Lemma 2

Tag 0CF4, permanent. Cite it as [Bourbaki, Tag 0CF4].

General Topology, chapter X, § 4. Approximation of continuous real-valued functions

Let $(P_n)$ be the sequence of polynomials without constant terms defined by

(1) \quad p_0(t) = 0, \quad p_{n+1}(t) = p_n(t) + \frac{1}{2} (t - (p_n(t))^2), \qquad n \geq 0.

In the interval $[0, 1]$ the sequence $(p_n)$ is increasing and converges uniformly to $\sqrt{t}$ .

To prove Lemma 2 it is enough to show that, for all $t \in [0, 1]$ , we have

(2) \qquad 0 \leq \sqrt{t} - p_n(t) \leq \frac{2 \sqrt{t}}{2 + n \sqrt{t}},

for (2) implies that

0 \leq \sqrt{t} - p_n(t) \leq 2/n

We prove (2) by induction on $n$ . It is true for $n = 0$ . If $n \geq 0$ it follows from the inductive hypothesis (2) that $0 \leq \sqrt{t} - p_n(t) \leq \sqrt{t}$ , hence $0 \leq p_n(t) \leq \sqrt{t}$ , and therefore from (1) we have

\sqrt{t} - p_{n+1}(t) = (\sqrt{t} - p_n(t)) (1 - \frac{1}{2} (\sqrt{t} + p_n(t))),

so that

\sqrt{t} - p_{n+1}(t) \geq 0

, and from (2)

\begin{align*} \sqrt{t} - p_{n+1}(t) &\leq \frac{2 \sqrt{t}}{2 + n \sqrt{t}} \left( 1 - \frac{\sqrt{t}}{2} \right) \\ &\leq \frac{2 \sqrt{t}}{2 + n \sqrt{t}} \left( 1 - \frac{\sqrt{t}}{2 + (n+1) \sqrt{t}} \right) \\ &= \frac{2 \sqrt{t}}{2 + (n+1) \sqrt{t}}. \end{align*}

Q.E.D.

If $X$ is not compact, the conclusion of Theorem 3 is not necessarily valid. For example, a continuous real-valued function on $\mathbf{R}$ which is bounded and not constant cannot be uniformly approximated in $\mathbf{R}$ by polynomials (cf. Exercise 6).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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Lemma 2

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