There seems to be a lot confusion on this notion of a domain of an operator $D(A)$ where $A$ is an operator.
Can someone use a simple example to illustrate exactly what this is?
Say, let $A$ be a linear operator s.t. $A(ax+y) = aAx + Ay$, then what is the domain of A?
For example, let $X=L^{2}[0,2\pi]$ and let $A=-\frac{d^{2}}{dx^{2}}$ on the domain $\mathcal{D}(A)$ consisting of all functions $f$ which are continuously differentiable on $[0,2\pi]$ and for which $f'$ is absolutely continuous on $(0,2\pi)$ with $f'' \in L^{2}[0,\pi]$. This is as large a domain as you can reasonable expect for $A$. This operator is not selfadjoint. It's adjoint is the restriction $A_{00}$ of $A$ $$ \mathcal{D}(A_{00}) = \{ f \in \mathcal{D}(A) | f(0)=f'(0)=0,\;f(2\pi)=f'(2\pi)=0 \} $$ Neither $A$ nor $A_{00}$ is selfadjoint, but $A_{00}$ is symmetric because $(A_{00}f,g)=(f,A_{00}g)$ for all $f,g \in \mathcal{D}(A_{00})$.
Consider the operator $A_{0}$ which is the restriction of $A$ to $$ \mathcal{D}(A_{0}) = \{ f \in \mathcal{D}(A) | f(0)=f(2\pi)=0 \} $$ This operator is selfadjoint and has a complete orthogonal basis of eigenvectors $$ \{ \sin(\frac{1}{2}x),\sin(x),\sin(\frac{3}{2}x)\,\sin(2x),\cdots \}. $$
Consider the operator $A_{p}$ which is the restriction of $A$ to periodic functions $$ \mathcal{D}(A_{p}) = \{ f \in \mathcal{D}(A) | f(0)=f(2\pi),\;f'(0)=f'(2\pi) \} $$ This operator is also selfadjoint with a complete orthogonal basis of eigenfunctions $$ \{ 1,\sin(x),\cos(x),\sin(2x),\cos(2x),\cdots \}. $$