I'm studying basic spectral theory from the book Elements of functional analysis by Hirsch and Lacombe, and I've encountered some difficulties in understanding the proof of the following theorem:
Theorem Let $E$ be a Banach space over $\mathbb{C}$ and suppose that $T \in L(E)$ (where $L(E)$ is the set of all bounded linear operators on $E$). Then the spectrum $\sigma(T)$ of $T$ is nonempty and $r(T) = \max \{|\lambda|: \lambda \in \sigma(T) \} $ (where $r(T) = \lim ||T^n||^{\frac{1}{n}}$).
I write here the beginning of the proof:
Let $\rho = \max \{|\lambda|: \lambda \in \sigma(T) \}$ and assume by contradiction $\rho<r(T)$. Take $\lambda \in \mathbb(C)$ s.t. $|\lambda|>r(T)$. So $\lambda I - T$ is invertible and $$ (\lambda I - T) ^{-1}= R(\lambda, T)= \sum_{n=0}^{\infty} \lambda^{-(n+1)}T^n,$$ where $R(\lambda, T)$ denotes the resolvent operator. Take $p \in \mathbb{N}$ and consider $$ \lambda^{p+1} R(\lambda, T) =\sum_{n=0}^{\infty} \lambda^{(p-n)}T^n. $$ Then write $\lambda$ as $\lambda = te^{i \theta}$, with $t>0$ and $\theta \in [0,2\pi] $. Then $$t^{p+1}e^{i(p+1)\theta}R(e^{i \theta}, T) = \sum_{n} t^{p-n} e^{i(p-n) \theta}T^n. $$ Here I cite: Integrating the result (with respect to $\theta$) from $0$ to $2\pi$, we obtain, by continuity of the Riemann integral with values in $L(E)$, $$ \int_{0} ^{2\pi}t^{p+1}e^{i(p+1)\theta}R(e^{i \theta}, T)d \theta = \\\sum_{n} \int_{0}^{2\pi} t^{p-n} e^{i(p-n) \theta}T^n d \theta = 2\pi T^p. $$ (The proof continues, but my question regards only this part that I've just written).
Now my question: In the last expression they integrate an expression in which appears an operator which takes value in an abstract Banach space, and they mention Riemann integral with values in $L(E)$. How should I interpret this phrase? Also, I would like to know why exchanging integral and summation sign is a legal operation. I know that this operation is allowed with standard Riemann integrals and absolutely convergent series, but as I said the context seems to be here more abstract.
What you need in order to understand the integral given there is the theory of the Banach space - valued Riemann integral. This is a more general integral as it is for functions $f: [a, b] \longrightarrow \mathbb{R}$. What one uses here for the construction are step functions $s: [a, b] \longrightarrow X$, which take values in a Banach space $X$. The usual construction of this integral is done by the BLT - theorem (Bounded linear transformation). See for example Reed/Simon 1 or here. It should be also noted that in the latest version of the book by Hirsch and Lacombe, there is a huge exercise on page 20, where this integral and its properties are constructed in eight steps. Note also that if $X$ is a Banach space, $L(X)$ is a Banach space and therefore the integral taken over bounded operators makes sense.
Now to your second question concerning the legitimation of interchanging the sum and the integral. Let $E$ be a Banach space, $T \in L(E)$ an operator and $\lambda \in \mathbb{K}$ such that $\lvert \lambda \rvert > r(T)$. Consider $r \in (r(T), \lvert \lambda \rvert )$. Then $r > r(T)$ and therefore there is an $n_0 \in \mathbb{N}$ such that $\|T^n \| \leq r^n $ for all $n \geq n_0$. Since $r < \lvert \lambda \rvert$, the series $\sum_{n = 0}^\infty \lambda^{- n - 1} T^n$ is absolutely convergent in the Banach space $L(E)$. Since the same assumptions are given in the proof you wrote in your question, it follows that your sum $\lambda^{p +1}R(\lambda, T)$ is absolutely convergent in $L(E)$. Now take $t \in (r(T), \infty)$ (note that $t$ is not just $>0$ as you wrote it, by the assumption on $\lambda$). It now follows easily by the Weierstraß-M-Test that the series $t^{p + 1} \mathrm e^{\mathrm i (p + 1) \theta} R(\mathrm e^{\mathrm i \theta}, T)$ is uniformly convergent (with respect to $\theta$) in $L(E)$. Here, you also have to use that the sum $\lambda^{p +1}R(\lambda, T)$ is convergent, which follows from absolute convergence in a Banach space. By the properties of the integral, interchanging the sum and the integral is then allowed.