How to show that the operator is densely defined. 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.
 A: By definition, the domain of $A^*$ is the space of those $v\in H$ such that $u\longmapsto \langle Au,v\rangle$ is bounded on $D(A)$. Given $u\in D(A)$, we have $u=\sum_{k=1}^m c_ke_k$. Then
$$\tag1
\langle Au,v\rangle=\sum_{k=1}^m\alpha_kc_k\langle e_k,v\rangle.
$$
This is bounded if and only if $\sum_k|\alpha_k\langle e_k,v\rangle|^2<\infty$. Indeed, if the sum is finite then we apply Hölder in $(1)$ to get
$$\tag2
|\langle Au,v\rangle|\leq  (\sum_k|c_k|^2 )^{1/2}\left(\sum_k|\alpha_k\langle e_k,v\rangle|^2\right)^{1/2}=\|u\|\,\left(\sum_k|\alpha_k\langle e_k,v\rangle|^2\right)^{1/2}.
$$
Conversely, if $|\langle  Au,v\rangle|\leq c\|u\|$ for all $u\in D(A)$, then taking $u=\sum_{k=1}^m \overline{\alpha_k\langle e_k,v\rangle}\,e_k$ we have 
\begin{align}
c\left(\sum_{k=1}^m|\alpha_k\langle e_k,v\rangle|^2\right)^{1/2}&=c\|u\|\geq|\langle Au,v\rangle|\\[0.3cm]
&= \left|\sum_j \sum_k|{\alpha_k}|^2\,\overline{\langle e_k,v\rangle}\,\overline{\langle v,e_j\rangle}\,\langle e_k,e_j\rangle\right|\\[0.3cm]
&= \left|\sum_j \sum_k|{\alpha_k}|^2\,\overline{\langle e_k,v\rangle}\,{\langle e_j,v\rangle}\,\langle e_k,e_j\rangle\right|\\[0.3cm]
&= \left| \sum_k|{\alpha_k}|^2\,|{\langle e_k,v\rangle}|^2\, \right|\\[0.3cm]
&=\sum_{k=1}^m|\alpha_k\langle e_k,v\rangle|^2.
\end{align}
It follows that $\sum_{k=1}^m|\alpha_k\langle e_k,v\rangle|^2\leq c^2$ for all $m$, and so 
$$\sum_{k=1}^\infty|\alpha_k\langle e_k,v\rangle|^2\leq c^2.$$ Thus $D(A^*)$ consists of those $v$ such that $\sum_k|\alpha_k\langle e_k,v\rangle|^2<\infty$. In particular $D(A)\subset D(A^*)$, so $D(A^*)$ is dense. 
