Let $f_n:[0,1]\rightarrow \mathbb{R}$ be continuous functions such that: $|f'_n(x)|\leq \frac{1}{\sqrt x}$ and $\int_0^1f_n(x)dx =0$ for every $n\in N$. Prove that exists a subsequence of $(f_n)$ which converges uniformly.
I know that I must apply Arzelá-Ascoli theorem. The problem then comes to show that the sequence $(f_n)$ is uniformly bounded and that it is equicontinuous.
What I have noted/done so far:
Since $|f'_n(x)|\leq \frac{1}{\sqrt x}$, if $x,y\in [0,1],x\neq y$, then
$$|f_n(y)-f_n(x)|=|\int_x^yf'_n(t)dt |\leq\int_x^y|f'_n(t)|dt\leq\int_x^y\frac{1}{\sqrt t}dt\leq\int_0^1\frac{1}{\sqrt t}dt = K$$
but I can't use this to solve the problem. Any help? Thanks in advance
Observe that, for all $n\in\mathbb{N}$ and every $x,y\in [0,1]$, $$ |f_n(y) - f_n(x)| \leq \left| \int_x^y \frac{1}{\sqrt{s}}\, ds \right| = 2 \left| \sqrt{y} - \sqrt{x}\right| \leq 2 \sqrt{|y-x|}, $$ so that the sequence $(f_n)$ is equicontinuous.
On the other hand, the condition $\int_0^1 f_n = 0$ implies that the sequence is equibounded. Namely, from the mean value theorem, for every $n$ there exists a point $x_n\in [0,1]$ such that $f_n(x_n) = 0$, so that $$ |f_n(x)| = |f_n(x) - f_n(x_n)| \leq \left| \int_{x_n}^x \frac{1}{\sqrt{s}}\, ds\right| \leq 2, \qquad \forall x\in [0,1]. $$