MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 9 down vote accepted

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.

share|cite|improve this answer

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

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.