This post is a review of some important results on series and power series in Banach spaces and algebras. Some knowledge about Banach spaces / algebras is presumed.

The following topics are covered:

Series In Banach Spaces

Let $X$ be a Banach space and let $(x_i)_{i \in \mathbb{N}}$ be a sequence in $X$. Define the sequence $ (s_n)_{n \in \mathbb{N}} $ by $$ \begin{equation} s_n = \sum_{i=1}^n x_i. \end{equation} $$ If $ (s_n)_{n \in \mathbb{N}} $ is convergent the limit is denoted by $ \sum_{i=1}^\infty x_i $.

Proposition: If $ (s_n)_{n \in \mathbb{N}} $ converges, then $ \lim_{i \to \infty} x_i =0 $.

Proof: Let $\varepsilon >0$. Since convergent sequences are Cauchy there is a $N \in \mathbb{N}$ so that $$ \begin{equation} | s_n - s_m | < \varepsilon \end{equation} $$ for all $n,m >N$. In particular with $m = n-1$: $$ \begin{equation} | x_n | = | s_n -s_{n-1} | < \varepsilon \end{equation} $$ for all $n > N+1$.

Definition: The sequence $(s_n)_{n \in \mathbb{N}}$ is called absolutely convergent if $\sum_{i=1}^\infty | x_i |< \infty$.

Proposition: Assume that $(s_n)_{n \in \mathbb{N}}$ converges absolutely. Then $(s_n)_{n \in \mathbb{N}}$ converges and the sequence $(\tilde{s}_n)_{n \in \mathbb{N}}$ defined by $\tilde{s}_n = \sum_{i}^n x_{\pi (i)}$ converges to the same limit for every bijective map $\pi : \mathbb{N} \to \mathbb{N}$.

Proof: Let $m,n \in \mathbb{N}$. Set $N = \min {m,n}$ and $M = \max {m,n }$. Then $ \begin{equation} |s_m -s_n | \leq \sum_{i=N+1}^M | x_i | \leq \sum_{i=N+1}^\infty | x_i | \to 0 \quad (N \to \infty). \end{equation} $ So $(s_n)_{n \in \mathbb{N}}$ is Cauchy and therefore convergent since $X$ is complete. Let $s$ be limit. Now let $\pi : \mathbb{N} \to \mathbb{N}$ be bijective. Let $\varepsilon >0$. Since convergent sequences are Cauchy there exist an $N \in \mathbb{N}$ with $ \begin{equation} \sum_{i=n}^m | x_i| <\varepsilon \end{equation} $ for all $m\geq n \geq N$. Let $k \in \mathbb{N}$ be so that ${1, \dots, N} \subset { \pi(1), \pi(2) \dots, \pi(k) }$. Note that $k\geq N$. Then for any $n \in \mathbb{N}$ with $n>k$: $ \begin{equation} | \tilde{s}_n - s_n | \leq \sum_{ i = N+1 }^n | x_i| < \varepsilon. \end{equation} $ And so $ \begin{equation} | \tilde{s}_n -s | = | \tilde{s}_n - s_n + s_n -s | \leq | \tilde{s}_n -s_n | + | s_n -s | \to 0 \quad (n \to \infty). \end{equation} $

Proposition: Let $r = \limsup_{n \to \infty} |x_n|^{1/n} $. Then

if $r<1$, then $(s_n)_{n \in \mathbb{N}}$ converges absolutely,
if $r>1$, then $(s_n)_{n \in \mathbb{N}}$ is divergent.

This is the so called root test.

Proof: Assume $r<1$. By definition $r$ is the supremum of all cluster points of the sequence. Now choose $\varepsilon>0$ so small, so that $\beta := r+ \varepsilon <1$. Then $\beta$ is strictly larger than any cluster point of the sequence and therefore there exist an $N \in \mathbb{N}$ so that $|x_n| < \beta ^n$ for all $n \geq N$. Since $\sum_{k=1}^\infty \beta^k <\infty$ it follows for $M \in \mathbb{N}$ with $M>N$ that $ \begin{equation} \sum_{n=1}^M | x_n | \leq \sum_{n=1}^N | x_n | + \sum_{n= N+1}^M \beta^n \leq \sum_{n=1}^N | x_n | + \sum_{n= N+1}^\infty \beta^n <\infty. \end{equation} $ And therefore $(s_n)_{n \in \mathbb{N}}$ converges absolutely. If $r>1$, then there is a $\varepsilon>0$ with $r-\varepsilon >1$. Since $r$ is the supremum of cluster points, this means that there is a cluster point $c$ with $c \geq r- \varepsilon >1$. But this means that there is a subsequence $ (|x_{n_k}|^{1/n_k})_{k \in \mathbb{N}}$ that converges to $c$. But then $(x_{n})_{n \in \mathbb{N}}$ is not a null sequence contradicting the convergence of $(s_n)_{n \in \mathbb{N}}$.

Power Series In Banach Spaces

For $a \in \mathbb{C}$ and $r \in [0, \infty]$. Define $ \begin{equation} D(a,r) := { z \in \mathbb{C} : | z-a| <r}. \end{equation} $

Proposition: Let $X$ be a complex Banach space and $(c_n)_{n \in \mathbb{N}}$ a sequence in $X$. Then there exist a unique $R \in [0, \infty]$, called the radius of convergence, so that $ \begin{equation} \sum_{n = 0}^\infty c_n z^n \end{equation} $ converges absolutely for all $z \in D(0,R)$ and uniformly (in $z$) on $\bar D(0,r)$ for all $r<R$, but diverges for all $z \notin \bar D (0,R)$. Furthermore $ \begin{equation} R = \frac{1}{\limsup_{n \to \infty} |c_n|^{1/n}} = \liminf_{n \to \infty} |c_n|^{-1/n}. \end{equation} $ Proof: Define a sequence $(x_n)_{n \in \mathbb{N}}$ in $X$ by setting $x_n = c_n z^n$. Then $ \begin{equation} \limsup_{n \to \infty} |x_n |^{1/n} = |z| \limsup_{n \to \infty } |c_n|^{1/n} = \frac{|z|}{R} \end{equation} $ which is $<1$ if $z \in D(0,R)$ and $>1$ if $z \notin \bar D (0,R)$. So the part about the absolute convergence / divergence and the uniqueness of $R$ follow from the root test. It is left to show the assertion on the uniform convergence. Let $r<R$ and $\varepsilon >0$ so that $r+ \varepsilon< R$. Then by the definition of the $\liminf$ there is a $M \in \mathbb{N}$ so that $r + \varepsilon \leq |c_n|^{-1/n}$ for all $n \geq M$. Then for $N>M$: $$ \begin{align} &\sup_{z \in \bar D(0,r)} | \sum_{n=0}^\infty c_n z^n - \sum_{n = 0}^N c_n z^n | \\ &= \sup_{z \in \bar D(0,r)} | \sum_{n=N+1}^\infty c_n z^n | \leq \sup_{z \in \bar D(0,r)} \sum_{n=N+1}^\infty |z|^n |c_n| \\ &\leq \sum_{n=N+1}^\infty \bigg( \frac{r}{r+ \varepsilon} \bigg)^n \to 0 \quad (N \to \infty). \end{align} $$

Power Series In Banach Algebras

Let $A$ be a (complex, unital) Banach Algebra and $(c_n)_{n \in \mathbb{N}}$ a sequence of complex numbers. Furthermore let $ \begin{equation} R = \frac{1}{\limsup_{n \to \infty} |c_n|^{1/n}} = \liminf_{n \to \infty} |c_n|^{-1/n} \end{equation} $ and $r(a) = \limsup_{n \to \infty } |a^n |^{1/n} $ be the spectral radius of $a \in A$ (note that $r(a) \leq |a|$).

Proposition: The series $ \begin{equation} \sum_{n=0}^\infty c_n a^n \end{equation} $ converges absolutely for all $a \in A$ with $r (a) < R$. Furthermore the convergence is uniform (in $a$) on $\bar B(0,r)$ (ball with radius $r$ in $A$) for any $r<R$.

Proof: Define a sequence $(x_n)_{n \in \mathbb{N}}$ in $A$ by $x_n = c_n a^n$. Then $$ \begin{align} &\limsup_{n \to \infty } |x_n|^{1/n} \leq \limsup_{n \to \infty } |c_n|^{1/n} |a^n|^{1/n} \\ & \leq \big( \limsup_{n \to \infty } |c_n|^{1/n} \big) \big( \limsup_{n \to \infty}|a^n|^{1/n} \big) = \frac{r(a)}{R}<1. \end{align} $$ The absolute convergence follows from the above using the root test. The assertion about the uniform convergence follows almost exactly as in the preceding proposition with $a$ taking the place of $z$ and $B(0,a)$ taking the place of $D(0,r)$ and further using that $|a^n| \leq |a|^n \leq r^n$ for all $a$ inside of the ball.

Remark: It is important to note that there is no divergence assertion in the above proposition. This is because it is possible that $0 \neq a \in A $ but $a^2=0$. Therefore it is possible that the above series converges for $a \in A$ with $|a| >R$.