A ball with slit at the radius is not $W^{1,1}$-extension domain 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
$$\int_{B(0,1)} Eu \partial_2 \phi dx = \int_{B_+} \partial_2 \phi dx= \int_{-1}^1   \phi(x,0) d x$$
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, $Eu\not\in W^{1,1}(B(0,1))$.
 A: If you are okay rotating your domain, so that the slit is the negative semiaxis instead of the positive, then the classical counterexample if the function written in polar coordinates $f(r,\theta)=r\theta$ where $0<r<1 $ and $-\pi<\theta<\pi$. One can show that this function is in $W^{1,\infty}$ but not Lipschitz continuous. If the domain where a $W^{1,1}$ extension domain, then you could extend $f$ to a  function $g$ in $W^{1,1}(R^2)$. There are several way to prove that this gives a contradiction. If you have seen traces, you could use trace theory to show that the restriction of $g$ to the upper half space has the same trace than the restriction of $g$ to the lower half space, but this is a contradiction since on the negative slit, the traces differ.
You could also use the theorem that say that a Sobolev function has a representative that is absolutely continuous on a.e. line parallel to the axes. The representative would have to be exactly $f$ on the unit ball, but again this is a contradiction since it is discontinuous across the negative slit.
You could also use the fact that in the unit ball the weak derivatives of $g$ are the standard derivatives of $f$ and so they are bounded. Thus the function is in $W^{1,\infty}(B)$ where $B$ is the unit ball. In particular it is in $W^{1,p}(B)$ for $2<p<\infty$. By Morrey’s theorem $f$ would have a Holder continuous representative, which is a contradiction again.
A: Define \begin{align*}
 u(x,y)= f(x)g(y)
\end{align*}
where $f\in C_c^\infty(0,1)$ such that $0\leq f\leq 1$
\begin{align*}
f(x)=\begin{cases}
1 & |x-\frac12|<\frac14\\
f(x)&\frac14\leq |x-\frac12|\leq\frac38\\
0 & |x-\frac12|>\frac38.
 \end{cases}
\end{align*}
\begin{align*}
g(y)=\begin{cases}
1 & y>0\\
0 &y\leq 0.
\end{cases}
\end{align*}
This is the same scenario as in OP. One can show that $u\in W^{1,1}(\Omega)$ with the weak derivatives in $\Omega$ given by  $\partial_y u=0$ and $\partial_xu = f'(x)g(y)$ a.e.
But $u$ cannot be extended as a function in $W^{1,1}(\Bbb R^2)$.
Indeed we have $u\not\in W^{1,1}(B(0,1))$ since $g'=\delta_0$ and hence
\begin{align*}
 \langle \partial_y u, \phi\rangle= \int_0^1\phi(x,0) dx=\langle f\otimes\delta_0, \phi\rangle. 
\end{align*}
This means $u$ is not even weakly differentiable on $B(0,1)$.
Another typical example is to consider the function with jumps, lying only the slit $\{ (x,0): x\geq0\}$, by passing to polar coordinates $(x,y)\equiv (r,\theta)$, as follows
\begin{align*}
u(r, \theta) 
&=\begin{cases}
1, & \theta\in (0,\frac{\pi}{2})\\
\frac12(1+\sin \theta), & \theta\in (\frac{\pi}{2}, \frac{3\pi}{2})\\
0 &\theta\in (\frac{3\pi}{2}, 2\pi).
\end{cases}
\end{align*}
which in cartesian coordiantes
\begin{align*}
\implies u(x,y) &=\begin{cases}
1 & x\geq0, y\geq0,\\
\frac12(1+\frac{y}{\sqrt{x^2+y^2} }) & x<0,\\
0 &x\geq0,y\leq 0.
\end{cases}
\end{align*}
This function has no jumps except for $\theta=0$.
By the same procedure, one can show that $u$ is not weakly differentiable on $B(0,1)$, but does on $\Omega.$
