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For $i=1,\ldots,n$, let $f_i\colon(0,\infty)\to(0,1)$ and $g_i\colon(0,\infty)\to(0,1)$ be $C^{\infty}$ functions.

Define $h\colon(0,\infty)\to(0,\infty)$ as follows: $$h(t)=\max_{i=1,\ldots,n} \frac{f_i(t)}{g_i(t)}\qquad \forall t\in(0,\infty).$$

I am wondering if the following is true:

The function $h$ is differentiable almost everywhere, and there exists $f,g\colon(0,\infty)\to(0,1)$ differentiable almost everywhere such that $h(t)=\frac{f(t)}{g(t)}$ for every $t\in (0,\infty)$.

I am pretty sure it is true, but I would like to have confirmation (and, if possible, a reference).

This question is quite related however it is not exactly the same. Still the answer of @Alex R. seems to confirm my belief that the affirmation is indeed true.

Note: I don't know if it helps, but the case I am interested in have the additional properties that $h$ is strictly decreasing, $\lim_{t\to 0}h(t)=\infty$ and $\lim_{t\to \infty}h(t)=1$.

idm
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    Might be heavier machinery than is required, but you should be able to show very easily that $h$ is locally Lipschitz and thus differentiable a.e. by Rademacher's theorem. – Anthony Carapetis Mar 29 '18 at 15:57
  • @AnthonyCarapetis Thanks for your comment. Indeed, $h$ should be locally Lipschitz and differentiable a.e. I think it follows from one of the theorems in the book of Clarke. I was however a bit in doubt regarding the existence of $f$ and $g$. – idm Mar 29 '18 at 16:05
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    The subproblem of writing $h$, or any function, as a ratio of two functions of the same regularity and with those bounds is not an issue: if you call $$u(x)=\begin{cases} 1&\text{if }x=0\ \frac{\arctan x}x&\text{if }x\ne 0\end{cases}$$ then $u$ is analytic on $\Bbb R$ and it satisfies $0< u\le 1$. You may then consider $f=\frac2\pi \arctan\circ h$, $g=\frac2\pi u\circ h$. Of course, this does not answer whether or not $h$ is differentiable a.e. –  Mar 29 '18 at 16:13
  • @G.Sassatelli Ok but your comment reduces my question to deciding wheather $h$ is a.e. differentiable or not, if I'm not mistaking. – idm Mar 29 '18 at 16:15
  • @idm Indeed, if I haven't stressed it enough. :) –  Mar 29 '18 at 16:16
  • @G.Sassatelli All in all, if I combine your comment with that of Anthony Carapetis, I should be able to prove that the statement is true. Or, am I missing something? – idm Mar 29 '18 at 16:19
  • @idm I'd say yes. Of course, there might be more intersting ways to write $h$ as a quotient. –  Mar 29 '18 at 16:43
  • @G.Sassatelli Indeed, but I'm only interested in existence so it is enough for me (and I am very happy because it saves me a lot of crazy computations). Do you wish to write an answer so that I can accept it? – idm Mar 29 '18 at 16:46
  • @AnthonyCarapetis After further small googling/thinking, in fact it is obvious that $h$ is locally (as these are smooth functions on an open interval so we may take any close intervals around the point). And sorry for my previous comment, I got mislead by Wikipedia who only stated the theorem for Lipschitz mappings. – idm Mar 29 '18 at 16:49

2 Answers2

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As it stands you may just assume $f_i(t)>0$, and define $$h(t):=\max_{1\leq i\leq n} f_i(t)\ .$$ For most $t_0$ there will be an $i_0$ such that $f_{i_0}(t_0)>f_i(t_0)$ for all $i\ne i_0$. It follows that $h(t)=f_{i_0}(t)$ for all $t$ in a full neighborhood $U$ of $t_0$, hence $h\in C^\infty(U)$.

I don't know about the control you have over the "critical set" $$C:=\bigl\{t>0\,\bigm|\,\exists i, j: \ i\ne j\ \wedge\ f_i(t)=f_j(t)\bigr\}\ .$$ If this set is, e.g., finite the function $h$ will be continuous and piecewise smooth. In order to see what's happening at points of $C$ consider the functions $f_1(t)=e^{t-1}$ and $f_2(t)=e^{1-t}$.

Christian Blatter
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  • Mhmm, but.. if $f_i=f_j$ for every $i,j$, then $h\in C^{\infty}((0,\infty))$. In which case $C=(0,\infty)$. On the other hand if $f_1>f_2>...>f_n$ everywhere, then $h = f_1$, $C=\emptyset$ and $h\in C^{\infty}((0,\infty))$. So I'm not sure how controlling the critical set $C$ will help me to understand the differentiability of $h$. – idm Mar 29 '18 at 16:29
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As @Christian Blatter noticed, the problem can be easily reduced to functions $h$ of the form $$h(t)=\max_{1\leq i \leq n} h_i(t)$$ where $h_i\colon (0,\infty)\to (0,\infty)$ is smooth ($h_i=f_i/g_i$ in OP). Now, as $h_i$ is smooth, it is Lipschitz continuous on every closed interval. It follows that $h$ is locally Lipschitz on $(0,\infty)$.

Now, @Anthony Carapetis suggested to use Rademacher's theorem which in facts, implies that $h$ is differentiable almost everywhere in $(0,\infty)$.

Finally, @G. Sassatelli noticed that one can always construct $f,g\colon (0,\infty)\to(0,1)$ with the same regularity as $h$ and such that $h(t)=\frac{f(t)}{g(t)}$ for every $t>0$. Take for instance $$f(t)=e^{-h(t)}, \qquad g(t)=\frac{e^{-h(t)}}{h(t)}.$$

idm
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