Spectral decomposition vs Taylor Expansion

Question

This question and the comments and answers it received encouraged me to ask this question, although I know that there will be some people who think that this belongs in the math forum. But I think that this topic is more relevant to mathematical physicists than to pure mathematicians.

Motivation: One of the answers to this question explains that if $f\colon\mathbb C\to\mathbb C$ is suitable function and $A\colon\text{Dom}(A)\subset H\to H$ is a suitable operator, we can define \begin{equation} f(A):=\int_{\mathbb C}f\,\mathrm{d}P_A \end{equation} where $P_A\colon B(\mathbb C)\to B(H)$ is a measure. However, it's much easier to understand the definition in terms of convergent series, e.g. in the case of the exponential or the logarithm. (In statistical physics, $S=k_B\langle\ln\rho\rangle$ is the entropy, when $\rho$ is the density operator.) That's why I'd like to know:

Is it also possible to write $f(A)$ in terms of a converging series when $f$ has a taylor expansion around some point?

In case that the answer is yes, I also wonder if there is a relatively easy way to see how the integral and the series are equivalent. (As far as I know, integrals - even $\int_{\mathbb C}f\,\mathrm{d}P_A$ - can be expressed as limit of some series, so maybe that would be a good starting point).

Examples: The expression \begin{equation} \sum_{n=0}^\infty \frac{1}{n!} A^n \end{equation} makes sense whenever $A$ is an element of a complete normed space and converges to $\mathrm{e}^A=\int_{\mathbb C}\text{exp}\,\mathrm{d}P_A$ when $A$ is suitable operator (source).

It is even known that \begin{equation} \left(\sum_{k=1}^N (-1)^{k+1}\frac{(A-\text{id})^k}{k}\right)_{N\in\mathbb N} \end{equation} converges to $\text{ln}(A)$ under certain circumstances (see here and here), so I was wondering if there is a general rule. That is, if we have \begin{equation} f(x)=\sum_{n=0}^{\infty}a_n(x-b)^n, \end{equation} in a neighborhood of $b$, is \begin{equation} f(A)=\sum_{n=0}^{\infty}a_n(A-b\cdot\text{id})^n? \end{equation}

Answer 1

If the operator $A$ belongs to $B(H)$ (the space of everywhere defined bounded operator on the Hilbert space $H$) and is normal: $$A^*A=AA^*$$ then it admits a spectral decomposition $$A = \int_{\mathbb{C}} z dP(z) = \int_{\sigma(A)} z dP(z)$$ and, with an obvious notation, $|\sigma(A)| \leq ||A|| <+\infty$.

In this case (and also in the general case where $A$ is unbounded (densely-defined, closed, normal)), $$f(A) := \int_{\sigma(A)} f(z) dP(z)$$ for every Borel measurable function $f: \sigma(A) \to \mathbb{C}$. In this case the answer is relatively easy.

Proposition.

Let $A \in B(H)$ be normal and consider $f: \Omega \to \mathbb{C}$ an analytic function on the open set $\Omega \subset \sigma(A) \subset \mathbb{C}$.

If $z_0 \in \Omega$ and the Taylor expansion of $f$ around $z_0$ $$f(z) = \sum_{n=0}^{+\infty} a_n (z-z_0)^n$$ has convergence radius $R+\epsilon$ for some $\epsilon>0$, and finally $$\sigma(A) \subset C_R(z_0):= \{ z\in \mathbb{C}\:|\: |z-z_0| <R\}$$ then $$f(A) = \sum_{n=0}^\infty a_n (A-z_0)^n$$ where the convergenge of the right-hand side is in the norm of $B(H)$.

PROOF. We start from the inequality $$||\int_{\mathbb{C}} g(z) dP(z)|| \leq ||g||_\infty$$ which is valid if $g$ is Borel-measurable and bounded. This inequality holds true even if $A$ is not bounded.

Exploiting that inequality we have $$\left|\left|\int_{\sigma(A)} \left[\sum_{n=0}^N a_n(\lambda -\lambda_0)^n - f(z)\right] dP(z)\right|\right| \leq \sup_{z\in \sigma(A)}\left|\sum_{n=0}^N a_n(\lambda -\lambda_0)^n - f(z)\right| \to 0$$ for $N\to +\infty$ since the convergence of the Taylor expansion is uniform in every compact in the convergence disk. Notice that $\sigma(A)$ is indeed a compact included in $C_{R+\epsilon}(z_0)$.

Using definition of $g(A)$, we therefore have that $$\int_{\sigma(A)} \sum_{n=0}^N a_n(\lambda -\lambda_0)^n dP(z) \to \int_{\sigma(A)} f(z) dP(z)$$ with respect to the norm of $B(H)$. In other words, if $N\to +\infty$ $$\sum_{n=0}^{+\infty} a_n (A-z_0I)^n = f(A)$$ in that topology. That is the thesis. QED

There are other results which can be proved analogously and which encompass the case of $A$ unbounded (densely defined, closed, and normal). If $\psi$ belongs to the projection space of $\int_{E} 1 dP(z)$, where $E \subset C_{R}(z_0)$ is a bounded Borel set (so that $\psi$ is an analytic vector of $A$), then $$f(A)\psi = \sum_{n=0}^\infty a_n (A-z_0)^n\psi$$ where now the convergence is in the Hilbert space norm.

It is difficult to produce a finer result.

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(As references I can quote my books https://doi.org/10.1007/978-3-319-70706-8 and https://doi.org/10.1007/978-3-030-18346-2)