# Existence of an extension operator $E: W_0^{s,p}(\Omega) \rightarrow W^{s,p}(\mathbb{R}^d)$

If $$\Omega$$ is an open Lipschitz domain with bounded boundary then their exists a continuous operator $$E: W^{s,p}(\Omega) \rightarrow W^{s,p}(\mathbb{R}^d)$$ where $$s\in (0,1)$$ such that $$Eu=u$$ on $$\Omega$$.

A similar result holds for integer order Sobolev spaces.

But in case of integer order sobolev spaces one always has the existence of a continous extension operator regardless of the regularity of $$\Omega$$ $$E: W_0^{k,p}(\Omega) \rightarrow W^{k,p}(\mathbb{R}^d)$$ i.e. the zero extension.

My question is if $$\Omega$$ is only open do we also have the existence of a continuous extension operator

$$E: W_0^{s,p}(\Omega) \rightarrow W^{s,p}(\mathbb{R}^d) \text{ where } s\in(0,1)$$

This is a very nice question. It appears to me that the answer is in general, no. Moreover, it appears to be an open problem determining geometric conditions on $$\Omega$$ for which such an extension holds. Essentially, this amounts to establishing fractional boundary Hardy inequality. Indeed, take $$u\in C^\infty_0(\Omega)$$ and let $$Eu(x) = \begin{cases} u(x), &\text{if } x\in \Omega,\\ 0, &\text{else} \end{cases}$$ i.e. extend by zero. Let $$0 and $$1\leq p <+\infty$$. Then $$\| Eu\|_{L^p(\mathbb R^n)}=\| u\|_{L^p(\Omega)}$$ and \begin{align*} [ Eu ]_{W^{s,p}(\mathbb R^n)}^p &= \int_{\mathbb R^n}\int_{\mathbb R^n} \frac{\vert Eu(x)-Eu(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx \\ &=\int_{\Omega}\int_{\Omega} \frac{\vert u(x)-u(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx + 2\int_{\mathbb R^n\setminus \Omega}\int_{\Omega} \frac{\vert u(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx\\ &= [ u ]_{W^{s,p}(\Omega)}^p+ 2\int_{\mathbb R^n\setminus \Omega}\int_{\Omega} \frac{\vert u(y)\vert^p}{\vert x- y \vert^{n+sp}}\, dy \, dx\\ &=[ u ]_{W^{s,p}(\Omega)}^p+ 2\int_{\Omega} \vert u(y)\vert^p w(y)\, dy \end{align*} where $$w(y) = \int_{\mathbb R^n\setminus \Omega}\frac{ dx}{\vert x- y \vert^{n+sp}}.$$

To prove $$\|Eu\|_{W^{s,p}(\mathbb R^n)} \leq C\|u\|_{W^{s,p}(\Omega)}$$ with $$C>0$$ depending only on $$n$$, $$s$$, $$p$$ (what we need to show to conclude $$E$$ is bounded), we must show that $$\int_{\Omega} \vert u(y)\vert^p w(y)\, dy \leq C \|u\|_{W^{s,p}(\Omega)}$$ for all $$u\in C^\infty_0(\Omega)$$. For $$\Omega$$ open, we have that $$w(y) \leq C \operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-sp}$$ (I have provided a proof of this fact below), so if one can show that $$\int_{\Omega} \frac{\vert u(y)\vert^p}{\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{sp}}\, dy \leq C \|u\|_{W^{s,p}(\Omega)} \tag{\ast}$$ for all $$u\in C^\infty_0(\Omega)$$ then we are done. This is the fractional boundary Hardy inequality which, as I mentioned before, is an open inequality in general.

Moreover, if $$\Omega=B_1$$ then I expect that we also have that $$w(y) \geq C \operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-sp}$$, so proving $$E$$ is bounded becomes equivalent to proving ($$\ast$$). However, it was proven by Dyda in A fractional order Hardy inequality that for a bounded Lipschitz domain (which $$B_1$$ is) the fractional boundary Hardy inequality holds if and only if $$sp>1$$. To read more, you could look at Fractional Hardy inequalities and visibility of the boundary by Ihnatsyeva, Lehrbäck, Tuominen, and Vähäkangas.

By the Layer-cake theorem, \begin{align*} w(y)&=(n+sp)\int_0^{+\infty} t^{n+sp-1}\vert \{x\in \mathbb R^n\setminus \Omega \text{ s.t. } \vert x-y\vert^{-1} >t\} \vert \, dt \\ &= (n+sp)\int_0^{+\infty}t^{n+sp-1} \vert ( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) \vert \, dt . \end{align*} When $$t>\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-1}$$ then $$( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) =\varnothing$$, so actually \begin{align*} w(y)&= (n+sp)\int_0^{\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-1}} t^{n+sp-1}\vert ( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) \vert \, dt . \end{align*} Hence, $$\vert ( \mathbb R^n\setminus \Omega ) \cap B_{t^{-1}}(y) \vert \leq Ct^{-n}$$, so $$w(y)\leq C\int_0^{\operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-1}} t^{sp-1} \, dt = C \operatorname{dist}(y,\mathbb R^n \setminus \Omega)^{-sp}.$$