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Given a function $f: \mathbb{R}\to\mathbb{R}$, we define its left modulus of continuity, $L(f): \mathbb{R} \times (0, \infty)\to [0,\infty]$ by

$$L(f)(x, e) := \sup \{d \ge 0 \,:\, f((x, x+d)) \subseteq [f(x) - e, f(x) + e]\} $$

Similarly define the right modulus of continuity, $R(f): \mathbb{R} \times (0, \infty)\to [0,\infty]$ by

$$R(f) (x, e) := \sup \{d \ge 0 \,:\, f((x-d, x)) \subseteq [f(x) - e, f(x) + e]\}$$

Suppose $f$ and $g$ are continuous functions such that $L(f) = L(g)$, $R(f) = R(g)$, and $f(0) = g(0) = 0$. Does it follow that either $f = g$ or $f = -g$?

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    $\begingroup$ If you were to replace the left/right modulus functions with a single two-sided modulus function, then the answer would be negative, in light of $f(x)=x$ and $g(x)=|x|$. $\endgroup$ – Joel David Hamkins Dec 21 '18 at 16:15
  • $\begingroup$ Yep, I realised that the single modulus was too messy, so I felt requiring both left and right was more natural. $\endgroup$ – James Baxter Dec 22 '18 at 4:41
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I think they are. Assume the contrary, then $f$ is not constant and therefore there exists a point $a$ such that $f$ is not constant neither on $[a,+\infty)$ nor on $(-\infty,a]$. Call such points admissible. The admissible points form an interval (possibly infinite). Without loss of generality $0$ is admissible point (else shift the argument and subtract the constant from $f$.)

Denote $L(f)(0,e)=A(e)$, $R(f)(0,e)=B(e)$. Clearly $A$ is decreasing function on $(0,\infty)$ and $|f(A(e))|=e$ (if $A(e)<\infty$), $|f|<e$ on $[0,A(e))$. The same holds for $g$, thus $g(A(e))=\pm f(A(e))$. I claim that the sign does not depend on $e$. Indeed, choose $e_1>0$ such that $B(e_1)<\infty$. Then analogously $f(-B(e_1))=\pm e_1, g(-B(e_1))=\pm e_1$. I claim that ${\rm sign}\, f(-B(e_1))/g(-B(e_1))={\rm sign}\, f(A(e))/g(A(e))$ for any positive $e$. Indeed, in the opposite case exactly one of the numbers $L(f)(-B(e_1),e+e_1)$, $L(g)(-B(e_1),e+e_1)$ equals to $B(e_1)+A(e)$. A contradiction. So, the sign is always the same.

We say that $0$ is a plus-point if this sign is always plus and 0 is minus-point if the sign is always minus. Analogously, any admissible point $a$ is a plus-point or minus-point. The next claim is that either all admissible points are plus-points or all of them are minus-points. It suffices to show that both sets of plus-points and minus-points are open. Assume the contrary: for example, 0 is a plus-point but there exists a sequence of minus-points $t_n\to 0$. Fix $e>0$ such that $A(e)>0$ and $A$ is continuous at $e$. This continuity yields that whenever $|f(s_n)| \to e$ or $|g(s_n)| \to e$ for a sequence $s_n\in [0,A(e)]$, we must have $s_n\to A(e)$.

Assume without loss of generality that $f(A(e))=g(A(e))=e$. Denote $b_n=L(f)(t_n,e-f(t_n))$. Then $t_n+b_n\leqslant A(e)$ and $f(t_n+b_n)=f(t_n)\pm (e-f(t_n))$, $g(t_n+b_n)=g(t_n)\mp (e-f(t_n))$ with opposite sign. In any case we have $|f(b_n+t_n)|\to e$, thus by aforementioned corollary of continuity of $A$ at $e$ we get $b_n\to A(e)$, but then one of functions $f, g$ becomes discontinuous at $A(e)$.

Now assume that all admissible points are plus-points but there exist points for which $f\ne g$. Denote by $\Omega$ the open set of admissible points for which $f\ne g$. $\Omega$ is non-empty but $0\notin \Omega$. Therefore $\Omega$ has a connected component $(\alpha,\beta)$ with $\alpha$ or $\beta$ admissible. Without loss of generality $\alpha=0$. Then for small $e$ we do not have $f(A(e))=g(A(e))$. A contradiction.

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    $\begingroup$ This will take awhile to digest, but i will get back to you eventually. $\endgroup$ – James Baxter Dec 22 '18 at 4:39
  • $\begingroup$ @James there were many bugs now fixed, but maybe some still remain. $\endgroup$ – Fedor Petrov Dec 22 '18 at 5:39
  • $\begingroup$ Hmm, everything seems to check out except the last line, Why is f(A(e)) =/= g(A(e)) for small e a contradiction? $\endgroup$ – James Baxter Dec 22 '18 at 7:33
  • $\begingroup$ Oh right, sorry. I misread something. It does work indeed. Thanks for the answer! $\endgroup$ – James Baxter Dec 22 '18 at 7:38
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    $\begingroup$ @JamesBaxter there are many partitions of $\mathbb{R}$ onto two subsets each of which have positive measure in any interval. Characteristic function of such a subset have left/right modulus of continuity (and essential modulus of continuity) equal to 0 for $e<1$ and equal infinity for $e\geqslant 1$. $\endgroup$ – Fedor Petrov Dec 23 '18 at 6:50

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