Assume $A$ is a densely-defined, positive, self-adjoint operator with compact resolvent on a Hilbert space $H$.
Also, assume that $D\subset H$ is a core domain for $A$; that is, $D$ is also dense in $\text{Dom(A)}$ with respect to the graph norm. Then, $\text{Range}(1+A)=H$.
Is it true that we can pick an eigenvector basis for $H$ to consist from elements in $D$? In that case, for example, we would have that if in addition $D\subset \text{Dom}(A^2)$, then $D$ would also be a core domain for $A^2$.
Thank you!
No, this is not true. Even more, for every operator satisfying your assumptions you find a core such that this core does not contain any eigenabsis.
To prove this, let $(\lambda_ n)_{n \in \mathbb{N}}$ be the eigenvalues of $A$ and $(e_n)_{n \in \mathbb{N}}$ be any eigenbasis. Note first that, by definition, $e_n \in \rm{ker} (A-\lambda_n)$, $n \in \mathbb{N}$. Thus it suffices to construct a core $\mathcal{D}$ of $A$ such that $$\rm{dim}(\mathcal{D}\cap\rm{ker} (A-\lambda_n))<\rm{dim}\,\rm{ker}(A-\lambda_n)$$ (the latter being finite since $A$ has compact resolvent) for some $n \in \mathbb{N}$. To this end, let me abbreviate $G:=(\mathcal{D}(A), \|\cdot \|_{gr})$ where $\|\cdot\|_{gr}$ denotes the graph norm of $A$, i.e. $G$ is a Banach space and $A$ is a bounded operator when viewed as a map from $G$ to $H$. Define the linear functional $f$ on $G$ through $$f(x):= \sum_{k=1}^\infty\lambda_k^2(x, e_{k}), \quad x \in \mathcal{D}(f):=\rm{span}\{e_n\mid n \in \mathbb{N}\}\subset G.$$ Note that $\mathcal{D}(f)$ is a core of $A$ and thus a dense subspace of $G$. Moreover, observe that $$\frac{|f(e_n)|}{\|e_n\|_{gr}}=\frac{\lambda_n^2}{1+\lambda_n}\to \infty, \quad n \to \infty.$$ Hence $f$ is an unbounded densely defined linear functional on the infinite-dimensional Banach space $G$. This necesitates that its kernel is dense in $G$. Now we set $\mathcal{D}:=\rm{ker}(f)$. Since $\mathcal{D}$ is dense in $G$, it is a core of $A$. Moreover, note that $f(e_n)=\lambda_n^2$ and thus $e_n \notin \mathcal{D}$, whenever $\lambda_n\neq0$ (which is the case for all but finitely many $\lambda_n$ as $A$ has compact resolvent). Hence, for those $n \in \mathbb{N}$, the inclusion
$$\mathcal{D} \cap \ker(A-\lambda_n) \subset \ker(A-\lambda_n)$$ must be strict and so the claim follows.
However, if you had a concrete exmaple in your mind when asking your question. There are a lot of 'canonical' cores which do satisfy your claim. For instance consider $$\mathcal{D}:=\bigcap_{n \in \mathbb{N}} \mathcal{D}(A^n);$$ the subspace of all $\mathcal{C}^\infty$-vectors of $A$. This is indeed a core of $A$ and is contains any eigenbasis of $A$. Moreover, in concrete examples this space is usually quite 'small', cf. e.g. the Dirichlet-Laplacian on compact interval. Many other cores one might think of are supersets of the core above.