Editing Total variation (section)

=====Proof of the equality=====
Under the conditions of the theorem, from the lemma we have:
:<math> \int_\Omega f\operatorname{div} \mathbf\varphi
= - \int_\Omega \mathbf\varphi\cdot\nabla f
\leq \left| \int_\Omega \mathbf\varphi\cdot\nabla f \right|
\leq \int_\Omega \left|\mathbf\varphi\right|\cdot\left|\nabla f\right|
\leq \int_\Omega \left|\nabla f\right| </math>
in the last part <math>\mathbf\varphi</math> could be omitted, because by definition its essential supremum is at most one.

On the other hand, we consider <math>\theta_N:=-\mathbb I_{\left[-N,N\right]}\mathbb I_{\{\nabla f\ne 0\}}\frac{\nabla f}{\left|\nabla f\right|}</math> and <math>\theta^*_N</math> which is the up to <math>\varepsilon</math> approximation of <math>\theta_N</math> in <math> C^1_c</math> with the same integral. We can do this since <math> C^1_c</math> is dense in <math> L^1 </math>. Now again substituting into the lemma:

:<math>\begin{align}
&\lim_{N\to\infty}\int_\Omega f\operatorname{div}\theta^*_N \\[4pt]
&= \lim_{N\to\infty}\int_{\{\nabla f\ne 0\}}\mathbb I_{\left[-N,N\right]}\nabla f\cdot\frac{\nabla f}{\left|\nabla f\right|} \\[4pt]
&= \lim_{N\to\infty}\int_{\left[-N,N\right]\cap{\{\nabla f\ne 0\}}} \nabla f\cdot\frac{\nabla f}{\left|\nabla f\right|} \\[4pt]
&= \int_\Omega\left|\nabla f\right|
\end{align}</math>
This means we have a convergent sequence of <math display="inline">\int_\Omega f \operatorname{div} \mathbf\varphi</math> that tends to <math display="inline">\int_\Omega\left|\nabla f\right|</math> as well as we know that <math display="inline">\int_\Omega f\operatorname{div}\mathbf\varphi \leq \int_\Omega\left|\nabla f\right| </math>. [[Q.E.D.]]

It can be seen from the proof that the supremum is attained when
: <math>\varphi\to \frac{-\nabla f}{\left|\nabla f\right|}.</math>

The [[Function (mathematics)|function]] <math>f</math> is said to be of [[bounded variation]] precisely if its total variation is finite.