Intuitively, why is the Gaussian the Fourier transform of itself?

Question

It's a standard exercise to find the Fourier transform of the Gaussian $e^{-x^2}$ and show that it is equal to itself. Although it is computationally straightforward, this has always somewhat surprised me. My intuition for the Gaussian is as the integrand of normal distributions, and my intuition for Fourier transforms is as a means to extract frequencies from a function. They seem unrelated, save for their use of the exponential function.

How should I understand this property of the Gaussian, or in general, eigenfunctions of the Fourier transform? The Hermite polynomials are eigenfunctions of the Fourier transform and play a central role in probability. Is this an instance of a deeper connection between probability and harmonic analysis?

Answer 1

The generalization of this phenomenon, from a probabilistic standpoint, is the Wiener-Askey Polynomial Chaos.

In general, there is a connection between orthogonal polynomial families in the Askey scheme and probability distribution/mass functions.

Orthogonality of these polynomials can be shown in an inner product space using a weighting function -- a weight function that typically happens to be, within a scale factor, the pdf/pmf of some distribution.

In other words, we can use these orthogonal polynomials as a basis for a series expansion of a random variable:

$$z = \sum_{i=0}^\infty z_i \Phi_i(\zeta).$$

The random variable $\zeta$ belongs to a distribution we choose, and the orthogonal polynomial family to which $\Phi$ belongs follows from this choice.

The deterministic coefficients $z_i$ can be computed easily by using Galerkin's method.

So, yes. There is a very deep connection in this regard, and it is extremely powerful, particularly in engineering applications. Strangely, many mathematicians do not know this relationship!

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See also: http://www.dtic.mil/cgi-bin/GetTRDoc?AD=ADA460654 and the Cameron-Martin Theorem.

Answer 2

There's a simple reason why taking a Fourier-like transform of a Gaussian-like function yields another Gaussian-like function. Consider the property $$\mathcal{T}[f^\prime](\xi) \propto \xi \hat{f}(\xi)$$ of a transform $\mathcal{T}$. We will call an invertible transform $\mathcal{F}$ "Fourier-like" if both it and its inverse have this property.

Define a "Gaussian-like" function as one with the form $$f(x) = A e^{a x^2}.$$ Functions with this form satisfy $$f^\prime(x) \propto x f(x).$$ Taking a Fourier-like transform of each each side yields $$\xi \hat{f}(\xi) \propto \hat{f}^\prime(\xi).$$ This is has the same form as the previous equation, so it is not surprising that its solutions have the Gaussian-like form $$\hat{f}(\xi) = B e^{b \xi^2}.$$

Answer 3

Yuval Filmus' answer here Gaussian fixed point Fourier transform proves this by using the stability property of the zero-mean normal distribution (which is of course very related to the CLT) to show that the higher (i.e. above 2nd-order) cumulants are $0$, so the Fourier transform is a (zero-mean) Gaussian:

『take $X$ to be a zero-mean Gaussian. We know that $(X+X)/\sqrt{2}$ is equidistributed with $X$. We immediately deduce that all higher-order cummulants are nil, and since the second cummulant is the variance, we get that the Fourier transform is $\exp(-\sigma^2x^2/2)$.』 - Yuval Filmus

Let me present the key ideas (reverting back to more conceptual analytic arguments at the end, instead of more probabilistic or computational like cumulants/Taylor series):

• probabilistic interpretation of Fourier transform: $\hat f(t)$ (for p.d.f. $f:\mathbb R \to [0,\infty)$ of a r.v. $X$) is the same as $\mathbb E[e^{itX}]$.
• Gaussian stability $\leadsto$ functional equation: For Gaussian $X$ (which the CLT "defines" to be the limiting distribution for a large class of distributions undergoing the iteration $X \leadsto \frac{X+\tilde X}{2}$; so naturally the Gaussian is stable under this iteration), we get $$\hat f(t) = \mathbb E[e^{itX}] = \mathbb E[e^{itX/\sqrt 2}]^2 = \hat f(\frac{t}{\sqrt 2})^2$$
• This variant of Paley-Wiener tells us that $\hat f$ extends to an entire function on $\mathbb C$. The only part important for us is that $\hat f(t)$ is real analytic in a small neighborhood of $t=0$.
• Note also $\hat f(0) = \mathbb E[e^0]=1$, and because the p.d.f. of $X$ is symmetric around $0$, $\hat f(t)$ is also symmetric around $t=0$.
• Consider $g(t) := \log(\hat f(t))$ on small neighborhood of $t=0$. Satisfies functional equation $g(z) = 2 g(\frac z{\sqrt 2})$. Also, $g$ is symmetric around $t=0$, satisfies $g(0)=0$, and is real-analytic.
• The previous sentence implies that we can define an analytic function $\Phi(t) := \frac{g(t)}{t^2}$ near $t=0$, which then satisfies the functional equation $\Phi(t) = \Phi(\frac t{\sqrt 2})$. The only analytic functions that satisfy this are constant functions (by the Identity Theorem, since $\Phi$ takes on the same value at $t_0, (\frac 1{\sqrt 2})t_0, (\frac 1{\sqrt 2})^2t_0, \ldots$ and at $0$ by continuity).
• So, tracing back, we conclude $\hat f(t) = e^{Ct^2}$.

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Further remarks: I think this is decently satisfactory. But in some situations, like the uncertainty principle in https://terrytao.wordpress.com/2021/01/23/246b-notes-2-some-connections-with-the-fourier-transform/#fo, or the generalization for general Holder conjugate exponents $p,q$ (i.e. $\frac 1p+\frac 1q=1$) instead of $(p,q)=(2,2)$ (found in e.g. Stein and Shakarchi's Complex Analysis $\S4$ Problem 1) which uses the "algebraic identity" $\xi,\eta\geq 0 \implies −\xi^p + \xi \eta \leq \eta^q$ (which plays the role of the algebraic identity "complete the square" used in the $(2,2)$ case: $-\xi^2 +2\xi \eta \leq \eta^2$ in the aforementioned Tao notes), there doesn't seem to be a more intuitive way than just the algebra.

The best algebraic explanation I think is what Terry Tao says in https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics,

Perhaps the one minor miracle, though, is that the Fourier transform of a Gaussian is again a Gaussian, although once one realises that the Fourier kernel is also an exponentiated bilinear form, this is not so surprising.

Answer 4

Let's take any function that satiafies $f(x)f(y) = f(\frac{x+y}{\sqrt{2}}) f(\frac{x-y}{\sqrt{2}})$. It is not hard to directly verify that the Fourier transform $\hat f$ must satisfy the same relation. Using the result of this question, we see that any function satisfying that relation is of the form $Ae^{bx^2}$. In particular, if the function satisfying the relation is bounded and decays at infinity, then $b< 0$ so that the function must be a Gaussian pdf times a constant.

Remark: It's a bit of a deep fact in probability, that there's only one distribution that's rotationally invariant by $\pi/4$ after self tensorization. That's hiding behind a lot of things, such as CLT.