Does the inclusion $A \subset B$ for closed densely defined operators $A$ and $B$ imply $A=B$? 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.
 A: If $A : \mathcal{D}(A)\subset\mathcal{H}\rightarrow\mathcal{H}$ is a densely-defined, closed, symmetric linear operator on a complex Hilbert space $\mathcal{H}$, then the adjoint $A^*$ is a densely-defined closed linear operator, and one has the graph inclusion $\mathcal{G}(A)\subseteq\mathcal{G}(A^*)$, and the following orthogonal decomposition in $\mathcal{H}\times\mathcal{H}$:
$$
  \mathcal{G}(A^*)=\mathcal{G}(A)\oplus\mathcal{D}_{-}\oplus\mathcal{D}_{+},
$$
where $\mathcal{D}_{\pm}$ are the restrictions of $A^*$ to $\mathcal{N}(A^*\pm iI)$, respectively. So, if $A$ is symmetric, but not selfadjoint, then $A$ has a proper closed extension $A^*$.
For example, define the operator $Lf=-if'$ on $\mathcal{D}(L)\subset L^2[0,1]$ consisting of functions $f \in L^2[0,1]$ that are equal a.e. to absolutely continuous functions $f$ with $f'\in L^2$ and $f(0)=f(1)=0$. $L : \mathcal{D}(L)\subset L^2\rightarrow L^2$ is a closed, densely-defined, symmetric operator that is not selfadjoint. The domain of $L^*$ is the same as that of $L$, except without the endpoint restrictions. Both $L$ and $L^*$ are closed and densely-defined, and $L^*$ is a proper extension of $L$.
A: The conjecture is not true.
As a counterexample, we choose $G=H=C([-1,1])$, $\mathcal D(A)=C^1([-1,1])$,
and $A$ as the derivative operator $Af = f'$.
$\newcommand\span{\rm span}$
$\newcommand\abs{\,\mathrm{abs}}$
Then it can be shown that $A$ is a closed operator, see here.
Also, $\mathcal D(A)$ is dense in $G$.
Let's construct an extension $B$ of $A$.
We set 
$$\mathcal D(B)= D(A)+\span(\abs)\subset C([-1,1])$$
(with $\abs(x) = |x|$)
which is strictly larger than $\mathcal D(A)$.
Now we can define $B$ via
$$
 B(f+\alpha\abs) = Af = f'.
$$
It remains to show that $B$ is a closed operator.
Let $g_n\in C^1([-1,1]),\alpha_n\in\mathbb R$ be sequences such that
$$
 g_n+\alpha_n\abs\to g\in C([-1,1]),
 \quad
 Ag_n = B(g_n+\alpha_n\abs) \to h\in C([-1,1]).
$$
Note that $g_n(0)\to g(0)$ as $n\to\infty$.
Using
$$
g_n(x) = g(0)+\int_0^x (Ag_n)(s) \mathrm ds
$$
it can be seen that $g_n$ is a Cauchy-Sequence in $C([-1,1])$.
Therefore there exists $g_0\in C([-1,1])$ such that $g_n\to g_0$.
Now (with $Ag_n\to h$) we can use the closedness of $A$ and conclude that
$Ag_0=h$ and $g_0\in C^1([-1,1])$. 
This implies $\alpha_n\to0$ and therefore $g_0=g$.
It follows that $g\in \mathcal D(B)$ and $Bg=h$.
To summarize, we have constructed operators $A,B$ satisfying the requirements from the question and with $A\subset B$, but $A\neq B$.
