If $\int_0^{2\pi} q = 0$, then $\lim_{n \to \infty} \int_0^{2\pi}p(x)q(nx) \, dx= 0$

Question

I'm learning about Fourier series and need help with the following exercise:

Let the functions $p, q \in L^1([0, 2\pi])$ be bounded and $2\pi$-periodic. If $\int_0^{2\pi} q = 0$, show that

$$\lim_{n \to \infty} \int_0^{2\pi}p(x)q(nx)\,dx = 0.$$

Hint: first consider the step function for $p$, then it's approximation.

I'm really lost here. I did go back several times through my notes and the only thing that I could relate this execise to is the following theorem

Let $f \in L^1([a, b])$. For every $\varepsilon > 0$, there exist a step function $\psi$ such that $\|f - \psi\|_{L^1[a, b]} < \varepsilon$.

The way I understand this is that we can approximate any integrable function on a close interval by a step function. Since $p \in L^1([0, 2\pi])$ by assumption, the above theorem holds. This is as far as I got.

Answer 1

The theorem you stated is definitely related to the question. Let $s$ be a step function so that $\| p-s\|_{L^1} <\epsilon$, then

$$\bigg|\int_0^{2\pi} p(x) q(nx)\,dx - \int_0^{2\pi}s(x) q(nx)\,dx\bigg|\le C \int_0^{2\pi}|p(x)-s(x)|\,dx <C\epsilon,$$

where $C$ is the bounded of $q$, that is $|q(y)|\le C$ for all $y\in [0,2\pi]$. Now if you have shown the theorem for step functions, then there is $N_\epsilon$ so that

$$\bigg|\int_0^{2\pi} s(x)q(nx) \bigg| <\epsilon$$

for all $n\ge N_\epsilon$. Then

$$\begin{split} \bigg|\int_0^{2\pi} p(x)q(nx)\,dx\bigg| &\le \bigg|\int_0^{2\pi} p(x) q(nx)\,dx - \int_0^{2\pi}s(x) q(nx)\,dx\bigg|+\bigg|\int_0^{2\pi} s(x)q(nx)\,dx \bigg| \\ &\le (C+1)\epsilon \end{split}$$

for all $n\ge N_\epsilon$. Thus the theorem is true for all $p$.

Answer 2

Your density lemma says it suffices to prove the result for a function of the form $q(x)=1_{[a,b]}(x)$ $[a,b] \subset [0,2\pi]$.

Now \begin{align*} \int _{0}^{2\pi} q(x) p(nx) d x &= \int _{0}^{2n\pi} \frac{1}{n}q(x/n)p(x) d x \\ &= \sum _{k=0}^{n-1} \int _{2(k)\pi}^{2(k+1)\pi}\frac{1}{n} q(x/n)p(x) d x \\ &= \sum _{k=0}^{n-1} \int _{0}^{2\pi}\frac{1}{n} q((x+2k\pi)/n)p(x+2k\pi) d x\\ &=\sum _{k=0}^{n-1} \int _{0}^{2\pi}\frac{1}{n} q((x+2k\pi)/n)p(x) d x\\ &=\sum _{k=0}^{n-1} \int _{0}^{2\pi}\frac{1}{n} \left (\sum _{k=0}^{n-1} q((x+2k\pi)/n)\right )p(x) d x \end{align*}

Now $$\frac{1}{n} \left (\sum _{k=0}^{n-1} q((x+2k\pi)/n)\right )=2\pi\frac{1}{2\pi n} \left (\sum _{k=0}^{n-1} q((x+2k\pi)/n)\right )\rightarrow \int _{0}^{2\pi} q(x)dx = 2\pi (b-a)$$

The last follows since this is a Riemann sum. And now noticing that $\frac{1}{n} \left (\sum _{k=0}^{n-1} q((x+2k\pi)/n)\right ) \leq 1$ we may apply dominated convergence and conclude that

$$\int _{0}^{2\pi} q(x) p(nx) d x \rightarrow \int_{0}^{2\pi} 2 \pi (b-a)p(x) dx = 0$$