Theorem 30.7.3 (Termwise Differentiation).label Let $E$ be a normed space over $K \in \RC$, $F$ be a complete locally convex 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)

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

  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): Let $\rho: F \to [0, \infty)$ be a continuous seminorm. For each $n \in \natz$ and $T \in L^{n}(E; F)$, let

\begin{align*}[T]_{L^n(E; F), \rho}&= \sup_{x \in B_E(0, 1)^n}\rho(Tx) \\ [T]_{L^n(E; L(E; F)), \rho}&= \sup_{x \in B_E(0, 1)^n}[Tx]_{L(E; F), \rho}\end{align*}

and

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

then $[S_{n}]_{L^n(E; L(E; F)), \rho}\le (n+1)[T_{n+1}]_{L^{n+1}(E; F), \rho}$. Since $(n+1)^{1/n}$ is convergent and $\{[T_{n+1}]_{L^{n+1}(E; F), \rho}\}_{1}^{\infty}$ is bounded,

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

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

(2): By the Theorem 30.4.7, 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 30.7.2, the proposed series converges uniformly on $B(a, r)$ for each $0 < r < R$. Thus by Theorem 30.2.8, $f$ 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 $D^{n}f$.$\square$

Post a Comment

Name:Email:
Please enter the tag of the current page (QM) to post the comment.
Tag: