# Property of Fejer kernel

Let $$F_n(x) = \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2$$ be the n-th Fejer-Kernel. Then $$\forall \epsilon > 0, r < \pi : \exists N \in \mathbb{N} : \forall n \ge N : \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t < \epsilon$$ The proof of this goes like this \begin{align*} \int_{[-\pi,\pi] \setminus [-r, r]} F_n(t) \, \mathrm{d} t & = \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{n} \left( \frac{ \sin(\frac{1}{2} n x ) } { \sin(\frac{1}{2} x ) } \right)^2 \, \mathrm{d} t \\ & \le \frac{1}{n} \int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2(\frac{1}{2}t)} \, \mathrm{d} t \\ & \le \frac{1}{n} \int_{[-\pi,\pi]} \frac{1}{\sin^2(\frac{1}{2}r)} \, \mathrm{d} t \qquad (*) \\ & = \frac{2\pi}{n} \frac{1}{\sin^2(\frac{1}{2}r)} \\ & < \epsilon \quad \textrm{for certain } ~ n \ge N. \end{align*} The step i marked with (*) is totally unclear to me, why could there replaced $t$ by $r$, because it need not to be the case that $\sin^2(0.5*r) < \sin^2(0.5*t)$ as i think?

To see the estimate (*), first note that for $0<r<x<\pi/2$ we have $$0<\sin r<\sin x<1$$ Hence for $x$ with $0<r<|x|<\pi$ $$0<\sin^2 \frac{r}{2}<\sin^2 \frac{x}{2}<1$$ that is $$0<\frac{1}{\sin^2 \frac{x}{2}}<\frac{1}{\sin^2 \frac{r}{2}}$$ and we get $$\int_{[-\pi,\pi] \setminus [-r, r]} \frac{1}{\sin^2\frac{x}{2}} dx \le \int_{[-\pi,\pi]\setminus[-r,r]} \frac{1}{\sin^2\frac{1}{2}r} dx \leq \int_{[-\pi,\pi]} \frac{1}{\sin^2\frac{1}{2}r} dt = \frac{2\pi}{\sin^2\frac{1}{2}r}$$
Note that you're integrating over the set $r < |x| \le \pi$.
Let $f(x) = \sin^2(\frac{1}{2}x)$. Differentiating, we get that $f'(x) = \frac{\sin(x)}{2}$. Now if $x \in (0,\pi]$, then $\sin(x) > 0$, and so $f$ is increasing on $(0,\pi]$. Thus $\sin^2(\frac{1}{2} r) < \sin^2(\frac{1}{2}x)$ for $0 < r < x \le \pi$. Similarly, we see that since $\sin(x) < 0$ for $x \in [-\pi, 0)$, so $f$ is decreasing on this interval. So if you pick some $-\pi \ge x > -r > 0$, it follows that $\sin^2(\frac{1}{2}r) < \sin^2(\frac{1}{2}x)$, which gives the desired inequality.