Let $\Omega$ be an open bounded subset of $\mathbb{R}^N$, working in the space $H_0^1(\Omega)$ with the inner product $$(u,v)_{H_0^1} = \int_\Omega \nabla u \cdot \nabla v$$ for $u\in H_0^1$ and $\mu(\{|\nabla u|>1\}) >0$, is there a way to define a function $v$ such that $$(u,v)_{H_0^1} = \int_{\{|\nabla u|>1\}} |\nabla u|^2. $$ So we want $\nabla v = \nabla u$ on $\{|\nabla u|>1\}$, and $\nabla v= 0$ on the set $\{|\nabla u|\leq 1\}$, but naively defining $v$ in this way will not guarantee that $v\in H_0^1(\Omega)$.

Edit: Here is the main problem I am trying to solve. Define $$E(u) = \sup_{v\in H_0^1} \Bigg\{\int_\Omega \nabla u\cdot \nabla v - \int_\Omega |\nabla v| \Bigg\}$$ I claim this is equal to $$E'(u) = \begin{cases} 0 & \text{if } |\nabla u|\leq 1 \text{ a.e.} \\ +\infty & \text{else } \end{cases}$$

By substituting $v = Cu$ for some $C \geq 0$, this is when $\int\nabla u\cdot\nabla v \leq \int |\nabla u||\nabla v|$ becomes an equality, the sup becomes $$E(u) = \sup_{C\geq 0} \Bigg\{C\bigg(\int_\Omega |\nabla u|^2- |\nabla u|\bigg) \Bigg\}$$

Now when taking $\sup_{v\in H_0^1}$, if we could somehow restrict the support of $v = Cu$ onto a subset $K \subset \{|\nabla u|>1\}$ (whenever $\mu(\{|\nabla u|>1\}) >0$, we require $\mu(K) >0$), we will get $$E(u) = \sup_{C\geq 0} \Bigg\{C\bigg(\int_K |\nabla u|^2- |\nabla u|\bigg) \Bigg\} = E'(u)$$ which is what is claimed.

Thank you very much!


1 Answer 1


There is generally no $v\in H^1_0$ with $\nabla v=\chi_{\{|\nabla u|>1\}}\nabla u$. Consider for example $u:(-1/2,2)\to\mathbb R$, $u(x)=\min\{1-x/2,1+2x\}$. Now $u\in H^1_0((-1/2,2))$ but there is no $v\in H^1_0$ with $v'=2\chi_{(-1/2,0)}$ by the fundamental theorem of calculus.

If $u\in C^1_0$ and $u$ is constant on $\{|\nabla u|=1\}$ then a truncation of $u$ works as $v$. You need to assume something like this before you can hope for the existence of $v$.

Another issue is that it is not clear if the vector field $\chi_{\{|\nabla u|>1\}}\nabla u$ has a potential in the first place. In dimension three a local requirement is that the curl of the vector field vanish. Writing $\phi=\chi_{\{|\nabla u|>1\}}$, we have formally $$ \nabla\times(\phi\nabla u)=\phi\nabla\times\nabla u+(\nabla\phi)\times\nabla u=(\nabla\phi)\times\nabla u. $$ Assuming the set $V=\{|\nabla u|>1\}$ and the function $u$ are sufficiently smooth, the condition $(\nabla\phi)\times\nabla u=0$ means that $\nabla u$ should be normal to $\partial V$. (The vector valued distribution $\nabla\phi$ is supported on $\partial V$ and is normal to it.) Thus $u$ should be constant on connected components of $\partial V$. This condition is very similar to the sufficient condition I gave in the previous paragraph.

You need to be more specific about your second question. I understand it like this: "Let $u\in H^1_0$ and $K\subset\{|\nabla u|>1\}$ with positive measure. Does there exist $v\in H^1_0$ with $(u,v)_{H^1_0}=\int_K|\nabla u|^2$?" The answer is yes: for some $\lambda\in[0,1]$ the choice $v=\lambda u$ works.

Edit: This is a remark regarding your edit. If $\phi$ is a cut-off function supported in $K$, a natural attempt is $v=\phi(u-c)$ for some constant $c$. One quickly ends up with the estimate $$ \int_\Omega\nabla u\cdot\nabla v-|\nabla v| \geq \int_\Omega\phi(|\nabla u|^2-|\nabla u|)+(u-c)\nabla u\cdot\nabla\phi-|u-c||\nabla\phi|. $$ But you can't in general choose $\phi$ so that the RHS would be positive since you can't increase $\int\phi$ arbitrarily but keep $\int|\nabla\phi|$ bounded. Something more clever is needed.

  • $\begingroup$ Thank you for the very helpful answer! For the last part, I knew the problem will be at the boundary of $\{|\nabla u| > 1\}$. So by relaxing the condition, I thought maybe there is some hope we could define $v$ on a smaller subset. I actually need $\nabla v$ to be supported on the set $\{|\nabla u| > 1\}$ and I will explain this in detail in my edit. $\endgroup$
    – Xiao
    Nov 17, 2014 at 7:52

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