How prove this limits $\lim_{n\to\infty}\int_{0}^{1}g(x)f'(nx)dx=0$ 
let $f'(x)$ be  continuation on $R$, and $f(x)=f(x+2\pi),\forall x\in R$, and such
  $|f'(x)|\le 1$,Now define $g(x)$,such for any $x_{1},x_{2}\in [0,1]$, we have
  $$|g(x_{1})-g(x_{2})|\le \sqrt{|x_{1}-x_{2}|}$$
show that
  $$I=\lim_{n\to\infty}\int_{0}^{1}g(x)f'(nx)dx=0$$

This problem is form analysis problem book excise (MIn hui Xie)
Maybe we can use Dominated convergence theorem,but I can't use this condition
$$|g(x_{1})-g(x_{2})|\le \sqrt{|x_{1}-x_{2}|}\Longrightarrow \left|\dfrac{g(x_{1})-g(x_{2})}{x_{1}-x_{2}}\right|\le \dfrac{1}{\sqrt{|x_{1}-x_{2}|}}$$
because $\dfrac{1}{\sqrt{|x_{1}-x_{2}|}}\in (0,+\infty),\forall x_{1},x_{2}\in [0,1]$,so $g$ is not Lipschitz continuity,Then I can't
 A: Let $[a] = \{ x : |x| \le |a| \}$ for any $a \in \mathbb{C}$.
$f'$ is continuous and periodic and $\int_{[t,t+2\pi]} f' = 0$ for any $t \in \mathbb{R}$, from the condition given.
$g$ is continuous, from the condition given.
[With $[0,1]$ being compact, the above conditions are all we actually need.]
Let $p,q \in \mathbb{R}$ such that $f' \in [p]$ on $\mathbb{R}$ and $g \in [q]$ on $[0,1]$
Also $g$ is uniformly continuous on $[0,1]$.
For any $ε > 0$:
  Let $a > 0$ such that $g(x+[a]) \subseteq g(x) + [ε]$ for any $x \in [0,1]$, by uniform continuity.
  For any $n \in \mathbb{R}$ such that $n \ge \max(\frac{2\pi}{a},\frac{2\pi}{ε})$:
    Let $c = \lfloor \frac{n}{2\pi} \rfloor$
    $\int_{[0,1]} g(x) f'(nx)\ dx$
    $ = \sum_{k=1}^c \int_{\frac{2\pi}{n}[k-1,k]} g(x) f'(nx)\ dx + \int_{[\frac{2\pi}{n}c,1]} g(x) f'(nx)\ dx$
    $ \in \sum_{k=1}^c \int_{\frac{2\pi}{n}[k-1,k]} \left( g(\frac{2\pi}{n}k) + [ε] \right) f'(nx)\ dx + (1-\frac{2\pi}{n}c) [pq]$ [because $\frac{2\pi}{n} < a$]
    $ = \sum_{k=1}^c \int_{\frac{2\pi}{n}[k-1,k]} [ε] f'(nx)\ dx + (1-\frac{2\pi}{n}c) [pq]$ [because $\int_{\frac{2\pi}{n}[k-1,k]} f'(nx)\ dx = 0$]
    $ \subseteq \sum_{k=1}^c \frac{2\pi}{n} [ε] [p] + \frac{2\pi}{n} [pq]$
    $ \subseteq [εp] + [εpq] = ([p]+[pq])ε$
Therefore $\int_{[0,1]} g(x) f'(nx)\ dx \to 0$ as $n \to \infty$
