I am trying to solve the following question:
Let $f_n$ be a sequence of continuous differentiable functions on $[a, b]$. Suppose that $f_n' \to g$ uniformly on $[a, b]$, and $(f_n(p))$ converges for some $p \in [a, b]$. Then there is a continuously differentiable function $f$ such that $f_n \to f$ uniformly on $[a, b]$ and $f'=g$.
I am able to make significant progress, but I am unable to complete the solution. Could you please help me finish it? Thank you.
My Work: Define $F:[a, b] \to \mathbb{R}$ by $F_n(x) = \int_p^x f_n'(t) dt = f_n(x) - f_n(p)$ $\hspace{1cm}$(by FTC 1).
By uniform continuity, $\int_p^x f_n'(t) dt \to \int_p^x g(t) dt.$ Therefore $\lim_n F_n(x)$ exists, and since $\lim f_n(p)$ exists, $\lim f_n(x)$ exists.
From here, I'm a little stuck.
Define, for each $x\in[a,b]$,$$f(x)=\lim_{n\to\infty}f_n(p)+\int_p^xg(t)\,\mathrm dt.$$Then $f$ is differentiable, $f'=g$, and $\lim_{n\to\infty}f_n(p)=f(p)$.
For each $x\in[a,b]$,\begin{align}\left|f(x)-f_n(x)\right|&=\left|\lim_{n\to\infty}f_n(p)+\int_p^xg(t)\,\mathrm dt-\left(f_n(p)+\int_p^xf_n'(t)\,\mathrm dt\right)\right|\\&=\left|\left(\left(\lim_{n\to\infty}f_n(p)\right)-f_n(p)\right)+\int_p^xg(t)-f_n'(t)\,\mathrm dt\right|\\&\leqslant\left|\left(\left(\lim_{n\to\infty}f_n(p)\right)-f_n(p)\right)\right|+\int_p^x\bigl|g(t)-f_n'(t)\bigr|\,\mathrm dt.\end{align}So, for each $\varepsilon>0$, take $N\in\Bbb N$ such that, if $n\geqslant N$,$$\left|\left(\left(\lim_{n\to\infty}f_n(p)\right)-f_n(p)\right)\right|<\frac\varepsilon2\text{ and }(\forall t\in[a,b]):\bigl|g(t)-f_n'(t)\bigr|<\frac\varepsilon{2(b-a)}.$$