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!