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Prove or disprove that $u=0$ a.e. on $\Bbb R^d$
Let $\Omega\subset\Bbb R^d$ be an open set. Let $k:\Bbb R^d\to [0,\infty)$ be measurable such that $0\in \operatorname{supp}k$. This implies that $\Omega\subset \Omega_k=\Omega+\operatorname{supp}k$. ...
7
votes
1
answer
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Level sets of weakly differentiable funtions
Let $C$ be a $C^1$ hypersurface in $R^n$ and let $u \in C^1(R^n)$. Suppose
$$\nabla u(x) \cdot \eta(x)=|\nabla u| \ \ \forall x\in C$$
where $\eta(x)$ is the normal vector to $C$ at $x$ ($\nabla u$ ...