The characteristic function of a bounded open subset of $\mathbb{R}^{n}$ is not in $W^{1,1}(\mathbb{R}^{n})$

Question

Let $E$ a bounded open subset of $\mathbb{R}^{n}$ , as i prove that $1_{E} \not\in W^{1,1}(\mathbb{R}^{n})$? I tried to use the proposition 9.3 (pg 267) in the book of Brezi's, but I could not!

Answer 1

This works for $W^{1,p}(\mathbb{R}^{n})$ if $p \in [1,\infty]$.

First, recall that if $u \in W^{1,p}(\mathbb{R}^{n})$ and we define $E_{0} = \{x \in \mathbb{R}^{n} \, \mid \, u(x) = 0\}$ and $E_{1}=\{x \in \mathbb{R}^{n} \, \mid \, u(x) = 1\}$, then $Du(x) = 0$ almost everywhere in $E_{0}$ and $E_{1}$.

Also recall that if $u \in W^{1,p}(\mathbb{R}^{n})$ and $Du = 0$ almost everywhere, then there is a constant $C \in \mathbb{R}$ such that $u(x) = C$ almost everywhere.

By way of contradiction, suppose there is a Lebesgue measurable set $E$ such that $\chi_{E} \in W^{1,p}(\mathbb{R}^{n})$. The first observation shows that $D\chi_{E}(x) = 0$ almost everywhere in $E$ and in $\mathbb{R}^{n} \setminus E$. In other words, $D \chi_{E}(x) = 0$ almost everywhere in $\mathbb{R}^{n}$. By the second observation, this proves there is a $C_{E} \in \mathbb{R}$ such that $\chi_{E}(x) = C_{E}$ almost everywhere in $\mathbb{R}^{n}$. Thus, if $C_{E} = 0$, then $|E| = 0$; otherwise, $C_{E} = 1$ and then $|\mathbb{R}^{n} \setminus E| = 0$.

In the case when $E$ is a bounded subset of $\mathbb{R}^{d}$, $|E| < \infty$ so the only possibility is $|E| = 0$.

Answer 2

Of course

For instance for $n=1$ Take $1_{[0,1]}$ its weak derivative is not a function but a measure

Indeed for $\phi\in C_0^\infty(\Bbb R)$ $$(1_{[0,1]}', \phi) = -(1_{[0,1]}, \phi')=-\int_0^1\phi'(x)dx= -\phi(1)+\phi(0)= -(\delta_1-\delta_0)(\phi)$$

Hence, $$\color{red}{1_{[0,1]}'= -\delta_1+\delta_0}$$ is not a function but a distribution

In general for a bounded set E in $\Bbb R^n$ the weak derivative of $1_{E}$ in $\mathcal{D}(\Bbb R^n)$ is not a function but it is a distribution instead.

Remark if you take $\Omega\subset E$ then the weak derivative of $1_{E}$ in $\mathcal{D}(\Omega)$ is the zero function. which means that $1_{E}\in W^{1,1}(\Omega).$

Answer 3

Hint: if $1_E\in W^{1,1}(\mathbb R^n)$, then for all $\phi\in C_c^{\infty}(\mathbb R^n)$, $$\int_E\partial_1\phi=\int_{\mathbb R^n}\partial_1\phi\cdot 1_E=-\int_{\mathbb R^n}\phi\cdot\partial_11_E,$$ therefore $$\left|\int_E\partial_1\phi\right|\leq\int_{\mathbb R^n}|\phi\cdot\partial_11_E|\leq C\|\phi\|_{\infty}.$$ If you can construct $\phi\in C_c^{\infty}(\mathbb R^n)$ that contradicts the last inequality, then you are done.