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. (1)

    $f \in C^{\infty}(B(a, R); F)$.

  2. (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. (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 $D^{n}f$.$\square$