Variant Of Inscribed Square Problem

Question

Looking at the inscribed square problem, I noticed that for some continuous curves $c:S^1\to\mathbf{R}^2$, one can prove that every continuous curve $c'$ with $\|c-c'\|_\infty<\epsilon$ contains an inscribed square.

For example, take $c$ to be the equilateral triangle (oriented like the letter $\Delta$) with bottom edge equal to $[0,1]\times\{0\}$. Then choose $0<a<1/2$ such that $1-a<a\sqrt{3}$ (meaning that a square sitting on $((a,0),(1,0))$ has its left upper vertex still inside the triangle) and define $f,g:[0,1]^2\to\mathbf{R}$ by letting $f(t,s)$ denote the signed distance of the upper left vertex of the square sitting on $((at,0),(1-as,0)$ from the curve $c$ (if it sits inside of $c$, make this term negative) and similar let $g(t,s)$ denote the signed distance of the upper right vertex from $c$. Now $f(0,s),g(t,0)>0$ but $f(1,s),g(t,1)<0$ for all $s,t\in[0,1]$ so there should exist a point $p\in[0,1]^2$ with $f(p)=g(p)=0$. (What's exactly the formal argument here?) And for $\epsilon$ sufficiently small, this works with $c'$ as well (where we must work with an appropriate parameterization of the bottom edge).

Which other results exist of this kind? Also, would it work if we don't assume that $\|c-c'\|_\infty<\epsilon$ but only that $c'$ lives in the $\epsilon$-tube around $c$ and there is a homotopy between $c$ and $c'$ inside this tube? And what features can I use to infer that two continuous functions $f,g:[0,1]^2\to\mathbf{R}$ have a common zero?

[EDIT] Yep, think one can show it for the homotopy case as well. Also, the title of that question is misleading, it's not a variant of the problem, more like a solution on a certain class of curves...

Answer 1

what features can I use to infer that two continuous functions $f,g:[0,1]^2\to\mathbf{R}$ have a common zero?

A general result is the main theorem of Part II at p. 86 of [CS].

Let $f:D\to P$ be a mapping of a disk into the plane, let $C$ be the boundary circle of $D$, and let $y$ be a point of the plane not on $fC$. If the winding number of $f|C$ about $y$ is not zero, then $y\in fD$; i.e. there is a point $x\in D$ such that $fx=y$.

But whereas the calculation of the winding number is a technical task, your special case follows from a known theorem in dimension theory.

Let $A$ and $B$ be disjoint subsets of a topological space $X$. A closed set $L\subset X$ is a *partition between $A$ and $B$ if there exist open sets $U,V\subset X$ such that $A\subset U$, $B\subset V$, $U\cap V=0$, and $X\setminus L = U\cup V$.

1.8.1. Theorem. Let $A_i$ and $B_i$, where $i = 1, 2, \dots, n$, be the subsets of the $n$-cube $I^n$ defined by the conditions $A_i = \{\{x_j\}\in I^n: x_i = 0\}$ and $B_i = \{\{x_j\}\in I^n: x_i = 1\}$, i.e., the pairs of opposite faces of $I^n$. If $L_i$ is a partition between $A_i$ and $B_i$ for $i = 1, 2,\dots, n$ then $\bigcap_{i=1}^n L_i\ne \emptyset.$

Proof. Let us consider open sets $U_i, W_i\subset I^n$ such that $A_i\subset U_i$, $B_i\subset W_i$, $U_i\cap W_i = \emptyset $ and $I^n\setminus L_i = U_i\cup W_i$ for $i = 1,2,\dots, n$. Since $(I^n\setminus W_i)\cap (I^n\setminus U_i)= I^n\setminus (U_i\cup W_i)=L_i$, the formulas

(1) $f_i(x)= \begin{cases} \frac 12\frac {\rho(x,L_i)}{ \rho(x,L_i)+ \rho(x,A_i)}+\frac 12\mbox{ for }x\in I^n\setminus W_i\\ -\frac 12\frac {\rho(x,L_i)}{ \rho(x,L_i)+ \rho(x,B_i)}+\frac 12\mbox{ for }x\in I^n\setminus U_i \end{cases},$

define for $i = 1,2,\dots, n$ a continuous function $f_i: I^n\to I$ [Here $\rho$ is the distance. AR]. Clearly, we have

(2) $f_i^{-1}(1/2) = L_i$, $f_i(A_i) =\{1\}$ and $f_i(B_i) =\{0\}$.

Assume that $\bigcap_{i=1}^n L_i =\emptyset$; it follows from the first part of (2) that the continuous mapping $f: I^n\to I^n$ defined by letting $f(x) = (f_1(x), f_2(x),\dots, f_n(x))$ for $x\in I^n$ does not assume the value $a = (1/2, 1/2,\dots , 1/2) \in I^n$. The composition $g: I^n\to I^n$ of the mapping $f$ and the projection $p$ of $I^n\setminus\{a\}$ from the point $a$ onto the boundary of $I^n$, i.e., onto the set $B =\bigcup_{i=1}^n (A_i\cup B_i)$, satisfies the inclusion $g(I^n)\subset B$; by the second and the third part of (2), we have $g(A_i)\subset B_i$, and $g(B_i)\subset A_i$. The last three inclusions show that $g(x)\ne x$ for every $x\in I^n$, which contradicts the Brouwer fixed-point theorem. Hence $\bigcap L_i\ne\emptyset$. $\square$

Now $f(0,s),g(t,0)>0$ but $f(1,s),g(t,1)<0$ for all $s,t\in[0,1]$ so there should exist a point $p\in[0,1]^2$ with $f(p)=g(p)=0$.

Now the required claim follows from the theorem for $n=2$, $L_1=g^{-1}(0)$, $U_1=g^{-1}(0,\infty)$, $W_1=g^{-1}(-\infty,0)$, $L_2=f^{-1}(0)$, $U_2=f^{-1}(0,\infty)$, and $W_2\in f^{-1}(-\infty,0)$.

References

[CS] W.G. Chinn, N.E. Steenrod. “First concepts of topology”, Random House, New York, 1966?.

[Eng] Ryszard Engelking, Dimension Theory, North-Holland, Amsterdam and Polish Scientific Publishers, Warsaw, 1978.