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.
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Related – Giuseppe Negro Dec 03 '17 at 11:25
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Not really. The post you're referring to discusses whether or not the old and new definitions of TV are equivalent (which are not) but I'm asking something different, namely how the new TV definition takes one a simpler form for smooth functions. – User32563 Feb 28 '18 at 11:42
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Ok, I retracted the close vote. – Giuseppe Negro Feb 28 '18 at 11:45
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I think it is a standard fact of L1 functions. The dual of L1 is Linfty and you are taking the sup of $\nabla f$ against the unit ball of $L^\infty $. – Giuseppe Negro Mar 01 '18 at 06:08
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$\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).
Giuseppe Negro
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Unfortunately it $div \phi$ and not $\phi$ itself that's been used as the integrand. – User32563 Apr 09 '18 at 21:13
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@Erfan: That's OK. Just do an integration by parts: $\int u(-\mathrm{div} \boldsymbol \phi), dx = \int \nabla u \cdot \boldsymbol \phi$ and now apply this answer. – Giuseppe Negro Apr 12 '18 at 09:55