Understanding Riesz's functional calculus

Question

I am studying Conway's Functional Analysis, where Riesz's functional calculus is introduced rather suddenly. Let $G \subset \mathbb C$ be open, $\mathcal A$ be a complex Banach algebra with identity $e$. The book states:

If $f: G \to \mathbb C$ is analytic and $\sigma(a) \subset G$, we will define an element $f(a) \in \mathcal A$ by $$f(a) = \frac{1}{2\pi i}\int_\Gamma f(z)(ze-a)^{-1}dz,$$

where $\Gamma$ is a suitable curve (I think further details are not necessary here, since this is common knowledge).

Question: what exactly is the object $\int_\Gamma f(z)(ze-a)^{-1}dz$? How can one understand this notation? For example, we can make sense of countable sums in a Banach space. But integration in a Banach space? I don't even have an intuition for it to be honest. Any answers, either rigorous or intuitive, are welcome.

Answer 1

There is nothing fancy going on here. You are still doing a line integral over the complex plane.

Something that is not emphasized (or even mentioned) when integration is taught, is the very asymmetric roles of the domain and the codomain. If you look at the definition of Riemann integral, what you do with the codomain is to multiply values of the function by complex numbers and add (which requires the codomain to be a vector space) and taking limits (which requires the codomain to be a topological space). So a Banach space is the perfect codomain to do Riemann integration.

Concretely, if $\Gamma$ is given by $\gamma(t)$, $t\in [0,1]$, then \begin{align} \int_\Gamma f(z)(ze-a)^{-1}dz&=\int_0^1f(\gamma(t))\,(\gamma(t)e-a)^{-1}\,\gamma'(t)\,dt\\[0.3cm] &=\lim_{n\to\infty}\frac1n\,\sum_{k=1}^nf\big(\gamma(\tfrac kn)\big)\,\big(\gamma(\tfrac kn)e-a\big)^{-1}\,\gamma'\big(\tfrac kn\big). \end{align} The limit exists when the argument is continuous, with the same proof as for the usual real-valued Riemann integral. Here, the continuity of $f$ and $\gamma$, together with the fact that $\Gamma$ is removed from the spectrum of $a$, guarantee that the integrand is continuous.

Answer 2

Martin Argerami has given an excellent answer on what the object of the mentioned integral is. I will add a bit more words on the very book you are reading since you can find information conveniently in Conway's book itself.

This is explicitly mentioned at the beginning of Section 4 of Chapter VII "The Riesz Functional Calculus" in Conway's book you are reading (A Course in Functional Analysis, 2nd edition, page 109, emphasis mine):

Let $G$ be an open subset of $\mathbb{C}$ and let $\mathscr{X}$ be a Banach space. If $f: G \rightarrow \mathscr{X}$ is analytic and $x^{*} \in \mathscr{X}^{*}$, then $z \mapsto\left\langle f(z), x^{*}\right\rangle$ is analytic on $G$ and its derivative is $\left\langle f^{\prime}(z), x^{*}\right\rangle$. By Exercise 4 of the preceding section, if $f: G \rightarrow \mathscr{X}$ is a function such that $z \mapsto\left\langle f(z), x^{*}\right\rangle$ is analytic for each $x^{*}$ in $\mathscr{X}^{*}$, then $f: G \rightarrow \mathscr{X}$ is analytic. These facts will help in discussing and proving many of the results below.

If $\gamma$ is a rectifiable curve in $G$ and $f$ is a continuous function defined in a neighborhood of $\{\gamma\}$ with values in $\mathscr{X}$, then $\int_{\gamma} f$ can be defined as for a scalar-valued $f$ as the limit in $\mathscr{X}$ of sums of the form $$ \sum_{j}\left[\gamma\left(t_{j}\right)-\gamma\left(t_{j-1}\right)\right] f\left(\gamma\left(t_{j}\right)\right) $$ where $\left\{t_{0}, t_{1}, \ldots, t_{n}\right\}$ is a partition of $[0,1]$. Hence $\int_{\gamma} f=\int_{0}^{1} f(\gamma(t)) d \gamma(t) \in \mathscr{X}$. It is easy to see that for every $x^{*}$ in $\mathscr{X}^{*},\left\langle\int_{\gamma} f, x^{*}\right\rangle=\int_{\gamma}\left\langle f(\cdot), x^{*}\right\rangle$.