Notes:Richard Sharp question

Problem

I wish to show:

• [math]2\Vert f\Vert_\infty \sum_{k:(k-nx)^2\ge n^2\delta^2}{}^nC_kx^k(1-x)^{n-k}\le 2\Vert f\Vert_\infty \frac{1}{n^2\delta^2}\sum_{k\eq 0}^n(k-nx)^2\ {}^nC_kx^k(1-x)^{n-k}[/math]

I am completely happy with the LHS, I understand the RHS is probably best written as:

• [math]2\Vert f\Vert_\infty \sum_{k\eq 0}^n\frac{(k-nx)^2}{n^2\delta^2}\ {}^nC_kx^k(1-x)^{n-k} [/math]

By hypothesis, for the summation on the LHS, [ilmath]k[/ilmath] is such that:

• [ilmath](k-nx)^2\ge n^2\delta^2\implies\frac{(k-nx)^2}{n^2\delta^2}\ge 1[/ilmath]

I cannot see how to get to the RHS

Supporting equations

Note that:

1. [math]\sum^n_{k\eq 0}{}^nC_kx^k(1-x)^{n-x}\eq 1[/math]
2. [math]\sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-1} \eq nx(1-x)[/math]
3. [math]\sum^n_{k\eq 0}k\ {}^nC_kx^k(1-x)^{n-1}\eq nx[/math] - probably not important

Possible solution

Starting from:

• [math]\sum_{\begin{array}{c}0\le k\le n\\k:\vert x-\frac{k}{n}\vert\ge\delta\end{array} } \vert f(x)-f(\tfrac{k}{n})\vert\ {}^nC_kx^k(1-x)^{n-k}[/math]

  [math]\le 2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:\vert x-\frac{k}{n}\vert\ge\delta\end{array} }{}^nC_kx^k(1-x)^{n-k}[/math] - by definition of the [ilmath]\Vert\cdot\Vert_\infty[/ilmath] norm

  [math]\eq2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:(k-nx)^2\ge \delta^2n^2\end{array} }{}^nC_kx^k(1-x)^{n-k} [/math] - by faffing about with absolute values

  [math]\le 2\Vert f\Vert_\infty \sum_{\begin{array}{c}0\le k\le n\\k:(k-nx)^2\ge \delta^2n^2\end{array} }\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}[/math] - as for such [ilmath]k[/ilmath] we see [math]\frac{(k-nx)^2}{\delta^2n^2}\ge 1[/math]

  [math]\le 2\Vert f\Vert_\infty \sum^n_{k\eq 0}\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}[/math] - because for the [ilmath]k[/ilmath] terms added we see [math]\frac{(k-nx)^2}{\delta^2n^2}\ {}^nC_kx^k(1-x)^{n-k}\ge 0[/math] - so it can only increase the summation's value

  [math]\eq \frac{2\Vert f\Vert_\infty}{\delta^2n^2} \sum^n_{k\eq 0}(k-nx)^2\ {}^nC_kx^k(1-x)^{n-k}[/math] - factoring

  [math]\eq \frac{2\Vert f\Vert_\infty}{\delta^2n^2}\cdot nx(1-x)[/math] (using result 2)

  [math]\le \frac{1}{4}n\frac{2\Vert f\Vert_\infty}{\delta^2n^2}[/math] - as [ilmath]x(1-x)\le\frac{1}{4} [/ilmath]

  [math]\eq \frac{\Vert f\Vert_\infty}{2n\delta^2}[/math]