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This question has already been addressed here and there's also a proof in Wikkipedia but to my frustration, I can't find any of these proofs to be coherent :| So I decided to come up with my own proof, albeit I admit I borrowed some ideas from here and there :) If you have the time, I'd be grateful if you checked my solution (particularly the bolded line) to see if it hits the nail on the head.

preliminary information:

Lemma 1. If $f_n \rightarrow f$ in $L^1$ then there exists a subsequence $f_{n_j}$ such that a.e $f_{n_j}\rightarrow f$.

Lemma 2 [The Dominated Convergence Theorem]. If $f_n$ is a sequence of integrable functions such that firstly, $f_n \rightarrow f$ a.e and secondly, there exists a function $g\in L^1$ such that a.e $\lvert f_n \rvert \leq g$ then $\int f = \lim_n \int f_n$.

(Both lemmas are proved in Folland's Real Analysis book.)

Problem. Define $$TV(u,\Omega):=\sup\left\{\int_\Omega u(x) \operatorname{div} \phi(x) \, \mathrm{d}x \colon \phi\in C_c^1(\Omega,\mathbb{R}^n),\ \Vert \phi\Vert_{L^\infty(\Omega)}\le 1\right\}$$ If $u\in W^{1,1}$ then $TV(u,\Omega) = \int_{\Omega} \lvert \nabla u(x) \rvert \, dx$.

Solution. By performing $n$ simple integrations by parts and summing up the results we easily see that $\int\limits_\Omega u\operatorname{div}\varphi = -\int_\Omega\nabla u\cdot\varphi$ therefore $$\int\limits_\Omega u\operatorname{div} \mathbf\phi = - \int\limits_\Omega \mathbf\phi\cdot\nabla u \leq \left| \int\limits_\Omega \mathbf\phi\cdot\nabla u \right|\leq \int\limits_\Omega \left|\mathbf\phi\right|\cdot\left|\nabla u\right|\leq \int\limits_\Omega \left|\nabla u\right|.$$ So in order to show that the supremum over $\phi$ of the leftmost term equals the rightmost term, it suffices to find a sequence $(\phi_n)_1^{\infty}$ such that $\int\limits_\Omega u\operatorname{div} \mathbf\phi_n$ tends to $\int\limits_\Omega \left|\nabla u\right| $.

Assume that $\nabla u \neq 0$ (otherwise the result follows trivially). Since $u\in W^{1,1}$ we have $ \nabla u \in L^1$ thus $\frac{-\nabla u}{\lvert \nabla u \rvert} \in L^1$ and since $C_c^1$ is dense in $L^1$ there is a sequence $(\phi_n)_1^{\infty} \subset C_c^1$ such that $\phi_n \rightarrow \frac{-\nabla u}{\lvert \nabla u \rvert}$ in $L^1$, and since the limit function $\frac{-\nabla u}{\lvert \nabla u \rvert}$ is a unit vector (function), we may assume $\phi_n$'s are unit vectors so that thay have norm 1.

By the first lemma above, there is a subsequence $(\phi_{n_j})$ such that a.e $ \phi_{n_j} \rightarrow \frac{-\nabla u}{\lvert \nabla u \rvert}$ thus $$ \nabla u \cdot \phi_{n_j} \rightarrow \frac{-\nabla u}{\lvert \nabla u \rvert }\cdot \nabla u \quad \text{a.e}.$$ Furthermore $$ \lvert \phi_{n_j}.\nabla u \rvert \leq \lvert \phi_{n_j} \rvert \lvert \nabla u \rvert \leq \lvert \nabla u \rvert \in L^1. $$ Henece both conditions of the second Lemma are met and this Lemma guarantees that we are eligible to take the $lim$ inside the integral: $$ \lim_j \int\limits_\Omega u\operatorname{div} \mathbf\phi_{n_j} = -\lim_j \int_{\Omega} \nabla u \cdot \phi_{n_j} = -\int_{\Omega} \lim_j \nabla u \cdot \phi_{n_j} = \int \frac{\nabla u}{\lvert \nabla u \rvert }\cdot \nabla u = \int_{\Omega} \lvert \nabla u \rvert. \quad \square $$

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You are assuming that $\nabla u\ne0$ everywhere in $\Omega$. It could be that $\nabla u=0$ on a set $E$ of positive measure and $\nabla u\ne0$ in $\Omega\setminus E$.

Update Assume that $\Omega$ is bounded. Define $$g(x):=\begin{cases}-\frac{\nabla u(x)}{\lvert \nabla u(x) \rvert}&\text{if } \nabla u(x)\ne 0,\\0 &\text{otherwise.}\end{cases}$$ Then $g\in L^1(\Omega)$ and so you can find $(\phi_n)_1^{\infty} \subset C_c^1$ converging to $g$ in $L^1$. Consider a smooth function $F:\mathbb{R}^N\to \mathbb{R}^N$ such that $F(y)=y$ if $|y|\le 2$ and $F(y)=0$ if $|y|\ge 4$ and take $g_{n_j}(x):=F(\phi_{n_j}(x))$. Then $|g_{n_j}(x)|\le 4$ and $g_{n_j}(x)\to g(x)$ since if $\phi_{n_j}(x)\to g(x)$, then $|\phi_{n_j}(x)|\le 2$ for all $j$ large enough, since $|g(x)|\le 1$. You can now continue as in your proof.

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  • $\begingroup$ I assumed $\nabla u \neq 0 $ so as to prevent the denominator of $\frac{\nabla u}{\lvert \nabla u \rvert}$ from vanishing. If I assume the gradient vanishes on set $ٍE$ of positive measure, does it affect the proof? Also, do you have any comments about the rest of the proof? Anything else wrong? Am I allowed to assume the $\phi_n$'s are unit vectors? $\endgroup$ – Erfan Jan 13 '18 at 16:04
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    $\begingroup$ I added more details to my reply. You don't need unit vectors, just bounded. $\endgroup$ – Gio67 Jan 13 '18 at 16:44

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