Lemma 7.4.1.label There exists a sequence of polynomials $\seq{p_n}\subset \real[x]$ such that $p_{n}(x) \upto \sqrt{x}$ uniformly on $[0, 1]$ as $n \to \infty$.
Proof. Let $p_{1} = 0$. For each $n \in \natp$ and $x \in [0, 1]$, define
\[p_{n+1}(x) = p_{n}(x) + \frac{x - p_{n}(x)^{2}}{2}\]
Assume inductively that $0 \le p_{n} \le \sqrt{t}$, then
\[p_{n+1}(x) = p_{n}(x) + \frac{x - p_{n}(x)^{2}}{2}\ge p_{n}(x) \ge 0\]
and
\begin{align*}\sqrt{x}- p_{n+1}(x)&= \sqrt{x}- p_{n}(x) - \frac{x - p_{n}(x)^{2}}{2}\\&= \sqrt{x}- p_{n}(x) - \frac{1}{2}(\sqrt{x}- p_{n}(x))(\sqrt{x}+ p_{n}(x)) \\&= (\sqrt{x}- p_{n}(x))\paren{1 - \frac{\sqrt{x}+ p_{n}(x)}{2}}\ge 0\end{align*}
for any $x \in [0, 1]$.
Since $\seq{p_n}\subset \real[x]$ is increasing on $[0, 1]$ with $p_{n}(x) \le \sqrt{x}$ for all $x \in [0, 1]$ and $n \in \natp$, there exists $f: [0, 1] \to [0, 1]$ such that $p_{n} \to f$ pointwise as $n \to \infty$. For each $x \in [0, 1]$,
\[f(x) = \limv{n}p_{n+1}(x) = \limv{n}p_{n}(x) + \frac{x - p_{n}(x)^{2}}{2}= f(x) + \frac{x - f(x)^{2}}{2}\]
so $f(x) = \sqrt{x}$. Since $p_{n} \upto f$ pointwise, $p_{n} \upto f$ uniformly as $n \to \infty$ by Dini’s Theorem.$\square$
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