Is $C_c^{\infty}$ dense in $X_0^{\alpha}$? While reading papers on fractional Laplacian, I always meet space $X_0^{\alpha}(\mathcal{C}_{\Omega})$ which is defined as following:
$$X_0^{\alpha}(\mathcal{C}_{\Omega})=\{z\in L^2(\mathcal{C}_{\Omega}): z=0 \text{ on }\partial_L\mathcal{C}_{\Omega}, \int_{\mathcal{C}_{\Omega}}y^{1-\alpha}|\nabla z(x,y)|^2\,dxdy<\infty\},$$
Where $\Omega$ is bounded domain in $\mathbb{R}^n$, $\mathcal{C}_{\Omega}=\{(x,y):x\in \Omega,y\in \mathbb{R}_+\}\subset \mathbb{R}^{n+1}_+$, and $\partial_L\mathcal{C}_{\Omega}$ is  the lateral boundary of $\mathcal{C}_{\Omega}$. And we equip $X_0^{\alpha}$ with norm 
$$\Vert z\Vert_{X_0^{\alpha}}^2=\int_{\mathcal{C}_{\Omega}}y^{1-\alpha}|\nabla z(x,y)|^2\,dxdy.$$
Then my question is: Is $C_c^{\infty}(\overline{\mathbb{R}^{n+1}_+})$ dense in $X_0^{\alpha}(\mathcal{C}_{\Omega})$, and how to prove it? I don't even know whether we have $\Vert \eta_{\varepsilon}*u-u\Vert_{X_0^{\alpha}}\to 0$, where $\eta_{\varepsilon}$ is the standard mollifier.
 A: Well, technically it's those functions $z \in C^\infty(\mathbb{R}^{n+1}_+)$ whose support are contained inside the region bounded by $\partial_L C_\Omega$, to use your notation. 
In that case, the answer is yes, and you do the mollification in the usual way (the following is a straightforward recounting of the mollification technique that you can find, for example, in the first chapter of Adams' book on Sobolev Spaces, slightly adapted for our case). 
Let $u \in X_0^\alpha(C_\Omega)$, and $\varphi$ be your favorite bump function supported on a ball of radius 1 in $\mathbb{R}^{n+1}$, say, and make your space the whole cylinder 
$$C_\Omega' = \{ (x,y): x \in \Omega, y \in \mathbb{R} \}$$
and extend $u$ by even reflection (for $y < 0$, let $u(x,y) = u(x,-y)$). Let $d(X) = d(x,\partial \Omega)$ where $X=(x,y)$. Define 
$$u(X,\epsilon) = \begin{cases}u(X) & \text{if} \,\,d>2\epsilon \\ 
                               0 &\text{otherwise}\end{cases}
$$
and then subsequently
$$u_\epsilon(X) = (u(X,\epsilon) *\frac{1}{\epsilon^n}\varphi(\frac{X}{\epsilon}))$$
From here, it's fairly simple to see that $u_\epsilon \in C^\infty_c(C_\Omega')$, and it remains only to show that
$$\int_{C_\Omega'} y^{1-\alpha} |\nabla u - \nabla u_\epsilon|^2 dx dy \rightarrow 0$$
as $\epsilon \rightarrow 0$. 
Notice that $y^{1-\alpha} dx dy$ is a nice bounded positive measure for $0 < \alpha < 2$ (which is always the case when you consider the extension for the fractional Laplacian), hence continuous functions of compact support are dense in $L^2(y^{1-\alpha} dx dy)$. Choose $n+1$ such nice functions $G=(g_1,\ldots,g_{n+1})$, such that $$\int_{C_\Omega'} |\nabla u - G|^2 y^{1-\alpha} dx dy  < \frac{\eta}{6}$$
Define $G_\epsilon(X)$ the same way we defined $u_\epsilon$. Now
$$|\nabla u - \nabla u_\epsilon| \leq |\nabla u - G| + |G - G_\epsilon| + |G_\epsilon - \nabla u_\epsilon|$$
It is clear (by simply using Fubini and doing the convolution first) that 
$$\int y^{1-\alpha} |G_\epsilon - \nabla u_\epsilon|^2 dx dy < \frac{\eta}{6}$$
so we are left only to consider the term in $|G_\epsilon - G|$. However, $G$ is uniformly continuous, hence for $\epsilon$ sufficiently small we have
$|G_\epsilon - G|^2 < \frac{\eta}{C}$, except possibly when $d(X) < 3\epsilon$. Since the support of $G$ is bounded, pick $C$ so that
$$\int y^{1-\alpha} |G_\epsilon - G|^2 dx dy < \frac{\eta}{6}$$
