Let $P_{\theta}: \Bbb{R}^2 \to \Bbb{R}$ the function $$P_{\theta}(x_1,x_2)=x_1\cos{\theta}+x_2\sin{\theta}, \text{ } \theta \in [0,\pi)$$ $P_{\theta}(x_1,x_2)$ is the orthogonal projection of $x=(x_1,x_2)$ onto the line through the origin in the direction $(\cos{\theta},\sin{\theta}).$
If $s>0$ and $x,y \in \Bbb{R}^2$ fixed, then how can I prove that the set $$A_s=\{\theta:|P_{\theta}(x-y)|<s \}$$ is an interval of length $2\arcsin{\frac{s}{|x-y|}}$?
Thank you in advance.
Let $u_{\theta}=(\cos\theta,\sin\theta)$. Since $P_{\theta}$ is linear in $(x_1,x_2)$, you can replace $x-y$ by some vector $v\in\mathbb{R}^2$, and then you want to know for which $\theta$ you have $|\langle u_{\theta},v\rangle|<s$, or $|\langle u_{\theta},\frac{v}{|v|}\rangle|<\frac{s}{|v|}$. By rotational invariance, we may assume that $\frac{v}{|v|}$ is the vector $(0,1)$, and now we are asking for which $\theta$ is it true that $|\sin\theta|<\frac{s}{|v|}$, and the answer is an inteval whose length is $2\arcsin \frac{s}{|v|}$, as required. However, this holds if we assume the domain of definition is $(-\pi/2,\pi/2)$. If it is $[0,\pi)$, then the set of $\theta$ is not and interval, but rather a union of two disjoint intervals.
For simplicity, assume (temporarily) that $x = 0.$ Then the vector $y - x$ projects onto a line through the origin as shown in the figure below.
If we put a line through the origin perpendicular to $y-x,$ the length of the projection is zero. Let $\theta_0$ be the angle of that line. Now construct a right triangle using $y - x$ as the hypotenuse, with a leg of length $\delta$ adjacent to the origin, the opposite angle is $\alpha = \arcsin\frac{\delta}{\lvert x - y\rvert}.$
Extending that leg to a line we get one of the lines at one of the angles $\theta_\min$ or $\theta_\max,$ which are the angles of the two lines onto which $y - x$ projects to a segment of length exactly $\delta.$
These two lines are rotated either clockwise or counterclockwise by an angle $\alpha$ from the line at angle $\theta_0.$ That is, their angles can be written $\theta_\min = \theta_0 - \alpha$ and $\theta_\max = \theta_0 + \alpha.$ On any line at an angle between those two extremes, $y - x$ will project onto a segment shorter than $\delta.$
If $x$ is not the origin, but we have the same vector $y - x,$ then the vector still projects onto a segment of length $\delta$ onto lines at angles $\theta_\min$ or $\theta_\max$ and onto a shorter segment on any line at any angle between those two. The diagram of the projection is just not quite as tidy as the figure above.
So regardless of which points $x$ and $y$ we choose, the range of angles at which $y - x$ (or $x - y$) projects onto a segment of length less than $\delta$ will be $\left(\theta_0 - \arcsin\frac{\delta}{\lvert x - y\rvert}, \theta_0 + \arcsin\frac{\delta}{\lvert x - y\rvert}\right).$ The size of this segment is $2\arcsin\frac{\delta}{\lvert x - y\rvert},$ where $\theta_0$ is the angle of a line orthogonal to $x - y.$
There is a small technical irregularity here, however, because in the paper you are reading, https://www.math.stonybrook.edu/~bishop/classes/math324.F15/book1Dec15.pdf, the domain of $\theta$ in the projection function $\Pi_\theta$ is only $[0,\pi).$ So what do we do when $0 \leq \theta_0 < \arcsin\frac{\delta}{\lvert x - y\rvert}$ or $\pi - \arcsin\frac{\delta}{\lvert x - y\rvert} < \theta_0 < \pi$? In those cases one of the angles $\theta_0 - \arcsin\frac{\delta}{\lvert x - y\rvert}$ or $\theta_0 + \arcsin\frac{\delta}{\lvert x - y\rvert}$ will be outside the given domain. An obvious answer is to split the set of $\theta$ values into two segments, one starting at $0$ and the other ending at $\pi,$ with total size $2\arcsin\frac{\delta}{\lvert x - y\rvert}.$ But technically that is not "a single interval" unless you identify $\pi$ with $0.$
The actual goal of this exercise, however, is actually just to show that $$\int_0^\pi \mathbf 1_{\Pi_\theta(x - y) < \delta} \; d\theta = 2\arcsin\frac{\delta}{\lvert x - y\rvert},$$ which is true whether the $\theta$ values all fall into one interval of size $2\arcsin\frac{\delta}{\lvert x - y\rvert}$ or into two intervals of combined size $2\arcsin\frac{\delta}{\lvert x - y\rvert}.$ The authors apparently did not think this distinction was worth pointing out.
Hint:
Write $(x_1-y_1)\cos\theta+(x_2-y_2)\sin\theta\,$ as $$\|x-y\|(\sin\varphi\cos\theta+\cos\varphi\sin\theta)=\|x-y\|\sin(\varphi+\theta),$$ so the condition on $\theta\,$ becomes $$|\sin(\varphi+\theta)|<\frac s{\|x-y\|}.$$ Can you continue?
