This question is from Stein-Shakarchi Real Analysis problem 1 in chapter 2. For a function $f$ integrable on $[0,2\pi]$:

(1) Prove $\int_0^{2\pi}f(x)e^{-inx}\mathrm{d}x\to 0$ as $n\to\infty$

(2) Use (1) to prove if $E$ is a measurable subset of $[0,2\pi]$, then $\int_E\cos^2{(nx+u_n)}\mathrm{d}x\to\frac{1}{2}m(E)$ as $n\to\infty$ for any sequence $(u_n)$.

I've done part (1), but for part (2), after split the integral using trig identity I have:$\int_E\cos^2{(nx+u_n)}\mathrm{d}x=\frac{1}{2}m(E)+\frac{1}{2}\int_E\cos{(2nx+2u_n)}\mathrm{d}x$. Then I tried to show the last part goes to $0$ as $n$ goes to infinity. But I cannot make any progress. In particular, I don't know how to use $E$ being a subset of $[0,2\pi]$. Any help is much appreciated.

up vote 0 down vote accepted

You can write $$ \int_E \cos{(nx+u_n)} = \Re \left( e^{-iu_n} \int_E e^{-inx} \, dx \right) = \Re \left( e^{-iu_n} \int_0^{2\pi} \chi_{E}(x) e^{-inx} \, dx \right). $$ $E$ is measurable, so $\chi_E$ is integrable, and you can apply the first part. ($e^{-iu_n}$ is bounded for real $u_n$, so does not affect convergence to zero.)

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.