26.6 Power Series

Definition 26.6.1 (Power Series).label Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $\bracsn{T_n}_{0}^{\infty}$ with $T_{n} \in L^{n}(E; F)$ for each $n \in \natz$, and $a \in E$, then the power series of $\bracsn{T_n}_{0}^{\infty}$ about $a$ is the function

\[f(x) = \sum_{n = 0}^{\infty} T_{n}(x - a)^{(n)}\]

defined on points on which the series converges.

Definition 26.6.2 (Radius of Convergence).label Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, and $\sum_{n = 0}^{\infty} T_{n}(x - a)^{(n)}$ be a power series about $a \in E$, then $R \in [0, \infty]$ be defined by^[1]

\[\frac{1}{R}= \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}\]

is the radius of convergence of $\sum_{n = 0}^{\infty} T_{n}(x - a)^{(n)}$. For each $0 < r < R$, the series converges uniformly and absolutely on $B_{E}(a, r)$.

Proof. For all $x \in B_{E}(a, r)$,

\[\sum_{n = 0}^{\infty} \normn{T_n(x - a)^{(n)}}_{F} \le \sum_{n \in \natz}\norm{T_n}_{L^n(E; F)}\norm{x - a}_{E}^{n} \le \sum_{n \in \natz}\norm{T_n}_{L^n(E; F)}r^{n}\]

For any $s \in (r, R)$, there exists $N \in \natp$ such that $\norm{T_n}_{L^n(E; F)}^{1/n}\le 1/s$ for all $n \ge N$. In which case,

\[\sum_{n = 0}^{\infty} \norm{T_n}_{L^n(E; F)}r^{n} \le \sum_{n = 0}^{N} \norm{T_n}_{L^n(E; F)}r^{n} + \sum_{n \ge N}\frac{r^{n}}{s^{n}}< \infty\]

As this estimate holds uniformly on $B_{E}(a, r)$, the series converges uniformly and absolutely on $B_{E}(a, r)$.$\square$

Remark 26.6.1.label In Definition 26.6.2, the radius of convergence appears to be an arbitrary lower bound on the domain of convergence. However, in the more specialised case of power series from $\complex$ to $\complex$ or in a Banach algebra, $R$ is the largest constant such that the series converges uniformly and absolutely on all $B(0, r)$ where $0 < r < R$. The lack of this ”maximum” claim is why the above statement is a definition.

Theorem 26.6.3 (Termwise Differentiation).label Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $f(x) = \sum_{n = 0}^{\infty} T_{n}(x - a)^{(n)}$ a power series about $a \in E$, and $R$ be its radius of convergence, then

(1)
$f \in C^{\infty}(B(a, R); F)$.
(2)
For each $x \in B(a, R)$ and $h \in E$,
\[Df(x)(h) = \sum_{n = 0}^{\infty} \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})\]
(3)
The radius of convergence of the above series is at least $R$.

Proof. (3): For each $n \in \natz$, let

\[S_{n}(x_{1}, \cdots, x_{n})(h) = \sum_{k = 1}^{n+1}T_{n+1}(((x_{1}, \cdots, x_{n}), h)^{\bracs{k}})\]

then $\norm{S_n}_{L^n(E; L(E; F))}\le (n+1)\norm{T_{n+1}}_{L^{n+1}(E; F)}$. Since $(n+1)^{1/n}$ is convergent and $\{||T_{n}||_{L^n(E; F)}^{1/n}\}$ is bounded,

\[\limsup_{n \to \infty}\norm{S_n}_{L^n(E; L(E; F))}^{1/n}\le \limsup_{n \to \infty}(n+1)^{1/n}\norm{T_{n+1}}_{L^{n+1}(E; F)}^{1/n}= \frac{1}{R}\]

so the radius of convergence of the proposed series is at least $R$.

(2): By the Theorem 26.4.6, for each $N \in \natp$,

\[D\braks{\sum_{n = 0}^N T_n(x - a)^{(n)}}(h) = \sum_{n = 0}^{N} \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})\]

By Definition 26.6.2, the proposed series converges uniformly on $B(a, r)$ for each $0 < r < R$. Thus by Theorem 26.2.8, $f$ is differentiable on $B(a, R)$ with

\[Df(x)(h) = \sum_{n = 0}^{\infty} S_{n}(x - a)(h) = \sum_{n = 0}^{\infty} \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})\]

(1): By (2), (3) applied inductively to $D^{n}f$.$\square$

Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.keyboard_return

Jerry's Digital Garden

26.6 Power Series

Direct References

Jerry's Digital Garden

Direct References