Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I do not understand one thing in an article on the blog of Terence Tao:

For instance, restricting a function $f: G \rightarrow \mathbb{C}$ to a subgroup $H$ causes the Fourier transform $\hat f$ to be averaged along the dual group $\widehat{H}$. In particular, restricting a function $f: \mathbb{R} \rightarrow \mathbb{C}$ to the integers (and renormalising it to become the measure $\sum_{n \in \mathbb{Z}} f(n) \delta_n$) causes the Fourier transform $\hat f: \mathbb{R} \rightarrow \mathbb{C}$ to become summed over the dual group $\mathbb{Z}^\perp = \mathbb{Z}$ to become the function $\sum_{m \in \mathbb{Z}} \hat f(\cdot+m)$. In particular, the zero Fourier coefficient of $\sum_{n \in \mathbb{Z}} f(n) \delta_n$ is $\sum_{m \in \mathbb{Z}} \hat f(m)$.

The thing that I do not understand is "to become the function $\sum_{m \in \mathbb{Z}} \hat f(\cdot+m)$."

Could someone give an explanation of this point?

share|improve this question
    
At first glance, I'd guess that it just means that if the Fourier transform of $f$ is $\hat f$, then the Fourier transform of $x\mapsto\sum\limits_{n\in\mathbb Z}f(n)\delta_n(x)$ is $\xi\mapsto\sum\limits_{m\in\mathbb Z}\hat f(\xi+m)$. –  Rahul Feb 8 '13 at 0:57

1 Answer 1

up vote 8 down vote accepted

I will present the same idea in a slightly less abstract way, i.e., without resorting to Pontryagin duality. This is sketched in Stein and Shakarchi's undergrad textbook, Fourier Analysis. I haven't Tao's post but I believe this is what you mean.

One assumes that the Fourier inversion formula holds everywhere, in particular by restricting oneself to the class of Schwartz functions (if this doesn't mean anything to you, don't worry -- just assume you're dealing with very 'nice' functions for the time being).

Basically, if you have a real valued function $f(x) : \mathbb{R} \rightarrow \mathbb{C}$, you can obtain the ``periodic version'' of $f$ by the formula $$f_p(x) = \sum_{n=-\infty}^\infty f(x+n).$$ (this is the formula you're having trouble with, i.e., the $f(\cdot+m)$ thing, I believe.) Notice that $f_p(x) = f_p(x+1)$, and this holds for all $x$, so $f_p$ is periodic of period 1. Also notice you require the function be nice to ensure that $f_p(x) < \infty$ everywhere ( the sum must converge).

Now what if you just took the discrete Fourier series of a real valued function? i.e., what if you had $$g(x) = \sum_{n=-\infty}^\infty \hat{f}(n)e^{2\pi i x n}?$$ Then $g$ is also periodic, of period 1. Turns out these two approaches give you the same function, i.e., $g(x) = f_p(x)$ for all $x \in \mathbb{R}$.

Proof: By the uniqueness theorem for Fourier transforms, i.e., if the Fourier coefficients of a sufficiently nice function agree, then the two functions agree almost everywhere. In other words, you want to show that the Fourier series of $f_p$ is that of $g$. By simple computation:

\begin{align} \int_{0}^{1} \left[ \sum_{n = - \infty}^{\infty} f(x + n) \right] e^{-2 \pi ixm} ~ d{x} &= \sum_{n = - \infty}^{\infty} \left[ \int_{0}^{1} f(x + n) e^{-2 \pi ixm} ~ d{x} \right] \quad (\text{By ‘niceness’.}) \\ &= \sum_{n = - \infty}^{\infty} \left[ \int_{n}^{n + 1} f(y) e^{-2 \pi i(y - n)m} ~ d{y} \right] \quad (\text{Change of variables.}) \\ &= \sum_{n = - \infty}^{\infty} \left[ \int_{n}^{n + 1} f(y) e^{-2 \pi iym} ~ d{y} \right] \\ &= \int_{- \infty}^{\infty} f(y) e^{-2 \pi iym} ~ d{y} \\ &= \hat{f}(m). \end{align} Thus, $$\sum_{-\infty}^\infty f(x+n) = \sum_{-\infty}^\infty \hat{f}(n)e^{2\pi i nx}.$$ Evalute this at $x = 0$, and you get $$\sum_{-\infty}^\infty f(n) = \sum_{-\infty}^\infty \hat{f}(n).$$ In other words, summing the Fourier transform (on the real line) of a function at the integers is the same thing as summing the function itself at the integers.

Does that help?

share|improve this answer
1  
snarski, I just corrected a few typesetting errors. :) –  Haskell Curry Feb 16 '13 at 18:00
1  
@Haskell Thanks! I'm looking through them; my LaTeX style is pretty old. I've been told I should use square brackets and such -- why? And what is the advantage of \quad vs. \tag and such? Thanks again :) –  snarski Feb 16 '13 at 18:01
2  
+1 nice explanation. As for LaTeX (without looking at the revisions), I remark that the limitations of MathJaX (and of its setup here) pretty much force users to use one style for Math.SE and another for actual documents. E.g., one cannot use \[ for delimiters here, and spacing around \colon looks (at least to me) worse than : when used in f\colon X\to Y (elsewhere it is the other way around). Oh well. –  user53153 Feb 16 '13 at 19:56

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.