# Interchange of limit and derivative [closed]

Suppose we have two sequences of functions, $$f_n(x)$$,$$g_n(x)$$., where the following relation holds $$\forall n, \hspace{5mm} \frac{d}{dx}f_n(x) = g_n(x)$$ Also $$\lim_{n \rightarrow \infty} f_n(x) = f(x)$$ We know that $$f_n$$ and $$f$$ are continuous functions. Under what conditions we are allowed to say $$\lim_{n \rightarrow \infty} g_n(x) = \frac{d}{dx}f(x)$$

1. A very simple sufficient condition is to assume each $$g_n$$ is Riemann-integrable and converges uniformly to some continuous function. In this case a proof can be given using integration and the Fundamental theorem of calculus.
2. A slightly weaker assumption, which requires a more careful argument is the following theorem (see Baby Rudin's book, theorem 7.17): Suppose $$f_n$$ is a sequence of differentiable functions on $$[a,b]$$, and which converges at some point $$x_0\in[a,b]$$. If $$g_n:=f_n'$$ converges uniformly on $$[a,b]$$ then $$f_n\to f$$ uniformly for some $$f$$ which is differentiable and for all $$x\in [a,b]$$, $$f'(x)=\lim\limits_{n\to\infty}g_n(x)$$.
Notice that in (2), we are no longer assume that $$f_n\to f$$, but that is part of our conclusion; we only assume convergence at one point. Also, we do not assume Riemann-integrability of the derivatives $$g_n:=f_n'$$.
1. Suppose $$U\subset\Bbb{C}$$ is open, and $$\{f_n\}_{n=1}^{\infty}$$ is a sequence of holomorphic functions on $$U$$ such that $$f_n\to f$$ for some function $$f$$ on $$U$$, such that the convergence is uniform on compact subsets of $$U$$. Then, $$f$$ is holomorphic and we also have that $$f_n'\to f'$$ uniformly on compact subsets of $$U$$.