# Why do we restrict unbounded linear operator to Banach spaces

Let $$(E, | \cdot|_E)$$ and $$(F, | \cdot|_F)$$ be Banach spaces. An unbounded linear operator $$A$$ from $$E$$ to $$F$$ is a linear map of the form $$A: D(A) \to F$$ where $$D(A)$$ is a subspace of $$E$$. Let $$A$$ be densely defined, i.e., $$D(A)$$ is dense in $$E$$. We define the adjoint $$A^\star: D(A^\star) \to E^\star$$ as follows. First, $$D (A^\star) := \{f \in F^\star \mid \exists c\ge 0, \forall x\in D(A), |\langle f, A x \rangle| \le c |x|_E\}.$$

Then $$D(A^\star)$$ is a subspace of $$F^\star$$. For each $$f \in D (A^\star)$$, we define a map $$g \in (D(A))^\star$$ by $$\langle g, x \rangle := \langle f, A x \rangle$$ for all $$x \in D(A)$$. By Hahn-Banach theorem, $$g$$ admits a linear continuous extension $$\tilde g \in E^\star$$. Because $$D(A)$$ is dense in $$E$$, such $$\tilde g$$ is unique. Then we define $$A^\star f:=g$$.

It follows from the construction that $$\langle A^\star f, x \rangle_{E^\star, E} = \langle f, Ax \rangle_{F^\star, F}, \quad \forall f\in D(A^\star), x\in D(A).$$

In Brezis's Functional Analysis and this Wikipedia page, unbounded linear operators are defined for Banach spaces, but the completeness property is not used in the construction. What is the motivation for such restriction?

In this specific case, you can always view $$A$$ as an operator $$\tilde A$$ between the completions $$\tilde E$$ and $$\tilde F$$. The dual spaces $$\tilde E^\ast$$ and $$\tilde F^\ast$$ are canonically isomorphic to $$\tilde E$$ and $$\tilde F$$, respectively, and you can check that $$\tilde A^\ast=A^\ast$$ under this identification. Thus there is really nothing won by allowing incomplete normed spaces here.