Obtaining a bound on the Bernstein approximation of Lipschitz functions I encountered the following excercise in a book:

Exercise: Given a Lipschitz continuous function $f$ on $[0,1]$, with Lipschitz constant $c$. 
Show that $|B_{n,f}(p) - f(p)|\le
 \frac{c}{2\sqrt{n}}$ for all $p\in[0,1]$, where $B_{n,f}$ is the
  Bernstein polynomial of order $n$ constructed from $f$.

The exercise also gives the following hint: 

$B_{n,f}(p) = \mathbf{E}[f(X/n)]$, where $X$ is a random variable with
  binomial distribution for $n$ trials with success probability $p$.

I got this far:

$$ \begin{align} \left\lvert B_{n,f}(p) - f(p)\right\rvert 
 = \left\lvert \mathbf{E}\left[f\left(\frac{X}{n}\right)\right] - f(p)\right\rvert  &= \left\lvert
 \mathbf{E}\left[f\left(\frac{X}{n}\right) - f(p)\right]\right\rvert
 \label{1}  \\ &\le  \mathbf{E}\left[\left\lvert
 f\left(\frac{X}{n}\right) - f(p)\right\rvert\right] & (|\cdot|\,\text{
 is convex}) \\ &\le  \mathbf{E}\left[c\left\lvert \frac{X}{n} -
 p\right\rvert\right] & (\text{Lipschitz continuity}) \\ &=c\cdot
 \mathbf{E}\left[\left\lvert \frac{X}{n} - p\right\rvert\right]
 \end{align} $$

So it remains to show that $\mathbf{E}\left[\left\lvert \frac{X}{n} - p\right\rvert\right] \le \frac{1}{2\sqrt{n}}$. I could not figure out how to proceed from this point. Any help will be appreciated.

The variance of $X$ is $np(1-p) \le n/4$. And since the variance is $E[|X-np|^2]$, it follows that 
$$E[|X/n -p|^2] =\frac1{n^2}E[|X-np|^2] \le \frac{1}{4n}$$
Then apply the Cauchy-Schwarz inequality.
