Define $P_{n+1}(x) = P_{n}(x)+\frac{x^{2} - P_{n}(x)^{2}}{2}$. Show $P_{n}$ converges uniformly on $[-1, 1]$ Let $P_{0} = 0$ and for $n = 0,1,2,...,$ define $P_{n+1}(x) = P_{n}(x)+\frac{x^{2} - P_{n}(x)^{2}}{2}$. Show that $P_{n}$ converges uniformly on $[-1, 1]$ and find its limit.
For proving that two claim we have to prove
Claim 1: 
$0 \leq P_{n}(x) \leq P_{n + 1}(x) \leq \vert x\vert \leq 1$ for $x\in [-1, 1]$
, proved easily.
Claim2: $0 \leq \vert x \vert - P_{n}(x) < \frac{2}{n + 1}$. This is difficult for me. I get a help form web note where they first prove 
$0 \leq \vert x \vert - P_{n}(x) \leq \vert x \vert \left(1 - \frac{\vert x \vert}{2}\right)^{n}$ and I understand the rest of the proof.
My question is where they get intention to prove $0 \leq \vert x \vert - P_{n}(x) \leq \vert x \vert \left(1 - \frac{\vert x \vert}{2}\right)^{n}$ ?
 A: Do it by summing: 
$P_{n+1}(x) = P_{n}(x)+\frac{x^{2} - P_{n}(x)^{2}}{2}$ 
so 
$\sum^n_{k=1}P_{k+1}(x)- P_{k}(x)=\sum^n_{k=1}\frac{x^{2} - P_{k}(x)^{2}}{2}\ge\sum^n_{k=1}\frac{|x|^{2} - P_{n}(x)^{2}}{2}$. 
Then, if $x\neq 0,$
$P_{n+1}(x)-P_1(x)\ge\frac{n|x|^2}{2}-\frac{n}{2}P_n(x)^2\Rightarrow$
$ |x|-P_n(x)\le\frac{2}{n}\cdot\frac{P_{n+1}(x)-P_1(x)}{|x|+P_n(x)}\le \frac{2}{n}\cdot\frac{|x|-P_1(x)}{|x|+P_n(x)}\le \frac{2}{n}\cdot\frac{|x|-P_1(x)}{|x|+P_1(x)}\le \frac{2}{n}.$
If $x=0,$ then $P_n(x)=0$ for all integers $n$ and the result is trivial: $0\le \frac{2}{n}$.
A: It appears that you already read a proof and would like to know the intention of certain step. For simplicity, let $y_n=P_n(x)\geq 0$ (assuming that claim 1 has been proven), one has $$y_n=y_{n-1}+\frac{x^2-y_{n-1}^2}2,n\geq 1.$$ Since $y_n$ is increasing and bounded above by $|x|$, it is natural to check if the limit of $y_n$ is $|x|$, so one may proceed to estimate $$|x|-y_n=(|x|-y_{n-1})\left(1-\frac{|x|+y_{n-1}}2\right)\leq (|x|-y_{n-1})(1-|x|/2),$$ hence by recursion $$|x|-y_n\leq (|x|-y_0)(1-|x|/2)^n=|x|(1-|x|/2)^n,~(*)$$ which also works when $n=0$.
Now consider the $n+1$ numbers: $n|x|/2,(1-|x|/2),\cdots,(1-|x|/2)$. Applying AM-GM inequality, one gets $$n/(n+1)\geq \sqrt[n+1]{(n|x|/2)(1-|x|/2)^n}$$
$$\Rightarrow |x|(1-|x|/2)^n\leq \frac 2 n(\frac n{n+1})(\frac n{n+1})^n<\frac 2 {n+1}.~(**)$$ Combining (*) and (**), $y_n\rightarrow |x|$ uniformly on $[-1,1]$.
A: The idea is simple. From $P_{n+1}(x) = P_{n}(x)+\frac{x^{2} - P_{n}(x)^{2}}{2}$ we obtain $$|x|-P_{n+1}(x) {= |x|-P_{n}(x)-\frac{x^{2} - P_{n}(x)^{2}}{2}\\=\left[|x|-P_n(x)\right]\cdot \left[1-{|x|+P_n(x)\over 2}\right]\\
}$$Since $0\le P_n(x)\le |x|$ (based on claim 1) we obtain $$0\le1-|x| \le 1-{|x|+P_n(x)\over 2}\le 1-{|x|\over 2}$$which leads to $$|x|-P_{n+1}(x) \le \left(|x|-P_{n}(x)\right)\left(1-{|x|\over 2}\right)$$and consequently

$$|x|-P_{n}(x)\le \left(|x|-P_0(x)\right)\left(1-{|x|\over 2}\right)^n=|x|\left(1-{|x|\over 2}\right)^n$$

