# Fejer Kernel is Unbounded

Statement: Given the Fejer Kernel $F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2$. Show that $F_n(x)$ is unbounded for $x=0$ as $n\rightarrow \infty$

• why don't you just compute the limit as x -> 0? Is that the right expression for the fejer kernel? wikipedia gives something else... – Tyler Mar 27 '14 at 3:47
• That's not the Fejer kernel. You're missing some squares. – Pedro Tamaroff Mar 27 '14 at 3:54

There is a big square missing: $$F_n(x) = \frac{1}{2n\pi}\left( \frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\right)^2\sim_0 \frac{1}{2n\pi}\left( \frac{\frac{nx}{2}}{\frac{x}{2}}\right)^2=\frac n{2\pi}$$
• How did you get those approximation? Also could have I used the fact that $F_n(x) = \frac{1}{n} \frac{1-\cos(nx)}{1-\cos(x)}$ and applied L'Hopital's rule twice to see that the limit at 0 is divergent as $n \rightarrow 0$? – nonameswereavailable Mar 27 '14 at 4:33
• I get them using the fact that $\sin$ has a derivative $=1$ on $x=0$. – mookid Mar 27 '14 at 4:41
When $a$ goes to $0$, $\sin(a) \simeq a$. Then replacing each sine by its argument, for $x$ going to $0$, you arrive to $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n$$ If you want to go further, you could use a Taylor expansion built at $x=0$ and obtain $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n-\frac{1}{12} \left (n-1)n(n+1\right) x^2+O\left(x^3\right)$$