Orthogonality of a domain and image of an operator in a Hilbert space

Question

Let A be an unbounded operator in a complex Hilbert space. $A: D(A)\subsetneq \mathcal{H} \rightarrow \mathcal{R}(A)$.

Could one have $D(A)^{\perp} = \mathcal{R}(A)$? Or is this not true for any unbounded operator in a complex Hilbert space?

Answer 1

Let $H_0$ be some infinite-dimensional Hilbert space, and let $A_0 : D(A_0) \subset H_0 \to H_0$ be your favorite unbounded surjective operator. You may even take $A_0$ to be closed and densely defined, e.g. take $H_0 = \ell^2$ realized as a space of functions from $\{1,2,3,\dots\}$ to $\mathbb{C}$, $D(A_0) = \{x \in \ell^2 : \sum_n n^2 |x(n)|^2 < \infty\}$ and $(A_0 x)(n) = n x(n)$.

Now let $H = H_0 \oplus H_0$, $D(A) = D(A_0) \oplus 0$, and $A(x,y) = (0, A_0 x)$. Then $A$ is closed and unbounded, and $D(A)^\perp = R(A) = 0 \oplus H_0$.