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 $S = T$.$\square$