6

I'm working out of Stein's Fourier Analysis (working with Riemann integrable functions), and I'm having trouble with problem 1:

Suppose $f$ is continuous and supported on $[-M,M] \subset \mathbb{R}$ such that the Fourier transform satisfies $|\hat{f}(\xi)| \leq \frac{B}{1+\xi^2}$ (of moderate decrease).

<p>Show that:</p>

<p>a) Let $L$ be such that $\frac{L}{2}$ and $\delta = \frac{1}{L} &gt; M$ </p>

<p>Show $f= \delta \sum \hat{f}(\delta n)e^{-2\pi i n \delta x}$ where $\hat{f}(\delta n)$ are the Fourier coefficients</p>

I don't have a problem with this, I used the fact that $\hat{f}$ is of moderate decrease to show that the series converges absolutely thus it converges uniformly to $f$

Now, my problem is with the part

b) Prove that if $F$ is continuous and of moderate decrease then 

<p>$\int^{\infty}_{-\infty} F({\xi}) d\xi = \operatorname{lim}_{\delta \to 0, \delta &gt; 0}  \delta \sum F(n\delta) $</p>

where all sums are taken from $-\infty$ to $\infty$

The book suggests approximating the integral by $\int^{N}_{-N} F$ and the series by $\delta \sum_{|n| \leq N/\delta} F(n\delta) $ then approximating the second integral by Riemann sums.

It isn't hard to see that for $ \epsilon > 0$ $\int_{|n| > N} F < \epsilon$ as $F$ is of moderate decrease

However, I'm stuck with what I should be doing now. Any help is appreciated.

The motivation here is that we want the Fourier inversion formula as a conclusion from the above two parts.

Lost
  • 2,095
  • What exactly do you mean by 'second integral'? The sum with the (finite) $\delta$ already is a Riemann sum. – Thomas Mar 25 '14 at 18:19
  • 1
    @Thomas That's what the book says. I wasn't sure what it meant either - maybe it means $\int^{N}_{-N}F$?

    As for the series, I'm confused about the $|n|\leq N/\delta$ bounds. How can we work with it?

     – Lost Mar 25 '14 at 23:25

1 Answers1

3

The sum is already a Riemann sum, I think the catch is just to formalize the infinite integration domain. There isn't much of anything to do beyond what's obvious. An integral like that is defined as a limit:

$$\int_{-\infty}^\infty=\lim_{a,b\to \infty}\int_{-a}^b$$

You have to do this here as well:

$$\lim_{a,b\to\infty}\lim_{\delta\to 0^+}\delta \left(\sum_{n=0}^{b/\delta} F(n\delta)+\sum_{n=1}^{a/\delta} F(-n\delta)\right) $$

Notice that I split the integral in two. We aren't allowed to use the same value for lower and upper integration boundary because that would be a Cauchy principal value, not a true integral.

Now this is a finite interval and you are formally allowed to treat it as a Riemann integral: $$\lim_{a,b\to\infty}\left(\int_0^b F(x)dx+\int_{-a}^0F(x)dx\right) $$

Because $|F(x)|\leq \frac{1}{1+x^2}$, these integrals converge and you are allowed to take the limit.

If you exchange the limits, you get infinite sums that you can bound by $n^{-2}$ so they converge as well.

As I said, it's all the matter of strictness, but the idea is already stated in your book.

orion
  • 15,781
  • I was uncertain about why the series was a Riemann sum as I haven't seen those show up much, but after some thought, I was able to see why it worked. Thanks for the answer. – Lost Apr 01 '14 at 04:47
  • Howcome we can interchange the limits $\lim_{a,b\rightarrow \infty}$ and $\lim_{\delta \rightarrow 0^{+}}$? How does using the identities above imply $ \int_{-\infty}^{\infty} F(\xi)d\xi = \lim_{\delta\rightarrow 0^+}\delta \sum_{\color{orange}{-\infty}}^{\color{orange}{\infty}} F(n\delta)$, because by continuity and moderate decrease I can see by Riemann integration/sums that $\int_{-N}^{N} F(\xi)d\xi = \lim_{\delta \rightarrow 0^+} \delta\sum_{-N/\delta}^{N/\delta} F(n\delta)$ the issue is as $N\rightarrow \infty$? – no lemon no melon Nov 02 '21 at 04:53