Show that $P_{n}(x)\rightrightarrows 0\qquad (x\in[0,1])$ 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!
 A: Given, $P_0(x) = \sqrt{x}$, and the recurssion $P_{n+1}(x)=\frac{1}{2}P_{n}^{2}(x)+(1-\sqrt{x})P_{n}(x)$, for $n \ge 0$.
We prove by induction, that $0 \le P_n(x) \le \dfrac{2\sqrt{x}}{2+n\sqrt{x}}$, $\forall \,x \in [0,1].$
The lower bound is easy to establish. As for the upper bound,
Since, $0 \le P_n(x) \le \sqrt{x} = P_{0}(x)$, (corresponds to case $n = 0$),
$ \implies P_{n+1}(x) = P_{n}(x)(\frac{1}{2}P_n(x) + (1-\sqrt{x})) \le \dfrac{2\sqrt{x}}{2+n\sqrt{x}}\cdot(1-\frac{\sqrt{x}}{2}) $
$\le \dfrac{2\sqrt{x}}{2+n\sqrt{x}}.\left(1-\dfrac{\sqrt{x}}{2+(n+1)\sqrt{x}}\right) = \dfrac{2\sqrt{x}}{2+(n+1)\sqrt{x}}$
Establishing our Induction hypothesis.
Hence, $P_n$ converges uniformly to $0$, on $[0,1]$. Q.E.D.
A: For simplicity, let we set $T_n(x)=P_n(x^2), T_0(x)=x$ in order to deal with the polynomial sequence:
$$T_{n+1}(x) = \frac{1}{2}T_n(x)^2+(1-x)T_n(x) = T_n(x)\left(1-x+\frac{1}{2}T_n(x)\right).$$
Obviously $T_n(x)$ is always positive over $(0,1)$. The behaviour in zero is always the same, hence it is worth to further set $T_{n}(x)=x U_n(x)$ in order to deal with:
$$U_{n+1}(x)=U_n(x)\left(1-x+\frac{x}{2}U_n(x)\right),\qquad U_0(x)=1.$$
Now it is not difficult to prove by induction that:
$$ U_n(x)\leq (1-x/2)^{n}, $$
hence:
$$ T_n(x) \leq x(1-x/2)^n, $$
where the maximum of the RHS occurs in $x=\frac{2}{n+1}$, giving:

$$ \sup_{x\in(0,1)}P_n(x)\leq \frac{2}{ne}.\tag{1}$$

A: 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
$$
P_n(x)\le\frac2{ne}\tag{3}
$$
A: To estimate the limit of $c = \lim\limits_{n\to+\infty}n\cdot\max\limits_{x\in[0,1]}|P_{n}(x)|$ the best constant for this inequality, $$|P_n(x)| \le \dfrac{c}{n}$$
Denote $T_n(x) = P_n(x^2)$, then $T_{n+1}(x) = T_n(x)\left(1-x + \frac{1}{2}T_n(x)\right)$, for $x \in [0,1]$
We can establish the estimate by induction or otherwise, $$\frac{2x}{1+exp\left({\dfrac{nx}{1-x}}\right)} \le T_n(x) \le \dfrac{2x}{1+e^{nx}}$$
We can have a short proof by induction on $n$:
The base case $n=0$ is obvious.
It remains to show that, $T_{n+1}(x) \le \dfrac{2x}{1+e^{nx}}\left(1-x+\dfrac{2x}{1+e^{nx}}\right) \le \dfrac{2x}{1+e^{(n+1)x}}$
and, $\dfrac{2x}{1+exp\left({\dfrac{(n+1)x}{1-x}}\right)} \le \dfrac{2x}{1+exp\left({\dfrac{nx}{1-x}}\right)}\left(1-x+\dfrac{2x}{1+exp\left({\dfrac{nx}{1-x}}\right)}\right) \le T_{n+1}(x)$
Which can be rearranged respectively to showing,
$\displaystyle \dfrac{e^x-1}{x} \le \dfrac{e^{(n+1)x}+1}{e^{nx}+1}$ and $\displaystyle\dfrac{e^{(n+1)h(x)}+1}{e^{nh(x)}+1} \le \dfrac{e^{h(x)}-1}{x}$ where, $\displaystyle h(x) = \dfrac{x}{1-x}$
Note that the Cauchy Schwarz inequality implies, the sequence $\dfrac{s^{n+1}+1}{s^n+1}$, is increasing for $n\ge 0$. Since, by Cauchy-Schwarz $(s^{n+1}+1)(s^{n-1}+1) \ge (s^n+1)^2 \implies \dfrac{s^{n+1}+1}{s^n+1} \ge \dfrac{s^n+1}{s^{n-1}+1}$, and taking the limits $n \to 0$ and $n\to \infty$ gives the bounds.
Thus, $\dfrac{s+1}{2} \le\dfrac{s^{n+1}+1}{s^n+1} \le s$, for $s > 1$ and $n\ge 0$.
Considering the two cases, where $s = e^x$ and $s=e^{h(x)}$ in the above inequality,
it suffices to show that $\dfrac{e^x+1}{2} \ge \dfrac{e^x-1}{x}$ and $\dfrac{e^{h(x)}-1}{x} \ge e^{h(x)}$, for $x\in (0,1)$. Both can be verified from the Taylor expansion of $e^x = 1+x+\frac{x^2}{2!}+\cdots$, thus establishing our bounds.
Now, $\displaystyle \sup\limits_{x \in [0,1]} \dfrac{2x}{1+e^x} = 2e^{-w} = 2(w-1)$, where $w$ is the unique solution to $e^{-x} = x-1$, i.e., $w = 1+W_0\left(\frac{1}{e}\right)$. Where, $W_0$ is the Lambert-W Function, and $1<w<2$.
So, $\sup\limits_{x \in [0,1]} \dfrac{2x}{1+e^{nx}} = \dfrac{2}{n}(w-1)$
i.e., $T_n(x) \le \dfrac{2}{n}(w-1)$
Similarly, $\displaystyle \sup\limits_{x \in [0,1]} \frac{2x}{1+exp\left({\dfrac{nx}{1-x}}\right)} = \displaystyle \sup\limits_{x >0} \dfrac{2x}{(1+x)(1+e^{nx})} = \frac{2}{n}(1+w_n)^2e^{-nw_n}$, where $w_n$ is the unique solution to $1+e^{-nx} = nx(1+x)$.
Denote, $\displaystyle v_n = nw_n \implies 1+e^{-v_n} = v_n\left(1+\frac{v_n}{n}\right)$. But, $0 < v_n < 2$, hence $v_n \to w$ as $n\to \infty$.
i.e., $\displaystyle 2(1+w_n)^2e^{-nw_n} {\to} 2e^{-w} = 2(w-1)$ as ${n \to \infty}$.
Therefore, $$\lim\limits_{n\to+\infty}n\cdot\sup\limits_{x\in[0,1]}|T_{n}(x)| = 2(w-1) = 2W_0\left(\frac{1}{e}\right)$$
