Consider the Fourier Series Approximation
$$f_n(x)=\sum_{|k|\leq n} c_k \phi_k(x)$$ where $$\phi_k(x)=\frac{1}{\sqrt{ 2 \pi}}e^{ikx}$$
I am given the following and asked to find what G(x,y) is.
I have for $x\in [0,2 \pi)$ that $$f_n(x)=\int_0^{2\pi}f_n(y)G(x,y)dy$$
I thought I could maybe use Parseval's relation but I would need absolute values for this to do any good I think. I also tried an integration by parts approach but that didn't get me anywhere either.
I appreciate any help you can give.
One can easily obtain:
$$\frac1{\sqrt{2\pi}}\sum c_k\int^{2\pi}_0 e^{iky}G(x,y)dy=\frac1{\sqrt{2\pi}}\sum c_k e^{ikx}$$
$$\sum c_k \left(\int^{2\pi}_0 e^{iky}G(x,y)dy-e^{ikx}\right)=0$$
It is natural to assume $$\int^{2\pi}_0 e^{iky}G(x,y) dy=e^{ikx}\qquad{(1)}$$
It is almost obvious that $G(x,y)=\delta(x-y)$, but let’s try to sort of prove it.
You can recognize the left hand side is related to the Fourier coefficients of $G(x,y)$’s Fourier series in $y$.
Let $$G(x,y)=\sum^\infty_{k=-\infty}A_k e^{iky}$$
By definition,
$$A_k=\int^\pi_{-\pi}e^{-iky}G(x,y)dy=e^{-ikx}$$ by $(1)$.
Hence, $$G(x,y)=\sum^\infty_{k=-\infty}e^{ik(y-x)}$$ which converges nowhere. This is indeed the ‘Fourier series’ of the Dirac delta $\delta(x-y)$.
We are now sure to say $$\color{red}{G(x,y)=\delta(x-y)}$$