# 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 hint - look at derivative of $P_n$. Commented Mar 10, 2019 at 1:52
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.
• Thanks a lot! In the third line of your proof $Q_{n}(x)=1-P_{n}(x)$ or $Q_{n}=1-P_{n+1}(x)$ ?? @Song
• I defined $Q_n(x)=1-P_{\color{red}n}(x)$ so that $1-P_{n+1}(x)=Q_{n+1}(x)=\frac12 (Q_n(x))^2+\frac{1-x^2}2$. Commented Mar 10, 2019 at 23:03