General Theory of Self-Reciprocal Functions in the Fourier Transform

Fixed Points of the Fourier Transform

In the following two articles, examples of self-reciprocal functions in the Fourier transform were provided:

• Fourier Transform of the Probability Density Function of the Normal Distribution (Gaussian Function)
• An Example of a Self-Reciprocal Function in Fourier Transform

The two examples are:

$$\exp [-\frac{t^2}{2}]$$

$$\frac{1}{\sqrt{|t|}}$$

These two functions are mapped to (except for a constant multiple) the same function under the Fourier transform.

Such functions are considered fixed points of the Fourier transform, and are referred to as self-reciprocal functions in the context of the Fourier transform.

There are various other such functions.

In this article, I will discuss the general theory related to such functions.

Unlike other articles, the following definition of the Fourier transform will be used here:

$$\mathcal{F}[f](\omega )=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e^{-i\omega t}dt$$

This is the third definition from Multiple Definitions of the Fourier Transform, used to ensure the Fourier transform is a unitary transform (avoiding constant multiples).

It will also be written as:

$$\hat{f}(\omega )=\mathcal{F}[f](\omega )$$

Repeated Application of the Fourier Transform

Let’s consider repeatedly applying the Fourier transform to a function $f(t)$.

After one Fourier transform, $f(t)$ is mapped as follows:

$$\mathcal{F}:f(t)\mapsto \hat{f}(\omega )$$

Here,

$$\mathcal{F}[f](\omega )=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e^{-i\omega t}dt$$

By substituting $s=-t$ in the integral, we get:

$$\mathcal{F}[f](\omega )=\frac{1}{\sqrt{2\pi }}{\int }_{s=\mathrm{\infty }}^{-\mathrm{\infty }}f(-s)e^{i\omega s}(-ds)=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(-s)e^{i\omega s}ds={\mathcal{F}}^{-1}[f(-t)]$$

The last transformation is the inverse Fourier transform.

From this, we can see that:

$${\mathcal{F}}^2[f](t)=f(-t)$$

Continuing further:

$${\mathcal{F}}^3[f](\omega )=\hat{f}(-\omega )$$

and

$${\mathcal{F}}^4[f](t)=f(-(-t))=f(t)$$

This type of operation, where something returns to its original form after four iterations, often appears in mathematics—such as multiplying by the imaginary unit $i$, or differentiating trigonometric functions.

To summarize this diagrammatically:

Returning to the main point, the important equation obtained here is:

$${\mathcal{F}}^4=1$$

Focusing specifically on even functions:

$${\mathcal{F}}^2=1$$

Eigenvalues of the Fourier Transform

Since the Fourier transform is a linear transformation, it is natural to consider its eigenvalues.

Let $\lambda$ be an eigenvalue of the Fourier transform $\mathcal{F}$, and $f$ be the eigenvector corresponding to $\lambda$ (in this case, it’s a function, so it’s called an eigenfunction).

By definition, we have:

$$\mathcal{F}[f]=\lambda f$$

Repeating this four times gives:

$${\mathcal{F}}^4[f]={\lambda }^{4}f$$

On the other hand, from the earlier analysis:

$${\mathcal{F}}^4[f]=1[f]=f$$

Thus:

$${\lambda }^4=1$$

Therefore:

$$\lambda =1,i,-1,-i$$

Later, I will show examples, but there are indeed non-zero functions $f$ that satisfy:

$$\mathcal{F}[f]=\lambda f$$

for $\lambda =1,i,-1,-i$, meaning that these are indeed eigenvalues of $\mathcal{F}$.

Let the eigenspaces corresponding to these eigenvalues be $W_1,W_i,W_{-1},W_{-i}$, respectively.

These eigenspaces are orthogonal to each other.

In fact, for $f\in W_a$ and $g\in W_b$, by Parseval’s theorem:

$$f\cdot g=\mathcal{F}[f]\cdot \mathcal{F}[g]=(af)\cdot (bg)=a\overline{b}(f\cdot g)$$

When $a\ne b$, $a\overline{b}\ne 0$, so we deduce that $f\cdot g=0$.

Thus, any function $f\in L^2(\mathbb{R})$ can be uniquely decomposed into these eigenspaces as:

$$f=f_1+f_i+f_{-1}+f_{-i}$$

Applying the Fourier transform repeatedly to this decomposition gives:

$$\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\end{array}\right)\left(\begin{array}{c}{f}_1\\ f_i\\ f_{-1}\\ f_{-i}\end{array}\right)=\left(\begin{array}{c}f\\ \mathcal{F}[f]\\ {\mathcal{F}}^2[f]\\ {\mathcal{F}}^3[f]\end{array}\right)$$

Solving this, we can express $f_1,f_i,f_{-1},f_{-i}$ as:

$$\left(\begin{array}{c}{f}_1\\ f_i\\ f_{-1}\\ f_{-i}\end{array}\right)=\frac{1}{4}\left(\begin{array}{cccc}1& 1& 1& 1\\ 1& -i& -1& i\\ 1& -1& 1& -1\\ 1& i& -1& -i\end{array}\right)\left(\begin{array}{c}f\\ \mathcal{F}[f]\\ {\mathcal{F}}^2[f]\\ {\mathcal{F}}^3[f]\end{array}\right)$$

Here, the $f_1$ component is the projection onto the eigenspace $W_1$, which is a self-reciprocal function.

Examples of Eigenfunctions Corresponding to Eigenvalues $1,i,-1,-i$

Example of an Eigenfunction for Eigenvalue $1$

This is precisely a self-reciprocal function, and a typical example is the Gaussian function mentioned at the beginning:

$$\exp [-\frac{t^2}{2}]$$

Differentiation in the Frequency Domain

One of the important properties of the Fourier transform is the frequency differentiation formula:

$$\mathcal{F}[tf(t)]=i\frac{d\hat{f}(\omega )}{d\omega }$$

This asserts that multiplying by $t$ in the time domain corresponds to differentiation in the frequency domain.

This can be derived by calculating as follows:

$$\frac{d\hat{f}(\omega )}{d\omega }=\frac{d}{d\omega }\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)e^{-i\omega t}dt=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)\left(\frac{d}{d\omega }{e}^{-i\omega t}\right)dt=\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}f(t)(-it{e}^{-i\omega t})dt=-i\frac{1}{\sqrt{2\pi }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}tf(t)e^{-i\omega t}dt=-i\mathcal{F}[tf(t)]$$

Below, we calculate Fourier transform of $t^n\exp [-\frac{t^2}{2}]$ using this formula. In general, these provide examples of eigenfunctions.

Fourier Transform of $t^n\exp [-\frac{t^2}{2}]$

$n=1$

$$\mathcal{F}[t\exp [-\frac{t^2}{2}]]=i\frac{d}{d\omega }\exp [-\frac{{\omega }^2}{2}]=-i\omega \exp [-\frac{{\omega }^2}{2}]$$

$n=2$

$$\mathcal{F}[t^2\exp [-\frac{t^2}{2}]]=i\frac{d}{d\omega }(-i\omega \exp [-\frac{{\omega }^2}{2}])=(1-{\omega }^2)\exp [-\frac{{\omega }^2}{2}]$$

$n=3$

$$\mathcal{F}[t^3\exp [-\frac{t^2}{2}]]=i\frac{d}{d\omega }((1-{\omega }^2)\exp [-\frac{{\omega }^2}{2}])=i(-3\omega +{\omega }^3)\exp [-\frac{{\omega }^2}{2}]$$

Example of an Eigenfunction for the Eigenvalue $-i$

From the result for $n=1$ above, we can see that

$$t\exp [-\frac{t^2}{2}]$$

is an eigenfunction corresponding to the eigenvalue $-i$.

Example of an Eigenfunction for the Eigenvalue $-1$

Although the result for $n=2$ cannot be directly applied, it is almost the answer when $n=2$.

Considering the Fourier transform of

$$f(t)=(t^2+\alpha )\exp [-\frac{t^2}{2}]$$

we obtain

$$\hat{f}(\omega )=-({\omega }^2-(1+\alpha ))\exp [-\frac{{\omega }^2}{2}]$$

Therefore, if we choose $\alpha$ such that $\alpha =-(1+\alpha )$, then $\hat{f}=-f$。

Solving this equation give $\alpha =-\frac{1}{2}$. Hence,

$$f(t)=(t^2-\frac{1}{2})\exp [-\frac{t^2}{2}]$$

is an eigenfunction corresponding to the eigenvalue $-1$.

Example of an Eigenfunction for the Eigenvalue $i$

Although the result for $n=3$ cannot be directly applied, it is almost the answer when $n=3$.

Considering the Fourier transform of

$$f(t)=(t^3+\beta t)\exp [-\frac{t^2}{2}]$$

We obtain

$$\hat{f}(\omega )=i({\omega }^3-(3+\beta )\omega )\exp [-\frac{{\omega }^2}{2}]$$

Therefore, if we choose $\beta$ such that $\beta =-(3+\beta )$, then $\hat{f}=if$.

Solving this equation gives $\beta =-\frac{3}{2}$. Hence,

$$f(t)=(t^3-\frac{3}{2}t)\exp [-\frac{t^2}{2}]$$

is an eigenfunction corresponding to the eigenvalue $i$.