$\mathbb{1}_{B_1(0)}$ doesn't have a $\partial_{x_i}$weak derivative in $\mathbb{R}^n$

Question

I want to prove that $\mathbb{1}_{B_1(0)}$ doesn't have a $\partial_{x_i}$weak derivative in $\mathbb{R}^n$. So I tried a proof by contradiction. So assume that $v$ is its weak derivative w.r.t $x_i$. Then we have that for any $\phi \in C_c^{\infty}(\mathbb{R}^n) $

$-\int_{\mathbb{R}^n}v \phi = \int \mathbb{1}_{B_1(0)} \phi_{x_i} = \int_{B_1(0)} \phi_{x_i}$.

So I think now I should construct a sequence of functions in $C_c^{\infty}(\mathbb{R}^n)$ so that they are uniformly bounded (by, say 1) and they converge to $0$ point-wise a.e. And try to get a contradiction. But I don't know what to do with $\int_{B_1(0)} \phi_{x_i}$ for such functions.

Answer 1

There is a formula for integration by parts in $\mathbb R^n$: $$\int_\Omega f_{x_i}g = \int_{\partial \Omega} fg \, \nu_i - \int_\Omega fg_{x_i}$$ where $\nu_i$ is the $i^{\text{th}}$ component of the outward facing unit normal. Using this with $f = \phi$ and $g = 1$ shows that $$\int_{B_1(0)} \phi_{x_i} = 0$$ for all $\phi$ supported in $B_1(0)$. Supposing such $v$ exists, we would then have $$\int_{\mathbb R^n} v \phi = \int_{B_1(0)} v \phi = 0$$ for any smooth $\phi$ with support in $B_1(0)$ and thus $v = 0$ a.e. in $B_1(0)$. Now taking $\phi$ with compact support in $\mathbb R^n \setminus B_1(0)$, we will also have $$\int_{\mathbb R^n \setminus B_1(0)} v\phi = \int_{\mathbb R^n} v\phi = \int_{B_1(0)} \phi_{x_i} = 0$$ and thus $v = 0$ a.e. in $\mathbb R^n \setminus B_1(0)$ as well. Hence $v = 0$ a.e. in $\mathbb R^n$. Thus we must have $$\int_{B_1(0)} \phi_{x_i} =0$$ for all $\phi \in C_c^\infty(\mathbb R^n)$ which is clearly absurd. Hence such $v$ does not exist.

Here I have twice used that fact that for open $\Omega \subseteq \mathbb R^n$, if $v \in L^2(\Omega)$ and $$\int_{\Omega} v\phi = 0$$ for all $\phi \in C_c^\infty(\Omega)$, then $v =0$ a.e. in $\Omega$.