The relation between the eigenspaces of $H_0$ and $H_0+H_1$

For the unperturbed Hamiltonian $H_0$, we have a set of eigenfunctions $\{ f_i \}$. For the perturbed Hamiltonian $H_1$, we have another set of eigenfunctions $\{g_i \}$.

Sometimes, it is not clear to me that the $f$'s and the $g$'s span the same Hilbert space.

For example, let $H_0 = \frac{1}{2} p^2 + \frac{1}{2}x^2$, and $H_1 = g \delta (x)$. It is not difficult to solve the eigenstates of $H_0+H_1$, at least numerically. But the question is, is it for sure that these eigenstates can be expanded by the harmonic oscillator eigenfunctions?

First of all, your example is flawed, because the "perturbation" you consider consists of a distribution, i.e. lies outside of the Hilbert space $L^2(\mathbb R)$. In this sense, one could formulate your problem in Hilbert space only (this is the domain of application of the famous Kato-Rellich theorem (*)), in which the full Hamiltonian is defined and is under certain conditions self-adjoint as is the non-perturbed one. In this case, both Hamiltonians have their seld-adjointness domains within the same Hilbert space.
However, since you bring in eigenvectors, then it is automatically natural to work with rigged Hilbert spaces, case in which both $H_0$ and $H_0 + H_1$ have the same space of distributions in which their generalized eigenvectors exist and are well-defined.
Example: the free particle Hamiltonian in 3D $\frac{\bf{p}^2}{2m}$ has the set of eigenvectors in the same $\left(C_0^{\infty}\left(\mathbb R^3\right)\right)'$ space as the Schroedinger H-atom (virtual particle) Hamiltonian $\frac{\bf{p}^2}{2m} + \frac{1}{r}$. You can check for yourself that there is a particular way in which these two sets of distributions are related one to another.
• The $\delta$ function does not define an operator so Kato-Rellich does not apply. – Keith McClary Nov 17 '17 at 0:46