Given an operator $T:D_1(T)\subset L^2 \rightarrow L^2$ and the same operator $T:D_2(T) \subset L^2 \rightarrow L^2$, such that the operator is both times self-adjoint and closed, with $D_1(T) \subset D_2(T)$ both being dense in $L^2$. Does this mean that for any $A \subset L^2$, the operator $T:A \rightarrow L^2$ is self-adjoint and closed, too?-if we have that $D_1(T) \subset A \subset D_2(T)$.
Let's denote the operator defined on $D_i(T)$ by $T_i$. Then since $T_1 \subset T_2$ we have $T_2^\ast \subset T_1^\ast$.
Now use that $T_i$ is self-adjoint to conclude that there is not much choice for $A$.
• so $D_1(T)=D_2(T)$? – user66906 Jun 30 '14 at 17:25