Let $X_0,X_1$ be Banach spaces. Let $A:D(A)\subseteq X_0\to X_1$ be a closable linear operator. Recall the definition of a core for such an operator:
A set $\mathcal D\subseteq D(A)$ is called a
core for $A$ if $\overline{A_{\mathcal D}}=\overline A$.
In the case of a bounded linear operator one has the result:
Let $A\in L(X_0,X_1)$, and $\mathcal D_0\subseteq X_0$ be a dense linear
subspace. Then $\mathcal D_0$ is a core for $A$.
My Question:
In the case that $A:D(A)\subseteq X_0\to X_1$ is a closed and densely defined linear operator, is there an analogous result which says that a (dense) subset $\mathcal D\subseteq D(A)$ is a core for A?
Best Answer
Let $A: D(A): \to X_1$ be some closed operator between Banach spaces and let $x\in X_0$ be some element not in $D(A)$. Define $$A': D(A)+\Bbb C \cdot x\to X_1\oplus_{\ell^1} \Bbb C, \qquad y+ \lambda x\mapsto (A(y), \lambda).$$
We will check that this is also a closed operator. Suppose $y_n +\lambda_n x$ converges and $(A(y_n), \lambda_n)$ also converges. It follows then that $\lambda_n$ converges (call the limit $\lambda$) and also that $A(y_n)$ converges. From $\lambda_n$ and $y_n+\lambda_n x$ converging you get that $y_n$ converges, call the limit $y$. By closedness of $A$ you get that $y\in D(A)$ and $A(y_n)\to A(y)$. So $$A'(y_n+\lambda_n x) = (A(y_n), \lambda_n) \to (A(y),\lambda) =A'(y+\lambda x)$$ verifying that $A'$ is closed.
If $i: X_1\to X_1\oplus_{\ell^1}\Bbb C$ is the inclusion note that $i\circ A$ is a closed operator. Putting it all together you have that $A'$ is a proper closed extension of the closed operator $i\circ A$. In particular $D(i\circ A) = D(A)\subseteq D(A')$ is densely defined, but $\overline{i\circ A}= i\circ A\neq A'$. As such $D(A)$ is not a core for $A'$.