Simultaneous Orthogonal basis for $L^2(\mathbb{R}^n)$ and $H^1(\mathbb{R}^n)$

Question

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.

Answer 1

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 $L^2$.

Answer 2

There is an abstract background which might help to explain the difference between the case of Sobolew spaces on bounded domains and unbounded ones. If $A$ is an unbounded self-adjoint operator on a Hilbert space, then its domain of definition has a natural structure of a Hilbert space and many (most?) Sobolew spaces arise in this way. For example, one can choose the Laplace operator on a domain (with suitable boundary conditions if it has a boundary). If the spectrum of the operator is discrete, then its eigenvectors will form a simultaneous orthogonal basis for the Hilbert space AND the Sobolew spaces. Simple examples are the Fourier bases for (higher dimensional) tori or Legendre polynomials for the case of an interval (in higher dimensions, a cube).

In your case, as pointed out above, the spectrum of the Laplace is not discrete so this does not apply. I can suggest two ways out.

1. You can replace the Laplacian operator with a Schrödinger type operator which will have discrete spectrum in many important situations. The simplest case leads to the Hermite functions as joint orthogonal basis. Of course, this changes the Sobolew space and only you can be the judge of whether the result is what you desire or require.

2. Even in the case of non discrete spectrum, there is a substitute, namely a spectral measure as in the spectral theorem for unbounded, self-adjoint operators in Hilbert space. Both this and the orthogonal basis of the discrete case (which is a special case of the former, of course) perform essentially the same task: to split the space into a sum of subspaces on each of which the operator is "constant". The former case is less preciae since, in contrast to the discrete one, the subspaces can be infinite dimensional and constancy is only approximate, i.e., up to a suitable error (the precise formulation is contained in the statement of the spectral theorem) but this suffices for many applications.

Answer 3

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.