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Definition. A function $f:\Bbb R\to\Bbb R$ will be called potentially continuous if there is a bijection $\phi:\Bbb R\to\Bbb R$ such that $f\circ \phi$ is continuous.

So one could say a potentially continuous (p.c.) function is "a continuous function with a mixed up domain". I was wondering whether there is an easy way to characterize such functions.


Some thoughts $\DeclareMathOperator{\im}{im}$

  • If the image $\im(f)$ is not connected (i.e. no interval), then $f$ is not p.c. because even mixing the domain cannot make $f$ satisfy the intermediate value theorem.
  • Bijective functions are always p.c. because we can choose $\phi=f^{-1}$. Every injective function with an open connected image is p.c. for a similar reason. However, only having a connected image is not enough, as e.g. there are bijections, but no continuous bijections $f:\Bbb R \to [0,1]$.
  • Initially I thought a function can never be p.c. if it attains every value (or at least uncountably many values) uncountably often, e.g. like Conways base 13 function. But then I discovered this: take a Peano curve like function $c$ (or any other continuous surjection $\Bbb R\to\Bbb R^2$) and only look at the $x$-component $c_x:\Bbb R\to\Bbb R$. This is a continuous function which attains every value uncountably often.
  • The question can also be asked this way. Given a family of pairs $(r_i,\kappa_i),i\in I$ of real numbers $r_i$ and cardinal numbers $\kappa_i\le\mathfrak c$ so that $\{r_i\mid i\in I\}$ is connected. Can we find a continuous function $f:\Bbb R\to\Bbb R$ with $|f^{-1}(r_i)|=\kappa_i$?
  • There is no continuous function which attains each real number exactly once except zero which is attained twice. So, e.g. the function $$f(x)=\begin{cases}x-1&\text{for $x\in\Bbb N$}\\x&\text{otherwise}\end{cases}$$ is not p.c., even though its image is all of $\Bbb R$.
M. Winter
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    (+1) Interesting question. My bet is that any function with a connected range is potentially continuous. – Jack D'Aurizio Oct 20 '17 at 21:43
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    @JackD'Aurizio I am pretty sure I found a counter-example. See the last item above. – M. Winter Oct 21 '17 at 00:16
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    And you are clearly right, the multiplicity of the elements of the range counts, too. Even more interesting. – Jack D'Aurizio Oct 21 '17 at 00:19
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    If $f^{-1}(0) = \{a,b\}$, and $f$ injective on $(a,b)$, then $f$ is strictly increasing or decreasing on $(a,b)$, so $f(a) \ne f(b)$, contradiction. – orangeskid Oct 21 '17 at 00:41
  • @orangeskid Yes :) I understand. I initially had no idea what you comment tried to show but I guess it was a proof to my "I am pretty sure"-part, right? – M. Winter Oct 21 '17 at 00:45
  • If $1< |f^{-1}(0)|< c$, then there exists a choice of sign $\pm$ and $\epsilon_0 > 0$, so that $|f^{-1}(\pm \epsilon)| > 1$ for all $0< \epsilon< \epsilon_0$. Just intermediate value. – orangeskid Oct 21 '17 at 01:13
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    It might be a good idea to first answer the question in the case when $\vert f^{-1}(x)\vert\le 2$ for all $x$ ... – Noah Schweber Oct 21 '17 at 02:03
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    My answers to https://math.stackexchange.com/questions/1441063/can-a-surjective-continuous-function-from-the-reals-to-the-reals-assume-each-val and https://math.stackexchange.com/questions/1439689/can-a-continuous-function-from-the-reals-to-the-reals-assume-each-value-an-even address some special cases of this, and suggest the question is fairly complicated to answer even in the case that all the fibers are finite. – Eric Wofsey Oct 21 '17 at 04:56
  • Do you think they have to be measurable? Any uses for it? – Felix B. Nov 12 '17 at 00:45
  • @FelixB. I think there are non-measurable bijections which are therefore p.c. For now I just thought this might be an interesting question without any specific usecase. – M. Winter Nov 12 '17 at 23:55
  • Generalizing your last example: If $n\in\mathbb N$ and $y\in\mathbb R$ so that $n\leq\left|f^{-1}(y)\right|<|\mathbb R|$ then there is a closed interval $I$ containing $y$ such that for all $y'\in I$, $2\lfloor n/2\rfloor\leq \left|f^{-1}(y')\right|.$ – Thomas Andrews Dec 14 '17 at 22:17
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    Is every Darboux function potentially continuous ? – Aditya Guha Roy Dec 18 '17 at 15:50
  • Given any potentially continuous function $f$ and any continuous $g:\mathbb R\to\mathbb R$, you'd have that $g\circ f$ is also potentially continuous. It seems unlikely that $f\circ g$ would be potentially continuous, in general, just because $f$ is, but I could be convinced otherwise. – Thomas Andrews Dec 19 '17 at 17:53
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    It would be interesting to find a pair $f,g$ of potentially continuous functions such that $f+g$ are not p.c, or are the p.c. functions a vector subspace of $\mathbb R^{\mathbb R}$? – Thomas Andrews Dec 19 '17 at 19:08
  • Also, if the continuum hypothesis is false, is it possible for $|\mathbb N|<|f^{-1}(x)|<|\mathbb R|$? – Thomas Andrews Dec 19 '17 at 20:14
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    @ThomasAndrews No, since $(f \circ \phi)^{-1}(x)$ is closed (because $f \circ \phi$ is continuous) hence Borel, so $|f^{-1}(x)| = |(f \circ \phi)^{-1}(x)|$ is either countable or $2^{\aleph_0}$. – Adayah May 28 '18 at 21:53
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    I don't think the second point of your second point is quite correct: what if $f$ is a bijection from $\mathbb R$ to $[0, 1]$? – Mees de Vries Jul 13 '18 at 08:07
  • If $f$ is a bijection from $\mathbb R$ to $[0, 1]$ then it is a fortiori a function from $\mathbb R$ to $\mathbb R$. I am responding to the claim that such a function, which is an injection with connected image, must be p.c. – Mees de Vries Jul 13 '18 at 09:00
  • @MeesdeVries you are right. I will edit my post. – M. Winter Jul 14 '18 at 12:39
  • If $f$ is a bijection from $\mathbb{R}$ to $[0,1]$ then it wouldn't be p.c.: no function with domain $\mathbb{R}$ and image $[0,1]$ can be one-to-one (in fact, there are infinitely many failures of injectivity near a point $x_0$ with $f(x_0) = 0$ or near a point with $f(x_0) = 1$). – Daniel Schepler Sep 25 '18 at 18:17
  • @Daniel Schepler: just to forestall any comments, of course you mean no continuous function with domain $\mathbb{R}$ and image $[0,1]$. I think a larger issue is that if $f \colon \mathbb{R} \to [0,1]$ is bijective then $f^{-1}$ does not extend to a bijection from $\mathbb{R}$ to $\mathbb{R}$, so the argument in the question about $f^{-1}$ won't work. – Carl Mummert Sep 25 '18 at 20:21
  • Really that point should say "bijections from $\mathbb{R}$ to $\mathbb{R}$ are always p.c.". – Carl Mummert Sep 25 '18 at 20:27
  • @CarlMummert I thought this was clear as I defined the term p.c. exclusively for function $f:\Bbb R\to\Bbb R$. I might change this if it is very unclear. – M. Winter Sep 25 '18 at 20:34
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    @NoahSchweber: in the case where all $\kappa$ are $\le 2$, there are four possibilities, depending on whether the function attains its infimum and whether it attains its supremum:$$\kappa \ge 1 \text{ on } (a,d),\ \kappa =2 \text{ on } \emptyset,\ -\infty \le a < d \le \infty$$ $$\kappa \ge 1 \text{ on } [a,d),\ \kappa =2 \text{ on } (a,b),\ -\infty < a < b \le d \le \infty$$ $$\kappa \ge 1 \text{ on } (a,d],\ \kappa =2 \text{ on } (c,d),\ -\infty \le a \le c < d < \infty$$ $$\kappa \ge 1 \text{ on } [a,d],\ \kappa =2 \text{ on } (a,b)\cup (c,d),\ -\infty < a < b \le c < d < \infty$$ –  Dec 27 '18 at 13:34
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    The composition of two continuous functions is continuous. If $f$ and $\phi$ are continuous, then $f\circ \phi$ is continuous therefore, and $f$ is potentially continuous. $\curvearrowright$ Each continuous function is potentially continuous. If $\phi$ is continuous, then $f$ is continuous. $f$ potentially continuous $\Leftrightarrow \forall x_0\in\mathbb{R}\colon \lim_{x\rightarrow x_0-}f(\phi(x))=\lim_{x\rightarrow x_0+}f(\phi(x))=f(\phi(x_0))$ – IV_ Dec 29 '18 at 20:23
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    To the editor: I've re-added the "set theory" tag since, even though set theory isn't *explicitly* part of the question, it's quite possible (in light of e.g. **descriptive set theory**) that the *answer* will involve set theory in a nontrivial way. While the tag is therefore hypothetical, it's (in my opinion) still appropriate. – Noah Schweber Dec 29 '18 at 20:48
  • @IV_ I think you're still misunderstanding the situation. Of course $f$ is potentially continuous via $\phi$ iff $\forall x_0\in\mathbb{R}:\lim_{x\rightarrow x_0^-}f(\phi(x))=\lim_{x\rightarrow x_0^+}f(\phi(x))=f(\phi(x_0))$ - the **definition** of $f$ being potentially continuous via $\phi$ is that $f\circ \phi$ is continuous, and that's exactly what you've written. So that doesn't help at all. – Noah Schweber Dec 29 '18 at 20:54
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    Also, while it's true that if $f$ is p.c. via a *continuous* $\phi$ then $f$ is actually continuous, that's the essentially-trivial case of the problem: *nowhere do we require $\phi$ to be continuous* (or monotonic, or remotely nice in any other way). – Noah Schweber Dec 29 '18 at 20:57
  • Maybe the definition helps to deduce consequences. – IV_ Dec 29 '18 at 20:58
  • Be $x_0\in\mathbb{R}$. If $\phi$ has a removable discontinuity or a jump discontinuity (discontinuities of the first kind) at $x_0$ with $\lim_{x\rightarrow x_0-}\phi(x)=y_-$, $\phi(x_0)=y$, and $\lim_{x\rightarrow x_0+}\phi(x)=y_+$, the following is sufficient for $f$: $r\in\mathbb{R}$, $f$ is continuous at all $r\notin\{y_-,y,y_+\}$, $\lim_{r\rightarrow y_-}f(r)=\lim_{r\rightarrow y_+}f(r)=f(y)$. – IV_ Dec 29 '18 at 22:03
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/87654/discussion-between-iv-and-noah-schweber). – IV_ Dec 29 '18 at 22:18
  • @Noah: While the username might not appear on the auto complete, you can ping editors just as well. – Asaf Karagila Dec 29 '18 at 22:57
  • for starters, if $f(\mathbb{R})$ is connected, I would try to find $\phi$ so that $f(\phi(x))$ is monotonic. Since $f(\phi(\mathbb(R)))$ is connected, that would essentially imply f is continuous – Nephanth Mar 11 '19 at 12:50
  • It's evident to see that a mapping is a p.c. function $f$ iff it is of the form $f = g \circ \psi$ for some continuous function $g$ and some bijective function $ \psi$. But what I think here is that since our $\psi$ can behave quite wildly, and our $g$ can be any continuous function, so there is not much hope of any neater desription. One can tweak the values of $\psi$ over any strict subset of the reals (whose complement in reals is uncountable [-- though not necessary] ) and then take any $g$ and apply it on the $\psi$. This also throws light on some claims proposed in some comments above. – Aditya Guha Roy Mar 17 '19 at 14:59
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    @Thomas Andrews, f and g is potentially continuous doesn't result in potentially continuous f+g. Let f(x)=x and g(x)=-x when x is irrational and g(x)=-x+1 when x is rational number. Both f and g is potentially continous since they're bijection but f+g is 0 and 1 respectively when x is irrational or rational number so it is not potentially continuous. – Zhaohui Du Aug 25 '19 at 10:43
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    The fact that this was a high visibility question for nearly two years, shows that *there is no easy way to characterize such functions.* – Moishe Kohan Oct 07 '19 at 21:04
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    @MoisheKohan nope. that logic would contradict the fact that there have been several famous problems open for tens of years that had easy proofs – mathworker21 Oct 08 '19 at 16:52
  • What about analogous case of "mixing-up" the codomain, where there $\exists$ a bijection $\phi:\mathbb{R} \to \mathbb{R}$ such that $\phi \circ f$ is continuous? Or even "mixing-up" both the domain and the codomain? – Geoffrey Trang Oct 10 '19 at 02:59
  • @M.Winter I was just reminded of this question. It looks like you have not asked it on MathOverflow; have you thought about doing so? I think it would be well-received there. – Noah Schweber May 14 '20 at 17:14

1 Answers1

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This should only be taken as a partial answer, since the results quoted are all "mod null", i.e. statements should only be understood to be true up to a set of measure zero. Probably a reader equipped with more descriptive set theory expertise would than me would know whether this can be upgraded to pointwise statements.

It is a result quoted, for example, in Brenier's "Polar Factorization and Monotone Rearrangement of Vector-Valued Functions", that if $(X,\mu)$ is a probability space (e.g a bounded domain in $\mathbb{R}^n$ with normalized Lebesgue measure), and $u\in L^p(X,\mu)$, then there exists a unique nondecreasing rearrangement $u^* \in L^p(0,1)$. Moreover, there exists a measure-preserving map $s$ from $(0,1)$ to $(X,\mu)$ such that $u\circ s = u^*$. In the case where X is understood to be a bounded subset of $\mathbb{R}^n$, this can be thought of as a rearrangement of the domain. (Brenier actually quotes the opposite result, but the paper he cites, by Ryff, gives both directions.)

By modifying $u^*$ on a null set, we can take $u^*$ to be lower-semicontinuous. In this case, it is apparent that $u^*$ is continuous iff it has no jump discontinuities.

In the case where $n>1$ one can also get some mileage out of Sobolev space theory. It follows from Theorem 0.1 of "A Co-area Formula with Applications to Monotone Rearrangement and to Regularity" by Rakotoson and Temam that if $u\in W^{1,p}(\Omega)$, then $u^* \in W^{1,p} (0,1)$. In particular, (up to choice of pointwise representative) $u^*$ is absolutely continuous, and therefore continuous. (This is not interesting in the case where $n=1$ since we would already be assuming that u is continuous.)

pseudocydonia
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    I offered a bounty but now i realize I am expected to evaluate all the answers to decide the award . I will likely need help. I will welcome any input on whether some answer deserves the bounty. I have not yet studied this A. – DanielWainfleet Oct 10 '19 at 04:50
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    This may be an answer to a better question than the one actually asked! As the question is stated, $\phi$ can be an arbitrary bijection without anything like measure-preserving properties. All $\phi$ is guaranteed to preserve is cardinality (of fibres, for example). – HTFB Oct 13 '19 at 21:28
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    This answer does not seem to shed any light on the original question. – Moishe Kohan Oct 14 '19 at 16:58