friends! I read that the algebra $\mathscr{L}(H,H)$ of the bounded operators on a Hilbert space $H$ is a $B^\ast$-algebra in the sense defined here . I easily verify all the properties except for the facts that for any $A\in\mathscr{L}(H,H)$ the self-adjoint exists in $\mathscr{L}(H,H)$, and that $\forall A\in\mathscr{L}(H,H)\quad \|AA^\ast\|=\|A\|^2$.
Thank you so much for any help!
First $\|A\|^{2} \le \|A^{\star}A\|$ holds because $$ \|Ax\|^{2}=(Ax,Ax)=(A^{\star}Ax,x) \le \|A^{\star}Ax\|\|x\| \le \|A^{\star}A\|\|x\|^{2}. $$ Consequently, $\|A\|^{2}\le \|A^{\star}A\|\le \|A^{\star}\|\|A\|$ gives $\|A\|\le \|A^{\star}\|$. Replacing $A$ by $A^{\star}$ in this inequality gives $\|A^{\star}\|\le\|A\|$; hence, $\|A\|=\|A^{\star}\|$. With that, $$ \|A\|^{2} \le \|A^{\star}A\| \le \|A^{\star}\|\|A\| = \|A\|^{2} \\ \implies \|A\|^{2}=\|A^{\star}A\|. $$