Let $A$ be a densely-defined, positive, self-adjoint operator with compact resolvent on a Hilbert space $H$. Then, $\text{Range}(1+A)=H$ and there is a basis for $H$ consisting of eigenvectors of $1+A$. Assume also that $D\subset H$ is a core domain for $A$; that is, $D$ is dense in $\text{Dom(A)}$ with respect to the graph norm.


Is it true that we can pick an eigenvector basis for $H$ to consist of elements in $D$ instead of $\text{Dom(A)}$?


Thank you in advance.



Diagonalizing selfadjoint operator on core domain
This sounds suspicious right away since the eigenvectors are what they are (nothing to choose here, unless you have degeneracies), but there is much choice for $D$ and we should be able to avoid eigenvectors.


For a concrete example, you can take $H=\ell^2$, $Ae\_n=ne\_n$, which is self-adjoint on its natural domain $D(A)=\{ x: \sum n^2 |x\_n|^2<\infty \}$. Then consider
$$
D= L( e\_1+e\_k/k^2: 1<k)
$$
($L$ = linear span).


This is a core of $A$: Clearly $e\_1+e\_k/k^2$ is close to $e\_1$ in graph norm for large $k$, and then $e\_1+e\_2/4-e\_1-e\_k/k^2$ is close to $e\_2/4$ etc. So all $e\_j$ are in the domain of the operator closure.


On the other hand, $D$ doesn't contain any of the eigenvectors $e\_j$.


