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Special sequences in a product metric space

by holy_toaster
Tags: metric, product, sequences, space, special
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holy_toaster
#1
Jul22-11, 01:19 PM
P: 32
Hi there,

I came across the following problem and I hope somebody can help me: I have some complete metric space [itex](X,d)[/itex] (non-compact) and its product with the reals [itex](R\times X, D)[/itex] with the metric [itex]D[/itex] just being [tex]D((t,x),(s,y))=|s-t|+d(x,y)[/tex] for [itex]x,y\in X; s,t\in R[/itex]. Then I have some sequences [itex]x_n\subset X[/itex], which converges to [itex]x[/itex] and [itex]t_n\subset R[/itex], which goes to infinity. These two give rise to sequences [itex](t_n,x)[/itex] and [itex](t_n,x_n)[/itex] in [itex]R\times X[/itex] which do not converge either, but nevertheless [tex]D((t_n,x),(t_n,x_n))\to 0[/tex] holds as [itex]n\to\infty[/itex]. Moreover I have now some continuous map [itex]f\colon R\times X\to X[/itex] and my problem is under what conditions does [tex]d(f(t_n,x),f(t_n,x_n))\to 0[/tex] hold as [itex]n\to\infty[/itex]?

I know that it does not hold in general as there are simple counterexamples and it does hold if [itex]f[/itex] is (globally) Lipschitz. But for my setting globally Lipschitz is quite restrictive, so I am looking for milder assumptions. Specifically I would be interested if it does hold when [itex]f[/itex] fulfills the following type of 'group condition': [tex]f(t+s,x)=f(t,f(s,x))[/tex] for all [itex]s,t\in R; x\in X[/itex]. I can not find a counterexample and cannot prove it either.

I know it's a quite specific problem, but I would be glad if somebody had an idea on that or could provide me with a source that helps because I am really stuck here with this.

Thanx.

PS: I think that [itex]X[/itex] is even a manifold and [itex]f[/itex] is smooth, but I don't think that makes much of a difference...
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zhentil
#2
Jul22-11, 10:05 PM
P: 491
I think it might be difficult to formulate something so generally. Let X be the real numbers, and then we have a function from the plane to the line. Make it even simpler: let f(t,x)=t*h(x). When is infinity times zero zero?

It's an interesting question, and it's clear that one can do better than Lipschitz under certain conditions. But your group property doesn't do it: f(t,x)=xe^t is a counterexample.
holy_toaster
#3
Jul23-11, 06:15 AM
P: 32
But your group property doesn't do it: f(t,x)=xe^t is a counterexample.
I see. I now thought that in general uniform continuity could do it.

But in the case of my group property: Maybe it would be enough if I additionally demanded that [itex]f(t_n,x_n)\to x[/itex] as [itex]n\to\infty[/itex]?


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