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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.

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    $\begingroup$ Question. In the third bullet, can you explain what is $\mathscr M$. Is it correct that, for every $f\in\mathscr M$ and $A,B\in\mathscr L (H,H)$, we have that $$f(A+B)=f(A)+f(B)\quad\text{and}\quad f(AB)=f(A)f(B)? $$ $\endgroup$ – Yiorgos S. Smyrlis Oct 2 '14 at 6:41
  • $\begingroup$ Thank you so much for your comment and your kindness. Exactly: I've edited to specify that. The issue is explained in an elegant and beautiful way by Tikhomirov in the appendix to Kolmogorov-Fomin's "Элементы теории функций и функционального анализа", but the English translation "Introductory Real Analysis"+"Elements of the Theory of Functions and Functional Analysis" doesn't contain it, as far as I know. $\aleph_1$ thanks! $\endgroup$ – Self-teaching worker Oct 2 '14 at 7:29
  • $\begingroup$ I have an idea... since any $A\in B(A_0)$ is the limit of a sequence of polynomials $\{p_n(A_0)=\sum_{k=0}^{m(n)} a_{n,k}A_0^k\}$ such that $p_n(A_0)\to A$ and $A^\ast=\lim_n \sum_{k=0}^{m(n)} \bar{a}_{n,k}A_0^k$, I think it would be sufficient to show that $f_M(A_0)=\overline{f_M(A_0)}$, i.e. that it's real, but I'm not able to see that... I know that the linear functional on a Hilbert space can be expressed by a scalar product: $\forall f\in H^\ast$ $\exists x_f\in H:\forall y\in H$ $f(y)=(y,x_f)$, but I don't think $B(A_0)$ is a Hilbert space... $\endgroup$ – Self-teaching worker Oct 2 '14 at 7:58
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    $\begingroup$ It is true that $f(B^*)=\overline{f(B)}$? $\endgroup$ – Yiorgos S. Smyrlis Oct 2 '14 at 10:27
  • $\begingroup$ I don't know: it isn't in the definition of $\mathscr{M}$... $\endgroup$ – Self-teaching worker Oct 2 '14 at 11:14
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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 $$ p(A_0)=\sum_{k=0}^na_kA_0^k\to A\quad\text{then}\quad p(A_0)=\sum_{k=0}^n\overline a_kA_0^k \to \overline{A}. $$

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  • $\begingroup$ I heartily thank you, dear Professor! Forgive me: I don't understand why $\|p(A_0)^2\|=\|p(A_0)\|^2$, nor why $\|A_0^2\|=\|A_0\|^2$, since my text, Kolmogorov-Fomin's Introductory Real Analysis, proves quite few properties of selfadjoint operators in chapter 4 (the theorem is from Tikhomirov's appendix in the Russian text). As to the third one, I have rewritten it in a clearer way; I see that $p(A_0)=p(A_0)^{\ast}$ and therefore their limits are selfadjoint too, but I'm not able to see why $\forall f_M\in\mathscr{M}\quad f_M(A)=\overline{f_M(A^{\ast})}$... :-( $\infty$ thanks!!! $\endgroup$ – Self-teaching worker Sep 28 '14 at 9:28
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    $\begingroup$ See the note I have added. $\endgroup$ – Yiorgos S. Smyrlis Sep 28 '14 at 11:55
  • $\begingroup$ Erased an uncorrect "proof" that I wrote. I notice that bullet 3 holds because $B(A_0)$ is a $B^\ast$-algebra in this sense: math.stackexchange.com/questions/953105/… $\endgroup$ – Self-teaching worker Oct 3 '14 at 17:58

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