Show that $P_{n}(x)\rightrightarrows 0\qquad (x\in[0,1])$

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.

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:

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

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 $$ P_n(x)\le\frac2{ne}\tag{3} $$