For $f\in C[0,1]$ show that for all $x\in[0,1]$ $\sum_{k=0}^nC_n^k x^k (1-x)^{n-x}(-1)^k f(\frac{k}{n})\to 0$ when $n\to \infty$
I don´t have a idea of how start with the proof, for one hand since our sum is the Bernstein polynomial with a extra $(-1)^k$ I think that must be $0$ since the Bernstein polynomial is a approach of our $f$ and therefore when we take all the terms of the approximation the error must be $0$.
For the other hand I need how fix the problem of the extra term $(-1)^k$ I think that we Can use the absolute value of our sum and find that these limit are $0$ for conclude that our original sum tends to $0$ when $n\to \infty$.
And basically I don´t know how I should continue with a formal proof of this.
I belive that I should calculate $$\lim_{n\to \infty}\left( \sum_{k=0}^nC_n^k x^k (1-x)^{n-x}(-1)^k f\left(\frac{k}{n}\right)\right)$$ or find $N\in \mathbb{N}$ such that $$\left|\sum_{k=0}^nC_n^k x^k (1-x)^{n-x}(-1)^k f\left(\frac{k}{n}\right)\right|<\varepsilon$$
Attempt Consider only the index $k\equiv 0\pmod 2$ then we should have $$\left| \sum_{k=0}^nC_n^{2k} x^{2k} (1-x)^{n-x} f\left(\frac{2k}{n}\right)\right|\leq \sum_{k=0}^{n}\left|f\left( \frac{2k}{n}\right) \right|C_{n}^{2k}x^{2k}(1-x)^{n-2k}$$ since the right sumand is at most $1$ then $$\left| \sum_{k=0}^nC_n^{2k} x^{2k} (1-x)^{n-x} f\left(\frac{2k}{n}\right)\right|\leq \sum_{k=0}^{n}\left|f\left( \frac{2k}{n}\right) \right|$$ but when $n\to \infty$ $\frac{2k}{n}\to 0$ and hence since $f(0)=p_n(1)=0$ $$\left| \sum_{k=0}^nC_n^{2k} x^{2k} (1-x)^{n-x} f\left(\frac{2k}{n}\right)\right|\leq \sum_{k=0}^{n}\left|f\left( \frac{2k}{n}\right) \right|=0$$ Finally the case when $k\equiv 1 \pmod{2}$ is analogous.