1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Isometry from R to itself.

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all isometries from the reals to itself.


    2. Relevant equations
    Well what we're basically doing is trying to find functions f: R -> R such that for any x, y in R, the property |f(x) - f(y)| = |x-y| holds.



    3. The attempt at a solution
    OK, so this shouldn't be too hard. It seems like you could just plug in 0 for one of the variables above and then you're basically done. But can't we prove differentiability as follows?

    [tex]|f'(y)| = \lim_{x \rightarrow y}\left|\frac{f(x) - f(y)}{x-y}\right| = \lim_{x \rightarrow y}\frac{|x-y|}{|x-y|} = 1.[/tex]

    This seems to immediately imply that the functions are defined by either i(x) = x + C or j(x) = - x + K for arbitrary real numbers C or K. Are there any holes I overlooked?
     
  2. jcsd
  3. Sep 21, 2009 #2

    Hurkyl

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Well, the equation

    [tex]
    \lim_{x \rightarrow y}\left|\frac{f(x) - f(y)}{x-y}\right| = 1.
    [/tex]

    doesn't actually imply

    [tex]
    \lim_{x \rightarrow y}\frac{f(x) - f(y)}{x-y} \in \{ 1, -1 \}
    [/tex]

    The best you can say is that the set of limit points of (f(x)-f(y))/(x-y) as x->y is a subset of {1, -1}.
     
  4. Sep 21, 2009 #3

    lurflurf

    User Avatar
    Homework Helper

    that would be true for continuous isometries
    say
    f(x)=f(y)+(x-y)g(x,y)
    all we would know is
    |g|=1
    we might have say
    g=-1 x rational
    g=1 x irrational
     
  5. Sep 21, 2009 #4
    I understand your counterexample for noncontinuous isometries, but this one is lipschitz and hence uniformly continuous, so is there a way to extend the original argument?

    *EDIT* The reason I pursued this approach in the original place is because I proved that [itex]|f(x) - f(y)| \leq (x-y)^2[/itex] for x,y in R implies that f is constant. In this case, you end up with |f'(y)| = 0, so we can just be rid of the absolute values and then recall the corollary to the mean value theorem. But it seems that I did not have to work with absolute values in the first place, since |a| = |b| implies that a = -b or a = b. However, I am aware of the concerns raised, so I am curious as to whether continuity - which we seem to have - can fix the original argument.

    Bah, upon closer examination of lurflurf's point, it doesn't matter whether f is continuous or not. Hmm well I think I have a better understanding of when to try something like this and when you can just substitute variables. Thanks.
     
    Last edited: Sep 21, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook