A continuous function which is not differentiable at a specified number of points.

Question

Consider functions $f:\Bbb R →\Bbb R$ with the property that $|f(x)-f(y)|\leq 4321|x-y|$ for all real numbers $x$ and $y$. Prove that there exists atleast one such f that is continuous but is non differentiable at exactly 2018 points and satisfies $\lim_{x→∞} (f(x))/|x| =2018.$ To solve this problem, I understand that the given $f$ is Lipchitz continuous and points of non differentiability can be created by producing bumps at required points. But I don’t understand the soul of the problem and the significance of $4321$ and $2018$ mentioned in the problem. Can someone help me please with the solution to this problem and explain the significance of the numbers mentioned? Thank you.

Answer 1

Here is one way:

Pick the points $1,...,2018$. The function $f$ will have zeros at each of these points (and more).

Let $f$ pass through $(1,0)$ and have slope $-2018$ for $x \le 1$.

Let $f$ be zero on $[1,2]$.

Then with $n \in \{2,4,6,...,2014,2016 \}$ and for $x \in [n,n+2]$, let $f$ be the straight lines joining the points $(n,0),(n+1,4321),(n+2,0)$.

Let $f$ pass through $(2018,0)$ and have slope $2018$ for $x \ge 2018$.

We see that $f$ is Lipschitz with rank $4321$, and for $x \le 1$, we have $f(x) = 2018(1-x)$ and for $x \ge 2018$ we have $f(x) = 2018(x-2018)$. It is straightforward to check that $\lim_{|x| \to \infty} {f(x) \over |x|} = 2018$.