Let $H,G$ be a Hilbert spaces and $A: \mathcal D (A) \subset G \to H \ $ be a densely defined operator with domain $\mathcal D (A) \ $. Assume there is an operator $B: \mathcal D (B) \subset G \to H$ which is an extension of $A$. i.e. $\mathcal D (A) \subset \mathcal D (B) \ $.
Assume both operators $A,B$ are closed. Can we conclude that $A=B$?
So far I tried the following: Let $x \in \mathcal D (B)$. Since $\mathcal D (B)$ is dense there is a sequence $(x_n)$ in both domains such that $x_n \to x$ and since $B$ is closed we have $\lim_n Bx_n=Bx$. Since every $x_n$ is also in the domain of $A$ we conclude that $(Ax_n)$ converges. Hence $x$ is in the domain of the closure of $A$ but this the same as the domain of $A$ since $A$ is closed.