Let $f:[0,1]\to \mathbb{R}$ is continuous Lipchitz in $[0,1] $with constant $L$ show that for $n\geq 1$ we have that $\forall x\in [0,1]$the inequality $|f(x)-B_n(f)|<\frac{L}{2 \sqrt{n}}$ holds where $B_n(f)$ is the Bernstein polynomial.
Attempt
Let us consider $p_{n,k}(x)=C_{n}^{k}x^k(1-x)^{n-k}$ and $p_n(x)=\sum_{k=0}^{n}f(\frac{k}{n})p_{n,k}(x)$.
Now notice that $f(x)=\sum_{k=0}^nf(x)p_{n,k}(x)$ and $p_n(x)=B_n(f)$ therefore our approach is $$|f(x)-p_n(x)|=\left|\sum_{k=0}^{n}\left(f(x)-f\left( \frac{k}{n}\right)\right)p_{n,k} \right| \leq \sum_{k=0}^n \left| f(x)-f \left( \frac{k}{n}\right)\right|$$ and for the condition of Lipchitz we have the following $$ |f(x)-p_n(x)| \leq L\sum_{k=0}^{n}\left| x-\frac{k}{n}\right|$$ I wish that $$\sum_{k=0}^{n}\left| x-\frac{k}{n}\right|< \frac{1}{2\sqrt{n}}$$ but I can´t bound this expression, any help, comment or hint was useful
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