Suppose $T$ is an operator with domain and range in the Hilbert space $\mathcal{H}$. The usual way of defining the adjoint $T^*$ of $T$ uses density of $dom(T)$. But cannot we use this same definition for general situation where the operator is not densely defined in the following way?
$dom(T^*):=\{ y\in \mathcal{H}:\text{the map }dom(T)\to \mathbb{C}:x\mapsto \langle Tx, y\rangle \text{ is continuous}\}$
Then by Riesz Representation theorem for the Hilbert space $\overline{dom(T)}$ there exists a unique $z\in \overline{dom(T)}$ for any $y\in dom(T^*)$. We'll define $T^*y=z$ for each such $y, z$.
This definition can be applied to any operators not necessarily the densely defined one and coincide with the usual notion for densely defined case. Isn't the case? If so, then why do we consider dense domain for defining adjoints?
EDIT: If we define $dom(T^*)$ as above then for any $y\in dom(T^*)$ there is a unique $z\in \overline{dom(T)}$ such that $\langle Tx, y\rangle = \langle x, z\rangle$ for all $x\in dom(T)$. There could be many such $z$ in $\mathscr{H}$ but exactly one such $z$ lies in $\overline{dom(T)}$, we'll define $T^*y$ to be that $z\in \overline{dom(T^*)}$. Is there any issue with this definition? Thank you.