Let $H$ be a separable complex Hilbert space with orthonormal basis $\{e_k; k \in\mathbb{N}\}$. Let $(\alpha_k)$ be a given sequence of complex numbers and let $A$ be the associated multiplication operator,$$ Au =\sum_{k=1}^{\infty}\alpha_k \langle u,e_k \rangle e_k $$with $D_A$ being the linear span of the orthonormal basis.
What is the adjoint of $A$?
Since,$$\langle Au,v \rangle = \left\langle \sum_{k=1}^{\infty}\alpha_k \langle u,e_k \rangle e_k,v \right\rangle =\sum_{k=1}^{\infty}\alpha_k \langle u,e_k \rangle \left\langle e_k,v \right\rangle=\left\langle u,\overline{\sum_{k=1}^{\infty}\alpha_k \langle e_k,v \rangle} e_k \right\rangle$$The adjoint of $A$ must be defined as$$A^{*}v={ \sum_{k=1}^{\infty}\overline{\alpha_k} \langle v,e_k \rangle}e_k; ~v\in D(A^{*})$$But the above argument is true as long as $A^{*}$ is densely defined. i.e. $D(A^{*})$, the domain of $A^{*}$ is dense in $H$, which I am having trouble to prove.