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.


1 Answer 1


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 $e_j$.

  • 1
    $\begingroup$ 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 $A$. $\endgroup$ Jun 5 at 17:07

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