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?



Transpose of unbounded operators between Banach spaces.
The transpose is **closed** but **it may not be densely defined**. For more info see Sec. 2.6 of



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> H. Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer Verlag, 2011
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Exercise 2.22 in this book describes a closed densely defined operator whose adjoint is not dense. Here it is.


Consider the Banach space $E=\ell^1$ with dual $E^\*=\ell^\infty$. Consider the densely defined operator


$$ A: D(A)\subset E\to E,$$


$$D(A)=\bigl\lbrace\; (u\_n)\in\ell^1;\;\; (nu\_n)\in \ell^1 \;\bigr\rbrace, \;\; A(u\_n)= (nu\_n).$$


Then $A$ is closed, densely defined, $A^\* $ is closed, but $ D(A^\*)$ is not dense. 


