Limit of a trigonometric integral

[hofhile]

Homework Statement

Excersice of Big Rudin: ch. 4, prob. 9.

If $$A\subset [0,2\pi]$$ and $$A$$ is measurable, prove that

$$\lim_{n\to\infty}\int_{A}\cos\,nx\,dx=0$$

Homework Equations

Bessel's inequality

The Attempt at a Solution

I give my solution, but I post because I think that my answer is completely wrong:

$$\chi_{A}\in L^{2}(T)$$ so $$c_{n}=\frac{1}{2\pi}\int_{A}e^{-int}\,dt$$ are the Fourier coefficient and by Bessel's inequality $$\sum_{n=-\infty}^{\infty}|c_{n}|^{2}<\infty$$ therefore $$|c_{n}|\to 0$$ and $$\lim_{n\to\infty}\int_{A}\cos\,nx\,dx=0$$.

Thank you.

 

[mighty2000]

Thank you for your solution. However, I believe there may be a mistake in your reasoning. Bessel's inequality states that the sum of the squares of the Fourier coefficients is finite, but it does not necessarily mean that each individual coefficient tends to zero. In order to prove that the limit of the integral is zero, we need to use a different approach.

First, we can rewrite the integral as follows:

\int_{A}\cos\,nx\,dx=\frac{1}{n}\int_{A}\sin\,nx\,d(nx)=\frac{1}{n}\int_{A}\sin\,nx\,dx

Then, using the substitution u=nx, we can rewrite it again as:

\frac{1}{n}\int_{nA}\sin\,u\,du

Since A is a measurable set, nA is also measurable. Now, we can use the Riemann-Lebesgue lemma to prove that the limit of the integral is zero. This lemma states that if f is an integrable function and \phi is a continuous function, then

\lim_{n\to\infty}\int_{a}^{b}f(x)\phi(nx)\,dx=0

In our case, f(x)=\sin\,x and \phi(x)=\chi_{nA}(x), where \chi_{nA}(x) is the characteristic function of nA. Since \chi_{nA}(x) is continuous, we can apply the Riemann-Lebesgue lemma and conclude that

\lim_{n\to\infty}\int_{A}\sin\,nx\,dx=0

Therefore, the original integral also tends to zero as n approaches infinity.

I hope this helps clarify the solution. Let me know if you have any further questions or concerns.