Can a densely defined closed operator have a closed extension? Let $X$ be a separable Banach space with two densely defined, closed unbounded operators $A_1 \subset A_2$ on it:
$$
  A_1: D(A_1) \to X, \quad A_2: D(A_2) \to X \\[8pt]
  D(A_1) \subseteq D(A_2) \subseteq X \\[8pt]
  \overline{D(A_1)} = \overline{D(A_2)} = X \\[8pt]
  \overline{A_1} = A_1, \quad \overline{A_2} = A_2 \\[8pt]
  A_1(x) = A_2(x) \quad \forall x \in D(A_1)
$$
Is it possible that $A_1 \neq A_2$?

I know that closed operators are defined by this condition:
$$
  x_n \to x \;\wedge\; Ax_n \to x \implies x\in D(A) \;\wedge\; Ax=y
$$
It seems to me, that $A_1 = A_2$ it doesn't follow from this condition. On the other hand, I couldn't find a counterexample and in my Quantum Mechanics class, we're always satisfied when we find any closed densely defined extension, so I always assumed other extensions just don't exist.
 A: In fact it is possible that $A_1\neq A_2$.  Here is an example.
Lemma.  Let $H=\ell ^2({\mathbb N})$.  Then there exists a bounded  operator $T:H\to H$, such that

*

*$T$ is injective,  and $T(H)$ is a proper dense subspace of  $H$,


*there exists a proper closed subspace $K\subseteq H$, such that $T(K)$ is also dense.
Proof.  Setting  $T((x_n)_n) = (x_n/n)_n$,  it is clear that $T$ satisfies all of the properties listed in (1).
Notice also that $T$ is self-adjoint.
Now let $y$ be any vector in $H\setminus T(H)$ and put $K=\{y\}^\perp$.  To see that $T(K)$ is dense, suppose by
contradiction that there is a nonzero vector  $z\in T(K)^\perp$.  This means that, for every
$x\in  K$  one has that
$$
  0 = \langle T(x), z\rangle  =   \langle x, T(z)\rangle ,
  $$
so
$$
  T(z) \in  K^\perp = (\{y\}^\perp)^\perp  = {\mathbb C}y.
  $$
Since $z$ is nonzero and $T$ is injective, $T(z)$ is nonzero and we deduce that $y$ is a multiple of $T(z)$,
contradicting the choice of $y$.  QED.

This said, let $A_2$ be the inverse of $T$, defined on $T(H)$, and let $A_1$ be the restriction of $A_2$ to $T(K)$.
Then $A_2$ is a closed operator since the its graph is (up to a flip of coordinates) the same as the
graph of the bounded operator $T$, and the same goes for $A_1$, being the inverse of  the restriction of $T$ to $K$.
