# Diagonalizing selfadjoint operator on core domain

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 $$\text{Dom(A)}$$?

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 $$D= L( e_1+e_k/k^2: 1 ($$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$$.
• Maybe a slightly better way of saying the same thing is as follows: Lemma: Let $H$ be a Hilbert space with ONB $\{e_n\}$. Then there is a dense subspace $D\subseteq H$ with $e_n\notin D$ for all $n$. Proof: Define $D$ as above. $\square$ To answer the actual question, apply this to the Hilbert space $(D(A),\|\cdot\|_A)$, with $e_n$ chosen as the eigenvectors of $A$. Jun 5 at 17:07