Q:

Integral of the Laplace Beltrami operator multiplied by the Heaviside function

Let $f \in W^{2,1}_p(U \times [0,T])$ where $U$ is a $C^2$ compact manifold. I'm having a really hard time understanding why the following computation is valid:

$\int_{U} H_\delta(f) \Delta_U f=-\int_U H_\delta'(f) {\vert \nabla f
 \vert }^2 (1)$

where $\Delta_U$ denotes the Laplace Beltrami operator and $H_\delta$ stands for a smooth approximation of the Heaviside function such that $\lim_{\delta \to 0^+} H_{\delta}=H$ in the sense of distributions.
I know that in any compact manifold $M$ it holds (as a consequence of the divergence theorem):
$\int_{M} f\Delta_M f=-\int_M {\vert \nabla f \vert }^2 (2)$
Although it seems that $(1),(2)$ are strongly related, I can't conclude $(1)$ at the end! Probably I'm missing something about $H_\delta$
I'm really stuck here so I'd appreciate any help.
Thanks in advance!

A:

You need to use (assume that $U$ has no boundary, of $f$ vanishes at the boundary)
$$\int_U \nabla g \cdot \nabla f d\mu = \int g \Delta f d\mu$$
with the chain rule  $\nabla (H_\delta (f)) = H_\delta '(f) \nabla f$. 

