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 or does there exist 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 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. 


As an attempt, 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 $L^2(\mathbb{R}^n)$ but not an orthogonal basis for $H^1(\mathbb{R}^n)$.


If I had instead a simultaneous orthogonal basis for $L^2(U)$ and $H^1(U)$, it would not be possible to extend outside $U$ by zero and other extensions would not preserve orthogonality.


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 $[0,T]\times \mathbb{R}^n$.


I have asked this on math.stackexchange [earlier](https://math.stackexchange.com/questions/683839/simultaneous-orthogonal-basis-for-l2-mathbbrn-and-h1-mathbbrn).



Simultaneous Orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$
If $u\_j$ is such a basis, from $(\nabla u\_j,\nabla u\_k)=0$ for all $j\not=k$ it follows $(\Delta u\_j,u\_k)=0$ which means $\Delta u\_j$ is a multiple of $u\_j$, and unfortunately $\Delta$ has no eigenvalues in $L^2$.


