Given a smooth bounded set $U\subset \mathbb{R}^n$, there is a simultaneous orthogonal basis for $L^2(U)$ and $H^1_0(U)$ by the existence of eigenvectors to the Laplacian in a bounded domain, which particularly requires boundedness for compactness of the solution operator of the corresponding elliptic problem. Is it possible to construct a simultaneous orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$ as well?
I thought it might be possible to use a basis for $L^2(U)$ where U is a cube and then by translations and dilations construct an orthogonal basis for $L^2(\mathbb{R}^n)$. I do not know if that will also be orthogonal basis for $H^1(\mathbb{R}^n)$ or if there will be some edge effects creating trouble.
I wanted to know because I was reading the existence of solutions to wave equations as given in Evans's book on Partial Differential Equations using the Galerkin Method and at one point it requires this simultaneous basis for $L^2(U)$ and $H^1_0(U)$, which is available only for bounded smooth domains $U$ and I wonder if that proof could be extended for existence in $[0,T]\times \mathbb{R}^n$.
Any help would be most welcome.
Possible Solution:
Looking at the answer to this question, I split $\mathbb{R}^n$ into the integer lattice $U_k:=U+k$ for $k\in\mathbb{Z}^n$, and where $U$ is the unit cube. For $L^2(U_k)$, there is an orthonormal basis $\{e_l^k; k\in\mathbb{Z}^n, l\in\mathbb{Z}\}$ which are also eigenvectors of the Laplacian $-\Delta$, and therefore, it also forms an orthogonal basis for $H_0^1(U_k)$. Now, we may extend each $e_n^k$ outside $U_k$ by $0$ so that it belongs to $H^1(\mathbb{R}^n)$. These $\{e_l^k; k\in\mathbb{Z}^n, l\in\mathbb{Z}\}$ form an orthonormal basis for $L^2(\mathbb{R}^n)$ and also an orthogonal basis for $H^1(\mathbb{R}^n)$.
Could someone confirm if this reasoning is correct or am I missing out some issue?