数分原理(Rudin)的一道证明题

Ex.7.23 Let $P_0 = 0, \; P_{n+1}(x) = P_n(x) + \small{\dfrac{x^2 -P_n^2(x)}{2}}.\quad$ Prove that $P_n\overset{_{[-1,1]}}{\rightrightarrows}|x|.$
$\qquad$(Stone-Weirestrass Th. can be proven without 1st prove Th.7.26)
$\qquad$Hint: Use the identity $\displaystyle{|x| -P_{n+1} = (|x|- P_n(x))\big(1-{\small{\frac{|x|+P_n(x)}{2}}}\big)}$ to prove that
$\qquad 0\le P_n(x)\le P_{n+1}(x) \le |x|\;(|x| < 1)$ hence $\displaystyle{|x| -P_n(x)\le |x|\big(1-{\small{\frac{|x|}{2}}}\big)^n < \small{\frac{2}{n+1}}}$
Proof From the recursive definition, we have
$\qquad |x| -P_{n+1}(x) = |x| - P_n(x) - {\small{\dfrac{x^2 -P_n^2(x)}{2}}} = (|x|- P_n(x))\big(1-{\small{\dfrac{|x|+P_n(x)}{2}}}\big).\qquad (\star)$
$\qquad$For $|x|< 1$, since $P_0 = 0$, by induction $(\star)$ implies $0\le P_n(x)\le P_{n+1}(x) \le |x|\;(\forall n)$ and so
$\qquad (\star)\implies 0\le |x| - P_{n+1}(x) \le (|x| -P_n(x))(1-\frac{|x|}{2})\le \cdots\le (|x| -P_{n-k})(1-\frac{|x|}{2})^{k+1}$.
$\qquad$Therefore $\quad 0\le |x| - P_n(x) \le |x|(1-\frac{|x|}{2})^n \le \frac{2}{n+1}\;(\forall n\in\mathbb{N}\,\forall x\in (-1,1)).\quad\square$
$\qquad$The last inequality comes from
$\qquad\qquad\quad \small{\varphi(t) = t(1-\frac{t}{2})^n,\,\varphi\,'(t) = (1- \frac{t}{2})^{n-1}(1-\frac{(n+1)t}{2}),\,\max\varphi([0,1]) = \varphi(\frac{2}{n+1}) < \frac{2}{n+1}}$