I've been told to prove the real-version of Riemann–Lebesgue lemma, which is: for $f$ integrable and $2\pi$ periodic: $$ \lim_{n\to\infty} \int_{0}^{2\pi} f(x)\cos(nx) \ dx = \lim_{n\to\infty} \int_{0}^{2\pi} f(x)\sin(nx) \ dx =0$$ Now, previously we proved this theorem for the complex version...

Let $f\in L^{p} (0, 2 \pi) (1\leq p < \infty.)$ How to show: $I_n = \int_{0}^{2\pi } \sin (nx) f (x) dx\to 0$ as $n\to \infty.$ (In other words, $f_n(x) =\sin (nx)$ converges to 0 weakly in $L^{p'}$) My Thoughts: I guess, I should prove for first for the dense class in $L^p.$ If suppos...

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 $p$, the...

If $f: \mathbb R \to \mathbb R$ is continuously differentiable periodic function of period $1$, then $$n\int_{0}^{1} f(x) \sin(2\pi nx)\mathrm dx \to 0 $$ as $ n\to\infty$.