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.