# Are continuous functions almost completely determined by their modulus of continuity?

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 $$f = -g$$?

• 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 $g(x)=|x|$. – Joel David Hamkins Dec 21 '18 at 16:15
• Yep, I realised that the single modulus was too messy, so I felt requiring both left and right was more natural. – James Baxter Dec 22 '18 at 4:41

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 $$A(e)<\infty$$), $$|f| on $$[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.

• This will take awhile to digest, but i will get back to you eventually. – James Baxter Dec 22 '18 at 4:39
• @James there were many bugs now fixed, but maybe some still remain. – Fedor Petrov Dec 22 '18 at 5:39
• Hmm, everything seems to check out except the last line, Why is f(A(e)) =/= g(A(e)) for small e a contradiction? – James Baxter Dec 22 '18 at 7:33
• Oh right, sorry. I misread something. It does work indeed. Thanks for the answer! – James Baxter Dec 22 '18 at 7:38
• @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 $e\geqslant 1$. – Fedor Petrov Dec 23 '18 at 6:50