Good day. The question is on proving the following relation ($\|\cdot\|$ here and on denotes $\ell\_2$ norm):
$$
\frac{dJ(f)}{df} = -\mathrm{div}\left(\frac{\nabla f}{\|\nabla f\|}\right) \qquad =: DJ,
$$
where $J(f)$ is the total variation norm:
$$
J(f) = \int \|\nabla f\| \, dx
$$


So what I was trying to do was an attempt to prove the Frechet derivative characterization (in other words, show that $DJ$ is the Frechet differential) 
$$
J(f + \delta f) - J(f) - \langle DJ, \delta f \rangle\_{L\_2} = o( \|\delta f \|\_{L\_2})
$$


I used the following fact (coming from Gauss-Ostrogradsky thm):
$$
\int f \,\mathrm{div}(\mathbf w) = -\int \langle \nabla f, \mathbf w \rangle
$$


So, 
$$ 
\int \left[ \| \nabla f + \nabla \delta f \| - \| \nabla f \| \right] - \int DJ \cdot \delta f = (\text{use the mentioned fact})
$$
$$
= \int \left[ \| \nabla f + \nabla \delta f \| - \| \nabla f \| - \langle \frac{\nabla f}{\|\nabla f\|}, \nabla \delta f \rangle \right] \le
$$
$$
\le \int \left[ \| \nabla f \| + \| \nabla \delta f \| - \| \nabla f \| - \langle \frac{\nabla f}{\|\nabla f\|}, \nabla \delta f \rangle \right] 
$$
$$
= \int \left[ \| \nabla \delta f \| - \langle \frac{\nabla f}{\|\nabla f\|}, \nabla \delta f \rangle \right],
$$
where I got stuck. Either I made a mstake above, or I don't know how to prove that the obtained is $o(\| \delta f \|\_{L\_2})$.


Any ideas or suggestions? This is just a curiosity question inspired by my friend's discussion. Thanks.



Fréchet derivative of the Total Variation norm
Not near zero, which is an issue (!), instead of using an inequality, write
\begin{eqnarray\*}
\| \nabla f+ \nabla\delta f \| - \|\nabla f\| &=& \frac{\| \nabla f+ \nabla\delta f \|^2 - \|\nabla f\|^2}{\| \nabla f+ \nabla\delta f \| +\|\nabla f\|}\\
&=& \langle \nabla\delta f , \frac{ 2\nabla f}{\| \nabla f+ \nabla \delta f \| +\|\nabla f\|}\rangle + \frac{\|\nabla \delta f\|^2}{\|\nabla f+ \nabla \delta f \| +\|\nabla f\|} 
\end{eqnarray\*}
**\* edited to be thorough\***
Next, 
\begin{eqnarray\*}
\frac{2\nabla f}{\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert }-\frac{\nabla f}{\left\Vert \nabla f\right\Vert } & = & \frac{\nabla f}{\left\Vert \nabla f\right\Vert }\left(\frac{\left\Vert \nabla f\right\Vert -\left\Vert \nabla f+\nabla\delta f\right\Vert }{\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert }\right)\\
 & = & \frac{\nabla f}{\left\Vert \nabla f\right\Vert }\left(\frac{2\left\langle \nabla\delta f,\nabla f\right\rangle -\|\nabla\delta f\|^{2}}{\left(\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert \right)^{2}}\right)\\
\left\langle \frac{2\nabla f}{\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert }-\frac{\nabla f}{\left\Vert \nabla f\right\Vert },\nabla\delta f\right\rangle & = & \left\langle \nabla\delta f,\frac{\nabla f}{\left\Vert \nabla f\right\Vert }\right\rangle ^{2}\frac{2\left\Vert \nabla f\right\Vert }{\left(\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert \right)^{2}}\\
 & - & \left\langle \nabla\delta f,\frac{\nabla f}{\left\Vert \nabla f\right\Vert }\right\rangle \frac{\left\Vert \nabla\delta f\right\Vert ^{2}}{\left(\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert \right)^{2}},
\end{eqnarray\*}
So altogether, provided $\min\left(\left\Vert \nabla f+\nabla\delta f\right\Vert ,\left\Vert \nabla f\right\Vert \right)>c>0$,
$$
\left\Vert \nabla f+\nabla\delta f\right\Vert +\left\Vert \nabla f\right\Vert -\left\langle \frac{\nabla f}{\left\Vert \nabla f\right\Vert },\nabla\delta f\right\rangle =O\left(\left\Vert \nabla\delta f\right\Vert ^{2}\right)
$$
Which is what you want, unless mistaken. I am not sure whether $L^2$ (instead of $W^{1,1}$ for example) was really part of your question. $J$ is not defined on $L^2$.


