## Isometry from R to itself.

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?

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

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?

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 Recognitions: Gold Member Science Advisor Staff Emeritus 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}.
 Recognitions: Homework Help 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

## Isometry from R to itself.

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 $|f(x) - f(y)| \leq (x-y)^2$ 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.