Does $H_0^1(\Omega)$ embed into $H_0^1(R^d)$?

Question

Given a domain $\Omega$ in $\mathbb{R}^d$ and a function $f\in H_0^1(\Omega)$, the closure of the test functions on $\Omega$, does the extension of f by 0 to all of $\mathbb{R}^d$ necessarily lie in $H_0^1(\mathbb{R}^d)$?

Answer 1

Yes. The big picture is that for we can glue two $H^1$-functions together using continuity.

Say let the extension be: $$ \widetilde{f}(x) = \begin{cases} f(x) & \text{ for } x\in \Omega, \\[2mm] 0 &\text{ for } x\in \mathbb{R}^d\backslash\Omega. \end{cases} $$ Apparently $\widetilde{f}\in L^2(\mathbb{R}^d)$. Now we want to find $\nabla \widetilde{f} = \big(\frac{\partial \widetilde{f}}{\partial x_i}\big)$, we don't know if this thing agrees with the weak derivative of $f$ in $\Omega$, and is 0 outside $\Omega$. We have to use the definition of the weak derivative: $$ \int_{\mathbb{R}^d} \frac{\partial \widetilde{f}}{\partial x_i} \phi = -\int_{\mathbb{R}^d} \widetilde{f}\frac{\partial\phi }{\partial x_i} \quad \text{for any }\phi\in C^{\infty}_c(\mathbb{R}^d), $$ now because of the integrability we can split the right hand side into two parts: $$ -\int_{\mathbb{R}^d} \widetilde{f}\frac{\partial\phi }{\partial x_i} = -\int_{\Omega} \widetilde{f}\frac{\partial\phi }{\partial x_i} -\int_{\mathbb{R}^d\backslash\Omega} \widetilde{f}\frac{\partial\phi }{\partial x_i} = -\int_{\Omega} f\frac{\partial\phi }{\partial x_i}. $$ Then use boundary condition of $f$: $$ -\int_{\Omega} f\frac{\partial\phi }{\partial x_i} = \int_{\Omega} \frac{\partial f}{\partial x_i}\phi - \int_{\partial\Omega} f\phi \nu_{x_i}\,dS = \int_{\Omega} \frac{\partial f}{\partial x_i}\phi. $$ Thus we have: $$ \int_{\mathbb{R}^d} \frac{\partial \widetilde{f}}{\partial x_i} \phi = \int_{\Omega} \frac{\partial f}{\partial x_i}\phi \quad \text{for any }\phi\in C^{\infty}_c(\mathbb{R}^d), $$ and this tells us that $$ \nabla \widetilde{f}(x) = \begin{cases} \nabla f(x) & \text{ for } x\in \Omega, \\[2mm] 0 &\text{ for } x\in \mathbb{R}^d\backslash\Omega. \end{cases} $$ Now since $\nabla f\in L^2(\Omega)$, $\nabla \widetilde{f}\in L^2(\mathbb{R}^d)$. The extension is valid.