# 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

$$\int_{B(0,1)} Eu \partial_2 \phi dx = \int_{B_+} \partial_2 \phi dx= \int_{-1}^1 \phi(t,0) d t$$ 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, $$Eu\not\in W^{1,1}(B(0,1))$$.

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.

• The point of the question is how to construct a concrete counterexample as in the diameter-case... The counter $\psi$ you provided clearly does not work: is it not a function in $W^{1,1}(\Omega)$. Indeed, because of the slit $x_1=1/2$, $\psi$ is not weakly differentiable in $\Omega$. Unless you are able to provide one, which is smooth near $x_1=1/2$. Nov 26, 2022 at 18:13
• I do not understand your problem. For the function I described the situation around the slit is locally at $x_1 > 1/2$ exactly the same as for your diameter counterexample. There is not supposed to be a jump of $\psi$ around $0$. You can even make $\psi$ to be zero around $0$, or, say, for $x_1 < 1/4$ if you like. Nov 26, 2022 at 20:04
• The way the function $\psi$ is described, cannot exist at all. Nov 26, 2022 at 23:02
• I'm sorry but I would think it does. Maybe you can describe your problem more precisely. I am saying that there is a bounded smooth function on the ball with a slit $\Omega$ whose value is $1$ in the region of the ball where $x_1 > 1/2$ and $x_2 > 0$, and $0$ in the region of the ball where $x_1 > 1/2$ and $x_2 < 0$. I do NOT claim that the function is smooth on $\overline\Omega$. Locally around the slit at $x_1 > 1/2$, the function looks precisely as the $\mathbb{1}_{B^+}$ in your original post which you seem to have no problems with being in the Sobolev space. Nov 27, 2022 at 11:07
• The first problem is the slit $x_1=1/2$. The second is, no smooth function is 0 on $x_2<0$ and 1 on $x_2>0$ Nov 27, 2022 at 18:32

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 $$\Omega.$$