We define the Banach space of functions of bounded variation on $\Omega\subseteq\mathbb{R}^n$ (assume as smooth a domain as we need) as all $u\in L^1(\Omega)$ for which $$\|u\|_{BV}:=\|u\|_1+\int|Du|<\infty$$ where the total variation is defined as $\int|Du|:=\sup_{\phi\in C^1_c(\Omega,\mathbb{R}^n),\|\phi\|_\infty=1}\int_\Omega u\nabla\phi dx$. The intuitive definition is that the distributional derivative $Du$ is a finite measure. I came across the following statement:
If $u$ is itself $C^1(\Omega)$, then then its variation is exactly the integral of the absolute value of its gradient: $\int|Du|=\int_\Omega|\nabla u|dx=\| |\nabla u| \|_1$, where $|\nabla u|=\sqrt{\nabla u\cdot\nabla u}$.
This makes sense considering the intuitive definition, though I am having trouble proving this fact from the definition.
The best I've come up with is if $u$ has two derivatives, then $\phi:=-\nabla u/|\nabla u|$ will be an admissible $C^1$ function which will give $\int|Du|\geq\||\nabla u|\|_1$. I cannot see how to prove the other inequality, nor how to even achieve this equality when $u$ is only $C^1$.
Use integration by parts, then the integrand becomes the product of our gradient with a test function bounded by 1. Although I would feel more comfortable if your test functions were required to have compact support.