# Functional calculus on compact self adjoint operators

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?

• 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. – Hermès Jul 18 at 11:06
• @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. – Guido A. Jul 18 at 11:15

## 1 Answer

• 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.