Good day, I am trying to solve an exercise in the course of distribution theory and fourier analysis. I am new to the matter of using distribution in calculating, and I am thankful for any help to solve the following question:
Consider the $2\pi$ -periodic function $f(x)$ defined on $[0,2\pi)$ by $f(x)=\frac{1}{2}(\pi-x)$. prove (by calculating the Fourier series) that in the sense of distributions $\sum_{n=1}^{\infty }\frac{\sin(nx)}{n}=f(x)$.
prove that-in the sense of distributions
$\sum_{n\in \mathbb{Z}}e^{inx}=2\pi\sum_{n\in \mathbb{Z}}\delta_{2\pi n}$ in $D^{'}(\mathbb{R})$ Where $\delta_{2\pi n}$ is the distribution $\phi \mapsto \phi (2 \pi n)$
Thanks.
For the second question I have tried the following: if one takes the distribution < sum e^{inx} , phi > = .... Then by taking the sum like this (-infinity,-1) & {0} & (1,infinity) With the notation of cos(nx) as a function of e^{inx} With the help of the derivative of f(x) the previous distribution may equal to <0,phi(x)>
Then I didn't know how to continue to get the second hand side (2 pi sum delta 2 pi n) Sorry