Show that the Fejér Kernel $F_n$ is a good kernel and $\int_{0}^{1}F_n(x)dx = 1$

Question

Given that the Fejér Kernel is defined as the Cesaro sum of the $k$-th Dirichlet kernels from wikipedia, it also notes that the closed form can be written as

$$ F_n(x) = \frac{1}{n}\left(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\right)^2.$$

However I don't really understand how to verify this. The textbook I am using got the closed form from convolution properties, but I am trying to show it without using anything related to convolution explicitly.

I think understanding the above form will help me understand how $\int_{0}^{1}F_n(x)dx = 1$. Please note that my background is only an introductory first semester to real analysis too.

Answer 1

If I understand correctly, you want to show that

${\displaystyle F_{n}(x)={\frac {1}{n}}\sum _{k=0}^{n-1}D_{k}(x),}$ where ${\displaystyle D_{k}(x)=\sum _{s=-k}^{k}{\rm {e}}^{isx}}$ can be written in closed form as $F_n(x) = \frac{1}{n}\left(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\right)^2$.

The way to do this is to use the geometric series formula twice. Once to write $D_k(x) = e^{-ikx}\frac{e^{(2k+1)ix}-1}{e^{ix}-1}$ and then on $F_n(x)$ to get the desired closed form.

Answer 2

Here's an easy way to do it,I hope this helped you

Let $$ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx $$

$$ a_{n}-a_{n-1}=\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \frac{\cos(2n-2)x-\cos 2nx}{\sin^2 x} dx=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x}{\sin x} dx $$

$$ a_{n}-a_{n-1}-(a_{n-1}-a_{n-2})=\int_{0}^{\frac{\pi}{2}} \frac{\sin(2n-1)x-\sin(2n-3)x}{\sin x} dx=2\int_{0}^{\frac{\pi}{2}} \cos(2n-2)x dx=0 $$

$$ a_{n}-2a_{n-1}+a_{n-2}=0 $$

So $ a_{n} $ is an arithmetic sequence,

$$ a_{n}=a+bn $$

We note

$$ a_{0}=0,a_{1}=\frac{\pi}{2} $$

we can get

$$ a=0,b=\frac{\pi}{2} $$

Thus

$$ a_{n}=\int_{0}^{\frac{\pi}{2}} \frac{\sin^2 nx}{\sin^2 x} dx=\frac{n\pi}{2} $$