Fejer Kernel is Unbounded

Question

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 $n\rightarrow \infty$

Answer 1

There is a big square missing: $$ F_n(x) = \frac{1}{2n\pi}\left( \frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\right)^2\sim_0 \frac{1}{2n\pi}\left( \frac{\frac{nx}{2}}{\frac{x}{2}}\right)^2=\frac n{2\pi} $$

Answer 2

Hint

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 $$F_n(x) = \frac{1}{n}\bigg(\frac{\sin(\frac{nx}{2})}{\sin(\frac{x}{2})}\bigg)^2\simeq n-\frac{1}{12} \left (n-1)n(n+1\right) x^2+O\left(x^3\right)$$