Flattening Convolution

Imagine someone tells you of a magical landscape where everywhere you go, the average elevation within 1km is always the same. Is there any way this landscape can have hills?

Essentially, I want to know whether there is a 2-dimensional real-valued function $f \ne f(p)=0$ that, when convolved with the unit disk ($|x|<1$), is constant (or without loss of generality, zero). Intuitively, this would mean that if you had a camera that blurred each point to a uniform disk, there would be a non-blank scene that would blur into a blank photo.

Formally, I'm curious whether there exists a function $f:\mathbb{R}^2\Rightarrow\mathbb{R}$ (not the constant function $f=0$) such that integration over any unit disk is 0: $$\forall p\in\mathbb{R}^2\quad\iint\limits_{|x-p|<1}f(x)\,\mathrm{d}A = \int\limits_0^{2\pi}\int\limits_0^1 f_p(r,\theta)\,r\,\mathrm{d}r\,\mathrm{d}\theta = 0$$ Where $f_p(r,\theta)=f(p_x+r\cos\theta,p_y+r\sin\theta)$. Or perhaps more simply $$f * I_\mathrm{disk}=0,\quad$$ Where $I_\mathrm{disk}$ is $1$ inside of the unit disk centered at the origin, and $0$ everywhere else, and $*$ is the convolution operator.

Using the Convolution Theorem

Work in progress - will convert away from the $e^{-\imath2\pi\omega t}$ style Fourier transform

According to this white paper, the Fourier transform ($\mathcal{F}$) of a unit disk is radially symmetric and equal to $$\mathcal{F}\{ I_\mathrm{disk} \}=F(\omega)=\frac{\sqrt{3}}{4|\omega|}J_1(2\pi |\omega|)$$ where $J_1$ is a Bessel function of the first kind. Following from the Convolution theorem we obtain the point-wise multiplication of $f$ and the disk in the Fourier domain: $$f * I_\mathrm{disk}=\mathcal{F}^{-1}\left\{ \mathcal{F}\{ f \} \cdot \mathcal{F}\{ I_\mathrm{disk} \} \right\}$$ Since the only way for $f * I_\mathrm{disk}$ to be the constant $0$ function is for its Fourier transform to be the same, $\mathcal{F}\{f\}$ can only have non-zero values where $\mathcal{F}\{ I_\mathrm{disk} \}=0$. From above, this means that $\mathcal{F}\{f\}$ takes on non-zero values specifically where $$|\omega|=\frac{j_{1,n}}{2\pi}$$ where $j_{1,n}$ are the zeros of $J_1$. Since we are assuming $f$ is real, we constrain $\mathcal{F}\{f\}(\omega)=\mathcal{F}\{f\}(-\omega)$ and thus:

$f$ satisfies the condition $f * I_\mathrm{disk}=0$ iff $f$ is a weighted sum of functions of the form $$f_{n,\hat{u},\phi}(\vec{v})=\cos(j_{1,n}(\vec{v} \cdot \hat{u})+\phi)$$ where $n$ selects the Bessel function zero, $\hat{u}$ is a unit vector indicating the direction of the oscillation, and $\phi$ is an arbitrary phase offset.

$f_p(\theta)\cos(\varphi + \theta)$
A density plot for an example function $f$ is shown on the left, with a unit disk (outline in red). Drag on the density plot to move the unit circle. To the right, a plot of $f_p(\theta)$ around the current center of the unit circle, and a weighted version with $\varphi$ based on the differential area formula below.

Constraints from differential area

Another interesting approach I've found is based on the fact that for any unit disks $A$ and $B$, their respective differences are equal. With the notation that for any area $C$, the integral of $f$ over $C$ is $A_f(C)=\iint_C f(x)\,\mathrm{d}A$, we get $A_f(A)=A_f(A\cap B)+A_f(A-B)=0$ and $A_f(B)=A_f(B\cap A)+A_f(B-A)=0$ we arrive at $$A_f(A-B)=A_f(B-A)$$ Intuitively, we can make $A$ and $B$ arbitrarily close so that we can identify constraints on the perimeter of any unit disk.

Given a unit disk $A$ with center $(0,0)$ and a unit disk $B$ with center $(d,0)$, we look at how their differences $B-A$, $A-B$ and their integrals behave as we change $d\to 0$. Parameterizing based on the angle $\theta$ from the center of $A$, we want to be able to measure $h(\theta)$ and obtain: $$A_f(B-A)=\int_{-\theta_0}^{\theta_0}\int_{1}^{h(\theta)}f(r,\theta)r\,\mathrm{d}r\,\mathrm{d}\theta$$ $$A_f(A-B)=\int_{-\theta_0+\pi}^{\theta_0+\pi}\int_{1}^{h(\theta)}f(r,\theta),\mathrm{d}r\,\mathrm{d}\theta$$ Where $\theta_0=\cos^{-1}\left(\frac{d}{2}\right)$ is the angle from the center of $A$ to where the boundaries of $A$ and $B$ intersect. It is important to note that $h(\theta)$ is the same for both the $A-B$ and $B-A$ cases. For $h(\theta)$, we use the law of cosines to obtain: $$1=h^2+d^2-2hd\cos\theta$$ This can be grouped into the polynomial $h^2-2hd\cos\theta+d^2-1=0$ and solved with the quadratic equation: $$h=\frac{2d\cos\theta\pm\sqrt{4d^2\cos^2\theta-4(d^2-1)}}{2}$$ and simplifying: \begin{align*} h&=d\cos\theta\pm\sqrt{d^2\cos^2\theta-d^2+1}\\ h&=d\cos\theta\pm\sqrt{d^2(\cos^2\theta-1)+1}\\ h&=d\cos\theta\pm\sqrt{1-d^2\sin^2\theta} \end{align*} Since we are interested in the far case, we are left with $$h(\theta)=d\cos\theta + \sqrt{1-d^2\sin^2\theta}$$ Wolfram Alpha shows that $\frac{\partial}{\partial d}h(\theta)=\cos\theta-d(\cdots)$ and thus at $d=0\ (A=B)$ we have $$h(\theta)=1, \qquad \frac{\partial}{\partial d}h(\theta)=\cos\theta$$ From the area integrals above, $A_f(B-A)-A_f(A-B)=0$ and we arrive at $$\int_{-\theta_0}^{\theta_0}\int_{1}^{h(\theta)}f(r,\theta)r\,\mathrm{d}r\,\mathrm{d}\theta - \int_{-\theta_0+\pi}^{\theta_0+\pi}\int_{1}^{h(\theta)}f(r,\theta),\mathrm{d}r\,\mathrm{d}\theta = 0$$ Since $\lim_{d\to 0}\theta_0=\pi/2$, and assuming $f$ is piecewise continuous, we get (VERIFY THIS): $$\frac{\partial}{\partial d}\left(A_f(B-A)-A_f(A-B)\right)=\\ \int\limits_{-\pi/2}^{\pi/2}f(1,\theta)\cos\theta\,\mathrm{d}\theta - \int\limits_{\pi/2}^{3\pi/2}f(1,\theta)\cos\theta\,\mathrm{d}\theta=0$$ Since $\cos$ is an even function, we get $$\int_{0}^{2\pi}f(1,\theta)\cos\theta\,\mathrm{d}\theta = 0$$ This can be extended for any angle $\varphi$ and any disk center (not just at the origin), leading to

$$\forall \varphi\in\mathbb{R},\quad \forall p\in\mathbb{R}^2,\quad \int_0^{2\pi}f_p(\theta)\cos(\varphi + \theta)\,\mathrm{d}\theta=0$$ Where $f_p(\theta)$ denotes $f(p_x+\cos\theta,p_y+\sin\theta)$, the unit circle with center $p$ parameterized by $\theta$, for all piecewise continuous functions $f$.

Coming soon

Or maybe these are exercises left to the reader?

  • Show that $\forall\varphi,\ \forall n\ge 2\in\mathbb{N},\quad \sum_{k=0}^{n}\cos(\frac{2\pi}{n}k+\varphi)=0$, and thus sums of functions of the form $f_p(\theta)=g(n\theta\mod{2\pi})$ i.e. $g$ repeated $n$ times around the circle where $g$ is piecewise continuous, satisfy the differential perimeter formula. Consider whether this is a necessary condition, now that we have methods for constructing $f$. If $n$ is prime, are these functions orthogonal? Could we decompose a function into these uniquely?
  • HTML5 canvas with density map of example $f$ functions
  • Add LaTeX formula references for clarity
  • Note 1D case is just any function periodic over a unit interval, $n$-dimensional unit hypercube case requires a function periodic in the same way.
  • Investigate sampling from distriutions of disk positions. Essentially, the integral should also hold not just for a uniform unit disk but for an (in)finite sum of unit disks.
  • Instead of unit disks, write down (simple?) extensions
  • Add drag hand to canvas


Please let me know below if you have any ideas, questions, or corrections about this problem. I'd love to hear new approaches or useful concepts that I haven't mentioned (as I may be unaware of them).

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