Beginner Fourier Problem and absolute convergence

Question

presumably as preparation for Fourier Analysis, I was given the following exercises:

1. Let $\sum_{n=1}^{\infty} a_{n}$ be a series which is absolute convergent and for $x\in\mathbb{R}$

   $$ f(x)=\sum_{n=1}^{\infty} a_{n}\sin(nx)$$

   Prove the equation

   $$a_{n}=\frac{1}{\pi}\int_{0}^{2\pi} f(x)\sin(nx)\text{d}x\qquad\forall n\in\mathbb{N}$$

2. Consider a sequence $ (a_{n})_{n\in\mathbb{N}}$ so that the series $$\sum_{n=1}^{\infty} n^{2k}a_{n}\qquad \forall n\in\mathbb{N}$$ is absolute convergent. Proof that for $x\in\mathbb{R}$ the function $$ f(x)=\sum_{n=1}^{\infty} a_{n}\sin(nx)$$ can be differentiated $2k$-times and it holds $$ f^{(k)}(x)=\sum_{n=1}^{\infty} b_{n}\sin(nx) $$ Determine the coefficient $b_{n}$.

I've spent quite a long time, but unfortunately I don't succeeded.

This is an outline what I've tried so far:

1. As it holds $$\left| a_{n}\sin(nx)\right| \leq \left| a_{n}\right|$$ and $\sum_{n=1}^{\infty}\left| a_{n}\right|$ is (absolute) convergent, Weierstrass M-Test gives us that $f(x)$ converge uniformly for $x\in\mathbb{R}$. Therefore $$ \frac{1}{\pi}\int_{0}^{2\pi} \left(\sum_{n=1}^{\infty} a_{n}\sin(nx)\right) \sin(nx)\text{d}x \\ = \frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \int_{0}^{2\pi} \sin^{2}(nx)\text{d}x \\ = \frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \left[ - \frac{\cos(nx)\sin(nx)}{2n} + \frac{x}{2}\right]^{2\pi}_{0} \\ =\frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \left[ - \frac{\sin(2nx)}{4n} + \frac{x}{2}\right]^{2\pi}_{0} \\ =\frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \cdot \left( - \frac{\sin(4\pi n)-4\pi n}{4n}\right) \\ =\frac{1}{\pi} \sum_{n=1}^{\infty} a_{n} \cdot \pi \\ = \sum_{n=1}^{\infty} a_{n} \overset{?}{=}a_{n}$$

2. I found no successful method, maybe Taylor's theorem for $\sin$ might be helpful, as it somehow matches to the setup of the exercise. $$f(x)=\sum_{n=1}^{\infty} a_{n}\sum_{i=0}^{\infty}(-1)^{i}\frac{(nx)^{2i+1}}{(2i+1)!}$$

Thanks in advance for your help.

Answer 1

1. The series $\sum_{n=1}^\infty a_n$ is absolutely convergent $\implies$ (by Weierstarss M-test) $f(x)=\sum_{n=1}^\infty a_n \sin(nx)$ converges uniformly. Applying the uniform convergence we get that $f$ is absolutely integrable on $[0,2\pi]$ : $$\int_{0}^{2\pi}|f(t)|dt=\int_{0}^{2\pi}|\sum_{n=1}^\infty a_n \sin(nt)|dt \le\sum_{n=1}^\infty|a_n|\int_{0}^{2\pi}|sin(nt)|dt=2\sum_{n=1}^\infty|a_n|<\infty$$

So we can compute it's Fourier coefficients.

Now rewrite, $f(x)=\sum_{n=1}^\infty a_n \sin(nx)=\sum_{n=1}^\infty\frac{a_n}{2i}(e^{inx}-e^{-inx})$

$\hat{f}(n)=\frac{1}{2\pi}\int_{0}^{2\pi} f(x) e^{-inx}dx=\frac{1}{2\pi}\int_{0}^{2\pi}\sum_{m=1}^\infty\frac{a_m}{2i}(e^{imx}-e^{-imx})(e^{-inx})dx=$$\frac{1}{2\pi}\sum_{m=1}^\infty\frac{a_m}{2i}\int_{0}^{2\pi}(e^{imx}-e^{-imx})(e^{-inx})dx$

$$\implies \hat{f}(n)=\frac{a_n}{2i} \frac{1}{2\pi}\int_{0}^{2\pi}dx=\frac{a_n}{2i}$$

Similarly, $$\hat{f}(-n)=-\frac{a_n}{2i}$$

So $a_n=\frac{a_n}{2}-(\frac{-a_n}{2})=i(\hat{f}(n)-\hat{f}(-n))=\frac{1}{\pi}\int_{0}^{2\pi} f(x)\sin(nx)dx$

2. $$\forall 0\le j \le2k,\sum_{n=1}^\infty |n^ja_n| \le \sum_{n=1}^\infty n^{2k}|a_n|<\infty \dots (*)$$ . Hence again Weierstrass M-test tells you that the series $\sum_{n=1}^\infty a_n\sin(nx)$ converges uniformly and also the series made of the derivatives $\sum_{n=1}^\infty na_n\cos(nx)$ converges uniformly. Thus the function defined as the sum $f(x)=\sum_{n=1}^\infty a_n\sin(nx)$ is diffrentiable and moreover, $f^{/}(x)=\sum_{n=1}^\infty na_n\cos(nx)$.

Keeping $(*)$ in mind keep on applying Weierstarss M-test and diffrentiability and computation of derivatives follow. It follows that $$ f^{(2k)}(x)= \begin{cases} \sum_{n=1}^\infty n^{2k} sin(nx) & \text{ if } 2k \equiv 0 \text{ mod 4} \\ \sum_{n=1}^\infty (-n^{2k}) sin(nx) & \text{ if } 2k \equiv 2 \text{ mod 4} \\ %-\sum_{n=1}^\infty n^{2k} cos(nx) & \text{ if } 2k \equiv 3 \text{ mod 4} \\% \end{cases}$$