The Fourier transform arises in many different ways. Historically, it first arose as a sort of limiting procedure for Fourier series. The goal was to extend the theory of Fourier series to functions which were aperiodic. It arose again in the context of Banach algebras as a special case of what is known as a Gelfand transform. The Fourier transform can also be arrived at by considering the spectral properties of the Laplace operator - this is quite similar in nature to the way it was discovered historically. Perhaps the most elegant approach to establishing the Fourier transform is as a lifted representation of characters on $\mathbb{R}$ to the $L^1(\mathbb{R})$ algebra. I will discuss this last characterization in an upcoming post.
Characterizing the Fourier Transform
Characterizing the Fourier Transform
The Gaussian - as any mathematician or physicst knows - plays an instrumental role in mathematics. It arises in the central limit theorem, in Brownian motion, as a Green's function for the heat equation and many other places. Particularly, the Gaussian, denoted $g$, plays a critical role in the theory of the Fourier transform as it is not only an eigenfunction of the Fourier transform (meaning $\mathcal{F}g=g$), but it is also the minimizer for the Heisenberg uncertainty product. In the literature, the role the Gaussian plays is viewed as a happy coincidence.
In this post, I give a new way to characterize the Fourier transform. This characterization is not equivalent to any of the ones above and is a bit unconventional in the following way. Typically, an operator is defined and, among other things, its spectral properties analyzed, including its eigenvectors. In this post, I present the following characterization for the Fourier transform. It is the integral transform $\mathcal{F}$ with kernel $\varphi$ which satisfies
- $\mathcal{F}g = g$,
- $\varphi(\omega, t) = f(\omega t)$ for some complex-valued $f$,
- $\varphi:\mathbb{R}^2\to\mathbb{C}$ is real analytic,
- If $\varphi = c + is$, where $c$ and $s$ are real-valued, then $c$ is even and $s$ is odd,
- $c$ and $s$ satisfy the same differential equation, and
- $\mathcal{F}$ is an isometry when restricted to a dense subspace of $L^2(\mathbb{R})$.
In essence, the Gaussian is taken to be the defining characteristic for the Fourier transform; the rest is added to ensure uniqueness.
Deriving the Fourier Transform
Since $\varphi$ is real analytic, we can express it as a power series which converges everywhere. In our first condition, note that both sides are real-valued. This particularly means that when we integrate $s$ against the Gaussian, we must get zero - else the left hand side would be complex in general. As such, we cannot hope to uncover what $s$ must be from property 1. Moreover, if we added any slowly growing odd function (e.g. $\omega t$) to $c$, the integration against the Gaussian would be unchanged. Thus to have uniqueness we must require that $c$ be even.
Thus we can restrict our attention to property 1 in the context of only $c$:
$$e^{-\frac{\omega^2}{2}} = \int_{-\infty}^{\infty} c(\omega t) e^{-\frac{t^2}{2}}\,dt.$$
Writing $c(\eta) = \sum_{n=0}^{\infty} c_n \eta^{2n}$ (note that we are using the assumption that $c$ is even), we get
$$e^{-\frac{\omega^2}{2}} = \int_{-\infty}^{\infty} \sum_{n=0}^{\infty} c_n(\omega t)^{2n} e^{-\frac{t^2}{2}}\,dt.$$
For now let us forsake rigor and simply interchange integral and summation:
$$e^{-\frac{\omega^2}{2}} = \sum_{n=0}^{\infty} 2 c_n \omega^{2n}\int_0^{\infty} t^{2n}e^{-\frac{t^2}{2}}\,dt.$$
Making a change of variable $z = \frac{t^2}{2}$, this becomes
$$e^{-\frac{\omega^2}{2}} = \sum_{n=0}^{\infty} 2 c_n \omega^{2n}\int_0^{\infty} ((2z)^{\frac{1}{2}})^{2n-1} e^{-z}\,dz = \sum_{n=0}^{\infty}2^{n+\frac{1}{2}} c_n\omega^{2n} \int_0^{\infty} z^{n-\frac{1}{2}} e^{-z}\,dz.$$
$$e^{-\frac{\omega^2}{2}} = \sum_{n=0}^{\infty} 2 c_n \omega^{2n}\int_0^{\infty} t^{2n}e^{-\frac{t^2}{2}}\,dt.$$
Making a change of variable $z = \frac{t^2}{2}$, this becomes
$$e^{-\frac{\omega^2}{2}} = \sum_{n=0}^{\infty} 2 c_n \omega^{2n}\int_0^{\infty} ((2z)^{\frac{1}{2}})^{2n-1} e^{-z}\,dz = \sum_{n=0}^{\infty}2^{n+\frac{1}{2}} c_n\omega^{2n} \int_0^{\infty} z^{n-\frac{1}{2}} e^{-z}\,dz.$$
This integral can be immediately recognized as the gamma function evaluated at $n+\frac{1}{2}$. One of the key properties of the gamma function is that for natural numbers $n$,
$$\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n} n!}.$$
Our expression then becomes
$$e^{-\frac{\omega^2}{2}}=\sqrt{2\pi}\sum_{n=0}^{\infty}\frac{(2n)!c_n}{2^nn!} \omega^{2n}.$$
Since $e^{-\frac{\omega^2}{2}}$ is an analytic function, we can write it as a power series and thus equate coefficients on both sides of the above equation:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2^n n!}\omega^{2n} = \sqrt{2\pi}\sum_{n=0}^{\infty} \frac{(2n)!c_n}{2^nn!} \omega^{2n}.$$
Hence $c_n = \frac{(-1)^n}{(2n)!}$, which gives
$$c(\eta) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\eta^{2n} = \frac{1}{\sqrt{2\pi}}\cos(\eta)$$
as desired. At this point, an application of Fubini-Tonelli justifies interchanging our limiting procedures. Alternatively, it can be deduced via uniform convergence of the power series for $c$.
Now that we have correctly deduced $c$, we are only tasked with determining $s$. To do so, we must consider what differential equation(s) $c$ solves. The most obvious differential equation that $\cos(\eta)$ solves is
$$\frac{d^2}{d\eta^2}\cos(\eta) = -\cos(\eta).$$
As such, we wish to find the odd solution to the equation
$$\frac{d^2}{d\eta^2} f = -f$$
to determine $s$. From basic differential equation theory, it is clear that $f(\eta) = C\sin(\eta)$, where $C$ is to be determined. The way by which to determine $C$ is by considering condition 6. For our purposes, we need only to pick an odd function in $L^2(\mathbb{R})$. The simplest such function is $f(t) = te^{-\frac{t^2}{2}}$.
Computing $\mathcal{F}f$, we get
\begin{eqnarray*}
\mathcal{F}f(\omega) &=& Ci\int_{-\infty}^{\infty} \sin(\omega t)t e^{-\frac{t^2}{2}}\,dt \\
&=& Ci\int_{-\infty}^{\infty} \frac{e^{i\omega t} - e^{-i\omega t}}{2i} t e^{-\frac{t^2}{2}}\,dt \\
&=& \frac{C}{2}\int_{-\infty}^{\infty} te^{-\frac{t^2}{2}+i\omega t}\,dt - \frac{C}{2}\int_{-\infty}^{\infty} t e^{-\frac{t^2}{2} - i\omega t}\,dt
\end{eqnarray*}
Employing the standard completing the square trick, we can rewrite these as $-\frac{t^2}{2} + i\omega t = -\frac{t^2}{2} + i\omega t + \frac{\omega^2}{2} - \frac{\omega^2}{2} = -\left(\frac{t}{\sqrt{2}} -i\frac{\omega}{\sqrt{2}}\right)^2 - \frac{\omega^2}{2}$ and similarly $-\frac{t^2}{2} - i\omega t = -\frac{t^2}{2} - i\omega t + \frac{\omega^2}{2} - \frac{\omega^2}{2} = -\left(\frac{t}{\sqrt{2}} + i \frac{\omega}{\sqrt{2}}\right)^2 - \frac{\omega^2}{2}$. Our integrals then become
$$\mathcal{F}f(\omega) = \frac{C}{2}e^{-\frac{\omega^2}{2}} \left(\int_{-\infty}^{\infty} t e^{-\left(\frac{t-i\omega}{\sqrt{2}}\right)^2}\,dt - \int_{-\infty}^{\infty} t e^{-\left(\frac{t+i\omega}{\sqrt{2}}\right)^2}\,dt\right)$$
Naively, one would make a change of variable of $z = t\pm i\omega$ but then our integral get shifted to one in the complex plane. To mitigate that, we simply note that our integrands are entire functions and thus have zero residues. So if we made contour integrals which were boxes with segments $[-R,R]$, $[-R\pm i\frac{\omega}{\sqrt{2}}, -R\mp i\frac{\omega}{\sqrt{2}}]$ (and their adjoining vertical segments), we would get a value of zero.
It is not hard to argue that the value of the contour integral along the vertical segments goes to zero as $R$ tends to infinity, thus the integral of our function along the real axis is equal to its integral on the shifted axis after a change of variable. This justifies the naive change of variable.
As such, we are left with evaluating
$$\mathcal{F}f(\omega) = \frac{C}{2}e^{-\frac{\omega^2}{2}}\left(\int_{-\infty}^{\infty} (t+i\omega) e^{-\frac{t^2}{2}}\,dt - \int_{-\infty}^{\infty} (t - i\omega) e^{-\frac{t^2}{2}}\,dt \right).$$
Making use of the oddness of $te^{-\frac{t^2}{2}}$, this simplifies immediately to
$$\mathcal{F}f(\omega) = -iC\omega e^{-\frac{\omega^2}{2}} \int_{-\infty}^{\infty} e^{-\frac{t^2}{2}}\,dt = -i\sqrt{2\pi}C\omega e^{-\frac{\omega^2}{2}}.$$
$$\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n)!\sqrt{\pi}}{2^{2n} n!}.$$
Our expression then becomes
$$e^{-\frac{\omega^2}{2}}=\sqrt{2\pi}\sum_{n=0}^{\infty}\frac{(2n)!c_n}{2^nn!} \omega^{2n}.$$
Since $e^{-\frac{\omega^2}{2}}$ is an analytic function, we can write it as a power series and thus equate coefficients on both sides of the above equation:
$$\sum_{n=0}^{\infty} \frac{(-1)^n}{2^n n!}\omega^{2n} = \sqrt{2\pi}\sum_{n=0}^{\infty} \frac{(2n)!c_n}{2^nn!} \omega^{2n}.$$
Hence $c_n = \frac{(-1)^n}{(2n)!}$, which gives
$$c(\eta) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\eta^{2n} = \frac{1}{\sqrt{2\pi}}\cos(\eta)$$
as desired. At this point, an application of Fubini-Tonelli justifies interchanging our limiting procedures. Alternatively, it can be deduced via uniform convergence of the power series for $c$.
Now that we have correctly deduced $c$, we are only tasked with determining $s$. To do so, we must consider what differential equation(s) $c$ solves. The most obvious differential equation that $\cos(\eta)$ solves is
$$\frac{d^2}{d\eta^2}\cos(\eta) = -\cos(\eta).$$
As such, we wish to find the odd solution to the equation
$$\frac{d^2}{d\eta^2} f = -f$$
to determine $s$. From basic differential equation theory, it is clear that $f(\eta) = C\sin(\eta)$, where $C$ is to be determined. The way by which to determine $C$ is by considering condition 6. For our purposes, we need only to pick an odd function in $L^2(\mathbb{R})$. The simplest such function is $f(t) = te^{-\frac{t^2}{2}}$.
Computing $\mathcal{F}f$, we get
\begin{eqnarray*}
\mathcal{F}f(\omega) &=& Ci\int_{-\infty}^{\infty} \sin(\omega t)t e^{-\frac{t^2}{2}}\,dt \\
&=& Ci\int_{-\infty}^{\infty} \frac{e^{i\omega t} - e^{-i\omega t}}{2i} t e^{-\frac{t^2}{2}}\,dt \\
&=& \frac{C}{2}\int_{-\infty}^{\infty} te^{-\frac{t^2}{2}+i\omega t}\,dt - \frac{C}{2}\int_{-\infty}^{\infty} t e^{-\frac{t^2}{2} - i\omega t}\,dt
\end{eqnarray*}
Employing the standard completing the square trick, we can rewrite these as $-\frac{t^2}{2} + i\omega t = -\frac{t^2}{2} + i\omega t + \frac{\omega^2}{2} - \frac{\omega^2}{2} = -\left(\frac{t}{\sqrt{2}} -i\frac{\omega}{\sqrt{2}}\right)^2 - \frac{\omega^2}{2}$ and similarly $-\frac{t^2}{2} - i\omega t = -\frac{t^2}{2} - i\omega t + \frac{\omega^2}{2} - \frac{\omega^2}{2} = -\left(\frac{t}{\sqrt{2}} + i \frac{\omega}{\sqrt{2}}\right)^2 - \frac{\omega^2}{2}$. Our integrals then become
$$\mathcal{F}f(\omega) = \frac{C}{2}e^{-\frac{\omega^2}{2}} \left(\int_{-\infty}^{\infty} t e^{-\left(\frac{t-i\omega}{\sqrt{2}}\right)^2}\,dt - \int_{-\infty}^{\infty} t e^{-\left(\frac{t+i\omega}{\sqrt{2}}\right)^2}\,dt\right)$$
Naively, one would make a change of variable of $z = t\pm i\omega$ but then our integral get shifted to one in the complex plane. To mitigate that, we simply note that our integrands are entire functions and thus have zero residues. So if we made contour integrals which were boxes with segments $[-R,R]$, $[-R\pm i\frac{\omega}{\sqrt{2}}, -R\mp i\frac{\omega}{\sqrt{2}}]$ (and their adjoining vertical segments), we would get a value of zero.
It is not hard to argue that the value of the contour integral along the vertical segments goes to zero as $R$ tends to infinity, thus the integral of our function along the real axis is equal to its integral on the shifted axis after a change of variable. This justifies the naive change of variable.
As such, we are left with evaluating
$$\mathcal{F}f(\omega) = \frac{C}{2}e^{-\frac{\omega^2}{2}}\left(\int_{-\infty}^{\infty} (t+i\omega) e^{-\frac{t^2}{2}}\,dt - \int_{-\infty}^{\infty} (t - i\omega) e^{-\frac{t^2}{2}}\,dt \right).$$
Making use of the oddness of $te^{-\frac{t^2}{2}}$, this simplifies immediately to
$$\mathcal{F}f(\omega) = -iC\omega e^{-\frac{\omega^2}{2}} \int_{-\infty}^{\infty} e^{-\frac{t^2}{2}}\,dt = -i\sqrt{2\pi}C\omega e^{-\frac{\omega^2}{2}}.$$
Since $f$ is actually an eigenfunction of $\mathcal{F}$ with eigenvalue $-iC\sqrt{2\pi}$, the only way for $\mathcal{F}$ to be an isometry on $L^2(\mathbb{R})$ is if $C = \pm \frac{1}{\sqrt{2\pi}}$. Thus we obtain the following functional form for $\varphi$:
$$\varphi(\omega t) = \frac{1}{\sqrt{2\pi}} e^{\pm i \omega t}$$
and thus the Fourier transform emerges. The $\pm$ is to be expected since the Fourier transform is only unique up to a sign in the exponent.
$$\varphi(\omega t) = \frac{1}{\sqrt{2\pi}} e^{\pm i \omega t}$$
and thus the Fourier transform emerges. The $\pm$ is to be expected since the Fourier transform is only unique up to a sign in the exponent.
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