# Ordering some functions without losing continuity

Let's consider the set $\mathbb R^{n}/S_n$, i.e. the quotient of $\mathbb R^{n}$ modulo permutations. An element $a \in \mathbb R^{n}/S_n$ is simply a $n$-tuple of real numbers (the order does not matter).

This set has clearly the quotient topology and it is metrizable, being the distance the so called "optimal matching distance": $$d(a,b)= \min_{\sigma \in S_n} \max_{1\le j \le n} \vert a_j-b_{\sigma(j)}\vert$$ for $a=\{a_1, \ldots , a_n\}$ and $b = \{b_1, \ldots , b_n\}$.

I know the following result holds true:

Theorem. Let $O:\mathbb R^{n}/S_n \to \mathbb R^{n}$ the map that "orders" the numbers of the n-tuple, i.e. $$\{a_1, \ldots , a_n\} \mapsto O(\{a_1, \ldots , a_n\}):=(\hat{a}_{1}, \ldots , \hat{a}_n)$$ where $\hat{a}_{1}\le \ldots \le \hat{a}_n$. Then $O$ it's Lipschitz continuous, i.e. $$\max_{i}\vert \hat{a}_i - \hat{b}_i\vert \le d(a,b)$$

Now take $n$ continuous functions, $\phi_j \colon [0,1] \to \mathbb {R}$. I would like to prove that there is a way to relabel the functions (without losing continuity) s.t. $$\phi_1(t) \le \phi_2(t) \le \ldots \le \phi_n(t) \le \phi_1(t) + 2\pi , \quad \forall t \in [0,1]$$

The functions $\phi_j$ have also this important property: for every $t \in [0,1]$ $$\phi_j(t) = \phi_j(t)+2k\pi, \quad k \in \mathbb Z$$ (indeed, they represent angular coordinates, therefore I can translate them up and down by multiple of $2\pi$). I would like to prove this property (but I am still not sure about the statement):

Given $n$ continuous functions $\phi_j \colon [0,1] \to \mathbb R$ ($j=1, \ldots n$) then there exist other $n$ continuous functions $\vartheta_j \colon [0,1] \to \mathbb R$ s.t.

1. $$\vartheta_1(t) \le \vartheta_2(t) \le \ldots \le \vartheta_n(t) \le \vartheta_1(t)+2\pi \qquad \forall t \in [0,1]$$
2. for every $t \in [0,1]$, for every $i=1, \ldots , N$ there exists $j \in \{1,\ldots , n\}$ and $k \in \mathbb Z$ s.t. $$\vartheta_i(t)=\phi_j(t)+2k\pi$$ ($j,z$ depend on $t$).

Do you think this last statemente is true? To prove it, I would like to use the theorem above: for every $\bar{t} \in [0,1]$ I consider the $(n+1)$-tuple given by $\{\phi_1(\bar{t}), \phi_2(\bar{t}), \ldots \le \phi_n(\bar{t}), \phi_1(\bar{t}) + 2\pi\}$: can I apply the theorem to conclude?

Example.

Consider the two functions $\phi_1(t)=-t$ and $\phi_2(t)=t$, defined on $[0,2\pi]$ (it's clearly the same of $[0,1]$, but it simplifies the calculations). Initially, there's no relationship between them. But if you put $$\vartheta_1 = \begin{cases} -t \qquad t \in [0,\pi] \\ t-2\pi \qquad t \in [\pi, 2\pi] \end{cases}$$ and $$\vartheta_2 = \begin{cases} t \qquad t \in [0,\pi] \\ -t+2\pi \qquad t \in [\pi,2\pi ] \end{cases}$$ then you can write $\vartheta_1(t) \le \vartheta_2(t) \le \vartheta_1(t)+2\pi$ for every $t \in [0,1]$. This is exacly what I want.

If you know any reference (book, article etc) which deals with this, please let me know. Thanks in advance for your help.

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If the order does not matter than you simply have a multiset, not a tuple. – Asaf Karagila Jun 14 '12 at 20:59

## 1 Answer

The question is not clear enough. What do you mean by "relabeling" the functions? If you are given $\phi_1(t)=t$ and $\phi_2(t)=1-t$, you may well decide to change the label (indices) and write instead $\phi_2(t)=t$ and $\phi_1(t)=1-t$. Of course, the functions don't become discontinuous from being called by another name. On the other hand, you don't get the ordering you want because for some values of $t$ the inequality between $\phi_1(t)$ and $\phi_2(t)$ goes one way, for others the opposite way.

Perhaps you want to introduce different functions: $\bar \phi_1(t)=\min(\phi_1(t),\phi_2(t))$ and $\bar \phi_2(t)=\max(\phi_1(t),\phi_2(t))$. This achieves $\bar\phi(t)\le \bar\phi_2(t)$ for all $t$. The functions are still continuous: the maximum of two continuous functions is continuous, and the minimum too. However, there is no reason to expect $\bar\phi_2(t)\le \bar\phi_1(t)+2\pi$ unless you have some information about your functions that involves $2\pi$. You stated no such assumptions in your post.

When $n>2$, the definition of $\bar \phi_k$ becomes a bit clumsy, but still works: $\bar \phi_k(t)=\min\limits_{I}\max\limits_{j\in I}\phi_j(t)$ where the minimum is taken over all sets $I\subset \{1,\dots,n\}$ of cardinality $k$.

An excellent current reference on maps into spaces like $\mathbb R^{n}/S_n$ is Q-valued functions revisited by De Lellis and Spadaro. Actually, their work emphasizes the more difficult issues involving $\mathbb R^{nQ}/S_{Q}$, i.e., the space of unordered $Q$-tuples of $n$-dimensional vectors.

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Thank you very much for your kind reply. I have just added some additional information and I've edited my post. Hope now the question is clear. Thanks for your help. – Romeo Jun 14 '12 at 21:29
@Romeo Counterexample: $\phi_1(t)=0$ for all $t$ and $\phi_2(t)=100t$. Now $\vartheta_1$ must be constant, and there is no way to choose $\vartheta_2$ continuously: it would be the same as continuous choice of argument on the entire circle. – user31373 Jun 14 '12 at 21:43
You are right. What if we add that $\phi_j$ are not constant but we require only that $\phi_j(0)=0$ for every $j=1, \ldots , N$? Thanks again. – Romeo Jun 14 '12 at 21:54
@Romeo Does not make a difference: take $\phi_1(t)=t/100$ and $\phi_2(t)=100t$. – user31373 Jun 14 '12 at 21:59
Consider for sake of simplicity $\phi_1(t)=t$ and $\phi_2(t)=2t$. Then for $t \in [0,2\pi]$ it's ok, we set $\vartheta_1=\phi_1$ and $\vartheta_2=\phi_2$. For $t \ge 2\pi$ just define $\vartheta_1 = 2t-2\pi$ and $\vartheta_2 = t+2\pi$ (clearly this works until $t=4\pi$, then we relabel again). Is this so different from taking $100t$ and $t/100$ (apart from calculations, I mean)? – Romeo Jun 14 '12 at 22:11