3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ What is $f$ in your question? $\endgroup$
    – Stefan Kohl
    Dec 11, 2013 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, 2013 at 19:20

1 Answer 1

1
$\begingroup$

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 $L^2$.

$\endgroup$
6
  • $\begingroup$ 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, 2013 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, 2013 at 19:32
  • $\begingroup$ @gron: I wrote above the explicit version. But dominated convergence would have been enough. $\endgroup$
    – username
    Dec 14, 2013 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, 2013 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, 2013 at 20:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.