TITLE: A smooth function such that the second derivative of its absolute value is a distribution of positive order
QUESTION [6 upvotes]: Let $f\in C^\infty(\mathbb R;\mathbb R)$ and let us define $g(x)=\vert f(x)\vert$. It is easy to verify that $g$ is locally Lipschitz-continuous function, but I would like to find an example of a smooth function $f$ where $g''$ is a distribution with positive order, that is a distribution which not a Radon measure.

REPLY [4 votes]: No, this is not possible. Clearly, the singular part of $g''$ results from points where $f(x)=0$ and $f'(x)\neq 0$. A given interval $[a,b]$ can contain at most countably many such points. Denote the set of these points by $X$. We need to show that $\sum_{x\in X} |f'(x)|$ is finite. Take any finite subset of $X$, ordered such that $a\le x_1<...<x_N\le b$. By Rolle's theorem, there is a zero of $f'$ between $x_i$ and $x_{i+1}$, hence
$$|f'(x_{i+1})|\le \int_{x_i}^{x_{i+1}} |f''(x)|\,dx.$$ It follows that
$$\sum_{i=1}^N |f'(x_i)|\le |f'(a)|+\int_a^b |f''(x)|\,dx.$$