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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?

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  • $\begingroup$ $|\sin u|\le |u|$ for all $u\in \mathbb R$ so it's obvious. Perhaps you meant something else. $\endgroup$ – zhw. Oct 7 '16 at 21:59
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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 $0$.

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