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?


The transpose is closed but it may not be densely defined. For more info see Sec. 2.6 of

H. Brezis: Functional Analysis, Sobolev Spaces and Partial Differential Equations, Springer Verlag, 2011

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.


In general, the transpose need not be densely defined. For an example see

S. G. Krein, Linear equations in Banach spaces. Birkhäuser, Boston, 1982.


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