18.1 Derivatives and Remainders

Definition 18.1.1 (Derivatives and Remainders). Let $E, F$ be TVSs over $K \in \RC$ and $\ch(E; F), \calr(E; F) \subset F^{E}$ be vector subspaces, then $\mathcal{HR}= (\ch(E; F), \calr(E; F))$ is a pair of derivatives and remainders if

  1. For any $T \in \ch$, if there exists $V \in \cn_{E}(0)$ and $r \in \calr$ such that $T|_{V} = r|_{V}$, then $T = 0$.

Definition 18.1.2 ($\mathcal{HR}$-Differentiability). Let $E, F$ be TVSs over $K \in \RC$, $\mathcal{HR}$ be a pair of derivatives and remainders, $U \subset E$ be open, $f: U \to F$ be a function, and $x_{0} \in U$, then $f$ is $\mathcal{HR}$-differentiable at $x_{0}$ if there exists $V \in \cn_{E}(0)$, $T \in \ch$, and $r \in \calr$ such that

\[f(x_{0} + h) = f(x_{0}) + Th + r(h)\]

for all $h \in V$. In which case, $T = D_{\mathcal{HR}}f(x_{0})$ is the unique element of $\ch$ satisfying the above, known as the $\mathcal{HR}$-derivative of $f$ at $x_{0}$.

Proof. Let $S, T \in \ch$, $r, s \in \calr$, and $V \in \cn_{E}(0)$ such that

\[f(x_{0} + h) - f(x_{0}) = Sh + r(h) = Th + s(h)\]

for all $h \in V$, then $(S - T)(h) = (s - r)(h)$. By (T), $S - T = 0$. Hence $S = T$.$\square$

Proposition 18.1.3. Let $E, F$ be TVSs over $K \in \RC$, $\mathcal{HR}$ be a pair of derivatives and remainders, $U \subset E$ be open, $f, g: U \to F$ be functions, and $x_{0} \in U$. If $f, g$ are $\mathcal{HR}$-differentiable at $x_{0}$, then for any $\lambda \in K$, $\lambda f + g$ is $\mathcal{HR}$-differentiable at $x_{0}$, and

\[D_{\mathcal{HR}}(\lambda f + g)(x_{0}) = \lambda D_{\mathcal{HR}}f(x_{0}) + D_{\mathcal{HR}}g(x_{0})\]

Proof. Let $V \in \cn_{E}(0)$ and $r, s \in \calr$ such that

\begin{align*}f(x_{0} + h) - f(x_{0})&= D_{\mathcal{HR}}f(x_{0})h + r(h) \\ g(x_{0} + h) - f(x_{0})&= D_{\mathcal{HR}}g(x_{0})h + s(h)\end{align*}
then
\[(\lambda f + g)(x_{0}+h) - (\lambda f + g)(x_{0}) = \underbrace{[\lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h)\]
$\square$