Extension of closed operator densely defined

Question

Let $X$ be a Banach space. Suppose $A:D(A)\subset X \longrightarrow X$ is a closed linear operator and $D(A)$ is dense in $X$. Prove that $A$ cannot be extended as a closed linear operator to any subspace $\mathcal{D} \supset D(A)$. The density of $D(A)$ is needed, since the restriction of a closed linear operator to a closed subspace is still a closed linear operator. Any hint is appreciated. Thanks.

Answer 1

A counterexample to your claim on Hilbert spaces would be the closure $\bar{A}$ of any non-self-adjoint but symmetric, densely defined and not bounded operator $A$ admitting a self-adjoint extension (which exists because you can construct symmetric operators with any two deficiency indices if I recall correctly, so in particular you can take $A$ with deficiency indices $(1,1)$). The not bounded part is optional, it's just so that we know the self-adjoint extension can't be bounded either, and thus the domain cannot be the whole space, in order to have a non-trivial example. Said self-adjoint extension is closed and thus must extend the aforementioned closure $\bar{A}$ at least, which exists because all symmetric operators are closable.

For a clear-cut example, the following operator from Example 1 in Chapters X of Reed and Simon's Fourier Analysis, Self-Adjointness (volume 2 of their series of books) has deficiency indices $(1,1)$: $$T : f \in H^1_0(0,1) \subset L^2(0,1) \mapsto i f' \in L^2(0,1)$$ whose adjoint is still $f \mapsto if'$ but with domain $H^1(0,1)$.