Eigenfunctions and spectrum of $T:H \to H^*$ where $H$ is a Hilbert space Let $H$ be a Hilbert space with dual $H^*$. Suppose $T:H \to H^*$ is a linear bounded symmetric operator. (We probably don't want to identify $H$ with $H^*$).
Can we talk about the eigenfunctions/eigenvalues/spectrum of such an operator?
I have in mind $T=-\Delta$ (Laplace-Beltrami on compact manifold) and $H=H^1(M)$. I am a little unclear how the weak Laplacian relates to the eigenfunctions of the Laplacian which we know are smooth functions. How does this work in the abstract case?

I think very likely the question you might wish to be asking includes more structure than the question you literally asked... based on your example of a Laplacian. That is, your Hilbert space $H$ is really a Sobolev space $H^1$ on some compact Riemannian manifold. Then, yes, the Laplacian maps $H^1$ to $H^{-1}$ continuously, and $H^{-1}$ is the Hilbert space dual to $H^1$. (Issues of complex conjugation are not the point here.) But this "full" Laplacian is not what we want for discussing eigenvectors. Rather, although $\Delta:H^1\to H^{-1}$ is continuous, it is not continuous on (for example) smooth functions given the $L^2$ topology. The restriction $T$ of the full $\Delta$ to smooth functions is a symmetric operator, but unbounded (=not continuous) in the $L^2$ topology. It does have (in this example) a unique self-adjoint extension $S$ (e.g., the Friedrichs extension) which maps its dense domain to $L^2$.
Indeed, this self-adjoint extension $S$ is still just a restriction of the full Laplacian.
That is, there are several different topologies in play, and a family $H^1\subset H^0=L^2\subset H^{-1}$ (sometimes called a Gelfand triple), and restrictions and extensions of the full (=distributional) $\Delta$.
The "eigenvectors" for this self-adjoint extension $S$ of the restriction of $\Delta$ lie in the domain inside $H^1$. But, no, $S$ is not defined on the whole $L^2$, and is not continuous in the $L^2$ topology. Still, it is continuous viewed as a restriction of $\Delta:H^1\to H^{-1}$, since the $H^1$ topology is finer, and the $H^{-1}$ is coarser than the $L^2$ topology.
