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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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 $$ p(A_0)=\sum_{k=0}^na_kA_0^k\to A\quad\text{then}\quad p(A_0)=\sum_{k=0}^n\overline a_kA_0^k \to \overline{A}. $$

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.