# 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.

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.
• Thanks I followed everything except that how did you come up eith this $|\langle Au,v\rangle|=\sum_{k=1}^m|\alpha_k\langle e_k,v\rangle|^2$ – gamma555 Mar 26 at 3:55