**Recall that:** A domain $\Omega\subset \mathbb{R}^d$ is an $W^{1,1}$-extension domain if there exists an operator $E:W^{1,1}(\Omega)\to W^{1,1}(\mathbb{R^d})$ and a constant $c= c(d,\Omega)>0$ such that $Eu|_\Omega= u$ and
$\|Eu\|_{W^{1,1}(\mathbb{R}^d)}\leq c\|u\|_{W^{1,1}(\Omega)}$ for all $u\in W^{1,1}(\Omega)$.

Let $\Omega= B(0,1)\setminus\{ (x_1,0): x_1\geq0 \}\subset \mathbb{R}^2$ be the unit ball off a radius.

**Question** how to show that $\Omega$ is not a $W^{1,1}$-extension domain?

Note that in the case where B is off the diameter, it is easier. Namely if $\Omega= B(0,1)\setminus\{ (x_1,0)\}= B(0,1)\cap\{ x_2=0\}\subset \mathbb{R}^2$.

It suffices to consider $u(x)= \mathbb{1}_{B_+}(x)$ with $B_+= B(0,1)\cap\{x_2>0\}$. In this case, one easily checks that $u\in W^{1,1}(\Omega)$. Assume $Eu$ an extension of $u$ to $\mathbb R^2$ exists

For any $\phi\in C_c^\infty(B(0,1)$, i.e., $\phi=0$ on $\partial B$, we have by integration by part that

$$\int_{B(0,1)} Eu \partial_2 \phi dx = \int_{B_+} \partial_2 \phi dx= \int_{-1}^1 \phi(t,0) d t$$ This means that $Eu$ is not weakly differentiable. In other words, any extension $Eu$ of $u$ to $\mathbb{R}^2$, is not weak differentiable on $B(0,1)$ a fortiori, $Eu\not\in W^{1,1}(B(0,1))$.