I have what is in essence a basic analysis question. 


To make working out a certain example a bit easier I found that I need to find existence of a function $f\in C^\infty(\mathbb{R})$ with the following properties: 


1. $f$ is increasing
2. $f(x)=0$ for all $x\leq 0$
3. $f(x)=1$ for all $x\geq 5$
4. $\frac{f(x)}{x}<1$ for all $x>0$
5. For any $y>-1$ the function $F(x,y)=\frac{f(x)}{f(x+f(x)y)}$ which is immediately well-defined and smooth for $x>0$ can be extended to a smooth function on $\{(x,y)| y>-1\}$.


To find a function $f$ that satisfies the first 4 conditions is relatively simple by tweaking one that looks like $e^{-\frac{1}{x^2}}$. So the main question is whether $f$ exists such that 5. is satisfied. After some discussion it seems like it should be possible even setting $F(x,y)=1$ for $x\leq 0$, but I have not been able to convince myself fully yet. The idea is to rewrite the denominator as $f(x(1+\frac{f(x)}{x}y))$ and use the fact that $\frac{f(x)}{x}$ vanishes to order $n$ at $x=0$ to deduce $n$ times differentiability of $F(x,y)$. Then, since $f^{(n)}(0)=0$ for arbitrary $n$ we find that smoothness. 


Any insights would be appreciated of course! 



Argument for differentiability of a certain quotient of smooth functions
$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$


It is straightforward, but a bit tedious, to construct a function $f\in C^\infty(\R)$ satisfying conditions 1--4 and such that $f(x)=e^{-1/x}$ for $x$ in a right open neighborhood $N$ of $0$ (see details at the end of this answer). Then 
\begin{equation}
 F(x,y)=e^{1/z-1/x}, 
\end{equation}
where $z:=x+ye^{-1/x}$; everywhere here, 
$y>-1$ and $x$ is a small enough positive real number (depending on $y$) such that $x$ and $z$ are both in $N$. Note that $z'\_x=1+ye^{-1/x}/x^2$ and $z'\_y=e^{-1/x}$, whence we have the crucial observation: 
\begin{multline\*}
 \Big(\frac1z-\frac1x\Big)'\_x
 =\frac1{x^2}-\frac1{z^2}\Big(1+e^{-1/x}\frac y{x^2}\Big)
 =\frac{(z-x)(z+x)}{x^2 z^2}-e^{-1/x}\frac y{x^2 z^2} \\
 =y e^{-1/x}\Big(\frac1{xz^2}+\frac1{zx^2}-\frac1{x^2z^2}\Big). 
\end{multline\*}


So, we see that 
\begin{align\*}
 F'\_x(x,y)&=F(x,y)p\_1(1/x,1/z,y,e^{-1/x})e^{-1/x},\\ 
 F'\_y(x,y)&=F(x,y)q\_1(1/x,1/z,y,e^{-1/x})e^{-1/x}
\end{align\*}
for some polynomials $p\_1$ and $q\_1$. So, by induction, all partial derivatives of $F(x,y)$ of orders $\ge1$ are of the form $F(x,y)p(1/x,1/z,y,e^{-1/x})e^{-1/x}$ for some polynomials $p$. 


Also, $|z-x|\ll e^{-1/x}$, $z\sim x$, $F(x,y)\to1$, which implies that all partial derivatives of $F(x,y)$ of orders $\ge1$ go to $0$; the convergence here is for $x\downarrow0$ uniformly over all $y$ in any compact subset of $(-1,\infty)$. This implies that $F\in C^\infty(\R\times(-1,\infty))$, with $F(x,y):=1$ for $x\le0$.




---


**Added:** details concerning the first sentence of this answer. For all real $x$,
let
\begin{equation}
 f(x) :=\int\_\R g(x + \ep u(x)t)Cu(t)\,dt,
\end{equation}
where
\begin{equation}
 g(x) := e^{-1/x} \ii{0 <x\le2}+[e^{-1/2} +(x-2)(1-e^{-1/2})]\,\ii{2 < x \le 3}+\ii{x > 3}
\end{equation}
\begin{equation}
 u(x) := \exp\Big\{-\frac1{(x - 1)(4 - x)}\Big\}\,\ii{1 < x < 4}, 
\end{equation}
$C := 1/\int\_\R u(x)\,dx$, $\ii{\cdot}$ is the indicator, and $\ep$ is a positive real number small
enough so that $1+4\ep u'> 0$ (whence $x+\ep u(x)t$ is increasing in $x \in\R$ for each $t$ in
the interval $(1,4)$). 


Then $f$ is in $C^\infty(\R)$, satisfies conditions 1--4, and $f(x)=e^{-1/x}$ for $x\in(0,1)$. In particular, $f$ is increasing -- because $g$ is so and $x+\ep u(x)t$ is increasing in $x \in\R$ for each $t\in(1,4)$. Also, $f(x)=e^{-1/x}<x$ for $x\in(0,1)$, $g\le1$ and hence $f\le1$ on $\R$, and so, $f(x)<x$ for $x>1$. That $f$ is $C^\infty$ on the interval $(1,4)$ follows because 
\begin{equation}
 f(x) =\int\_\R g(y)Cu\Big(\frac{y-x}{\ep u(x)}\Big)\,\frac{dy}{\ep u(x)}
\end{equation}
for $x\in(1,4)$. 


