# boundness about Fejer kernel

let $F_N(x)$ denote the $N^\text{th}$ Fejer kernel.$$F_N(x)=\frac{1}{N}\frac{\sin^2(\frac{Nx}{2})}{\sin^2(\frac{x}{2})} \forall x\in[-\pi,\pi]/\{0\}$$ $$F_N(0)=N$$

now we have to show that there is some fixed $C>0$ such that for all $x\in[-\pi,\pi]$. $$F_N(x)\leq \min\{N,\frac{C}{Nx^2} \}$$

I have proved that $F_N(x)\leq \frac{\pi^2}{Nx^2}$ for all $x \in[-\pi,\pi]\setminus\{0\}$

but how to prove from here?? or there is some other way ??

• Just evaluate the limit at $x=0$. That is the only point left. – tattwamasi amrutam Feb 13 '18 at 16:33

## 1 Answer

The Dirichlet kernel is $$D_N(x) = \sum_{n=-N}^{N}e^{inx}$$ The Fejer kernel $F_N$ is the average of the Dirichlet kernels $D_0,D_1,\cdots,D_{N-1}$. $$F_N(x) = \frac{1}{N}\sum_{n=0}^{N-1}D_n(x).$$ So \begin{align} |F_N(x)|&= \left|\frac{1}{N}\sum_{n=0}^{N-1}\sum_{k=-n}^{n}e^{ikx}\right|\\ & \le \frac{1}{N}\sum_{n=0}^{N-1}\sum_{k=-n}^{n}1 \\ & =\frac{1}{N}\sum_{n=0}^{N-1}(2n+1) \\ & =\frac{1}{N}\left(2\frac{N(N-1)}{2}+N\right) = N. \end{align}