# Extension of unbounded symmetric operator ranges equal implies trivial extension

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.

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 $$C$$.
• and shouldn't it be $i$ instead of $-i$? Nov 5, 2020 at 2:14
• Added explanation. It can be $i$ or $-i$. Nov 5, 2020 at 5:26