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Define $(p_n)_n$ recursively by $p_0(x) = 0$ and $p_n+1(x) = p_n(x) + frac12(x - p_n(x)^2)$. Prove that $p_n(x) rightarrow sqrtx$ uniformly for $0 leq x leq 1$ as $n rightarrow infty$.

I'm not sure where to go with this. I can show inductively that $forall n, 0 leq p_n(x) leq 1$ for $0 leq x leq 1$, and I think that $p_n(x)$ is increasing with $n$, but I have no idea whether this helps me. I know that I need to show that

$forall epsilon>0: exists N in mathbbN: forall n geq N: forall xin [0,1]: big|p_n(x) - sqrtxbig| < epsilon$

Edit: I just had a thought. I'm given a version of Dini's theorem that reads

Let $X$ be a compact metric space, $(f_n)_n$ a sequence of real-valued continuous functions on $X$, $f$ a continuous function on $X$ such that:

(i) $f_n rightarrow f$ pointwise on $X$, and

(ii) $f_n(x) geq f_n+1(x) $ for all $x in X, n in mathbbN$.

Then: $f_n rightarrow f$ uniformly on $X$.

Obviously, (ii) is not true for my sequence, but what if I took $f_n(x) = -p_n(x)$ and $f(x) = - sqrtx$? Then I could prove (ii) to be true, I know that $[0,1]$ is compact, and $f_n(x)$ is real-valued. The only part I would need to prove is (i), which is where I get confused. Correct me if I'm wrong, but after that I could say that since $(-p_n)_n rightarrow -sqrtx$, then $(p_n)_n rightarrow sqrtx$. Any advice?

By the recursion formula, we have
$$p_n + 1(x) - sqrtx = (p_n(x) - sqrtx)left[1 - frac12(p_n(x) + sqrtx)right].$$
Use this recursion $n$ times, it follows that
$$p_n(x) - sqrtx = (p_0(x) - sqrtx)prod_k = 0^n - 1left[1 - frac12(p_k(x) + sqrtx)right]. tag1$$

We shall show that for each $k$, it holds that
$$0 leq 1 - frac12(p_k(x) + sqrtx) leq 1 - frac12sqrtx. tag2$$
It is easy to see that $p_n(x) geq 0$ for $x in [0, 1]$ (as you stated, it in fact holds that $0 leq p_n(x) leq 1$, which can be proved inductively), hence the right inequality of $(2)$ holds. To show the left side inequality, notice that
beginalign
& 2 - p_k(x) - sqrtx \
= & 2 - p_k - 1(x) - frac12x + frac12p_k - 1^2(x) - sqrtx \
= & frac12(p_k - 1(x) - 1)^2 + frac32 - frac12x - sqrtx \
geq & frac12(p_k - 1(x) - 1)^2 geq 0.
endalign
Therefore $(2)$ holds. Consequently, we can obtain an upper bound of the right hand side of $(1)$:
beginalign
& |p_n(x) - sqrtx| \
= & sqrtx prod_k = 0^n - 1left| 1 - frac12(p_k(x) + sqrtx)right| \
leq & sqrtxleft(1 - frac12sqrtxright)^n tag3
endalign

Since the right side of $(3)$ converges to $0$ uniformly as $n to infty$, the result follows.