Bourbaki

Exercise 20, § 1

Functions of a Real Variable, II, 83

Show that as nn increases indefinitely the polynomial

fn(x)=∫0x(1−t2)n dt∫01(1−t2)n dt f_n(x) = \frac{\int_0^x (1-t^2)^n\,dt}{\int_0^1 (1-t^2)^n\,dt}
tends uniformly to −1-1 on every interval [−1,−ε][-1, -\varepsilon], and tends uniformly to +1+1 on every interval [ε,+1][\varepsilon, +1], where ε>0\varepsilon > 0 (note that ∫01(1−t2)n dt≥∫01(1−t)n dt\int_0^1 (1-t^2)^n\,dt \geq \int_0^1 (1-t)^n\,dt). Deduce that the polynomial gn(x)=∫0xfn(t) dtg_n(x) = \int_0^x f_n(t)\,dt tends uniformly to ∣x∣|x| on [−1,+1][-1, +1], which provides a new proof of the Weierstrass theorem (Gen. Top., X, p 313, prop. 3).

Solution

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Transcribed from the printed volume. Not translated.