A problem on Left Hilbert algebra.

Question

I got stuck with the following problem while going through Section 10.1 from the book 'Lectures on von Neumann Algebras' by Strătilă and Zsidó.

Let $\mathfrak{A}$ be a complex algebra with involution, which is also endowed with a scalar product $\langle\cdot | \cdot\rangle$. We denote by $\xi\mapsto\xi^{\text{#}}$ the involution in $\mathfrak{A}$ and by $\mathscr{H}$ the Hilbert space obtained by the completion of $\mathfrak{A}$. We denote by $\mathfrak{A}^2$ the vector space generated by the elements of the form $\xi\eta,\,\xi,\,\eta\in\mathfrak{A}$. One says that $\mathfrak{A}$ is a left Hilbert algebra if

1. $\mathfrak{A}\ni\eta\mapsto\xi\eta\in\mathfrak{A}$ is continuous, for any $\xi\in\mathfrak{A}$.
2. $\langle\xi\eta_1|\eta_2\rangle=\langle\eta_1|\xi^{\text{#}}\eta_2\rangle$ for any $\xi,\,\eta_1,\,\eta_2\in\mathfrak{A}$.
3. $\mathfrak{A}^2$ is dense in $\mathfrak{A}$.
4. $\mathscr{H}\supseteq\mathfrak{A}\ni\xi\mapsto\xi^{\text{#}}\in\mathscr{H}$ is a preclosed antilinear operator.

In accordance with $1$, for any $\xi\in\mathfrak{A}$, one define $L_{\xi}\in\mathscr{B}(\mathscr{H})$ by the formula $L_{\xi}(\eta)=\xi\eta,\,\eta\in\mathfrak{A}$.

Problem: Prove that $I\in\overline{\{L_{\xi}:\xi\in\mathfrak{A}\}}^{so}$, where $I$ is the identity map on $\mathscr{H}$ defined by $I(\eta)=\eta,\,\eta\in\mathfrak{A}$.

The authors say that it follows from property $3$, but I am not getting how to argue that. Thanks in advance for any help.

Answer 1

Let us denote $\mathscr{A}=\{L_{\xi}:\xi\in\mathfrak{A}\}$. Fix $\eta\in\mathscr{H}$ and let $\Large\chi$$=\overline{\mathscr{A}\eta}$. Then as $\Large\chi$ and $\Large\chi$$^{\perp}$ both are invariant under $\mathscr{A}$, so the orthogonal projection $P_{\Large\chi}$ on the subspace $\Large\chi$ is in $\mathscr{A}'$. We want to show that $\eta\in\Large\chi$. Now for all $\xi\in\mathfrak{A}$, \begin{equation*} \begin{split} L_{\xi}\big(I-P_{\Large\chi}\big)\eta=\big(I-P_{\Large\chi}\big)L_{\xi}(\eta)=0,\\ \text{ i.e. } {L_{\xi}}^*\big(I-P_{\large\chi}\big)\eta=L_{\xi^{\text{#}}}\big(I-P_{\Large\chi}\big)\eta=0,\\ \text{ i.e. } \langle \zeta,{L_{\xi}}^*\big(I-P_{\Large\chi}\big)\eta\rangle=0 \text{ for all } \zeta\in\mathscr{H},\\ \text{ i.e. } \langle L_{\xi}(\zeta),\big(I-P_{\Large\chi}\big)\eta\rangle=0 \text{ for all } \zeta\in\mathscr{H}. \end{split} \end{equation*} But as $\mathfrak{A}^2\subseteq \{L_{\xi}(\zeta):\xi\in\mathfrak{A},\zeta\in\mathscr{H}\}$ is dense in $\mathfrak{A}$, hence it is dense in $\mathscr{H}$. Therefore $\big(I-P_{\Large\chi}\big)\eta=0$, i.e. $\eta\in\Large\chi$.