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Consider a proof from a textbook on Harmonic Analysis:

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Note that $\mathcal{S}(ℝ)$ denotes the Schwartz Space.

Question 1: Why does the top left formula in the proof start out as:

$$ \int_0^1 \left( \sum_{m ∈ ℤ} \phi(x+m) \right) \mathrm{e}^{-2 \pi i n x} \,\mathrm{d}x? $$

Shouldn't the integrand be $\int_{-\infty}^{\infty}$?

Question 2: What justifies the leap from

$$ \sum_{m ∈ ℤ} \int_0^1 \phi(x+m) \mathrm{e}^{-2 \pi i n x} \,\mathrm{d}x = \sum_{m ∈ ℤ} \int_m^{m+1} ϕ(y) \mathrm{e}^{-2 \pi i n y} \,\mathrm{d}y ? $$

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  • $\begingroup$ (1) Is just the computation of the $n$-th Fourier coefficient. Recall that the $n$-th Fourier coefficient of a $1$-periodic function $f$ is the integral $\int_{0}^{1} f(x) \mathrm{e}^{-2\pi inx}\,\mathrm{d}x$. (2) This is a change of variables (let $y=x+m$) plus the observation that $\mathrm{e}^{-2\pi inm} = 1$. $\endgroup$
    – Xander Henderson
    Mar 3, 2018 at 22:12
  • $\begingroup$ But how do we know that $f$ is 1-periodic? $\endgroup$ Mar 3, 2018 at 22:12
  • $\begingroup$ $\phi((x+m )+ 1) = \phi( x + (m+1))$. Both terms are in series. $\endgroup$
    – Xander Henderson
    Mar 3, 2018 at 22:15
  • $\begingroup$ Out of curiosity, what book is this from? $\endgroup$
    – Xander Henderson
    Mar 3, 2018 at 22:34
  • $\begingroup$ The book is called Harmonic Analysis: From Fourier to Wavelets by Pereyra and Ward. $\endgroup$ Mar 4, 2018 at 12:00

4 Answers 4

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Question 1: Define the function $f$ by $$ f(x) := \sum_{m\in\mathbb{Z}} \phi(x+m). $$ Observe that $$ f(x+1) = \sum_{m\in\mathbb{Z}} \phi((x+1)+m) = \sum_{m\in\mathbb{Z}} \phi( x + (m+1) ) = \sum_{m'\in\mathbb{Z}} \phi(x+m') = f(x), $$ where $m' = m+1$. Thus $f$ is a $1$-periodic function. Since $f$ is a $1$-periodic function, we may compute its $n$-th Fourier coefficient using the usual integration, i.e. $$ \hat{f}(n) = \int_{0}^{1} f(x) \mathrm{e}^{-2\pi inx}\,\mathrm{d}x= \int_{0}^{1} f(x) \mathrm{e}^{-2\pi inx}\,\mathrm{d}x = \int_{0}^{1} \sum_{m\in\mathbb{Z}} \phi(x+m) \mathrm{e}^{-2\pi inx}\,\mathrm{d}x, $$ which answers your first question.

Question 2: This is a fairly straight-forward change of variables. Let $y = x+m$. Then (using the usual shorthand) $\mathrm{d}y = \mathrm{d}x$, which gives us \begin{align} \sum_{m\in\mathbb{Z}} \int_{0}^{1} \phi(x+m) \mathrm{e}^{-2\pi inx} \,\mathrm{d}x &= \sum_{m\in\mathbb{Z}} \int_{m}^{m+1} \phi(y) \mathrm{e}^{-2\pi in(y-m)} \,\mathrm{d}y && (\text{change of variables}) \\ &= \sum_{m\in\mathbb{Z}} \int_{m}^{m+1} \phi(y)\, \mathrm{e}^{-2\pi iny}\mathrm{e}^{2\pi inm}\, \mathrm{d}y \\ &= \sum_{m\in\mathbb{Z}} \int_{m}^{m+1} \phi(y) \mathrm{e}^{-2\pi iny}\, \mathrm{d}y. && (\text{since $\mathrm{e}^{2\pi ik} = 1 \forall k\in\mathbb{Z}$}) \end{align} This seems to answer your second question.

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  • $\begingroup$ Is there a definition of the nth Fourier coefficient that doesn't depend on knowing the periodicity of the function in question? Or does the definition of a Fourier coefficient always depend on the specific periodicity of the function in question? If so, what if the function isn't periodic? Does that mean there's no Fourier coefficient definable? $\endgroup$ Mar 4, 2018 at 11:59
  • $\begingroup$ @user1770201 The basic answer to your question is that when we talk about Fourier coefficients, we are talking about the values of the Fourier transform of a function defined on the circle (a periodic function). There is a lovely theory that generalizes this idea to locally compact abelian groups; under this generalization, the Fourier coefficients are related to the inverse Fourier transform. However, the term "Fourier coefficient" is typically only meaningful for periodic functions. In other groups, the objects are a little different, with different language used to describe those ideas. $\endgroup$
    – Xander Henderson
    Mar 5, 2018 at 1:59
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Elias Stein (famous analyst) tells a story about this theorem (folklore that I am going to mess up). He says (he gives names to at least one of these people, but I don't remember their names) a [math] professor has a class where he is talking to some of his three students. He asks the students, "Ok, given a Schwartz function, how can one get a periodic function?"

Student A says, "We can take the Fourier transform $\hat{\phi}$, apply it to integers getting $\hat{\phi}(n)$ and then form the trigonometric series $\sum_{n \in \mathbb{Z}}\hat{\phi}(n)e^{2\pi i n x}$." The Teacher says, "Good."

Student B says, "We can take the function $\phi$ and the value at $x$ should be the value gotten by adding the value of $\phi$ at all points on the real line an integer distance from $x$ getting $\sum_{n \in \mathbb{Z}}\phi(x+n)$." The Teacher responds well.

Student C (who is presumably Poisson), says, "These are both good... and they are equal!"

So, what we see that the idea is that we start with a function $\phi$ that is a Schwartz function on the real line. We then create a periodic function of period $1$ using both these formulas and the theorem says that they are equal. So, when the proof says that it just needs to make sure that they have the same Fourier coefficients, it is talking about these periodic functions that we were given have same Fourier coefficients because they are continuous. That is why in your first question they are only integrating on the interval $[0,1]$.

They are both continuous because $\phi$, being a Schwartz function, decays faster than any polynomial and so does its Fourier transform. In particular, there is a constant $C$ such that $\hat{\phi}(n) \leq \frac{C}{n^2}$ and $|\phi(y)| \leq \frac{C}{(1+y)^2}$ for all $n \in \mathbb{Z}$ and all $y\in\mathbb{R}$. Both of these bounds means that the the series converge uniformly and so the periodic functions that we made are continuous so their Fourier coefficients determine what function they are and we can interchange the integral and summation that you were wondering about.

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  • $\begingroup$ Where you wrote "We can take the function $f\, [\ldots]$" did you mean "We can take the function $\phi\, [\ldots] \text? \qquad$ $\endgroup$ Mar 4, 2018 at 0:01
  • $\begingroup$ Oops.... missed that mistake. $\endgroup$
    – user357980
    Mar 4, 2018 at 0:18
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there is lot easier than that

consider the fourier series expansion for the 1-periodic function

$$ \lfloor x\rfloor = x - \frac{1}{2} + \frac{1}{\pi} \sum_{k=1}^\infty \frac{\sin(2 \pi k x)}{k} $$

and get this into the Euler sum formula

and use Euler sum formula plus integration by parts

$$ \sum_{n=-\infty}^{\infty}f(n) = - \int_{-\infty}^{\infty}f'(x)\lfloor x] $$

and you have the Poisson sum formula, very easily

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This was meant to be a comment in reply to a comment in Xander Henderson's nice answer, but it got too long.

The abelian group $A = \{x\}$ of the domain of complex-valued functions $f(x)$ we are thinking about is always understood in the context of the problem, and then the Fourier transform $\mathcal F\colon f(x)\mapsto \hat f(\xi)$ yields a complex-valued function $\hat f(\xi)$ on the "dual" group $\hat A = \{\xi\}$.

Also it's very common that the abelian groups $A$ and $\hat A$ have topological structure (because we want to talk about convergence), or in other words a notion of "open sets," and usually a rather descript one at that (locally compact is often assumed, as well as Hausdorff), or even smooth structure if we want to study the relationship between the operator $\mathcal F$ and differential operators, or use said relationship as (a collection of) tools for something else we're working on.

When passing from a rapidly decaying function $f(x)$ on $\mathbb R$ to its periodization, $f_{per}(x) = \sum_{n=-\infty}^\infty f(x-n)$ (say as a tool to prove the Poisson summation formula, for example), we consider $f_{per}(x)$ as a function on the circle ($S^1$). Then we need to keep in mind that although

  • $\hat{\mathbb R} = \mathbb R$, hence $\hat f(\xi)$ is a function of $\xi\in\mathbb R$
  • $\hat{S^1} = \mathbb Z$, so $\hat{f_{per}}(\xi)$ is a function of $\xi\in \mathbb Z$. So we usually use the psychologically convenient variable $n$ instead of $\xi$ and write $\hat{f_{per}}(n)$.
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