Let $E,F$ be two Banach Spaces and $A : D(A) \subset E \to F$ be a linear unbounded operator which is densely defined. Now, we would like to define $A^{*}$ as the adjoint of $A$. Let $A^{*} : D(A^{*}) \subset E^{*} \to F^{*}$ be unbounded operator. The domain of $A^{*}$ is defined as $D(A^{*}) = \{ v \in F^{*} \, | \, \exists c \geq 0 \forall u \in D(A) \ni |\langle v, Au \rangle_{F^{*},F} | \leq c ||u||\}$.
Here, $E^{*}$ and $F^{*}$ are the dual spaces of $E$ and $F$ respectively.

Remark 16
If A is a bounded operator, then $A^{*}$ is also a bounded operator.

So, to see Remark 16, the book says that it is clear that $D(A^{*}) = F^{*}$. My idea is since we know by default that $D(A^{*}) \subset F^{*}$. So I want to claim that $F^{*} \subset D(A^{*})$. This is my attempt :

Take $v \in F^{*}$. Since $A$ is bounded linear operator, then we have :
$\forall u \in D(A), |\langle v, Au \rangle|_{F^{*},F}\leq ||v||_{F^{*}}||Au||_{F}\leq ||v||_{F^{*}}||A||_{\mathcal{L}(E,F)}||u||_{E}$. Choose $c = ||v||_{F^{*}}||A||_{\mathcal{L}(E,F)} \geq 0$ and thus $v \in D(A^{*})$.
My question : Is my reasoning correct? Also, I use Cauchy-Schwarz to obtain $|\langle v, Au \rangle|_{F^{*},F}\leq ||v||_{F^{*}}||Au||_{F}$. Can I use Cauchy-Schwarz Inequality here for the dual pairing?

Any help will be much appreciated!

  • 1
    $\begingroup$ I don’t know much about unbounded operators, but what you did looks good! Observe that if $E$ is a Banach space, the dual pairing $\langle \cdot , \cdot \rangle : E^* \times E \to \mathbb{C}$ is given by $\langle f , \xi \rangle = f(\xi)$. Thus, since by definition of the operator norm one has: $ |f(\xi)| \leq \| f \| \| \xi \| $, it makes perfect sense to use Cauchy-Schwarz as you did. $\endgroup$ – Alonso Delfín Nov 14 '18 at 7:07
  • 1
    $\begingroup$ thank you so much for your constructive comments! $\endgroup$ – Evan William Chandra Nov 15 '18 at 2:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.