I am learning about adjoints of unbounded operators on infinite-dimensional Hilbert spaces. I understand that the domain of such an adjoint (to operator $A$, from Hilbert space $H$ onto itself) is defined as
$$D_A* = \{v \in H |\,\exists w: w\cdot u = v \cdot Au, \forall u \in D_A\}$$
Question: Why is $D_A^*$ not simply all of $H$, in the case where $H$ is separable?
Consider the standard construction for the existence of $w$, given any $v \in H$:
Define $w = (v\cdot A e_1)e_1 + (v\cdot Ae_2)e_2 \dots$
where $e_1, e_2 \dots$ is an orthonormalized Schauder basis of separable Hilbert space $H$.
A simple check will confirm that, given any vector $u \in D_A$, $$ w\cdot u = (v\cdot Ae_1)(e_1\cdot u) + (v\cdot Ae_2)(e_2\cdot u) … = v\cdot Au$$
All your steps go through informally, this means that in finite dimensions or if you don't have to bother about anything being unbounded, this is actually the case. In infinite dimensions, a problem arises. We know that $D_A$ cannot be the whole space (by the theorem of Hellinger-Toeplitz) and it can only be dense in $H$.
Assuming that $D_A$ is dense in $H$, it may be possible to find a Schauder basis such that $Ae_j$ is well-defined for all $j$. But now, in your last calculation you actually pull $A$ inside a limit. In your last step, consider
$$A u = A \lim_{n \to \infty} \sum_{j=0}^n (e_j \cdot u) e_j \neq \lim_{n \to \infty} A \sum_{j=0}^n (e_j \cdot u) e_j = \sum_{j=0}^\infty (e_j \cdot u) A e_j, $$
in general because being unbounded for a linear operator is the same as failing to be continuous. It is sometimes useful to write out the limit like this so you see what can go wrong with unbounded/not continuous operators.
Edited: Answer given was wrong, now it should be correct.