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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 Benning 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 ? –  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. – 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 – tbrugere 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. –  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