# Lipschitz map problem

Let $$f : \mathbb{R}\rightarrow \mathbb{R}$$ be a function such that $$| f(x)-f(y)| \leq 4321|x-y|$$ for all $$x,y \in \mathbb{R}$$. Then which one of the following is true.

1. $$f$$ is always differentiable.
2. There exists at least one such $$f$$ that is continuous and satisfies $$\displaystyle{\lim_{x \to \pm\infty}\frac{f(x)}{|x|}} = \infty$$.
3. There exists at least one such $$f$$ that is continuous but not differentiable at exactly $$2018$$ points and satisfies $$\displaystyle{\lim_{x \to \pm\infty}\frac{f(x)}{|x|}} = 2018$$.
4. It is not possible to find a sequence $$\{x_n\}$$ of reals such that $$\displaystyle{\lim_{n \to \infty} x_n = \infty}$$ and further satisfying $$\displaystyle{\lim_{n \to \infty}|\frac{f(x_n)}{x_n}|} \leq 10000$$.

My argument: It is known that a Lipschitz map is uniformly continuous and differentiable almost everywhere. Thus option one is eliminated. Also since $$\frac{f(x)}{x}$$ is bounded, option 2 and 4 can be eliminated. But I cannot establish the necessary argument for option 3.

• For (3), get a function $g(x) = 2018 |x|$. The behavour on infinity is ok, now you are free to make whatever you want with this function on any compact. E.g. look what happens with the function $| g(x) -1|$. Oct 27, 2021 at 9:55

To elimnate 1) you have to give an example. $$f(x)=|x|$$ will do.
$$f(x)=c \sum\limits_{k=1}^{2018} \frac {|x-k|} {2^{k}}$$ satsifies 3) if $$c$$ is chosen so that $$c \sum\limits_{k=1}^{n} \frac {1} {2^{k}}=2018$$.
• @debabratachakraborty I just built $f$ on the fact that $|x-c|$ is differentiable at points except $c$. So our $f$ is differentiable at all points except $1,2,...,2018$. We can make the limit of $\frac {f(x)} x$ equal to any number by choosing $c$ appropriately. Oct 27, 2021 at 12:40