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


2 Answers 2


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.

  • $\begingroup$ 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 $x_1=1/2$. $\endgroup$
    – Guy Fsone
    Nov 26, 2022 at 18:13
  • $\begingroup$ 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. $\endgroup$
    – Hannes
    Nov 26, 2022 at 20:04
  • $\begingroup$ The way the function $\psi$ is described, cannot exist at all. $\endgroup$
    – Guy Fsone
    Nov 26, 2022 at 23:02
  • $\begingroup$ 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. $\endgroup$
    – Hannes
    Nov 27, 2022 at 11:07
  • $\begingroup$ 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 $x_2>0$ $\endgroup$
    – Guy Fsone
    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 $\Omega.$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.