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Find all functions $f : \mathbb{R} \rightarrow \mathbb{R}$ such that for all $I$ open bounded interval it follows $f(I)$ is also an open bounded interval of same length as $I$.

It's easy to see $f(x)=\pm x + c, c \in \mathbb{R}$ are solutions, but I cannot prove/disprove these are all.

Intuitively, it seems that $|f(x) - f(y)|=|x - y|$ should stand for all $x, y$, following the same length condition.

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Hint: $f$ must be monotone, so $(f(x)-f(y))/(x-y)$ should be the same for all $y \ne x$.

EDIT: To show $f$ is monotone: if $a < b < c$, $f((a,c)) = f((a,b)) \cup f((b,c)) \cup \{f(b)\}$. Now the length of $f((a,c))$ is the length of $(a,c)$, which is the sum of the lengths of $(a,b)$ and $(b,c)$, and thus the sum of the lengths of $f((a,b))$ and $f((b,c))$. The only way the union of two open intervals (plus a point) can have length equal to the sum of the lengths of the intervals is that those intervals are disjoint.

Now if $f$ was not monotone, there would be some $x < y < z$ such that $f(y)$ is not between $f(x)$ and $f(z)$: either $f(y) \ge \max(f(x), f(z))$ or $f(y) \le \min(f(x), f(z))$. Suppose $f(x) \le f(z) \le f(y)$ (the other cases will be similar). Then taking $a < x < y < b < z < c$, $f((a,b))$ and $f((b,c))$ both include $f(z)$ and are thus not disjoint, contradiction.

Robert Israel
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  • How can I prove $f$ monotone? –  Jan 02 '18 at 09:10
  • @Eugen Covaci : with $a \le b \le c$, we can't have $f(a)<f(c)<f(b)$ or $f(c)<f(a)<f(b)$ (because we'd have $|f(c)-f(a)| < |f(b)-f(a)|$, or $|f(c)-f(a)| < |f(c)-f(b)|$, absurd). Hence $f(a) < f(b) \Rightarrow f(a)<f(b) \le f(c)$ (resp. with $>$). I can't see how to do it without taking $x<y<z$ and considering the different cases... – charmd Jan 02 '18 at 10:08
  • @charMD Why is $|f(c)-f(a)| < |f(b)-f(a)|$ absurd? –  Jan 02 '18 at 16:34
  • Because this would imply $|c-a| < |b-a|$ i.e. $c-a < b-a$, i.e. $c<b$ – charmd Jan 02 '18 at 16:41
  • @charMD $|f(c)-f(a)| < |f(b)-f(a)|$ doesn't mean $|f((a,c))| \lt |f((a,b))| $ –  Jan 03 '18 at 17:44
  • @Eugen Covaci Oh, you are right, I don't know then, sorry – charmd Jan 03 '18 at 20:08