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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 $$ \lim_{n \rightarrow \infty} g_n(x) = \frac{d}{dx}f(x) $$

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  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 $g_n:=f_n'$.


As a fun fact, we have the following wonderful theorem in complex analysis (I know your question is mainly focused on the real case, but I thought it would be nice to contrast with the nice behavior in the complex case)

  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 $U$.
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