Subalgebra generated by selfadjoint operator $A_0\in\mathscr{L}(H,H)$

Question

Let $\mathscr{L}(H,H)$ be the Banach algebra of bounded operators defined on a complex Hilbert space $H$ and let $B(A_0)$ be the subalgebra generated by the selfadjoint operator $A_0$, i.e. $\overline{\text{Span}(A_0)}$. I would like to show that:

• $B(A_0)$ is commutative, but I only see that $\text{Span}(A_0)$ is commutative and am not sure of what happens at its frontier;

• $B(A_0)$ is regular, in the sense that $\forall A\in B(A_0)\quad \|A^2\|=\|A\|^2$;

• $B(A_0)$ is symmetric, i.e. for all $A\in B(A_0)$ there is a $B\in B(A_0)$ such that, for all $f_M\in\mathscr{M}$ (where $\mathscr{M}$ is the set of all non-trivial continuous multiplicative linear functionals$^1$ $B(A_0)\to\mathbb{C}$), $f_M(A)=\overline{f_M(B)}$ where the overline means the complex conjugation, and $B=A^{\ast}$ precisely is the selfadjoint operator of $A$.

I study by myself and my text does not give a detailed introduction to Banach algebras. Could anybody help me with a proof or a link to one?

I $\infty$-ly thank you!

$^1$ Continuous multiplicative linear functionals are defined as the continuous linear functionals, belonging to the dual space $B(A_0)^\ast$, such that $\forall A,B\in B(A_0)\quad f_M(AB)=f_M(A)f_M(B)$. $\mathscr{M}$ can be identified with the set of all non-trivial maximal ideals of $B(A_0)$: for all non-trivial maximal ideal $M\subset B(A_0)$ there is one and only one $f_M$ such that $\ker f_M=M$ and for any $f_M$ its kernel is a non-trivial maximal ideal. Cfr. pp. 521-523 here.

Answer 1

Every element of $B(A_0)$ is the limit of polynomials of the form $$ p(A_0)=\sum_{k=0}^na_kA_0^k,\quad n\in\mathbb N,\,\,a_k\in\mathbb C. $$ Hence the first and second bullets hold.

Note that $$ \|B\|=\sup_{\|x\|=\|y\|=1}(x,By), $$ hence, if $A$ is self-adjoint, then $$ \|A^2\|=\sup_{\|x\|=\|y\|=1}(x,A^2y)=\sup_{\|x\|=\|y\|=1}(Ax,Ay)\ge \sup_{\|x\|=1}(Ax,Ax)=\|A\|^2, $$ and as $\|B^2\|\le\|B\|^2$, for all operators, then $\|A^2\|=\|A\|^2$.

For the third bullet, if $$ p(A_0)=\sum_{k=0}^na_kA_0^k\to A\quad\text{then}\quad p(A_0)=\sum_{k=0}^n\overline a_kA_0^k \to \overline{A}. $$