Consider the series of function $\{ f_n : \mathbb R \to \mathbb R\}_{n=1}^\infty$ and suppose each $f_n$ is $C^1$ and $f_n(0)=0.$
Prove that if $f_n'(x)$ uniformly converges to $g(x)$ on arbitrary compact sets, then $f_n(x)$ uniformly converges to $\displaystyle\int_0^x g(t) dt$ on arbitrary compact sets.
I have to ideas (i)(ii) to solve this problem, but I 'm not sure whether (ii) is correct or not.
Solution (i)
Let $[a,b]\subset \mathbb R$ be an arbitrary compact set and let $\epsilon >0$.
I can choose $c>0$ s.t. $-c<a, b <c.$
Since $f' _n \to g$ uniformly on arbitrary compact sets, there exists $N\in \mathbb N$ s.t. $$n \geqq N \Rightarrow |f_n'(x)-g(x)|<\epsilon \mathrm{\ for \ } \forall x\in [a,b].$$ When $n \geqq N,$ \begin{align} \left| f_n (x)-\int_0^x g(t) dt \right| &=\left| \int_0^x \big(f_n'(t)-g(t)\big)dt\right|\\ &\leqq \left| \int_0^x |f_n'(t)-g(t)| dt\right|\\ &\leqq \left| \int_0^x \epsilon dt\right| \\ &=|x| \epsilon \\ &<c\epsilon \mathrm{\ for\ } \forall x \in [a,b]. \end{align} Thus $f_n(x)$ uniformly converges to $\displaystyle\int_0^x g(t) dt$ on $[a,b]$.
Solution(ii)
Let $[a,b]\subset \mathbb R$.
I have $f_n(x)=\displaystyle\int_0^x f_n'(t) dt+ f_n (0)=\int_0^x f_n'(t) dt$ on $[a,b]$.
Letting $n\to \infty,$ $\displaystyle\lim_{n\to \infty}f_n(x)=\lim_{n\to \infty} \int_0^x f_n'(t) dt=\int_0^x \lim_{n\to \infty}f_n'(t) dt=\int_0^x g(t) dt$ on $[a,b]$.
Thus $f_n(x) \to \displaystyle\int_0^x g(t) dt$ on $[a,b]$.
In (ii), I think that I only proved $$\forall x\in [a,b]\ ; \ f_n(x) \to \int_0^x g(t) dt.$$ This seems not to be uniform convergence but to be pointwise convergence.
Isn't (ii) good ? And does (i) make sense ?