Proving the uniform convergence of a the recursively sequence given as $P_{n+1}=P_{n}(x)+\frac{1}{2}(x^{2}-P_{n}(x)^{2})$. Given a sequence of algebraic polynomials $\lbrace P_{n}:[0,1] \to \mathbb{R}\rbrace_{n=0}^{\infty}$ defined recursively as $P_{0}(x)=1$ and for every $n \in  \mathbb{N}$; $$P_{n+1}=P_{n}(x)+\frac{1}{2}(x^{2}-P_{n}(x)^{2}).$$ 
Prove the sequence $\lbrace P_{n} \rbrace_{n=0}^{\infty}$ is uniformly convergent to $|x|$ for $x \in [-1,1]$.
My attemp to the proof goes as follow:
As we are given the limit in the exercise, Im looking for a numerical sequence $h_{n} \to 0$ such for every $n \in \mathbb{N}$ and $x \in [0,1]$ (or $x \in [-1,1]$ , im not sure which is the correct interval) we have that
$$|P_{n}(x)-|x|| \leq h_{n}.$$
As $P_{n}$ is defined recursively for every $n \in \mathbb{N}$, I have to prove the inequality given above inductively. This is, to prove
$$|P_{1}(x)-|x|| \leq h_{n}.$$
Suppose $|P_{n}(x)-|x|| \leq h_{n}$, then prove
$$|P_{n+1}(x)-|x|| \leq h_{n}.$$ As
$$P_{1}(x)-|x|=\frac{x^{2}}{2}-|x|$$, Im stucked at finding the correct sequence $\lbrace h_{n} \rbrace$ mentioned before. Any help with this proof will be aprecciated. Thanks
 A: Note that
$$
1-P_{n+1}(x)=\frac12(1-P_n(x))^2+\frac{1-x^2}2
$$ so by defining $Q_n(x)=1-P_n(x)$, we have $Q_{n+1}(x)=\frac12 Q_n^2(x)+\frac{1-x^2}2$. We can further observe inductively that $Q_n(\pm 1)=0$ for all $n\ge 1$, thus if we let $R_n(x)=\frac{Q_n(x)}{1-x^2}$, then it follows
$$
R_{n+1}(x)=\frac{1-x^2}2R_n^2(x)+\frac12.
$$ Note that $R_n\ge \frac 12>R_0=0$ for all $n\ge 1$. By induction, we can see that $R_n\le 1$ implies $R_{n+1}\le 1$. Since $R_0=0$, we have $R_n\le 1$ for all $n$. Also by induction, we can see that $R_n\le R_{n+1}$ implies $R_{n+1}\le R_{n+2}$, which implies $R_n\le R_{n+1}$ for all $n\ge 0$ by induction hypothesis. These facts together imply that for each $x$, $\{R_n(x)\}_{n\ge 1}$ is a bounded increasing sequence, converging pointwise to $R(x)\in [\frac12,1]$ satisfying
$$
R(x)=\frac{1-x^2}2 R^2(x)+\frac 12\implies R(x)=\frac{1\pm\sqrt{1-(1-x^2)}}{1-x^2}.
$$ This gives $R(x)=\frac1{1+ |x|}$ or $R(x)=\frac1{1-|x|}$, but since $R(x)\le 1$, it follows that $R(x)=\frac{1}{1+|x|}$. Now, we invoke Dini's theorem to verify that $R_n$ converges uniformly to $R$ on $[-1,1]$. This readily implies that $P_n(x) =1-(1-x^2)R_n(x)$ converges uniformly to $P(x)=|x|$ as desired.
