Let $f:[0,1] \to \mathbb{R}$ be a continuous function. Prove that $$\lim_{n \to \infty} \frac{1}{2^n}\sum_{k=0}^n(-1)^k {n\choose k}f\left(\frac{k}{n} \right)=0$$
I know that $f$ is uniformly continuous and I tried to get some inequalities for the terms $f\left(\frac{k}{n} \right)$. For all $\epsilon>0$, we have $|f(0)-f\left(\frac{1}{n} \right)|<\epsilon, \dots, |f\left(\frac{n-1}{n} \right)-f(1)|<\epsilon$ when $n$ is large enough. Then I tried to apply these for the sum in the statement and squeeze it to $0$, but I only got to prove it is less than $\frac{n}{2},$ which doesn't go to $0$.
This sum is $$2^{-n}\sum_{k=0}^n(-1)^k\left\{\binom{n-1}{k-1}+\binom{n-1}k\right\}f\left(\dfrac{k}{n}\right) =\\=2^{-n}\sum_{k=0}^{n-1}(-1)^k\binom{n-1}k\left\{f\left(\dfrac{k}{n}\right)-f\left(\dfrac{k+1}{n}\right)\right\}$$ By uniform continuity, for any $\epsilon$, when $n$ is large enough then $|f(k/n)-f((k+1)/n)|<\epsilon$ for all $k$. The whole sum is bounded in absolute value by $$\epsilon 2^{-n}\sum_{k=0}^{n-1}\binom{n-1}k=\frac{\epsilon}2.$$