y\right|\tag{1}$$ for all $x,y\in\mathbb{R}$. Now assume that for all $\epsilon>0$ exists a $C=C\left(\epsilon\right)>0$ such that $$\left|f\left(x\right)-f\left(y\right)\right|\leq C\left(\epsilon\right)\left|x-y\right|\tag{2}$$ for all $x,y\in\mathbb{R}$ such that $\left|x-y\right|>\epsilon$. My question is:
are these definitions equivalent?
It is obvious that $\left(1\right)\Rightarrow\left(2\right)$ but I'm not sure about the other implication. I tried to prove it by absurd but I get nothing.
I assume you mean to write $|x-y|<\epsilon$ at the end there. The definitions are not equivalent in general. We call the functions satisfying $(2)$ locally Lipschitz continuous. If your space is nice enough, i.e. locally compact, then we have $f$ is locally Lipschitz iff $f$ is Lipschitz on every compact subset of the space.