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.


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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.