Let $f:R\to R,x\mapsto|x|$ and $T_f\in\mathcal D'(R)$ the regular distribution of $f$. I want to calculate the second derivative of $T_f$. By definition: $$\frac{d^2}{dx^2}T_f[\varphi]=T_f\left[\left(\frac{d^2}{dx^2}\right)^*\varphi\right]=T_f\left[\frac{d^2}{dx^2}\varphi\right]=\int_R|x|\frac{d^2}{dx^2}\varphi(x)dx,$$ because $\frac{d^2}{dx^2}$ is self-adjoint. As far as I know you can't simply swap the derivatives though, becuase $|x|\notin C^2(R)$.
Instead, I wanted integrate from $(-\infty,0)$ and $(0,\infty)$ (as $\{0\}$ is a null set of $R$): $$\frac{d^2}{dx^2}T_f[\varphi]=-\int_{(-\infty,0)}\frac{d}{dx}|x|\frac{d}{dx}\varphi(x)dx-\int_{(0,\infty)}\frac{d}{dx}|x|\frac{d}{dx}\varphi(x)dx$$ However, this should yield $0$: $$\frac{d^2}{dx^2}T_f[\varphi]=\int_{(-\infty,0)}\frac{d}{dx}\varphi(x)dx-\int_{(0,\infty)}\frac{d}{dx}\varphi(x)dx=0,$$ as $\varphi$ is compactly supported.
However, if we visualize the distribution, one would expect to get the Heaviside step function $\theta$, whose distribution has the Dirac-Delta function $\delta_0$ as derivative.
Is my intuition or above calculation wrong?
You fell at the last hurdle! Check the contribution of the endpoints of the integrals at 0.