TITLE: Simultaneous Orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: I your aim is to apply the Galerkin method, you do not need simultaneous orthonormal basis.
An inspection of Evans’ proof shows that you need a sequence of linear maps $(P_n)_{n \in \mathbb{N}}$ such that 

for each $m \in \mathbb{N}$, the range of $P_n$ is finite-dimensional,
for each $u \in L^2 (U)$,
$$
  \lim_{m \to \infty} \Vert P_m (u) - u \Vert_{L^2 (U)} = 0
$$
there exists $C > 0$ such that for each $m \in \mathbb{N}$ and $u \in L^2 (U)$,
$$
  \Vert P_m (u) \Vert_{L^2 (U)} \le  C \Vert u \Vert_{L^2 (U)},
$$ 

for each $u \in H^1_0 (U)$,
$$
  \lim_{m \to \infty} \Vert P_m (u) - u \Vert_{H^1_0 (U)} = 0
$$

there exists $C > 0$ such that for each $m \in \mathbb{N}$ and $u \in H^1_0 (U)$,
$$
  \Vert P_m (u) \Vert_{H^1_0 (U)} \le  C \Vert u \Vert_{H^1_0 (U)}.
$$

(These assumptions are in fact redundant in view of the uniform boundedness principle and the density of $H^1_0 (U)$ in $L^2 (U)$.)
To construct such a family, consider a function $\eta \in C^1 (\mathbb{R}^n;[0, 1])$ such that $\eta = 0$ in $\mathbb{R}^n \setminus B_1 (0)$ and $\eta = 1$ in $B_{1/2} (0)$, set $\eta_\ell (x) = \eta (x/\ell)$ and define the linear map
$$
  Q_\ell (u) = \eta_\ell u.
$$
Observe that 
 - for each $\ell \in \mathbb{N}$, $Q_\ell (L^2 (\mathbb{R}^n)) \subset L^2 (B_{\ell})$ and $Q_\ell (H^1_0 (\mathbb{R}^n)) \subset H^1_0 (B_{\ell})$
 - for each $u \in L^2 (U)$,
$$
  \lim_{\ell \to \infty} \Vert Q_\ell (u) - u \Vert_{L^2 (U)} = 0
$$
 - for each $\ell \in \mathbb{N}$ and $u \in L^2 (U)$,
$$
  \Vert Q_\ell (u) \Vert_{L^2 (U)} \le \Vert u \Vert_{L^2 (U)},
$$ 
- for each $u \in H^1_0 (U)$,
$$
  \lim_{\ell \to \infty} \Vert Q_\ell (u) - u \Vert_{H^1_0 (U)} = 0
$$
 - for each $\ell \in \mathbb{N}$ and $u \in H^1_0 (U)$,
$$
  \Vert Q_\ell (u) \Vert_{H^1_0 (U)} \le  (1 + \Vert \nabla \eta \Vert_{L^\infty}) \Vert u \Vert_{H^1_0 (U)}.
$$
In a second step, consider a basis $(w_{k, \ell})_{k \in \mathbb{N}}$ of eigenvectors of $-\Delta$ in $B_{\ell}$ with Dirichlet boundary conditions, define
$$
  Q_{k, \ell} (u)= \sum_{j = 1}^k (w_{k, j}, Q_\ell (u)) w_{k, j},
$$
and observe that
- for each $k, \ell \in \mathbb{N}$, $Q_\ell$ has finite range,
 - for each $u \in L^2 (U)$,
$$
  \lim_{\ell \to \infty} \lim_{k \to \infty} \Vert Q_{k, \ell} (u) - u \Vert_{L^2 (U)} = 0
$$
 - for each $k, \ell \in \mathbb{N}$ and $u \in L^2 (U)$,
$$
  \Vert Q_{k,\ell} (u) \Vert_{L^2 (U)} \le  \Vert u \Vert_{L^2 (U)},
$$ 
- for each $u \in H^1_0 (U)$,
$$
  \lim_{\ell \to \infty} \lim_{k \to \infty} \Vert Q_\ell (u) - u \Vert_{H^1_0 (U)} = 0
$$
 - for each $k, \ell \in \mathbb{N}$ and $u \in H^1_0 (U)$,
$$
  \Vert Q_{k, \ell} (u) \Vert_{H^1_0 (U)} \le  (1 + \Vert \nabla \eta \Vert_{L^\infty}) \Vert u \Vert_{H^1_0 (U)}.
$$
Since $H^1_0 (\mathbb{R}^n)$ and $L^2 (\mathbb{R}^n)$ are separable, conclude with a diagonal argument.