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?

## Best Answer

In general, bits and pieces of operator theory can be done in incomplete normed spaces, too, but it's usually a hassle and people don't care to check how much of the theory goes through without completeness. In practice, this is of limited interest.

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.