Consider the function $$f(x,\vec{v}):=g(I_t+\nabla\cdot(I\vec{v}))$$ where $I=I(\vec{x},t)$ is the image intensity function, $g:\mathbb{R}\to[0,\infty)$ is continuous and non negative, $\vec{v}\in H^1(\Omega)\times H^1(\Omega)$. Here $\Omega=[0,1]^2$. It is assumed that $I\in C^2(\Omega\times[0,\infty)), I_t\in L^2(\Omega),I_x,I_y\in L^\infty(\Omega)$. I have to show that $f$ is Lipschitz wrt $\vec{v}$ provided $g$ is uniformly Lipschitz in every closed ball of radius $r$, $\bar{B(r)}\subseteq\mathbb{R}$ with Lipschitz constant $L_r$. This is what I have tried so far. \begin{align*} |f(x,\vec{v})-f(x,\vec{v}_0)|&=|g(I_t+\nabla\cdot(I\vec{v}))-g(I_t+\nabla\cdot(I\:\vec{v}_0))|\\ &\le L_r|I_t+\nabla\cdot(I\vec{v})-I_t+\nabla\cdot(I\vec{v}_0)|\\ &\le L_r|\nabla\cdot\{I(\vec{v}-\vec{v}_0)\}|\\ &\le L_r|\nabla I\cdot(\vec{v}-\vec{v}_0))|+L_r|I\nabla\cdot(\vec{v}-\vec{v}_0))| \end{align*} Using the given hypothesis, we can show that that the first part satisfies the bounds $$|\nabla I\cdot(\vec{v}-\vec{v}_0))|\le\sqrt 2CL_r\|\vec{v}-\vec{v}_0\|$$ where $C=\max\{|I_x|^2,|I_y|^2\}$. I am stuck with the second part. How to get the required bound for the second part ? Do we need any additional requirement for $I$ ? Any help will be appreciated. Thank You.

  • $\begingroup$ Lipschitz from which space to which space? The map is unbounded from $H^1$ to itself. It's Lipschitz from $H^1$ to $L^2$ - just estimate the second term in your last line. $\endgroup$ Commented Nov 12, 2018 at 15:43
  • $\begingroup$ @HansEngler, I didn't get you exactly. To get the last line I used $\nabla\cdot (I\vec{w})=\nabla I\cdot\vec{w}+I\nabla\cdot\vec{w}$ and used triangle inequality. $\endgroup$
    – creative
    Commented Nov 12, 2018 at 15:48
  • 1
    $\begingroup$ What is the norm $\| \cdot \|$ on the right hand side of your last equation? $\endgroup$ Commented Nov 12, 2018 at 15:55
  • $\begingroup$ @HansEngler I think $L^2$ norm. $\endgroup$
    – creative
    Commented Nov 12, 2018 at 16:03
  • $\begingroup$ There also ought to be a norm on the left hand side of the inequality. What is that norm? $\endgroup$ Commented Nov 12, 2018 at 17:51

1 Answer 1


I doubt that this statement is correct. For a counterexample, consider $g(x) = |x|^p$ with large positive $p$ and $I_t = 0, I = 1$. Then $$ f(x,\vec{v}) = |\nabla \cdot \vec{v}|^p $$ If $p > 2$, this expression need not even be integrable over $\Omega$, since $\nabla \cdot \vec{v}$ is only in $L^2(\Omega)$.

  • $\begingroup$ What if we assume $1\le p<2$ ? $\endgroup$
    – creative
    Commented Nov 13, 2018 at 4:01
  • $\begingroup$ I have edited the question. I forgot to provide some more information. Can you edit your answer accordingly ? Sorry for the inconvenience and you can assume $1\le p < 2$. $\endgroup$
    – creative
    Commented Nov 13, 2018 at 4:39

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