# A continuous selection from a countable family of uniformly Lipschitz functions is Lipschitz

Conider a countable collection of positive Lipschitz functions $\varphi_n : \mathbb{R} \to \mathbb{R}$ such that there is a positive real number $K$ which works as a Lipschitz constant for every $\varphi_n$. Clearly, these Lipschitz functions can intersect each other, and let us look at a branch of this family. I want to say that the said branch (any branch of the family) is also Lipschitz, but not completely sure, because the intersections could be quite ugly. It also seems to me intuitively that the nature of the intersections of the $\varphi$'s should not matter, because they share the same Lipschitz constant, but I could use some opinion on this.

By a "branch of this family", I mean a continuous function $\varphi$ such that for every real number $x$, we have some $\varphi_k$ such that $\varphi (x) = \varphi_k (x)$. As a special example, you can consider $\varphi (x) = \inf_k \varphi_k (x)$. That will be the "lowest branch".

• Could you explain what you mean by "a branch of this family"? – Jose27 Nov 17 '15 at 15:50
• @Jose27 Thanks for your comment, edited! – novice Nov 17 '15 at 15:55
• By the way, your example does not necessary work, since $\inf_k \varphi_k(x)$ need not be attained. – user147263 Nov 27 '15 at 8:16

This is true. The problem becomes more transparent if we use the following characterization of $K$-Lipschitz functions: $f$ is $K$-Lipschitz if and only if both $Kx-f(x)$ and $Kx+f(x)$ are nondecreasing.
Thus, it suffices to prove that any branch of a countable family of nondecreasing functions $\psi_n$ is also nondecreasing.
Suppose that $f$ is a branch such that $f(a)>f(b)$ for some $a<b$. For each $y\in ( f(b), f(a))$ define $$\xi(y) = \inf \{x\in [a,b]: f(x) \le y\}$$ Note that $f(\xi(y)) = y$ and that $$\xi(y_1)>\xi(y_2)\quad \text { when } y_1<y_2\tag{1}$$
For each $y$ as above, there is $n$ such that $f(\xi(y))=\psi_n(\xi(y))$. Since there are only countably many $n$, same $n$ will be used for two distinct values $y_1,y_2$. But then
$$\psi_n(\xi(y_i)) =f(\xi(y_i)) = y_i,\quad i=1,2$$ which in view of $(1)$ implies $\psi_n$ cannot be nondecreasing.