# Confusion about the closure of $\mathcal{D}(\Omega)$ in $H^s(\mathbb{R}^n)$

On page 77 of his book Strongly Elliptic Systems and Boundary Integral Equations, McLean defines the Sobolev space $$\tilde{H}^s(\Omega)$$ as $$\tilde{H}^s(\Omega)=\text{closure of}\ \mathcal{D}(\Omega)\ \text{in}\ H^s(\mathbb{R}^n),$$ where $$\mathcal{D}(\Omega)=C_{\rm{comp}}^\infty(\Omega)=\{u:u\in C^\infty_K(\Omega)\ \text{for some}\ K\Subset\Omega\},$$ where $$C_K^\infty(\Omega)=\bigcap_{r\ge0}C_K^r(\Omega),$$ where $$C_K^r(\Omega)=\{u\in C^r(\Omega):\rm{supp}\,u\subseteq K\},$$ where $$C^r(\Omega)=\{u:\partial^\alpha u\ \text{exists and is continuous on}\ \Omega\ \text{for}\|\alpha|\le r\}.$$ Also, Theorem 3.30 on page 92 says that if $$\Omega$$ is a Lipschitz domain, then $$H^s(\Omega)^*=\tilde{H}^{-s}(\Omega)$$ where $$H^s(\Omega)=\{u\in\mathcal{D}(\Omega)^*\,|\,u=U|_\Omega\ \text{for some}\ U\in H^s(\mathbb{R}^n)\},$$ where $$H^s(\mathbb{R})=\{u\in\mathcal{S}^*(\mathbb{R}^n)\,|\,\mathcal{J}^su\in L_2(\mathbb{R}^n)\},$$ where $$\mathcal{S}(\mathbb{R}^n)$$ is the Schwarz space of rapidly diminishing functions. Theorem 3.30 claims that $$\tilde{H}^{s}(\Omega)$$ is a set of distributions. Considering the fact that virtually any other Sobolev space introduced in the book is defined as a set of distributions, this makes sense to me. However, what strikes me weird is the fact that $$\tilde{H}^{s}(\Omega)$$ is the closure of $$\mathcal{D}(\Omega)$$, which is a set of infinitely differentiable functions. It don't see why elements of $$H^s(\Omega)^*$$ would need to satisfy this requirement. Sure enough, $$\tilde{H}^s(\Omega)$$ is not $$\mathcal{D}(\Omega)$$, but its closure in $$H^s(\mathbb{R}^n)$$, which leads to me to the question, what exactly is the closure of $$\mathcal{D}(\Omega)$$ in $$H^s(\mathbb{R}^n)$$? What elements does this construct contain?

Lastly, I want to mention, that I am aware that the notation $$\partial u$$ is sometimes used to denote derivatives defined in a weak sense, however, unless I missed it, no such thing was ever mentioned for the elements of $$C^r(\Omega)$$.

Edit: The Bessel potential of order s is given by $$\mathcal{J}^su(x)=\int_{\mathbb{R}^n}(1+|\xi|^2)^{s/2}\hat{u}(\xi)\,e^{i2\pi\xi\cdot x}\,d\xi\quad\text{for}\ x\in\mathbb{R}^n,$$ where the Fourier transfrom $$\hat{u}$$ is given by $$\hat{u}(\xi)=\int_{\mathbb{R}^n}e^{-i2\pi\xi\cdot x}u(x)\,dx\quad\text{for}\ \xi\in\mathbb{R}^n,$$ for $$u\in L_1(\mathbb{R}^n)$$.

• I am not sure I understand what you are asking. Notice that elements of $\mathcal{D}(\Omega)$ can be extended by zero to become a compactly supported smooth function on $\mathbb{R}^n$, and so can be identified as elements of $H^s(\mathbb{R}^n)$. This letter set has a topology (induced by the Hilbert space inner product/norm) and the closure can be taken in this sense. In other words, you are looking at $H^s$ functions that can be obtained as a limit of $\mathcal{D}(\Omega)$ functions, in the $H^s$ topology. – Willie Wong Apr 1 at 0:53
• @WillieWong I think that this is actually the answer that I was looking for. Is it correct that by extended by zero, you mean $u(x)=0$ for $x\in\Omega^c$? Also, is my understanding correct these limits may not even be continuous functions anymore? I have to admit that I have an engineering background and so these things may not be so clear to me as to, say your average physics/math student :) – Felix Crazzolara Apr 2 at 7:29
• Yes, that's what I meant by extended by zero. The limits may not be continuous in general, for $s$ not too large (otherwise you have Morrey's inequality/embedding). But the elements in $\tilde{H}^s$ has to "vanish" in some sense on the boundary of $\Omega$. So for example, for $s > \frac12$, the characteristic function of $\Omega$ is not in $\tilde{H}^s(\Omega)$. – Willie Wong Apr 2 at 12:49
• @WillieWong Thank you for clarifying. Thank you also for mentioning the issue that the elements of $\tilde{H}^s$ may have to vanish on the boundary. This is actually very relevant for me. – Felix Crazzolara Apr 2 at 14:48