A restriction of a symmetric operator such that the range of (operator)+i is the same

Question

I have this problem and I really can't see how to do it.

Suppose that $C$ is a symmetric operator, $A\subset C$ and that $\operatorname{Ran}(C+i)=\operatorname{Ran}(A+i)$. Prove that $C=A$.

Where $A\subset C$ means that $\operatorname{Dom}(A)\subset \operatorname{Dom}(C)$ and $Ax=Cx$, for all $x\in \operatorname{Dom}(A)$.

I tried this

For any $y \in R(C + i)$ we have $x\in \operatorname{Dom}(C)$ and $z \in \operatorname{Dom}(A)$ such that $$(C + i)z = (A + i)x = y$$ Since $A ⊂ C$,we have also $(C + i)x = y$.

If I could prove that $C + i$ is injective, it must hold that $z = x \in \operatorname{Dom}(A)$, this implies that $\operatorname{Dom}(C) ⊆ \operatorname{Dom}(A)$ and therefore $A = C$.

Answer 1

Lemma: The operator $C+i$ is injective: If $(C+i)x=0$, then $$i\|x\|^2=\langle ix,x\rangle=\langle -Cx,x\rangle=-\langle x,Cx\rangle=-\langle x,-ix\rangle=-i\|x\|^2,$$ hence $x=0$.

(More generally, all eigenvalues of a symmetric operator are real by the same argument.)

Solution of the original problem: Let $x\in\mathrm{Dom}(C)$ be arbitrary. Since $\mathrm{Ran}(C+i)=\mathrm{Ran}(A+i)$, there is a $z\in \mathrm{Dom}(A)$ such that $(A+i)z=(C+i)x$. By assumption, $(A+i)z=(C+i)z$, and the above lemma yields $x=z\in\mathrm{Dom}(A)$.