Where my mistake in find : $\lim_{n\to +\infty}\int_{\mathbb{R}}\sin^{2} (nx)\Phi (x)dx≠0$

Question

I don't know where is my mistake to solve this two limits :

Prove that :

$$\lim_{n\to +\infty}\int_{\mathbb{R}}\sin (nx)\Phi (x)dx=0$$

$$\lim_{n\to +\infty}\int_{\mathbb{R}}\sin^{2} (nx)\Phi (x)dx≠0$$

For all : $\Phi \in D(\mathbb{R})$

Mean that : $\Phi (x)≠0\implies |x|<a,a>0$

My solution is :

In first limite $y=nx$ we get :

$$\lim_{n\to +\infty}\frac{1}{n}\int_{\mathbb{R}}\sin (x)\Phi (\frac{x}{n})dx=\lim_{n\to +\infty}\frac{1}{n}\int_{[-a,a]}\sin (x)\Phi (\frac{x}{n})dx≤\lim_{n\to +\infty}\frac{sup_{[-a,a]}|\Phi (x)|}{n}\int_{[-a,a]}\sin (x)dx$$

And its clearly go to zero

But when I applied in second limit give me also zero ? Why ?

$$\lim_{n\to +\infty}\frac{1}{n}\int_{\mathbb{R}}\sin^{2} (x)\Phi (\frac{x}{n})dx=\lim_{n\to +\infty}\frac{1}{n}\int_{[-a,a]}\sin^{2} (x)\Phi (\frac{x}{n})dx≤\lim_{n\to +\infty}\frac{sup_{[-a,a]}|\Phi (x)|}{n}\int_{[-a,a]}\sin^{2} (x)dx=0$$ ??

Answer 1

When you make the change of variable $y= nx$ you should integrate over $[-na,na]$, not $[-a,a]$. Same mistake is leading to a wrong answer in the second question.

To find the correct limit for $\int \sin^{2} (nx) \Phi(x)dx$ write $\sin ^{2}(nx)=\frac 1 2 (1-\cos (2nx))$ and show that $\int \cos (2nx) \Phi (x) dx \to 0$. So we get the answer as $ \frac 1 2 \int \Phi(x)dx$.

The fact that $\int \sin (nx) \Phi(x)dx $ and $\int \sin (nx) \Phi(x)dx $ both tend to $0$ is called Riemann Lebesgue Lemma. See https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma