I'm working on a project for a class on Dynamical systems, and I've come to a point where even the professor was unable to help me. He excused me from having to prove it directly, but out of my own curiosity, and because I think it will help the later steps, I've come here to see if anyone can prove it.
We are using the skinny baker map, defined as: \begin{equation} B(x,y) = \begin{cases} (\frac{1}{3}x, 2y) & \text{if 0 $\leq y \leq \frac{1}{2}$}\\ (\frac{1}{3}x+\frac{2}{3}, 2y-1) & \text{if $\frac{1}{2} < y \leq 1$}\\ \end{cases} \end{equation}
Now, this is the theorem it is asking us to prove:
$Theorem$: Let B denote the skinn baker map, and let $d>0$. Assume there is a set of points $\{{x_0, x_1, ..., x_{k-1}, x_k=x_0}\}$ such that each coordinate of B$(x_i)$ and $x_i$ differs by less than $d$ for $i=0,1,...,k-1$. Then there is a periodic orbit $\{z_0,z_1,...,z_{k-1}\}$ such that $|x_i-z_i|<2d$ for $i=0,1,...,k-1$.
So far, to solve this, I have shown that B$(S_i)$ lies across $S_{i+1}$, where $S_i$ is the $3d$ by $2d$ rectangle centered at $x_i$. This allows me to conclude by a corollary that B$^k$ has a fixed point in $S_0$. That was provided in the hint.
From here, I really just don't know how to show that the periodic orbit exists with the given inequality $|x_i-z_i|<2d$ for $i=0,1,...,k-1$. I'm really just struggling with the inequality part.
Thank you for any help, and I probably won't be able to figure this out on my own.