Assume that $\{f_n\}$ is a sequence of complex-valued functions defined on $[0, 1]$ such that $f_n\rightarrow f$ uniformly, $f'_n$ is continuous for all $n$, and $f'_n\rightarrow g$ uniformly.
I want to prove $g$ is actually the derivative of $f$. When each $f_n$ is a real-valued function, the statement can be proved by applying the Mean Value Theorem. But I have no idea to deal with the case that each $f_n$ is complex-valued. Does anyone have any idea for proving ?
$f_n(x)=f_n(0)+\int_0^{x} f_n'(t)dt$. Letting $n \to \infty$ we get $f(x)=f(0)+\int_0^{x} g(t) dt$ (by uniform convergence). Note also that $g$ is continuous by uniform convergence. This implies that $f$ is differentiable and $f'(x)=g(x)$. [There is no difference between the real case and the complex case in this argument].