A ball with slit at the radius is not $W^{1,1}$-extension domain

Question

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, $Eu\not\in W^{1,1}(B(0,1))$.

Answer 1

If $\Omega$ was a $W^{1,1}$-extension domain, then restrictions of $C_c^\infty(\mathbb{R}^n)$ functions to $\Omega$ would be dense in $W^{1,1}(\Omega)$ since they are dense in $W^{1,1}(\mathbb{R}^n)$. But this cannot be since restrictions of smooth functions to $\Omega$ must be smooth across the slit, too, so they can never approximate (even smooth) functions on $\Omega$ which have, for example, a jump across the slit. This argument also applies to the 'sliced' domain as an alternative to your calculation.

Indeed, pick a smooth bounded function $\psi$ on $\Omega$ which is, say, $0$ for $x_1 > 1/2$ and $x_2 > 0$, and $1$ for $x_1 > 1/2$ and $x_2 < 0$. Suppose we could approximate $\psi$ in $W^{1,1}(\Omega)$ by restrictions $\psi_n$ of $C_c^\infty(\mathbb{R}^n)$ functions to $\Omega$. Then $\psi_n$ converges to some $\Psi$ in $W^{1,1}(B(0,1))$, too, since the slit is a nullset, and $\Psi = \psi$ almost everywhere. Hence $\nabla \Psi = 0$ on the connected set $B(0,1) \cap \{ x_1 > 1/2 \}$, so $\Psi$ must be constant. But then it cannot coincide with $\psi$ almost everywhere.

Answer 2

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>0, y\geq0,\\ \frac12(1+\frac{y}{\sqrt{x^2+y^2} }) & x<0,\\ 0 &x>0,y\leq 0. \end{cases} \end{align*}

This function has no jumps except on the segment $\theta=0$.

By the same procedure, one can show that $u$ is not weakly differentiable on $B(0,1)$, but does on $\Omega.$