# 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.

## 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

## Feedback

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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