# Show Fejer kernel on the real line is good, without using trignometric integrals.

This is from page 163 of Stein's Fourier Analysis.

Fejer kernel on the real line is defined by

$$\mathcal{F}_R(t) = R\left(\frac{\sin(\pi t R)}{\pi t R}\right)^2$$

When $t=0$, $\mathcal{F}_R(t)=R$.

I want to show that $\lim_{R\to 0} \mathcal{F}_R(t) = 0$. However, consider

$$\int_{-\infty}^\infty\mathcal{F}_R(t) dt=\int_{-\infty}^\infty \frac{\sin(\pi t R)^2}{\pi^2 t^2 R} dt=\frac{1}{\pi^2 R}\int_{-\infty}^\infty\frac{\sin(\pi t R)}{t^2}dt\le\frac{1}{\pi^2 R}\int\frac{1}{t^2} dt$$

This is not good enough. I also tried $$\int_{-\infty}^\infty\mathcal{F}_R(t) dt=\int_{-\infty}^\infty \frac{\sin(\pi t R)^2}{\pi^2 t^2 R} dt=\int\frac{\sin^2(u)}{u^2}du$$

But I suppose I do not have to resort to trigonometric integral. Any hint?

• $|\sin u|\le |u|$ for all $u\in \mathbb R$ so it's obvious. Perhaps you meant something else. – zhw. Oct 7 '16 at 21:59

Why do you want to show that $\lim_{R\to 0} F_R (t)=0$? That's not a property for good kernels. But if you do want to show it, it's easy since $\frac{\sin x}{x} \to 1$ as $x \to 0$ and so the function inside the square is bounded near $0$ and $R\to 0$, so the whole goes to $0$.