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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.

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  • $\begingroup$ What is $f$ in your question? $\endgroup$ – Stefan Kohl Dec 11 '13 at 21:48
  • $\begingroup$ @Stefan Kohl, basically $f$ is a function of (several) arguments. Since my question is not a certain problem with well-known setting, I would say that we can add any assumptions which are necessary. $\endgroup$ – agronskiy Dec 13 '13 at 19:20
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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 $L^2$.

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  • $\begingroup$ The restriction on $\nabla f\neq 0$ is necessary. In fact, if $\nabla f=0$ on a set of positive measure, then there will be a problem. However in general we can use the subdifferential since $J(f)$ is a convex functional. $\endgroup$ – Kelei Wang Dec 13 '13 at 3:46
  • $\begingroup$ Thanks for your answers, that's definitely a good idea, since it simplifies a bit, but I still can't see how this helps to prove that the final integral is $o( \| \delta f \|_{L_2})$. The dominated convergence doesn't help here a lot, does it? $\endgroup$ – agronskiy Dec 13 '13 at 19:32
  • $\begingroup$ @gron: I wrote above the explicit version. But dominated convergence would have been enough. $\endgroup$ – username Dec 14 '13 at 18:15
  • $\begingroup$ @AthanagorWurlitzer I'm starting to get the point. The difference between what I stated (the Frechet derivative definition for $L_2$, which is $o( \| \delta f \|_{L_2})$) and what you have shown (the same for $o(\| \nabla\delta f \|)$ is motivated by the fact that the proper space to work in is the Sobolev's one, not $L_2$. Am I right with that? $\endgroup$ – agronskiy Dec 14 '13 at 21:30
  • $\begingroup$ @gron: truly speaking, BV is a better idea, because of the inherent problem with $L^{1}$ not being reflexive. $\endgroup$ – username Dec 16 '13 at 20:17

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