Consider the following fragment from Takesaki's book "Theory of operator algebras II" chapter VI "Left Hilbert algebras" (p6):
I am trying to understand how lemma 1.9 implies that $\mathfrak{A}'$ is a $*$- algebra. In particular, I am unable to justify why $$\xi \in \mathfrak{A}' \implies \xi^\flat \in \mathfrak{B}'$$ which is necessary to see that the involution on $\mathfrak{A}'$ is well-defined.
Why is this the case?
Thanks in advance for any help!
Let $\xi\in\mathfrak A'$. We need to show that $\xi^\flat$ is right-bounded. By assumption, there exists a constant $C>0$ such that $\lVert \pi_l(b)\xi\rVert\leq C\lVert b\rVert$ for all $b\in\mathfrak A$. For $a,b\in\mathfrak A$ we have $$ \langle \pi_l(a)\xi^\flat,b\rangle=\langle \xi^\flat,a^\sharp b\rangle=\langle (a^\sharp b)^\sharp,\xi\rangle=\langle b^\sharp a,\xi\rangle=\langle a,\pi_l(b)\xi\rangle. $$ Since $\mathfrak A$ is dense in $\mathfrak H$, it follows that $$ \lVert \pi_l(a)\xi^\flat\rVert=\sup_{\substack{b\in\mathfrak A\\\lVert b\rVert\leq 1}}\lvert \langle \pi_l(a)\xi^\flat,b\rangle\rvert=\sup_{\substack{b\in\mathfrak A\\\lVert b\rVert\leq 1}}\lvert\langle a,\pi_l(b)\xi\rangle\rvert\leq \lVert a\rVert\sup_{\substack{b\in\mathfrak A\\\lVert b\rVert\leq 1}}\lVert \pi_l(b)\xi\rVert\leq c\lVert a\rVert. $$ Hence $\xi^\flat$ is right-bounded (and $\pi_r(\xi^\flat)=\pi_r(\xi)^\ast$).