User dejan govc - MathOverflow

Answer by Dejan Govc for A Brouwer fixed point theorem on finite sets

2011-10-14T13:48:59Z

I believe the following is a counter-example:

$f: \lbrace -1,0,1\rbrace^2 \to \lbrace -1,0,1\rbrace^2$

$\forall x:$

$ f(-1,x) = (1,x)$

$f(0,x)=(1,x)$

$f(1,x)=(0,x)$

What is known about Ulam's problem of metric spaces with isometric squares?

2011-10-07T12:17:15Z

Background

In the book Problems in Modern Mathematics, S. Ulam asks the following question:

Suppose $A$ and $B$ are metric spaces, such that $A^2$ and $B^2$ equipped with the 2-metric $d((x_1, y_1),(x_2, y_2)) = \sqrt{d(x_1, x_2)^2 + d(y_1, y_2)^2}$ are isometric. Does it follow that $A$ and $B$ must also be isometric themselves?

The answer to this general question is negative. Take for example the space of the rational numbers $\mathbb{Q}$ and the space $\mathbb{Q} \sqrt{2}$ of rational multiples of $\sqrt{2}$. These two spaces are obviously not isometric, but their squares are, isometry being just the rotation by $\frac{\pi}{4}$. See http://www.ams.org/journals/proc/1971-029-03/S0002-9939-1971-0278262-5/S0002-9939-1971-0278262-5.pdf for the proof.

Question

What is the status of this problem for slightly less general metric spaces? I am especially interested in the following cases:

What is known about this problem for complete metric spaces?
What is known for finite metric spaces?

A good survey if it exists would be more than welcome.

Answer by Dejan Govc for Topological space with some conditions

2011-09-06T01:40:17Z

I believe such a space cannot exist for the following reason:

Suppose it does. By the second requirement, there should be an unbounded function $f: X \to (0, \infty)$, which is bounded on every compact set. This means we have countably many points $x_1, x_2, x_3, \dots$ such that for each $n$ the inequality $f(x_n)\ge n$ holds. The singletons $\lbrace x_n\rbrace$ are finite sets, therefore compact. By the first requirement, the set $\lbrace x_n| n\in\mathbb{N}\rbrace$ is therefore relatively compact and so its closure must be compact. But then $f$ is unbounded on a compact set, which is a contradiction.

Answer by Dejan Govc for Interpolating between piecewise linear functions, with a family of smooth functions

2011-08-23T16:07:15Z

Interesting problem! This is not an actual answer, just a comment. I believe some minor condition may be missing, since it seems to me that the problem as posed does not necessarily have a solution. Set for example $F:[0,2)\to\mathbb R$, $F|_ {[0,1)}\equiv 1$, $F|_ {[1,2)}\equiv 0$ and $G:[0,2)\to\mathbb R$, $G|_ {[0,1)}\equiv 2$, $G|_ {[1,2)}\equiv 1$. These two functions satisfy conditions 1-4 but a smooth family you're looking for would have to contain some functions discontinuous at point $x=1$ because of condition 2 which asks that the family be strictly increasing. Maybe the inequalities for $L$ should be strict? Or maybe you should add to $E$ those points where $F(x)$ equals the left or right limit of $G$ there? Maybe I'm just missing something. In the latter case, I kindly ask the moderators to delete this comment.

Answer by Dejan Govc for Can a continuous, nowhere differentiable function have specified "shape" at every point?

2011-07-14T15:21:33Z

I'm new here, so I hope my answer is of any use and not too late.

I was wondering: Wouldn't it be perhaps more natural to consider limits of the form

$\lim_{y\to x}\frac{f(y)-f(x)}{\phi(y)-\phi(x)}$?

If for example we take $f(x) = |x|$ for $x\in \bf{R}$ and $\phi = f$, the "derivative" would be equal to one everywhere, which makes sense, since $f$ and $\phi$ are really the same, therefore their shape should be the same, right?

By the previous definition which uses $\phi(y-x)$ in the denominator, we would get limits like

$\lim_{y\to x}\frac{|y|-|x|}{|y-x|}$

which exists only for $x = 0$, even though our intuition tells us the shape is supposed to be the same everywhere. Any comments?

Comment by Dejan Govc

2012-03-24T23:36:34Z

+1. Beautifully written.

Comment by Dejan Govc

2011-11-14T15:14:31Z

Not in characteristic 2. =)

Comment by Dejan Govc

2011-11-02T21:30:49Z

A similar example I find interesting is to take the unique positive solution of $x^x=x+1$. Then you can easily check that $x^{(x^{(x^x)})} =(x^x)^{(x^x)}$ using the defining equation and the usual exponentiation rules. Although the transcendentality doesn't seem obvious anymore ...

Comment by Dejan Govc

2011-10-16T01:59:10Z

It is a beautiful problem, I must say.

Comment by Dejan Govc

2011-10-16T01:58:24Z

I have thought about it some more and $d(f(x_0),x_0)\leq 1$ won't always work, but I am quite sure you can prove that there always exists a point $x_0$ such that $d(f(x_0),x_0)\leq C$ where $C$ is a constant depending only on the dimension $d$. This is because you can extend your function by piecewise linear interpolation to a continuous function on $[-n,n]^d \subseteq \mathbb{R}^d$ and then use the original Brouwer fixed point theorem obtaining a fixed point. Now choose the closest lattice point. This point has the property we seek (this follows from your property and piecewise linearity).

Comment by Dejan Govc

2011-10-15T16:50:03Z

I am wondering ... Maybe it is still possible to get a version of Brouwer's fixed point theorem. The condition $f(A(x,y))\subseteq A(f(x,y))$ looks a lot like continuity. If you equip $X$ with the Euclidean metric, it just says that whenever $d(x,y) \leq 1$ it follows that $d(f(x),f(y)) \leq 1$. My intuition suggests that it should still be possible to prove that there is an almost-fixed point $x_0$ in the sense that $d(f(x_0),x_0) \leq 1$. This certainly works for the one-dimensional analog of the problem ($X = \lbrace 1,2, ..., n \rbrace $).

Comment by Dejan Govc

2011-10-08T16:28:42Z

Max Euwe showed using the Thue-Morse sequence that the rules of chess in his time did not exclude the possibility of an infinite game. I'd say that's interesting mathematically.

Comment by Dejan Govc

2011-10-08T10:45:03Z

Thank you! After reading the first article, it seems the problem for complete metric spaces is still wide open.

Comment by Dejan Govc

2011-07-14T20:31:56Z

Oh, I think I get it now. You're looking for a function that is locally of the same shape as $\phi$ is around zero. My answer is probably a bit off-topic then ...