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Homework Help: 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


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    Well, the equation

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

    doesn't actually imply

    \lim_{x \rightarrow y}\frac{f(x) - f(y)}{x-y} \in \{ 1, -1 \}

    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


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    that would be true for continuous isometries
    all we would know is
    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
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