Let us say we have the indicator function $\chi_{\{|x|\leq 1\}}$ in $\mathbb{R}^2$.
How can I write out the weak derivative of this indicator function?
Is it $\delta_{|x|=1}$? Or it should be vector valued measure like $\bigg(\frac{\partial}{\partial x_1} \chi_{{|x|\leq 1}},\frac{\partial}{\partial x_2} \chi_{{|x|\leq 1}}\bigg)$, but now $\frac{\partial}{\partial x_1} \chi_{{|x|\leq 1}}$ is something that depends on the value of $x_2$.
Best Answer
My edit of the other answer (which is wrong) was rejected, so I'll copy it with correction and due credit to its original author.
The weak gradient of the characteristic function of a domain $\Omega$ with a smooth boundary is the vector-valued measure $\nu(x)d\sigma(x)$ where $\nu(x)$ is the inward unit normal at $x\in\partial\Omega$ and $d\sigma$ is the surface measure.
Why so? By the divergence theorem. Recall that by definition, the weak gradient satisfies $$ \int \nabla u\cdot \varphi = -\int u\operatorname{div}\varphi $$ for every compactly supported smooth vector field $\varphi$. With $u=\chi_\Omega$ and $\nabla u$ as above this is exactly the divergence theorem.