A smooth function such that the second derivative of its absolute value is a distribution of positive order 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.
 A: 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
$$\sum_{i=1}^N |f'(x_i)|\le |f'(a)|+\int_a^b |f''(x)|\,dx.$$
A: This is not an answer but rather a perhaps non-obvious example. A natural try is a smooth function which changes signs very often like $f(x)=h(x)\sin(\pi/x)$ with, e.g., $h(x)=\exp(-1/x)$ and $h(x)=0$ for $x<0$ .  The idea is that the second derivative would contain (in some sense) a sequence $c_n \delta_{1/n}$ which might prevent order zero.
However, for a regular distribution of sign changes in points $x_n\to 0$ this will not work, because by partial integration
$$
\langle g'',\varphi\rangle =\int \varphi''(x) |f(x)| dx =\sum_{n=1}^\infty (-1)^n \int_{x_n}^{x_{n+1}} \varphi''(x) f(x)dx$$
$$ = \sum_{n=1}^\infty (-1)^n \left(\left[\varphi'(x)f(x)-\varphi(x)f'(x)\right]_{x_n}^{x_{n+1}}+\int_{x_n}^{x_{n+1}} \varphi(x) f''(x)dx\right).$$
Since $x_n$ are zeros of $f$, the critical terms involving $\varphi'$ disappear and one can estimate $\langle g'',\varphi\rangle$ by the sup-norm of the test function $\varphi$, i.e., $g''$ has order $0$.
I don't know whether this phenomenon can be made a proof because the set of sign changes of a smooth function can probably be very wild.
