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!
Best Answer
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.