Let $H=(H,(\cdot,\cdot)_H)$ be Hilbert space and $A:D(A) \subset H \longrightarrow H$ be a self-adjoint operator, not necessarily bounded.
Question. Is true that $\text{Ker}(A)^{\perp}=\text{Range}(A)$?
Note that, given any $v \in \text{Range}(A)$ there exist $u \in D(A)$ such that $A(u)=v$. Hence, if $w \in \text{Ker}(A)$ is arbitrary, then since $A$ is self-adjoint we obtain
$$
0=(A(w),u)_H=(w,A(u))_H=(w,v)_H
$$
that is, $\text{Ker}(A)^{\perp}$. Thus, $ \text{Range}(A) \subset \text{Ker}(A)^{\perp}$. But, the reverse inclusion holds? If so, would this be in a sense Fredholm Alternative for self-adjoint operators? Or is there a Fredholm Alternative for self-adjoint operators? I only know the version for compact operators.
Best Answer
If $T$ is densely defined with domain $\mathcal{D}(T)$ then $T^*$ is closed and $$y \perp \text{Range}(T)\iff \forall x\in \mathcal{D}(T):\langle Tx,y\rangle=0 \iff y\in\mathcal{D}(T^*) \land T^*(y) = 0$$ This proves $$\text{Range}(T)^\perp = \text{Ker}(T^*)$$ and proves, inter alia that $\text{Ker}(T^*)$ is a closed subspace because the orthogonal complement to any set is a closed subspace.
If $T$ is densely defined and closable then $T^*$ is densely defined (and as before closed). Additionally, $T^{**} = \overline{T}$ the closure of $T$ obtained by taking the closure of the graph of $T$. From this we have $$\text{Range}(T^*)^\perp = \text{Ker}(T^{**})=\text{Ker}(\overline{T})$$