Let $X$ and $Y$ be Banach spaces, and let $L : X \rightarrow Y$ be a unbounded operator with dense domain $\operatorname{dom}(L)$. We can then talk about the transposed operator

$L' : \operatorname{dom}(L') \subset Y' \rightarrow X' : y' \rightarrow y'(T\cdot)$

whose domain is given by those functionals $y'$, such that the term $y'(T\cdot)$, initially defined on $\operatorname{dom}(L)$, has bounded extension to all of $X$. If $L$ is closed and densely defined, then it is standard to show that $L'$ is closed, too. But if what the density of the domain of transpose? The proof by Reed and Simons seems in the Hilbert space case seems to use specific Hilbert space techniques.

**Question:** Suppose $L$ is a closed densely-defined operator between Banach spaces. Is it transpose a closed densely-defined operator, too?