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.
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.
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 $f$.
Or maybe these are exercises left to the reader?
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).