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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.

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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.

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  • $\begingroup$ Thanks I followed everything except that how did you come up eith this $|\langle Au,v\rangle|=\sum_{k=1}^m|\alpha_k\langle e_k,v\rangle|^2$ $\endgroup$ – gamma555 Mar 26 at 3:55
  • $\begingroup$ And also the part where used Holder inequality. $\endgroup$ – gamma555 Mar 26 at 4:09
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    $\begingroup$ I added a little more detail, but I wouldn't know what else to say. $\endgroup$ – Martin Argerami Mar 26 at 4:23
  • $\begingroup$ Thanks, it makes sense perfectly now. $\endgroup$ – gamma555 Mar 26 at 5:00

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