Derivative Normed Vector Spaces

Question

Could someone provide an explanation for Proposition 3.9 below. The author states "The proposition is a simple application of Theorem 20.6" where Theorem 20.6 is the chain rule. However, I am uncertain as to how the chain rule is being applied here.

Answer 1

Hope this isn't too late... the solution is really just a chain rule application and basic linear algebra. Let $\iota_k: F_k \to F$ be defined by $\iota_k(\xi) = (0, \dots, \xi, \dots, 0)$, with $\xi$ in the $k^{th}$ slot. Also, define $\pi_k: F \to F_k$ by $\pi_k(\xi_1, \dots, \xi_m) = \xi_k$.

Here's some terminology: $\iota_k$ is often called the canonical injection of $F_k$ into $F$, and $\pi_k$ is called the canonical projection of $F$ onto $F_k$ (you can check that $\iota_k$ is injective and $\pi_k$ is surjective so these really are suitable names). These functions are continuous linear transformations (continuity is guranteed if you're working with finite-dimensional spaces).

Recall that if $V$ and $W$ are normed vector spaces and $T: V \to W$ is a continuous linear transformation, then for every $a \in V$, $DT(a)$ exists and equals $T$. Now, onto your question. Note that the statement $f = (f_1, \dots, f_m)$ is a convenient but loose way of saying \begin{align} f = \sum_{k=1}^m \iota_k \circ f_k. \tag{*} \end{align}

Suppose first that each $f_k$ is differentiable at $a$. Since each $\iota_k$ is a continuous linear map, it is differentiable everywhere; hence the composite map $\iota_k \circ f_k$ is also differentiable at $a$ by the chain rule. By (*), we have that $f$ is a sum of functions which are differentiable at $a$; hence is differentiable at $a$ as well, and its derivative is given by \begin{align} Df(a) &= \sum_{k=1}^m D(\iota_k \circ f_k)(a) \\ &= \sum_{k=1}^m D \iota_k(f_k(a)) \circ D f_k(a) \\ &= \sum_{k=1}^m \iota_k \circ Df_k(a). \end{align} This assertion is often written as $Df(a) = \left( Df_1(a), \dots, Df_m(a) \right)$.

Conversely, suppose that $f$ is differentiable at $a$. Let $k \in \{1, \dots, m\}$ be arbitrary; we want to show $f_k$ is differentiable at $a$. To do this, note that $\pi_k$ is a continuous linear map and hence everywhere differentiable. Since we have the identity $f_k = \pi_k \circ f$, it follows by the chain rule that $f_k$ is differentiable at $a$. This completes the proof.