Let $(\phi_k)_k$ be some laplacian-orthonormal basis in $L^2(D)$ where $D\subset\mathbb{R}^d$ is bounded and with smooth boundary. Is it true that $(\phi_k)_k$ is also an orthogonal basis of $\dot{H}^s(D)$ where $s$ is a real number (possibly negative)?
Note: $\dot{H}^s(D)$ is the homogeneous Sobolev space, defined by
$$||f||_{\dot{H}^s}^2 = \sum_k \lambda_k^s\langle\phi_k,f\rangle^2 $$
where $\lambda_k$ are the eigenvalues of the laplacian.
It is clear that $\{\phi_k\}_k$ is a orthogonal set both in $L^2$ and in $\dot{H}^s$. The question is whether it is complete in the latter space.
The answer is affirmative. We need to prove that $\langle f, \phi_k\rangle_{\dot{H}^s}=0$ for all $k$ implies that $f=0$. Now, $$ 0=\langle f, \phi_k\rangle_{\dot{H}^s}= \lambda_k^s \langle f, \phi_k\rangle_{L^2}, $$ and $\lambda_k\ne 0$ for all $k$ (actually $k=0$ is an exception, as $\lambda_0=0$; however, typically you define $\dot{H}^s$ in such a way as to avoid this difficulty. Here the question is unclear and scarce, so I don't feel like entering into details). Therefore $\langle f, \phi_k\rangle_{L^2}=0$ for all $k$, with this caveat for $k=0$, and since $\phi_k$ is a complete system in $L^2$ we conclude that $f=0$.