# To verify Lipschitz continuity of the given function $f$.

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.

• 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. Commented Nov 12, 2018 at 15:43
• @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. Commented Nov 12, 2018 at 15:48
• What is the norm $\| \cdot \|$ on the right hand side of your last equation? Commented Nov 12, 2018 at 15:55
• @HansEngler I think $L^2$ norm. Commented Nov 12, 2018 at 16:03
• There also ought to be a norm on the left hand side of the inequality. What is that norm? Commented Nov 12, 2018 at 17:51

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 $$L^2(\Omega)$$.
• What if we assume $1\le p<2$ ? Commented Nov 13, 2018 at 4:01
• 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 $1\le p < 2$. Commented Nov 13, 2018 at 4:39