TITLE: Show that $P_{n}(x)\rightrightarrows 0\qquad (x\in[0,1])$
QUESTION [7 upvotes]: Let $P_{0}(x)=\sqrt{x}$,   $P_{n+1}(x)=\frac{1}{2}P_{n}^{2}(x)+(1-\sqrt{x})P_{n}(x)$,Prove that
$$ P_{n}(x)\rightrightarrows 0\qquad (x\in[0,1]) $$(the double arrows means Uniform convergence)
my idea:it can deduce that 
$$ P_{n+1}(x)\leq P_{n}(x)$$
and 
$$ 0\leq P_{n}(x)\leq \sqrt{x}$$,also for any fixed $x$,I find that $ \displaystyle\lim_{n\to\infty}P(x)=0$,but I don't know how to prove that Uniform convergence,if we can prove that,does there exist a constant $c>0$ which makes
$$\lim_{n\to+\infty}n\cdot\max_{x\in[0,1]}|P_{n}(x)|=c$$
or not ?
My friend told me that we can prove 
$$ P_{n}\leq \frac{2}{n+1}$$ at frist,but I don't know how to do it..
Thank you in advance!

REPLY [4 votes]: If $P_n(x)\le\sqrt{x}$, then
$$
\begin{align}
\frac{P_{n+1}(x)}{P_n(x)}
&=\frac12P_n(x)+1-\sqrt{x}\\
&\le1-\frac12\sqrt{x}\tag{1}
\end{align}
$$
Inequality $(1)$ implies that $P_{n+1}(x)\le\sqrt{x}$ as well. Therefore, by induction and the fact that $1+x\le e^x$, we have
$$
\begin{align}
P_n(x)
&\le\sqrt{x}\left(1-\frac12\sqrt{x}\right)^n\\
&\le\sqrt{x}e^{-n\sqrt{x}/2}\\
&=\frac2n\cdot(n\sqrt{x}/2)e^{-n\sqrt{x}/2}\tag{2}
\end{align}
$$
Since $xe^{-x}\le\frac1e$, $(2)$ implies
$$
P_n(x)\le\frac2{ne}\tag{3}
$$