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Let $b > 0 \in \mathbb R$, a continuous function $f : [0, b] \to \mathbb R$, and a periodic and continuous function $p : \mathbb R \to \mathbb R$ of period $T = 1$.

  1. If $p \geq 0$ on $\mathbb R$ et $\int_0^1 p(t) dt = 1$, prove that $$\lim_{n \to \infty} \int_0^b f(t)p(nt) dt = \int_0^bf(t) dt. $$

I was told that we should use Riemann/Darboux sums, is there any another way to do it ? If not how to do it with Riemann sums ?

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  • $\begingroup$ Are you familiar with Lebsgue integration? $\endgroup$ Feb 22 at 20:20
  • $\begingroup$ No sorry, only Riemann sums and other well known properties $\endgroup$
    – Kilkik
    Feb 22 at 20:21
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This is what I would do:

$1)$ Prove that the statement is true for $p(x)=e^{2\pi i x}$.

$2)$ Prove that it is true for trigonometric polynomials $$p(x)=\sum\limits_{j=-m}^ma_je^{2\pi i jx} \quad with \quad\sum\limits_{j=-m}^ma_j=1,$$ ie, for any element belonging to the span of $\{e^{2\pi inx}:n\in\mathbb{Z}\}$ with norm equal to 1.

$3)$ Use that $\{e^{2\pi inx}:n\in\mathbb{Z}\}$ is an orthonormal complete system, which implies that $\{e^{2\pi inx}:n\in\mathbb{Z}\}$ is dense in $L^2(\mathbb{T})$. In other words, any function $p$ fulfilling the hypothesis can be approximated by a sequence of trigonometric polynomials that satisfy the original property.

Note: I solved the problem assuming that $f$´s domain was $[0,1]$ instead of $[0,b]$, but this should be easy to fix.

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