Existence of function with prescribed values?

Question

Does there exist an infinitely differentiable function $f: \mathbb{R} \rightarrow \mathbb{R}$ equal to $|x|$ when $x \in \mathbb{Z}$?

Answer 1

Yes, but I think it may be somewhat laborious to prove.

Consider the function $$ \varphi(x) = \begin{cases} 0 & \text{if }x\le 0, \\ e^{-1/x} & \text{if } x>0. \end{cases} $$ This is obviously continuous everywhere and infinitely differentiable at all nonzero points, and continuous at $0$. It is fairly easily shown to be differentiable at $0$. To show that it is infinitely differentiable at $0$ is the more onerous part.

Supposing this to be shown, consider the function $$ \chi(x) = c\int_{-\infty}^x\varphi(u)\varphi(1-u)\,du. $$ This is equal to $0$ when $x\le 0$, equal to $1$ when $x\ge 1$ (for the right value of $c>0$), and infinitely differentiable. Multiply it by $x$ and you get a function equal to $|x|$ when $x\ge1$ and to $0$ when $x\ge0$.

In a similar way, construct a function equal to $|x|$ when $x\le-1$ (you can use $x\mapsto\chi(-x)\cdot|x|$) and to $0$ when $x\ge0$.

Then add those two functions together.