Defining the Fourier transform on LCA groups

My previous post said that all the familiar variations on Fourier transforms—Fourier series analysis and synthesis, Fourier transforms on the real line, discrete Fourier transforms, etc.—can be unified into a single theory. They’re all instances of a Fourier transform on a locally compact Abelian (LCA) group. The difference between them is the underlying group.

Given an LCA group G, the Fourier transform takes a function on G and returns a function on the dual group of G. We said this much last time, but we didn’t define the dual group; we just stated examples. We also didn’t say just how you define a Fourier transform in this general setting.

Characters and dual groups

Before we can define a dual group, we have to define group homomorphisms. A homomorphism between two groups G and H is a function h between the groups that preserves the group structure. Suppose the group operation is denoted by addition on G and by multiplication on H (as it will be in our application), saying h preserves the group structure means

h(x + y) = h(x) h(y)

for all x and y in G.

Next, let T be the unit circle, i.e. complex numbers with absolute value 1. T is a group with respect to multiplication. (Why T for circle? This is a common notation, anticipating generalization to toruses in all dimensions. A circle is a one-dimensional torus.)

Now a character on G is a continuous homomorphism from G to T. The set of all characters on G is the dual group of G. Call this group Γ. If G is an LCA group, then so is Γ.


The classical Fourier transform is defined by an integral. To define the Fourier transform on a group we have to have a way to do integration on that group. And there’s a theorem that says we can always do that. For every LCA group, there exists a Haar measure μ, and this measure is nice enough to develop our theory. This measure is essentially unique: Any two Haar measures on the same LCA group must be proportional to each other. In other words, the measure is unique up to multiplying by a constant.

On a discrete group—for our purposes, think of the integers and the integers mod m—Haar measure is just counting; the measure of a set is the number of things in the set. And integration with respect to this measure is summation.

Fourier transform defined

Let f be a function in L¹(G), i.e. an absolutely integrable function on G. Then the Fourier transform of f is a function on Γ defined by

\hat{f}(\gamma) = \int_G f(x)\, \gamma(-x) \, d\mu

What does this have to do with the classical Fourier transform? The classical Fourier transform takes a function of time and returns a function of frequency. The correspondence between the classical Fourier transform and the abstract Fourier transform is to associate the frequency ω with the character that takes x to the value exp(iωx).

There are multiple slightly different conventions for the classical Fourier transform cataloged here. These correspond to different constant multiples in the choice of measure on G and Γ, i.e. whether to divide by or multiply by √(2π), and in the correspondence between frequencies and characters, whether ω corresponds to exp(±iωx) or exp(±2πiωx).

Unified theory of Fourier transforms

You can take a periodic function and analyze it into its Fourier coefficients, or use the Fourier coefficients in a sum to synthesize a periodic function. You can take the Fourier transform of a function defined on the whole real line and get another such function. And you can compute the discrete Fourier transform via the FFT algorithm.

Is there a general theory that unifies all these related but different things? Why yes, yes there is.


Everything in the opening paragraph is simply a Fourier transform, each in a different context. And the contexts correspond to groups. Specifically, locally compact Abelian groups.

Some of these groups are easier to see than others. Clearly the real numbers with addition form a group: the sum of two real numbers is a real number, etc. But where are the groups in the other contexts?

You can think of a periodic function as a function on a circle; the function values have to agree at both ends of an interval, so you might as well think of those two points as the same point, i.e. join them to make a circle. Shifting along an interval, wrapping around if necessary, corresponds to a rotation of the circle, and rotations form a group. So analyzing a periodic function in to a set of Fourier coefficients is a Fourier transform on the circle.

You can think of a set of Fourier coefficients as a function on the integers, mapping n to the nth coefficient. Synthesizing a set of Fourier coefficients into a periodic function is a Fourier transform on the group of integers.

What about a discrete Fourier transform (DFT)? If you have a function sampled at m points, you could think of those points as the group of integers mod m. Your sampled points constitute a function on the integers mod m, and the DFT is a Fourier transform on that group.

Note that the DFT is a Fourier transform in its own right. It’s not an approximation per se, though it’s nearly always used as part of an approximation process. You can start with a continuous function, approximate it by a finite set of samples, compute the DFT of these samples, and the result will give you an approximation to the Fourier transform of the original continuous function.

What about functions of several variables? These are functions on groups too. A function of two real variables, for example, is a function on R², which is a group with (vector) addition.

Dual groups

A Fourier transform takes a function defined on a group and returns a function defined on the dual of that group. I go into exactly what a dual group is in my next post, but for now, just note that the Fourier transform takes a function defined on one group and returns a function defined on another group.

The dual of the circle is the integers, and vice versa. That’s why the Fourier transform of a function on the circle is an infinite set of Fourier coefficients, which we think of as a function on the integers. The Fourier transform of the function on the integers, i.e. a set of Fourier coefficients, is a function on the circle, i.e. a periodic function.

The dual group of the real numbers is the real numbers again. That’s why the Fourier transform of a function on the real line is another function on the real line.

The integers mod m is also its own dual group. So the DFT takes a set of m numbers and returns a set of m numbers.

Locally compact Abelian (LCA) groups

What do locally compact and Abelian mean? And why do we make these assumptions?

Let’s start with Abelian. This just means that the group operation is commutative. When we’re adding real numbers, or composing rotations of a circle, these operations are commutative.

Locally compactness is a more technical requirement. The circle is compact, and so are the integers mod m. But if we restricted out attention to compact groups, that would leave out the integers and the real numbers. These spaces are not compact, but they’re locally compact, and that’s enough for the theory to go through.

It turns out that LCA groups are a sort of theoretical sweet spot. Assuming groups are LCA is general enough to include the examples we care about the most, but it’s not so general that the theory becomes harder and the results less powerful.

More connections

This post relates Fourier series (analysis and synthesis) to Fourier transforms (on the real line) by saying they’re both special cases of Fourier analysis on LCA groups. There are a couple other ways to connect Fourier series and Fourier transforms.

You can take the Fourier transform (not Fourier series) of a periodic function two ways: by restricting it to one period and defining it to be zero everywhere else, or by letting it repeat forever across the real line and taking the Fourier transform in the sense of generalized functions. You can read more about these two approaches in this post.

Fourier transform of a function on a graph

What is a Fourier transform at its core? An expansion of function in terms of eigenfunctions of the Laplacian. For a function on the real line, the Laplacian is simply the second derivative. The functions mapped to multiples of themselves by taking second derivatives are sines and cosines of various frequencies. A Fourier series is a change of basis, using as basis vectors those functions who behave the simplest under the second derivative.

The Fourier transform of a function on a graph is also a change of basis, expanding a discrete function in terms of eigenvalues of the Laplacian, in this case the graph Laplacian.

The Fourier transform of a function f, evaluated at a frequency ω, is the inner product of f with the eigenfunction exp(2πiωt).

\hat{f}(\omega) = \langle f, \exp(2\pi i \omega t) \rangle = \int_{-\infty}^\infty f(t) \exp(-2\pi i \omega t) \, dx

The inner product of two complex functions f and g is the integral of the product of f and the conjugate of g. Conjugation is why exp(2πiωt) became exp(-2πiωt).

The Fourier transform of a discrete function f on a graph, evaluated at an eigenvalue λi, is the inner product of f (i.e. the vector of values of f at each node) with the eigenvector associated with λi.

\hat{f}(\lambda_i) = \langle f, v^*_i \rangle = \sum_{j=1}^N f(j) v_i^*(j)

Here the inner product is a discrete sum rather than an integral. As before, we take the complex conjugate of the second item in the product.

The eigenvectors associated with the smallest eigenvalues of the graph Laplacian are analogous to low frequency sines and cosines. The eigenvalue components corresponding to nearly vertices in a graph should be close together. This analogy explains why spectral coordinates work so well.

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Discrete Laplace transform

The relationship between the discrete Laplace transform and discrete Fourier transform is not quite the same as that between their continuous counterparts.

Continuous Fourier and Laplace transforms

The continuous versions of the Fourier and Laplace transforms are given as follows.

Fourier transform:

{\cal F}(f)(\omega) = \int_{-\infty}^\infty \exp(-i\omega x) f(x)\, dx

Laplace transform:

{\cal L}(f)(s) = \int_0^\infty \exp(s x) f(x)\, dx

The Fourier transform is defined several ways, and I actually prefer the convention that puts a factor of 2π in the exponential, but the convention above makes the analogy with Laplace transform simpler. There are two differences between the Fourier and Laplace transforms. The Laplace transform integrates over only half the real line, compared to the entire real line for Fourier. But a variation on the Laplace transform, the Bilateral Laplace transform integrates over the entire real line. The Bilateral Laplace transform at s is simply the Fourier transform at xis. And of course the same is true for the (one-sided) Laplace transform if the function f is only non-zero for positive values.

I’ve encountered the Fourier transform more in application, and the Laplace transform more in teaching. This is not to say the Laplace transform isn’t used in practice; it certainly is used in applications. But the two transforms serve similar purposes, and the Laplace transform is easier to teach. Because the factor exp(-sx) decays rapidly, the integral defining the Laplace transform converges for functions where the integral defining the Fourier transform would not. Such functions may still have Fourier transforms, but the transforms require distribution theory whereas the Laplace transforms can be computed using basic calculus.

Discrete Fourier and Laplace Transforms

There’s more difference between the discrete versions of the Fourier and Laplace transforms than between the continuous versions.

The discrete Fourier transform (DFT) approximates the integral defining the (continuous) Fourier transform with a finite sum. It discretizes the integral and truncates its domain. The discrete Laplace transform is an infinite sum. It discretizes the integral defining the Laplace transform, but it does not truncate the domain. Given a step size η > 0, the discrete Laplace transform of f is

{\cal L}_\eta(f)(s) = \eta \sum_{n=0}^\infty \exp(-sn\eta) f(n\eta)

The discrete Laplace transform isn’t “as discrete” as the discrete Fourier transform. The latter takes a finite sequence and returns a finite sequence. The former evaluates a function at an infinite number of points and produces a continuous function.

The discrete Laplace transform is used in applications such as signal processing, as well as in the theory of analytic functions.

Connection with the z-transform and generating functions

If η = 1 and z = exp(-s), the discrete Laplace transform becomes the z-transform of the values of f at non-negative integers. And if we replace z with 1/z, or equivalently set z = exp(s) instead of z = exp(-s), we get the generating function of the values of f at non-negative integers.

z-transforms are common in digital signal processing, while generating functions are common in combinatorics. They are essentially the same thing.

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Visualizing the DFT matrix

The discrete Fourier transform (DFT) of length N multiplies a vector by a matrix whose (j, k) entry is ωjk where ω = exp(-2πi/N), with j and k running from 0 to – 1. Each element of the matrix is a rotation, so if N = 12, we can represent each element by an hour on a clock. The angle between the hour hand and minute hand corresponds to the phase of the matrix entry. We could also view each element as a color around a color wheel. The image below does both.

The matrix representing the inverse of the DFT is the conjugate of the DFT matrix (divided by Nf, but we’re only looking at phase here, so we can ignore this rescaling.) The image below displays the DFT matrix on the left and it’s inverse on the right.

Taking the conjugate amounts to making all the clocks run backward.

The DFT is often called the FFT. Strictly speaking, the FFT is an algorithm for computing the DFT. Nobody computes a DFT by multiplying by the DFT matrix, because the FFT is faster. The DFT matrix has a lot of special structure, which the FFT takes advantage of to compute the product faster than using ordinary matrix multiplication.

By the way, there are Unicode characters for clock times on the hour, U+1F550 through U+1F55B. I created the image above by writing a script that put the right characters in a table. The colors have HSL values where H is proportional to the angle and S = L =0.8.

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Relating Fourier series and Fourier transforms

Fourier series and Fourier transforms may seem more different than they are because of the way they’re typically taught. Fourier series are presented more as a representation of a function, not a transformation. Here’s a function on an interval. We can write it as a sum of sines and cosines, just as we can write a function as a sum of powers in a power series. There’s not much emphasis on the coefficients per se. They appear inside a sum, but don’t get much attention on their own.

Fourier transforms, on the other hand, are presented as genuine transforms. Here’s a function, and here’s its transform, another function. One’s a function of time, the other a function of frequency. Or maybe both are presented as representations of the same function in two different domains, the time domain and the frequency domain.

You could think of the Fourier series as a kind of transform, taking a periodic function and mapping it to an infinite sequence, the Fourier series coefficients. And you could think of the Fourier transform as being a kind of continuous set of coefficients for representing a function, if you interpret the inversion theorem the right way.

Here are a couple connections between Fourier series and Fourier transforms. Start with a function f on an interval and compute its Fourier series. The Fourier series is periodic, so we could think of f as periodic, even though we only care about f on the interval. Instead, let’s think of extending f to be 0 everywhere outside the interval. Now we take the Fourier transform of f. The Fourier series coefficients are the Fourier transform of f evaluated at integer arguments.

Now let’s go back to thinking of f as a periodic function. What would it’s Fourier transform look like? In classical analysis, you can’t do that. Periodic functions have Fourier series but they don’t have Fourier transforms because the integral defining the latter does not converge. But by the magic of tempered distributions, we can indeed take the Fourier transform of a periodic function. The result is a weighted sum of delta distributions at each integer, and the coefficient of the delta distribution at n is the nth Fourier series coefficient.

The proof of the claim in the previous paragraph is simple once you understand the sha function Ш. Start with a function f defined on a unit interval and extended to be zero outside that interval. Convolving f with Ш make a periodic function f*Ш extending f. The Fourier transform of a convolution is the product of the convolutions. The Fourier transform of f is simply its classical Fourier transform F. The Ш function is it’s own Fourier transform, so the transform of f*Ш is FШ. Multiplying a function by Ш samples that function, and the samples of F are the Fourier coefficients of the Fourier series of f*Ш, the periodic extension of f.

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An example of coming full circle

Here’s an interesting line from Brad Osgood:

Isn’t it a little embarrassing that multibillion dollar industries seem to depend on integrals that don’t converge?

In context, he’s not saying that huge companies are blithely using bad math. Some are, but that’s not what he’s getting at here. His discussion is an example of coming full circle, where experts and novices come to the same conclusion for different reasons.

The divergent integrals Osgood refers to are Fourier transforms of certain functions. A beginner might not notice that said integrals don’t converge. An expert knows that the calculations are justified by a more sophisticated theory. Someone in-between would have objections. Experts can be casual, not because they’re ignorant of technical difficulties but because they’ve mastered these difficulties. [1]

The expert in Fourier analysis has all the technicalities in the back of his or her mind. Often these don’t need to be explicitly exercised. You can blithely go about using formal calculations that aren’t justified by the classical theory.

But the expert doesn’t entirely come full circle, not in the sense of walking in circles in the woods. It’s more like winding around a parking garage, coming back to the same (x, y) location but one level up. Sometimes the expert needs to pull out the technical machinery to avoid an error the beginner could fall into. The theory of tempered distributions, for example, doesn’t justify every calculation a novice might try.

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[1] In a nutshell, here’s the theory that justifies apparently sloppy calculations with Fourier transforms. The key is to view the function you want to transform not as a function on the real line but as a tempered distribution, a linear functional on the space of smooth, rapidly decaying test functions. A function acts on a test function by forming their product and integrating. Then use Parseval’s theorem from the classical theory as the definition in this new context, moving the transform operation from the original function to the test function. Simple, right?

The Dirac comb or Sha function

shaThe sha function, also known as the Dirac comb, is denoted with the Cyrillic letter sha (Ш, U+0428). This letter was chosen because it looks like how people visualize the function, a long series of vertical spikes. The function is called the Dirac comb for the same reason. This function is very important in Fourier analysis because it relates Fourier series and Fourier transforms. It relates sampling and periodization.  It’s its own Fourier transform, and with a few qualifiers discussed later, the only such function.

The Ш function, really the Ш distribution, is defined as

sha(x) = \sum_{n=-\infty}^\infty \delta(x-n)

Here δ(xn) is the Dirac delta distribution centered at n. The action of δ(xn) on a test function is to evaluate that function at n. You can envision Ш as an infinite sequence of spikes, one at each integer. The action of Ш on a test function is to add up its values at every integer.


The product of Ш with a function f is a new distribution whose action on a test function φ is the sum of f φ over all integers. Or you could think of the distribution as a sort of clothesline on which to hang the sampled values of f, much the way a generating function works.


Next let’s look at a function f that lives on [0, 1], i.e. is zero everywhere outside the unit interval. The convolution of f with δ(xn) is f(xn), i.e. a copy of f shifted over to live on the interval [n, n+1]. So by taking the convolution with Ш, we create copies of f all over the real line. We’ve made f into a periodic function. So instead of saying “the function f extended to create a periodic function” you can simply say f*Ш.

Fourier transform

Now let’s think about the Fourier transform of Ш. The Fourier transform of δ(x) is 1, i.e. the function equal to 1 everywhere [1].  (The more concentrated a function is, the more spread out its Fourier transform. So if you have an infinitely concentrated function δ, its Fourier transform is perfectly flat, 1. You can calculate the transform rigorously, this this is the intuition.) If you shift a function by n, you rotate its Fourier transform by exp(-2πinω). So we can compute the transform of Ш:

Fourier transform of sha = \sum_{n=-\infty}^\infty \exp(-2\pi i n \omega)

This equation only makes sense in terms of distributions. The right hand side does not converge in the classical sense; the individual terms don’t even go to zero, since each term has magnitude 1. So what kind of distribution is this thing on the right side? It is in fact the Ш function again, though this is not obvious.

To see that the exponential sum is actually the Ш function, i.e. that Ш is its own Fourier transform, we need to back up a little bit and define Fourier transform of a distribution. As usual with distributions, we take a classical theorem and turn it into a definition in a broader context.

For absolutely integrable functions, we have

\int_{-\infty}^ \infty \hat{f}(x) \, \varphi(x) \, dx = \int_{-\infty}^ \infty f(x) \, \hat{ \varphi }(x) \, dx

where the hat on top of a function indicates its Fourier transform. Inspired by the theorem above, we define the Fourier transform of a distribution f to be the functional whose action on a test function φ is given below.

 \hat{f} : \varphi \mapsto \int_{-\infty}^ \infty f(x) \, \hat{ \varphi }(x) \, dx

As we noted in a previous post, the integral above can be taken literally if f is a distribution associated with an ordinary function, but in general it means the application of the linear functional to the test function.

As a distribution, exp(-2πinω) acts on a test function φ by integrating against it. From the definition of a (classical) Fourier transform, this gives the Fourier transform of φ evaluated at n. So the Fourier transform of Ш acts on φ by summing the values of φ’s Fourier transform over all integers. By the Poisson summation formula, this is the same as summing the values of φ itself over all integers. Which is the same as applying Ш. So the Fourier transform of Ш has the same effect on test functions as Ш. In other words, Ш is its own Fourier transform.


We haven’t been explicit about where our test functions come from. We require that xn φ(x) goes to zero as x goes to ±∞ for any positive integer n. These are called functions of rapid decay. And the distributions we define as linear functionals on such test functions are called tempered distributions.

The Ш distribution is essentially unique. Any tempered distribution with period 1 that equals its own Fourier transform must be a multiple of Ш.

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[1] All Fourier transform calculations here use the convention I call (-1, τ, 1) in these notes on various definitions. This may be the most common definition, though there are several minor variations in common use.

Sinc and Jinc sums

In the previous post, we looked at an elegant equation involving integrals of the sinc function and computed the corresponding integrals for the jinc function.

\int_{-\infty}^\infty \mbox{sinc}(x) \, dx = \int_{-\infty}^\infty \mbox{sinc}^2(x) \, dx = \pi

It turns out the analogous equation holds for sums as well:

\sum_{n=-\infty}^\infty \mbox{sinc}(n) = \sum_{n=-\infty}^\infty \mbox{sinc}^2(n) = \pi

As before, we’d like to compute these two sums and see whether we can compute the corresponding sums for the jinc function.

The Poisson summation formula says that a function and its Fourier transform produce the same sums over the integers:

\sum_{n=-\infty}^\infty f(n) = \sum_{n=-\infty}^\infty \hat{f}(n)

Recall from the previous post that the Fourier transform of sinc is the function π box(π x) where the box function is 1 on [-1/2, 1/2] and zero elsewhere. The only integer n with πn inside [-1/2, 1/2] is 0, so the sum of sinc(n) over the integers equals π. A very similar argument shows that the sum of jinc(n) over the integers equals its Fourier transform at 0, which equals 2.

Let tri(x) be the triangle function, defined to be 1 – |x| for -1 < x < 1 and 0 otherwise. Then the Fourier transform of tri(x) is sinc2(π ω) and so π tri(π x) and sinc2 are Fourier transform pairs. The Poisson summation formula says the sum of sinc2 over the integers is the sum of π tri(π x) over the integers, which is π.

I don’t know the Fourier transform of jinc2 and doubt it’s easy to compute. Maybe the sum could be computed more easily without Fourier transforms, e.g. using contour integration.

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Sinc and Jinc integrals

The sinc function is defined by sinc(x) = sin(x)/x. Philip Woodward introduced the name of the function in 1952, saying it “occurs so often in Fourier analysis and its applications that it does seem to merit some notation of its own.”

Here’s an elegant equation involving the integrals of the sinc function:

\int_{-\infty}^\infty \mbox{sinc}(x) \, dx = \int_{-\infty}^\infty \mbox{sinc}^2(x) \, dx = \pi

When I ran across this recently I wondered two things: How hard is it to compute these two integrals? What are the corresponding results for the jinc function? The jinc function is analogous to sinc, but using a Bessel function in place of sine: jinc(x) = J1(x)/x.

The Fourier transform of the box function, the function box(x) that is 1 on the interval [-1/2, 1/2] and zero everywhere else, is sinc(π ω). (That’s one of the reasons sinc comes up so often in Fourier analysis, as Woodward observed.) So the Fourier transform of sinc(x) is π box(π x). The integral of a function is the value of its Fourier transform at zero, so sinc integrates to π. [1]

By Plancherel’s theorem, the integral of sinc2(x) is the integral of it’s Fourier transform squared, which equals π.

[There are several conventions for defining the Fourier transform. Here I’m using what I call the (-1, τ, 1) definition in these notes. See that page for other conventions and how to convert between them.]

Now for the jinc function. It also has a simple Fourier transform: f(ω) = 2 √(1 – (2πω)) for |x| < 1/2π and zero otherwise. As above, we can compute the integral of jinc over the real line by evaluating its Fourier transform at 0, which equals 2.

Also as above, the integral of jinc2 is the integral of its Fourier transform squared, which works out to 8/3π.

Update: See the next post for the analogous relations for sums.

Related posts:

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[1] You may have a couple objections to this calculation. I found the Fourier transform of the box function was sinc, then concluded that the transform of sinc is the box function. But applying the Fourier transform twice doesn’t give you the original function back, right? When you transform f(x) twice you get  f(-x), but the functions involved here are even, so  f(-x) =  f(x).

OK, but you may still have another objection: the sinc function does not have bounded L1 norm, so you can’t just take it’s Fourier transform. True, but you can justify the transform in terms of L2 theory or distribution theory.

Fourier analysis notes

There are six or eight ways to define a Fourier transform. The differences in the various conventions are minor, but they lead to differences in the basic results. So whenever you look up a result, you have to make sure the reference’s definition matches the one you’re expecting. Or maybe you re-derive the result. This is good exercise, but it’s a distraction when you’re in the middle of working on something else.

This has annoyed me periodically since shortly after I learned what a Fourier transform was. I’ve thought about making a Rosetta stone of sorts for Fourier transforms, listing the basic formulas using each of the various conventions, and now I finally did it. See these notes:

Related: Fourier transform results under various conventions

Generalized Fourier transforms

How do you take the Fourier transform of a function when the integral that would define its transform doesn’t converge? The answer is similar to how you can differentiate a non-differentiable function: you take a theorem from ordinary functions and make it a definition for generalized functions. I’ll outline how this works below.

Generalized functions are linear functionals on smooth (ordinary) functions. Given an ordinary function f, you can create a generalized function that maps a smooth test function φ to the integral of fφ.

There are other kinds of generalized functions. For example, the Dirac delta “function” is really the generalized function δ that maps φ to φ(0). This is the formalism behind the hand-wavy nonsense about a function infinitely concentrated at 0 and integrating to 1. “Integrating” the product δφ is really applying the linear functional δ to φ.

Now for absolutely integrable functions f and g, we have

\int_{-\infty}^\infty \hat{f} g = \int_{-\infty}^\infty f \hat{g}

In words, the integral of the Fourier transform of f times g equals the integral of f times the Fourier transform of g. This is the theorem we use as motivation for our definition.

Now suppose f is a function that doesn’t have a classical Fourier transform. We make f into a generalized function and define its Fourier transform as the linear function that maps a test function φ to the integral of f times the Fourier transform of φ.

More generally, the Fourier transform of a generalized function f is is the linear function that maps a test function φ to the action of f on the Fourier transform of φ.

This allows us to say, for example, that the Fourier transform of the constant function f(x) = 1 is 2πδ, an exercise left for the reader.

The Heisenberg uncertainty principle for ordinary functions says that the flatter a function is, the more concentrated its Fourier transform and vice versa. Generalized Fourier transforms take this to an extreme. The Fourier transform of the flattest functions, i.e. constant functions, are multiples of the most concentrated function, the delta (generalized) function.

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Teach yourself Fourier analysis in two weeks

From William Thompson (Lord Kelvin), 1840:

I had become filled with the utmost admiration for the splendor and poetry of Fourier. … I asked [John Pringle] Nichol if he thought I could read Fourier. He replied ‘perhaps.’ He thought the book a work of most transcendent merit. So on the 1st of May … I took Fourier out of the University Library; and in a fortnight I had mastered it — gone right through it.


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The opening chord of "A Hard Day’s Night"

The opening chord of the Beatles song “A Hard Day’s Night” has been something of a mystery. Guitarists have tried to reproduce the chord with limited success. Turns out there’s a good reason why they haven’t figured it out: the chord cannot be played on a guitar alone.

Jason Brown has digitally analyzed the chord using Fourier analysis and determined that there must have been a piano in the recording studio playing along with the guitars. Brown has determined what notes each member of the Beatles were playing.

I heard Jason Brown’s story on the Mathematical Moments podcast. In addition to the chord discussed above, Brown talks about other things he has discovered about the Beatles and about the relationship between music and math in general. Unfortunately, Mathematical Moments does not make it easy to link to individual episodes. Here is a link to a PDF file of show notes with the audio embedded. The file is slow to download, and your PDF viewer may not support it. Here’s a link directly to just the MP3 audio file.

The Mathematical Moments podcast also does not make it obvious that you can subscribe to the podcast; they only provide links to individual episodes with fat PDF files. However, you can subscribe by using the URL

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