Is there a densely defined closed operator $A$ such that $codim(D(A))=1$?

Question

Let $X$ be a Banach space. Is there a closed operator $ A:D(A)\subset X \rightarrow X $ such that $ D(A)\neq \overline{D(A)}=X $ and $ codim (D(A))=1$ ?

Thank you.

Answer 1

The graph $G(A) = \{(x,A(x)) \; : \; x \in D(A)\}$ is a closed subspace of $X^2$. Take some $w \in X \backslash D(A)$. Then $G(A) \oplus \text{span}((x,0))$ is a closed subspace of $X^2$ and is the graph of a linear operator $B$. By the Closed Graph Theorem, $B$ is bounded. Now if $x_n \in D(A)$ with $x_n \to w$, we have $(x_n, A(x_n)) \to (w, 0)$, and the assumption that $w \notin D(A)$ contradicts the assumption that $A$ is a closed operator.