$f_n(x)$ is a sequence of continuous differentiable functions on $[0,1]$ such that $f_n(0)=0$, $|f^{\prime}_n(x)|\le x^{-1/2}$ a.e., and there exists $h$ such that $\lim_{n\to\infty}f^{\prime}_n(x)=h(x)$ for all $x$. Prove that there is an absolutely continuous function $f$ such that $f_n(x)$ converges to $f$ uniformly on $[0,1]$.
I think $f=\int_0^xh(t)dt$, and we know that $h$ is integrable by given conditions. Also, $f$ is absolutely continuous by fundamental theorem of calculus. $f_n(x)=\int_0^xf_n^{\prime}(t)dt$ as $f_n$ is continuous differentiable and $f_n(0)=0$. Our goal is to show that $\sup_{x\in[0,1]}|\int_0^xf_n^{\prime}(t)-h(t)dt|\to0$ as $n\to\infty$.
Thanks for any hint.
For each $x\in [0,1]$,
$$|f_n(x) - f(x)| = \left|\int_0^x f_n'(t) - h(t) dt\right| \le \int_0^x |f_n' - h| \le \int_0^1 |f_n' - h|.$$ $$\Rightarrow \sup_{x\in [0,1]} |f_n(x) - f(x)|\le \int_0^1 |f_n' - h|.$$ The Lebesgue dominated convergence theorem applied to the sequence $\{f_n'\}$ (note $|f_n'(x)| \le x^{-1/2}$ and $x^{-1/2}$ is integrable) implies that
$$\int_0^1 |f_n' - h| \to 0$$
as $n\to \infty$. Thus
$$\sup_{x\in [0,1]} |f_n(x) - f(x)| \to 0$$
as $n\to \infty$ and so the convergence is uniform.