Under what conditions on a function $f: \mathbb{R} \to \mathbb{R}$ can we say that given any real numbers $x,y$ with $x<y$ if $f(x) \ne f(y)$ then there is a sub-interval $S_{(x,y)}$ of $(x,y)$ such that $f$ behaves injective on $S_{(x,y)}$ ?
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$\begingroup$ Unfortunately continuity is not enough : (The devil here is) the Weierstrass function. $\endgroup$– Aditya Guha RoyCommented May 7, 2017 at 3:59
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1$\begingroup$ Differentiablity is of course enough. $\endgroup$– Claudio GorodskiCommented May 7, 2017 at 12:57
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$\begingroup$ @ClaudioGorodski I think differentiability over the whole domain is necessary (just think about the step functions). $\endgroup$– Aditya Guha RoyCommented May 7, 2017 at 13:53
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$\begingroup$ In general, I think absolute continuity is enough, though it may not be the strongest and the best solution to it. Also it may have some lacks. $\endgroup$– Aditya Guha RoyCommented May 7, 2017 at 13:54
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$\begingroup$ Ok if you (anyone) suspects the source : I say I never read it being true. I actually was thinking about it because I felt that if this was true for certain (though peculiarly) general functions then many things were true for many general categories of functions (one such thing is : mathoverflow.net/questions/269064/… ) , however unfortunately it is not so general that it holds with the sufficiency of continuity, but I don't know what to tell about absolute continuity. $\endgroup$– Aditya Guha RoyCommented May 9, 2017 at 11:29
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