Some context tangential to the question: I'm currently preparing for a final on an introductory course in functional analysis, which consists of a short presentation. Looking through these notes, I have found an elementary consequence of the spectral theorem for compact and self adjoint operators which seems interesting to investigate.

The author gives a version of functional calculus for these type of operators (namely Corollary 3.10 and the subsequent observation, which includes equation 3.11). Paraphrasing (and focusing only on the case over $\mathbb{R}$ in which the Hilbert space is separable, which is the one I'm interested in),

Theorem. Let $A \in \mathscr{L}(H)$ be a compact self adjoint operator on a separable Hilbert space. Given an orthonormal basis of eigenvectors $\{e_n\}_{n \geq 1}$ with corresponding eigenvalues $\{\alpha_n\}_{n \geq 1} \subset \mathbb{R}$. There exists a Banach algebra continuous homomorphism $$ f \in B(\{\alpha_n\}_{n \geq 1},\mathbb{R}) \mapsto f(A) \in \mathscr{L}(H) $$ via $$ f(A)(x) := \sum_{n \geq 1}f(\alpha_n)\langle e_n,x \rangle e_n, $$ which sends $\mathsf{1}$ to $id_{H}$ and $id_{\{\alpha_n\}_{n \geq 1}}$ to $A$.

In particular, we have

Remark: given $A$ as above, and $ z \not \in \{0\} \cup \{\alpha_n\}_{n \geq 1}$, then $$ (A-zI)^{-1}(x) = \sum_{n \geq 1}\frac{1}{\alpha_n - z}\langle e_n,x \rangle e_n. $$

I have a couple of questions motivated by the former results:

  • It is my impression that the theorem should be generalizable to continuous real valued functions by precomposing with the restriction $f \mapsto f|_{\{\alpha_n\}_n}$, which if I am not mistaken, is a continuous Banach algebra homomorphism. Is this the case?

  • What are the relations between the spectrum of $A$ and $f(A)$? By a direct verification it seems that we have $f(\sigma(A)) \subset \sigma(f(A))$.

  • What are (if any) some sufficient conditions to guarantee that $f(A)$ is self adjoint and/or compact?

  • Are there any available references which treat functional calculus in the specific case of compact self adjoint operators?

  • 1
    $\begingroup$ You can look at Methods of Mathematical Physics, Functional Analysis (Vol.1) by Reeds & Simon, Theorem VII.1. Note that, for the specific case of compact self-operator defined over your all Hilbert (the reference above treats the general case of densely-defined operator), there's (I believe) another book by Simon called Analysis of Operator. To check. $\endgroup$ – Hermès Jul 18 at 11:06
  • $\begingroup$ @Hermès this seems like what I was looking for. Many thanks! I'll wait for some more responses, but in any case I think this should be an answer rather than a comment. $\endgroup$ – Guido A. Jul 18 at 11:15
  • Yes, that restriction map is a Banach algebra homomorphism. For this, you want to check that it is linear and that $(fg)|_{\{a_n\}} = f|_{\{a_n\}} g|_{\{a_n\}}$. Both of these properties are straightforward.
  • In fact, one has that $\sigma(f(A)) = f(\sigma(A))$. For the inclusion you say that you don't have, assume that $\lambda \not \in f(\sigma(A))$ and define $g(x) = (f(x) - \lambda)^{-1}$. Then check that $g(A) = (f(A) - \lambda)^{-1}$ so that $\lambda \not \in \sigma(f(A))$.
  • $f(A)$ is always self-adjoint. This is an easy computation with the definition since $$\langle f(A)x, y \rangle = \sum_{n \geq 1} \langle e_n, x \rangle \langle e_n,y \rangle f(a_n) = \langle x, f(A)y \rangle.$$ One condition for compactness of $f(A)$ is as follows. Note that if $P$ is a polynomial with $0$ constant term then $P(A)$ is easily seen to be compact. Since the map $f \mapsto f(A)$ is continuous, if $P_n \to f$ in the $\sup$-norm then $P_n(A) \to f(A)$ in operator norm which implies that $f(A)$ is also compact. Hence $f(A)$ is compact whenever $f$ is a continuous function on the spectrum such that $f(0) = 0$ by an application of Stone-Weierstrass.
  • It seems you already got some good literature recommendations to check in the comments whilst I was in the process of writing this answer.

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