Prove that $\{ f_n \}$ has a subsequence that converges uniformly in $[0,1]$ I have problems with exercise:
Let $f_n : [0,1] \longrightarrow{} \mathbb{R}$ Let R be a sequence of continuous functions such that $f_n$is Riemann integrable in $[0, 1]$ Supposethat for all $n \in \mathbb{N}$ we have $| {f_n}^{\prime}(t)  | \leq{} t ^{-1/2}$ for all $t$ and $\displaystyle\int_{0}^{1} f_n (t) dt =0 $. Prove that $\{ f_n \}$ has a subsequence that converges uniformly in $[0,1]$
My attempt:
$\left| F_n(x) - F_n(y) \right| = \left| \displaystyle\int_x^y f_n(t) \, dt  \right|  \le M|y-x|$. We have that $M$ is unifor bound of $f_n$.
How can I apply the Arzelá-Ascoli theorem?
 A: $\sqrt x$ is a continuous function on $[0,1]$ and hence it is uniformly continuous. Choose $\delta>0$ such that $|\sqrt y -\sqrt x| <\epsilon/2$ for $|x-y|<\delta$. Then $|f_n(x)-f_n(y)|=|\int_x^{y}f_n'(t) dt| \leq |\int_x^{y}t^{-1/2} dt|=2|\sqrt y -\sqrt x| <\epsilon$ for $|x-y|<\delta$.
Note also that $|f_n(x)| \geq |f_n(y)|-|\int_x^{y}f_n'(t) dt| >|f_n(y)|-2$ for all $x,y$. Can you show from this and the hypothesis $\int_0^{1} f_n(t)dt=0$ to that $(f_n)$ is uniformly bounded?
A: To show that the sequence $f_n$ is bounded, simply write
$$
f_n(t) = f_n(0) + \int\limits_0^t f_n'
$$
and notice that
$$
\left|\int\limits_0^t f_n'\right| \leq 2\sqrt{t} \leq 2.
$$
If the sequence $f_n(0)$ were not bounded, then it would possess a subsequence such that $f_{n_k}(0) > 2k$ and so $f_{n_k}(t) > 2k - 2 = 2(k-1)$ and this would violate the hypothesis that the integral of each $f_n$ is zero.
Again, by the fundamental theorem of calculus,
$$
|f_n(y) - f_n(x)| \leq 2 \left| \sqrt{y} - \sqrt{x} \right|,
$$
and since the square-root is a continuous function and $[0, 1]$ is compact, you may invoke the Heine-Cantor theorem to conclude that this function is uniformly continuous on $[0, 1]$ and conclude the exercise with Ascoli's theorem.
