Uniform Convergence of Sequence of Functions and their derivatives

Question

The question: Let $I$ be a set which $[\alpha, \beta] \subset I$. Suppose that $f_{n}:I \rightarrow \mathbb{R}$ is a sequence of continuously differenciable functions that has the following properties:

i. The sequence, $f_n$ converges point wise on $I$ to $f$

ii.The derived sequence $f'_{n}$ converges uniformly on $I$ to the function $g$

iii. $f: I \rightarrow \mathbb{R}$ is continuously differentiable and $f'(x) = g(x)$ for all x $\in I$.

Show the sequence $f_{n}: [\alpha, \beta] \rightarrow \mathbb{R}$ converges uniformly to $f: [\alpha, \beta] \rightarrow \mathbb{R}$

I am not sure what to do here. Do I have to use the fundamental theorem of calculus eventually? Can someone give me some hints on how to solve this problem please?

Thank you very much!! I really appreciate it!

Answer 1

Yes, by the fundamental theorem of calculus for continuously differentiable functions, $f_n(x)=f_n(\alpha)+\int_\alpha^x f_n'(t)\,dt$ and $f(x)=f(\alpha)+\int_\alpha^x g(t)\,dt$ for each $x\in[\alpha,\beta],$ so

$\begin{align}|f_n(x)-f(x)|&=\left|f_n(\alpha)-f(\alpha)+\int_\alpha^x f_n'(t)-g(t)\,dt\right|\\ &\leq |f_n(\alpha)-f(\alpha)|+\left|\int_\alpha^x f_n'(t)-g(t)\,dt\right|\\ &\leq |f_n(\alpha)-f(\alpha)|+(\beta-\alpha)\max_{t\in[\alpha,\beta]}|f_n'(t)-g(t)|. \end{align}$

Everything in the last line goes to $0$ as $n\to\infty$, with no dependence on $x$.