# What functions can be made continuous by “mixing up their domain”?

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 $$\Bbb R$$.
• (+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
• @JackD'Aurizio I am pretty sure I found a counter-example. See the last item above. – M. Winter Oct 21 '17 at 0:16
• 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 0:41
• Is every Darboux function potentially continuous ? – adityaguharoy Dec 18 '17 at 15:50
• @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

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.)
• 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