> [!theorem] > > Let $U \subset \cx$ be open, and $\seq{f_n} \subset C^1$ with each $f_n: U \to \cy$. If $f_n \to f$ [[Pointwise Convergence|pointwise]] and $Df_n \to g$ converges [[Uniform Convergence|uniformly]], then $f$ is differentiable with $Df = g$. > > *Proof*. Let $x \in U$ and $V \subset U$ be an open neighbourhood of $x$. For any $m, n \in \nat$, by the mean value theorem, > $ > \begin{align*} > &\abs{f_n(x + h) - f_m(x + h) - (f_n(x) - f_m(x))} \\ > &\le \abs{h} \sup_{y \in V}\abs{Df_n(y) - Df_m(y)} > \end{align*} > $ > Let $\varepsilon > 0$, then there exists $N \in \nat$ such that > $ > \norm{Df_m - Df_n}_u, \norm{Df_n - g}_u < \varepsilon \quad \forall m, n \ge N > $ > and sending $m \to \infty$ yields > $ > \begin{align*} > &\abs{f_n(x + h) - f(x + h) - (f_n(x) - f(x))} \\ > &\le \limv{m}\abs{h} \sup_{y \in V}\abs{Df_n(y) - Df_m(y)} \\ > &\le \abs{h} \cdot \varepsilon > \end{align*} > $ > for all $n \ge N$ thanks to uniform convergence. > > By the mean value theorem again, > $ > \abs{f_n(x + h) - f_n(x) - Df_n(x)h} \le \abs{h}\sup_{x + h \in V}\abs{Df(x + h) - Df(x)} > $ > Since each $f_n \in C^1$, there exists $\delta > 0$ such that > $ > \abs{Df(x + h) - Df(x)} < \varepsilon \quad \forall h: \abs{h} < \delta > $ > Combining the two above inequalities and $\norm{Df_n - g}_u < \varepsilon$, we have > $ > \begin{align*} > &\abs{f(x + h) - f(x) - g(x)h} \\ > &\le \abs{f(x + h) - f(x) - Df_n(x)h} + \abs{h} \cdot \varepsilon \\ > &\le \abs{f(x + h) - f(x) - (f_n(x + h) - f_n(x))} + 2\abs{h} \cdot \varepsilon \\ > &\le 3\abs{h} \cdot \varepsilon > \end{align*} > $ Let $U \subset \real$