When the total variation of a function equals the integral of its gradient

Question

As you probably know, for any function $u\in L^1_{loc} (\Omega)$ the total variation is defined as $$\text{TV}(u,\Omega)= \sup \, \bigg\{ -\int_{\Omega} u\, div \phi \, dx : \phi \in C_c^{\infty} (\Omega,\mathbb{R}^N), \, \lvert \phi (x) \rvert \leq 1\, \forall x\in \Omega \bigg \}. $$ If also $u \in C^1(\Omega)$, by a simple 'integration by parts' and considering the definition of weak derivative in the Sobolev Space $W^{1,1}$ we easily derive $$ -\int_{\Omega} u\, div\phi\, dx = \int_{\Omega} \phi.\nabla u\, dx. $$ However, the text I'm reading claims that here, the $\sup$ over all $\phi$ with $\lvert \phi \rvert \leq 1$ is $$ \text{TV} (u,\Omega)=\int_{\Omega} \lvert \nabla u \rvert\, dx. $$ But to my frustration, I can't show this. Any help will be appreciated.

Answer 1

$\newcommand{\fbold}{\mathbf f} \newcommand{\phibold}{\boldsymbol \phi}$ Set $\fbold:=\nabla u$. You know that $\fbold\in L^1(\Omega; \mathbb R^n)$ and you are asking whether $$\tag{1} \sup\left\{ \int_{\mathbb R^n} \fbold\cdot \phibold\, dx\ \Big|\ \phibold\in L^\infty(\Omega; \mathbb R^n),\ \|\phibold\|_\infty\le 1\right\}=\|\fbold\|_1.$$ This is true, and it is a consequence of the relation $(L^1(\Omega;\mathbb R^n))^\star = L^\infty(\Omega;\mathbb R^n)$, where the duality pairing is given by the integral in (1).