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Prove that for the function $f\in C([0,1])$ $\forall x\in[0,1]:$ $\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}(-1)^kf(\frac {k}{n}) \to0$, $n\to \infty$. I think we can use a Bernstein polynomial, but how?

Ihor
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  • This looks like a Riemann-Lebesgue lemma kind of thing: what's a function which is equal to $(-1)^k f(k/n)$ at $x=k/n$? I can think of one: take $g_k=f h_k$ where $h_k$ is a cosine of amplitude $1$ and period $2/n$. So your function is a Bernstein polynomial for $g_k$. – Ian Jun 02 '15 at 16:52
  • Maybe. But I believe that this sum can be estimated in any way. – Ihor Jun 02 '15 at 17:05
  • Let $S_n$ be your sum and define $R_n = 2 \sum_{k=1}^{\lceil n/2 \rceil} {n \choose 2k-1} x^{2k-1} (1-x)^{n-2k+1} f((2k-1)/n)$ and $Q_n = \sum_{k=0}^n {n \choose k} x^k (1-x)^k f(k/n)$. (Here $R_n$ is supposed to be twice the sum of the terms with odd $k$ in $S_n$, please pardon me if I made an error in setting up the expression for it.) Then $S_n= Q_n - R_n$. That seems to suggest that if you can get a probabilistic representation of $R_n$ then you may be able to finish the proof by following http://en.wikipedia.org/wiki/Bernstein_polynomial#Proof – Ian Jun 02 '15 at 17:54
  • I tried use formula $\lim_{n\to \infty}{\sum_{k=0}^n \binom{n}{k}x^k(1-x)^{n-k}f(\frac {k}{n})= f(x)$, but couldn't do this. – Ihor Jun 02 '15 at 17:57
  • That isn't enough: you basically need to show that what I called $R_n$ also converges to $f(x)$. – Ian Jun 02 '15 at 17:58
  • Therein problem. – Ihor Jun 02 '15 at 18:11
  • Perhaps we can use Cantor's theorem? – Ihor Jun 02 '15 at 18:14
  • Which Cantor's theorem? Do you mean Heine-Cantor, i.e. that we can assume $f$ is uniformly continuous? – Ian Jun 02 '15 at 18:41
  • Yes, because $[0,1]$ is compact. – Ihor Jun 03 '15 at 08:44

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