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I have this question that relates the Fejer theorem with the Fourier Transform. Any help would be appreciated.

If $f$ is of moderate decrease then

$$\int_{-R}^{R}\left(1-\frac{|\xi|}{R}\right) \hat{f}(\xi)\exp(2\pi i x \xi) d\xi =(f*F_R)(x)$$

where $${F}_R = R \left(\frac{\sin\pi t R}{\pi tR}\right)^2$$ if $t\neq0$ and $${F}_R = R,$$ if $t=0$

Show that $F_R$ is a family of good kernels as $R \rightarrow \infty,$ and therefore that the integral tends uniformly to $f(x)$ as $R \rightarrow \infty$.

I know the three properties for showing that something is a good kernel. Also I know the theorem saying that if something is a good kernel then $(f*F_R)(x)$ converges uniformly to $f$ if $f$ is continuous everywhere. Just how would I go about proving that it is a good kernel using what I have? Should I substitute the $F_R$ in for the $\hat{f}$ in the integral to the left or should I work with the definition of a convolution. I tried looking for other posts about the topic but couldn't find any. Please help me with some starting steps as I always ask for. Thanks!

user110842
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1 Answers1

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This is an old post, but I've just come across this exercise and have solved it so I put the solution here for anyone who may need help with this exercise from Stein and Shakarchi's Fourier Analysis Exercise 9, Chapter 5. So the most difficult part with this problem is showing the first property, that is, the integral of the kernel is equal to $1$. Then the second property, that $\int |F_R (t)|dt$ is bounded, and the third, for any $\eta >0,$ $\int_{|t|\gt \eta} |R| (\frac{\sin \pi tR}{\pi tR})^2dt \to 0$ as $R \to \infty$ are obvious (for the third property, just see that the integral of $\frac{1}{t^2}$ is bounded if $t$ is bounded away from $0$.

To show the first part, we need to use the famous integral $\int_0^\infty \frac{\sin x}{x}dx=\pi/2$, which is also given as Exercise 12, Chapter 3 of the book.

Now, consider $R>0$ (I think we're supposed to restrict $R>0$ here otherwise we get $-1$). Then since the integral (Lebesgue or Riemann) is invariant under a change of the integrand at a point, we can simply consider the integral of the kernel to be given as $\int \frac{\sin^2 \pi tR}{\pi^2 t^2R}dt$, which makes sense since it is absolutely convergent. Now make the change of variables $\pi tR=u$, then $dt=\frac{1}{\pi R}du$. And we get the integral to be $\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{\sin^2 u}{u^2}du=2/\pi \int_0^\infty \frac{\sin^2 u}{u^2}du.$ And now the rest is just integration by parts. $\int_0^\infty \frac{\sin^2 u}{u^2}du=-\frac{1}{u}\sin^2 u\Big|_0^\infty +\int_0^\infty \frac{1}{u}2\sin u\cos u du=0+2\int_0^\infty \frac{\sin2u}{2u}du=\int_0^\infty \frac{\sin v}{v}dv=\pi/2.$ Note that to get $0$ in the first integration by parts, we used the fact that $\frac{\sin u}{u}$ has limit $1$ as $u \to 0$ and then used another change of variables $2u=v$ from which $du=1/2 dv$ to get rid of $2$ in front of the integral.

nomadicmathematician
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