Bourbaki

Exercise 20, § 1

Functions of a Real Variable, II, p. 32

Now prove that, when n 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 in every interval (−1, −ε), and tends uniformly to +1 in every interval (ε, +1), where ε > 0 (notice 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| in (−1, +1), which gives a new proof of the Weierstrass th. (TG, X, p. 37, prop. 3).

Solution

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Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.