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