Do eigenfunctions of elliptic operator form basis of $H^k(M)$?

Question

We know that the eigenfunctions of the Laplacian on a compact manifold $M$ form a countable basis of $H^1(M)$. If $L$ is a $2k$-order elliptic operator, do the eigenfunctions of $L$ form a basis for $H^k(M)$? References/more detail would be appreciated. Thanks.

Answer 1

Yes, if your operator is self-adjoint and coercive in the sense that $$ \langle Lu,u\rangle \geq \alpha \|u\|_{H^k}^2 - C \|u\|_{L^2}^2, $$ where $\alpha>0$. This includes in particular strongly elliptic operators.

More generally, let $V$ and $H$ be Hilbert spaces, with the compact and dense embedding $V\hookrightarrow H$. We embed $H$ into $V'$ through the identification $H\eqsim H'$. Let $L:V\to V'$ be a bounded linear operator satisfying $$ \langle Lu,v\rangle = \langle u,Lv\rangle, \qquad\textrm{for}\quad u,v\in V, $$ and $$ \langle Lu,u\rangle \geq \alpha \|u\|_{V}^2 - C \|u\|_{H}^2, \qquad\textrm{for}\quad u\in V, $$ with $\alpha>0$. Then

• By the Riesz representation theorem, $L+C\,\mathrm{id}:V\to V'$ is invertible.
• The inverse $R=(L+C\,\mathrm{id})^{-1}|_{H}$, restricted to $H$, as an operator $R:H\to H$, is compact.
• There is an orthonormal basis $\{u_n\}$ of $H$, consisting of the eigenvectors of $L$. This is an application of the spectral theory of compact operators.
• The eigenvalues are real and of finite multiplicity, and satisfy $\lambda_n> -C$ and $\lambda_n\to\infty$ (follows from the Hilbert-Schmidt theory). This and the preceding item can also be derived directly working with $L$ by variational methods.
• The functions $\{u_n\}$ form also a basis of $V$, which is orthogonal with respect to the inner product $\langle L\cdot,\cdot\rangle + C\langle\cdot,\cdot\rangle_{H}$.

Let me elaborate on the last point a bit since this seems to be the heart of the question. The orthogonality follows from $$ \langle Lu_n,u_m\rangle + C\langle u_n,u_m\rangle_{H} = \langle \lambda_nu_n,u_m\rangle + C\delta_{nm} = (\lambda+C)\delta_{nm}. $$ Completeness of $\{u_n\}$ in $V$ follows similarly from the completeness of $\{u_n\}$ in $H$. Letting $v\in V$, we have $$ \langle Lu_n,v\rangle + C\langle u_n,v\rangle_{H} = \langle \lambda_nu_n,v\rangle + C\langle u_n,v\rangle_{H} = (\lambda_n+C) \langle u_n,v\rangle_{H}. $$ So if $v$ is orthogonal to each $u_n$ with respect to the new inner product in $V$ then $v=0$.