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.