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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 2.2 of 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.

1

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 2.2 of 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.