Let $H$ be a Hilbert space and let $A : D(A) \subset H \to H$ be an essentially self-adjoint operator. Let $\overline A$ be the unique self-adjoint extension of $A$.
Question: Is it true that $\ker(A) = \ker(\overline A)$?
I don't really see any reason for this to be true, so I tried constructing a counterexample... but somehow this is not so easy to do!
For instance, I tried starting with the negative Laplacian $\Delta = - \frac{d^2}{dx^2}$ on $L^2(S^1)$ with domain $\mathrm{dom}(\Delta) = C^\infty(S^1)$. This is an essentially self-adjoint operator and the closure $\overline \Delta$ is easily understood if we look to the frequency domain. On $\ell^2(\mathbb{Z})$, we find $\overline \Delta$ is conjugate to the diagonal operator $D$ on $\ell^2(\mathbb{Z})$ given by $D(a_n) = (n^2a_n)$ and with domain $\{ (a_n) \in \ell^2(\mathbb{Z}) : (na_n) \in \ell^2(\mathbb{Z})\}$. From this, we conclude that $\Delta$ and $\overline \Delta$ have the same one-dimensional kernel. $$\ker(\Delta) = \ker(\overline \Delta) = \{\text{constant functions}\} \subset L^2(S^1).$$ Next I tried to replace $\Delta$ by its restriction $\Delta_S$ to some smaller subspace $S \subset C^\infty(S^1)$ such that $S$ intersects trivially with the constant functions, but still manages to be a core for $\Delta$, i.e. has $\overline \Delta_S = \overline \Delta$. However, every subspace $S$ like this I try turns out to give a non essentially self-adjoint $\Delta_S$. For instance, taking $S = \{f \in C^\infty(S^1) : f(p) = 0\}$ for some point $p$ fails. So does taking $S$ to be the smooth functions vanishing on a whole neighbourhood of some point $p$.
Any thoughts?
Let $B$ be selfadjoint with $\ker B = \operatorname{span}\{x_0\}$. Denote by $\|\cdot\|_B$ the graph norm with respect to $B$ on $D(B)$. Choose a $\|\cdot\|_B$-dense subspace $D\subset D(B)$ such that $D(B) = D + \operatorname{span}\{x_0\}$. Such a subspace may arise as the kernel of a non-$\|\cdot\|_B$-continuous linear functional on $D(B)$ (Note: This is non-constructive. As far as I know, such a functional could not be contructed, so far. However, the existence of such functionals follows from the axiom of choice. See here: https://en.wikipedia.org/wiki/Discontinuous_linear_map#A_nonconstructive_example). Now, define $A := B|D$. Then $A$ is essentially selfadjoint with $\overline A = B$ and $\ker A = \{0\}$.