Why is the Spectrum of an Operator Used as the Domain in Continuous Functional Calculus?

Question

I'm currently working to grasp the concepts of (continuous) functional calculus, aiming to prove the spectral theorem for bounded self-adjoint operators as outlined in "Introduction to Hilbert space and the theory of spectral multiplicity" by Paul Halmos and "Quantum Theory for Mathematicians" by Brian Hall. The specific definition of continuous functional calculus I'm exploring is as follows:

Theorem 8.3 ("Quantum Theory for Mathematicians" - Brian Hall):
Let $A \in \mathcal{B}(H)$ be a self-adjoint operator. Then there exists a unique bounded linear map from $\mathcal{C}(\sigma(A); \mathbb{R})$ to $\mathcal{B}(H)$, denoted as $f \mapsto f(A)$, such that when $f(\lambda)=\lambda^{m}$, we have $f(A)=A^{m}$. This mapping, denoted as $f \mapsto f(A)$, where $f\in \mathcal{C}(\sigma(A);\mathbb{R})$, is commonly referred to as the (real-valued) functional calculus for the operator $A$."

My questions are mainly concerned with the choice of domain of $f$:

1. Why is the domain of $f$ restricted to $\sigma(A)$ (the spectrum of $A$)? Why not simply use $\mathbb{R}$ as the domain, given that $A$ is self-adjoint and $\sigma(A) \subset \mathbb{R}$?

2. In Halmos' book, for a polynomial $p(\lambda)=\sum_{j=0}^{n}\alpha_{j}\lambda^{j}$ and an operator $A$ on a Hilbert space, he defines $p(A)$ to be $p(A)= \sum_{j=0}^{n}\alpha_{j} A^{j}$. To me, this seems acceptable, assuming that the domain and codomain of $p(A)$ and $A$ are the same. However, am I correct that when considering more complex functions like Borel or holomorphic functions, we need to be cautious about our choice of domain? Unfortunately, this aspect is not generally explained.

Update: The proof makes use of the Stone-Weierstrass theorem, i.e. that polynomials are dense in $C(K;\mathbb{R})$, where $K$ is a compact set. Thus we could use any compact $K \supset \sigma(A)$. However, since $\sigma(A)$ is compact, we wouldn't gain anything as pointed out in the answers below.

Answer 1

The point is that $f(A)$ is determined by the properties/behaviour of $f$ on $\sigma(A)$. In particular, $f$ does not even need to be defined for $\mathbb R\setminus\sigma(A)$.

This becomes more clear when we consider the case in which $H$ is finite-dimensional: For a symmetric matrix $A$, which has a diagonalization of the form $Q^\top A Q = \operatorname{diag}(\lambda_1, \ldots)$, you would define said map $f$ according to $$f(A):= Q\operatorname{diag}(f(\lambda_1), f(\lambda_2), \ldots)Q^\top.$$ Note that $f|_{\{\lambda_1, \ldots\}}$ (i.e. $f$ on $\sigma(A)$) is all that you need to know about $f$ in order to define $f(A)$.

Comment for question 2. The definition is consistent for polynomial $f$, but extends to more general functions as long as they are defined of $\sigma(A)$. Note that the theorem states that if there is a function $f$ defined on $\sigma(A)$, you may define (!) $f(A)$ accordingly, in a way that is consistent with the case in which $f$ is a polynomial. The fact that $f(A)$ is a definition cannot be stressed enough.

Answer 2

1. The proof of the theorem has $C(\sigma(A);\Bbb{R})$ mapping one-to-one onto the self-adjoints of the norm-closed subalgebra generated by $A$. Consequently, $C(\sigma(A); \Bbb{C})$ maps one-to-one onto the norm-closed subalgebra generated by $A$. You can extend the mapping artificially to any superset of $\sigma(A)$ by mapping $f$ to the same operator as $f\rvert_{\sigma(A)}$ maps to. If you do this you get a kernel of all functions $f$ that are zero on $\sigma(A)$.
2. Since $A$ is bounded, the domain is always the entire Hilbert space $H$. The codomain is a subspace of $H$ so you can take powers like that. The largest class to which the functional calculus extends is bounded Borel-measureable functions on $\sigma(A)$. The image is the von-Neumann algebra generated by $A$. If you want to apply the calculus to a holomorphic function defined on an open neighborhood of $\sigma(A)$ that's fine, because $\sigma(A)$ is compact so the function is still bounded on $\sigma(A)$.

Answer 3

To complement the other answers, if you decide to use $C(\mathbb R)$ instead of $C(\sigma(A))$, you are unnecessarily disallowing valid functions. For an example consider $$A=\begin{bmatrix}1&0\\0&2\end{bmatrix}$$ and the function $f(t)=1/t$.