Extension of unbounded symmetric operator ranges equal implies trivial extension

Question

This is a problem I have encountered in my work and studies in operator theory and functional analysis.

We take a Hilbert space $H$. We take a symmetric (possibly unbounded) operator $C$ which extends $A$, $A \subseteq C$. We are given that $\text{Range}(A+i)=\text{Range}(C+i)$. I need to prove $C=A$, or basically that the domains are equal $D(A)=D(C)$.

Here are the definitions I have used. If $T$ is a densely defined linear operator on a Hilbert space $H$, the domain $D(T^*)$ is the set of $\phi \in H$ for which there is a $\eta \in H$ with $$ \langle T\psi,\phi \rangle = \langle \psi,\eta \rangle $$ for all $\psi \in D(T)$. For each such $\phi \in D(T^*)$ we define $T^* \phi = \eta$, and $T^*$ is called the adjoint of $T$. A densely-defined operator is said to be symmetric if $\langle T\phi,\psi \rangle = \langle \phi,T\psi \rangle$ for all $\phi,\psi \in D(T)$, and in this case $D(T) \subseteq D(T^*)$ and $T=T^*$ on $D(T)$ and $T^*$ is said to extend $T$. A symmetric operator is self-adjoint iff $D(T)=D(T^*)$ and thus $T=T^*$.

So $C$ is densely defined but $A$ may not be. In fact, I have no idea how to do this. I do not know how to use the fact that $C$ is symmetric. I would appreciate any help with this. I thank all helpers.

Answer 1

Let $\gamma \in \mathscr D(C)$ but not $\in \mathscr D(A)$.

Then there is an $\alpha \in \mathscr D(A)$ such that $(A-i) \alpha = (C-i)\gamma$.

Then $(C-i)(\alpha - \gamma) =0$, which implies
$$((\alpha - \gamma),C(\alpha - \gamma))=i((\alpha - \gamma),(\alpha - \gamma))$$ which cannot hold for symmetric $C$.