# Remark 16 in Chapter 2 from Brezis - Functional Analysis, Sobolev Spaces, and Partial Differential Equations.

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!

• 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. – Alonso Delfín Nov 14 '18 at 7:07
• thank you so much for your constructive comments! – Evan William Chandra Nov 15 '18 at 2:20